Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Mnemonic Devices - Free WonderKids-e
Document related concepts
Nanofluidic circuitry wikipedia , lookup
Hydrogen-bond catalysis wikipedia , lookup
Double layer forces wikipedia , lookup
Chemical reaction wikipedia , lookup
Water splitting wikipedia , lookup
Lewis acid catalysis wikipedia , lookup
Stoichiometry wikipedia , lookup
Transition state theory wikipedia , lookup
Artificial photosynthesis wikipedia , lookup
Click chemistry wikipedia , lookup
Bioorthogonal chemistry wikipedia , lookup
Photoredox catalysis wikipedia , lookup
Photosynthetic reaction centre wikipedia , lookup
Strychnine total synthesis wikipedia , lookup
Theory of solar cells wikipedia , lookup
Nickel–cadmium battery wikipedia , lookup
Nickel–metal hydride battery wikipedia , lookup
Gaseous detection device wikipedia , lookup
Metalloprotein wikipedia , lookup
History of electrochemistry wikipedia , lookup
Electrolysis of water wikipedia , lookup
Evolution of metal ions in biological systems wikipedia , lookup
Transcript
Electrochemistry 1 III) Oxidation – Reduction Reaction for Metal with Non-Metal – driven by Electron Transfer (and possibly Formation of Gas), (Electron transfer reactions can also take place between two non-metals, and the compound formed is not ionic. A sure sign of that happening is the presence of oxygen, O 2 ( g ) , as a reactant or product): e. g. 1) 2 Na( s) + Cl2 ( g ) → 2 NaCl( s) 2) 2 Mg( s) + O2 ( g ) → 2 MgO( s) 3) 2 Al( s) + Fe2 O3( s) → 2 Fe( s ) + Al2 O3( s) 4) a) 2 HCl( aq ) + NaCO3( aq ) → CO2 ( g ) + H 2 O( l ) + NaCl( aq ) The fourth driving force is in play in this reaction because a gas is formed. Another example: b) 2 HCl( aq ) + Zn( s) → H2 ( g ) + ZnCl( aq ) Oxidation is the process whereby a substance in a chemical reaction losses one or more electrons; Reduction is the process whereby a substance gains one or more electrons. Oxidation-Reduction is used to describe any reaction in which electrons are transferred from one reactant to another reactant. Oxidizing agents remove electrons from reducing agents, causing the reducing agent to oxidized (loss of electrons) and the oxidizing agent to be reduced (gain of electrons). Substance oxidized by providing electron(s) = reducing agent; Substance reduced by accepting electron(s) = oxidizing agent Mnemonic Devices • • • • OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons, not to be confused with oxidation number) LEO says GER: Loss of Electrons Oxidation, Gain of Electrons Reduction RED CAT AN OX: REDuction CAThode and ANode OXidation ReduCtion has a C(athode) but no A(node), and OxidAtion has a A(node) but no C(athode) An oxidation number is the charge that an atom appears to have when the electrons in each bond it is participating in are assigned to the more electronegative of the two atoms involved in the bond. The following set of operational rules are used to determine oxidation numbers: Group # 1 2 3 - 12 13 14 15 16 Elements All Oxidation # +1 H: +1 in combination with nonmetals; -1 in combination with metals All +2 Ag +1 Zn +2 Other elements in group have more than one possible oxidation # Al +3 C and Si +4 or −4 Other elements in group have more than one possible oxidation # N and P −3 Other elements in group have more than one possible oxidation #, such as N: NH 3 , N 2 O, NO, N 2 O3 , NO 2 , N 2 O 4 , N 2 O5 , HNO3 −3 +1 +2 +3 +4 +4 +5 +5 O and S −2 O: Exceptions include OF2, since F is more electronegative than O ( ), and BaO2 ( ), due to 2), O's O.N. is the structure of the peroxide ion, which is [O-O] . In superoxides (such as Other elements in group have more than one possible oxidation # 17 All −1 18 All 0 Rule # 1: The oxidation # of a free element is always 0 Rule # 2: The oxidation # of a monoatomic ion equals the charge of the ion Rule # 3: The oxidation # in a polyatomic ion equals the charge of the ion Rule # 4: The sum of the oxidation numbers of all the atoms in a neutral compound is 0 (1) Balancing Redox Equations using Oxidation Number Method: e. g.: To balance the equation S( s ) + HNO3( aq ) → SO2 ( g ) + NO( g ) + H2O( l ) , follow the steps below: 0 (i) +1+5− 2 +4 − 2 +2 − 2 +1 − 2 Assign oxidation numbers to all atoms: S( s) + HNO3( aq ) → SO2( g ) + NO( g ) + H2O( l ) (ii) Electrochemistry 2 Identify the element oxidized and the element reduced: In this reaction, sulfur changes from 0 to +4, a change of +4 (oxidation). Nitrogen changes from +5 to +2, a change of –3 (reduction). +4 (iii) Connect the atoms that change oxidation numbers: (iv) -3 Choose coefficients that make the total increase in oxidation number equal the total decrease in oxidation number: 3 * (+4) =12 0 +1+5− 2 +4 − 2 +2 − 2 +1 − 2 S( s) + HNO3(aq ) → SO2( g ) + NO( g ) + H2O(l ) +1+5− 2 0 +4 − 2 +2 − 2 +1 − 2 S( s) + HNO3( aq ) → SO2( g ) + NO( g ) + H2O( l ) 4 * ( - 3) = - 12 +1+5− 2 0 +4 − 2 +2 − 2 +1 − 2 3 S( s) + 4 HNO3(aq ) → 3SO2( g ) + 4 NO( g ) + H2O(l ) (i) Balance the remaining elements by inspection, and then check the final equation: 0 +1+ 5− 2 +4 − 2 +2 − 2 +1 − 2 3 S( s) + 4 HNO3( aq ) → 3SO2 ( g ) + 4 NO( g ) + 2 H2 O( l ) (2) Balancing Redox Equations using the Ion-Electron or Half-Reaction Method: Ini this method, the oxidation and reduction processes are divided into separate equations called half-reactions that are balanced separately. Example: Balance the net ionic equation for the reaction of iron (III) chloride with tin (II) chloride. Solution: Step 1. Separate the equation into two half equations and balance the charge in each by adding : (oxidation) (reduction) Step 2: Adding the two resulting equations back together, we get a balanced equation: Note: Sometimes a reaction that occurs in an acidic water solution can be balanced by adding H2 O and H + to either side of the equation as necessary. In these cases, you should add H2 O to the side of the equation that needs O atoms and H + to the side that needs H atoms. Similarly, you can use OH − and H2 O to balance reactions in basic solutions Electrochemistry 3 Electrochemistry 4 Electrochemistry 5 Electrochemistry Example: Complete and balance the chemical equation for the dissolution of copper (II) sulfide in aqueous nitric acid: CuS ( s ) + NO3−( aq) → Cu 2+ ( aq ) + SO42(−aq ) + NO( g ) 6 Electrochemistry 7 Electrochemistry 8 Electrochemistry 9 Exercises: (1) Complete and balance the following equations for reaction taking place in acidic solution. + 2+ 2− → VO( aq ) + SO4( aq) (a) VO( aq ) + SO2( g ) (b) Br 2 ( l ) + SO → Br ( −aq ) + SO 2(g ) − (c) Cr 2 O 72aq ) + Np 4+ ( aq ) 2− 4 ( aq ) → Cr (3aq+ ) + NpO 2+ 2 ( aq ) − 2+ → CO2( g ) + Mn( aq ) (d) HCOOH( aq) + MnO4( aq ) (e) Hg 2 HPO 4(s) + Au ( s ) + Cl (−aq ) → Hg (l ) + H 2 PO 4−( aq ) + AuCl − 4 ( aq ) (2) Complete and balance the following equations for reaction taking place in basic solution 2− − → CrO4( aq ) + Br( aq ) (a) Cr (OH ) 3( s ) + Br2( aq) 2− 2− → Zr( s ) + SO4( aq ) (b) ZrO(OH ) 2( s ) + SO3( aq) − − → Pb( s ) + Re O4( aq ) (c) HPbO2( aq) + Re ( s ) − 4− → XeO6( aq) + Xe( g ) (d) HXeO4( aq) 2− → N 2( g ) + CO( g ) (e) N 2 H 4( aq ) + CO3aq ) (3) The following balanced equations represent reactions that occur in aqueous acid. Break them down into balanced oxidation and reduction half-equation. − 2+ → 2 Fe( aq ) + 4 H 2 O(l (a) 2 H 3 O( aq ) + H 2 O2( aq ) + 2 Fe( aq ) + 3 − 2+ − → 2Mn( aq) + 5HSO4( aq ) (b) H 3 O( aq ) + H 2 O(l ) + 2MnO4( aq ) + 5SO2( aq ) − + − → 4ClO2( g ) + Cl( aq) + 6 H 2 O(l ) (c) 5ClO2( aq ) + 4 H 3 O( aq ) (4) Nitrous acid ( HNO 2 ) disproportionates in acidic solution to nitrate ion ( ( NO3− ) a nitrogen oxide (NO). Write a balanced equation for this reaction. Answers: + 2+ 2− → 2VO( aq) + SO4( aq ) (1) (a) 2VO( aq ) + SO2( g ) − 2− + → 2 Br( aq ) + SO4( aq ) + 4 H 3 O( aq ) (b) Br2(l ) + SO2( g ) + 6 H 2 O(l ) 2− 4+ + 3+ 2+ → 2Cr( aq ) + 3NpO2( aq ) + 3H 2 O(l ) (c) Cr2 O7 aq ) + 3Np( aq ) + 2 H 3 O( aq) − + 2+ → 5CO2( g ) + 2Mn( aq ) + 14H 2O(l ) (d) 5HCOOH( aq) + 2MnO4( aq ) + 6 H 3 O( aq ) → 6 Hg (l ) + 3H 2 PO4−( aq ) + 2 AuCl4−( aq ) + 3H 2 O(l ) (e) 3Hg 2 HPO4( s ) + 2 Au ( s ) + 8Cl(−aq) + 3H 3O(+aq ) − 2− − → 2CrO4( aq) + 6 Br( aq ) + 8H 2 O(l ) (2) (a) 2Cr (OH ) 3( s ) + 3Br2( aq ) + 10OH ( aq) 2− 2− → Zr( s ) + 2SO4( aq ) + H 2 O(l ) (b) ZrO(OH ) 2( s ) + 2SO3( aq) − − − → 7 Pb( s ) + 2 Re O4( aq ) + H 2 O(l ) + 5OH ( aq ) (c) 7 HPbO2( aq ) + 2 Re ( s ) − − 4− → 3 XeO6( aq ) + Xe( g ) + 6 H 2 O(l ) (d) 4 HXeO4( aq) + 8OH ( aq) → N 2( g ) + 2CO( g ) + 4OH (−aq ) (e) N 2 H 4( aq) + 2CO32(−aq ) 2+ 3+ → Fe( aq ) + e (3) (a) Oxidation: Fe( aq ) + − − → 4 H 2 O(l ) Reduction: H 2 O2( aq ) + 2 H 3 O( aq ) + 2e − + → HSO4( aq) + 3H 3 O( aq ) + 2e (b) Oxidation: 5H 2 O(l ) + SO2( aq ) − + − − 2+ → Mn( aq ) + 12H 2 O(l ) Reduction: MnO4( aq ) + 8H 3 O( aq ) + 5e − → ClO2( g ) + e (c) Oxidation ClO2( aq ) − → Cl(−aq ) + 6 H 2 O(l ) Reduction: ClO2−( aq ) + 4 H 3 O(+aq ) + 4e − → NO3−aq ) + 2 NO( g ) + H 3 O(+aq) (4) 3HNO2 aq ) Electrochemistry 10 Oxidation – Reduction Reactions can be further classified into 4 different types as follows: 1. Combustion Reaction – Many chemical reactions that involve oxygen produce energy (heat) so rapidly that a flame results. There are many combustion reactions, most of which are used to provide heat or electricity for homes or businesses or energy for transportation. Some examples: e. g. 1) Combustion of propane (used to heat some rural home): 2) Combustion of gasoline (used to power cars and trucks): C3 H 8( g ) + 5O2( g ) → 4 H 2O( g ) + 3CO2( g ) 2C8 H18(l ) + 25O2( g ) → 18 H 2O( g ) + 16CO 2( g ) 3) Combustion of coal used to generate electricity: C ( s ) + O2 ( g ) → CO2 ( g ) 4) Many direct combination reactions are combustion reactions, but not all combustion reactions are necessarily direct combination reactions because there are two reactants that combine, but more than one product is formed. Consider the combustion reaction in which gasohol burns in a car’s engine. Gasohol is a type of fuel that is made from regular unleaded gasoline and alcohol. It is intended to burn a little more cleanly in order to release fewer pollutants. The alcohol in gasohol burns according to the equation: C2 H5OH + 3O2 → 2CO2 + 3H2 O 2. Synthesis (Combination) Reaction (reactants are elements) – This is an important reaction to produce synthetic materials: e. g. 1) Synthesis of water: 2 H 2 ( g ) + O2 ( g ) → H 2 O( l ) 2) Synthesis of carbon dioxide: C( s) + O2 ( g ) → CO2 ( g ) N 2 ( g ) + O2 ( g ) → 2 NO( g ) 3) Synthesis of nitrogen monoxide: 4) Fe + N 2 + 4 H2 + 2 S + 4 O2 → Fe( NH4 ) 2 ( SO4 ) 2 Above reactions all involve oxygen: that is why they are called oxidation – reduction reaction. These first two reactions are called combustion reactions because they produce flames. There are many synthesis reactions that do not involve oxygen: e. g. 5) Synthesis of sodium chloride: 2 Na (s) + Cl 2( g ) → 2 NaCl also (s) 6) Synthesis of magnesium fluoride: Mg( s) + F2 ( g ) → MgF2 ( s) 7) A simple molecule or element is added to another molecule: HCl + C5 H10 → C5 H11Cl 3. Decomposition Reaction (products are elements) – These are usually accomplished by heating or by the application of an electric current: → 2 H 2( g ) + O2( g ) e. g. 1) Decomposition of water: 2H 2O(l ) electric current 2) Decomposition of mercury (II) oxide: 2HgO( s) → 2Hg (l ) + O2( g ) heat 3) Decomposition of sodium chloride: 2 NaCl(l ) → 2 Na(l ) + Cl2( g ) electric current 4) Decomposition – a large molecule decomposes into elements or smaller molecules: C12 H 22O11 →12C + 11H 2O heat Examples of decomposition reactions that result in at least one compound as a product are: 5) 2 KClO3 → 2 KCl + 3O2 There are Decomposition Reactions where there is no oxidation number change (therefore cannot be called redox reactions) and are called Metathetical Reactions. e. g. 6) CaCO3 → CaO + CO2 4. Single Displacement – A single type of anion is exchanged between two cations in contrast with Double Displacement Reaction in Precipitation Reaction or in Acid-Base Neutralization Reaction. Both metals and non-metals (such as halogens) participate in Single Displacement Reactions: e. g. 1) Zn( s ) + 2 HCl( aq ) → H 2( g ) + ZnCl 2( aq ) ; 2) 2 AgNO3 + Zn → 2 Ag + Zn( NO3 ) 2 3) Cl 2 + 2 KBr → Br2 + 2 KCl Industrial important single-replacement reactions use carbon (C) to displace metals from metal oxides in the refining process. For example, Fe 2 O 3 is refined into iron in the reaction: 4) 2 Fe 2 O 3 + 3C → 4 Fe + 3CO 2 Electrochemistry 11 Activity Series of Metals (i) Very Active Metals - Li, K, Na, Rb, Cs, Ca, Sn, Ba : react with water in Single-Replacement reaction to produce H 2 (and lot of heat, so hydrogen may ignite) (ii) Active Metals - Mg, Al, Zn, Mn, Pb, Ni, Ti, Cr, Fe, Cd, Sn, Co : Do not react with water but with acids in Single-Replacement reaction to produce H 2 . (iii) Inactive Metals - Cu , Ag , Pt , Au : Do not undergo Single-Replacement reactions with water or acids. • Cu & Ag may react with conc. Nitric Acid to produce nitrogen oxides but not hydrogen; • Au may react with a mixture, called aqua regia, of three parts conc. HCl and one part conc. HNO 3 . Note: Only the active and inactive metals will displace each other from compounds. The very active metals listed above do not displace other metals. Although the very active metals are certainly reactive enough to result in such displacement, their very activity causes them to react preferentially with water instead. (iv) Nonmetals such as halogens also participate in Single Displacement reactions. Halogens have their own activity series (can be predicted based on the concept of electronegativity): Most reactive >… fluorine >chlorine > bromine > iodine .> Least reactive Examples: (i) Nonmetal to displace another nonmetal -(ii) Nonmetal to displace a metal -------------Metal – Name, Electronegativity, & symbol Activity Series of Metals The activity series lists metal elements according to their ability to react, which is used to determine which element replaces another in a single replacement reaction. An element can replace any element that is below it in the list. Although hydrogen is not a metal, it is included in the series because it acts like a metal in replacement reactions. Reducing Agents: Metals metal ions Hydrocarbons Oxidizing Agents: , may act as either oxidizing or reducing agents. 1. Potassium .…..0.8… K 2. Barium ….… 0.9. …Ba 3. Sodium ….…. 0.9… Na 4. Lithium ……...1.0…, Li 5. Strontium .…...1.0...,...Sr 6. Calcium …..…1.0…,.Ca 7. Magnesium.…1.2…, Mg 8. Aluminum .….1.5…, Al 9. Titanium …….1.5 Ti 10. Manganese …1.6…, Mn 11. Zinc ……….....1.6…,.Zn 12. Chromium 1.7 , Cr 13. Cadmium …...1.7….,Cd 14. Iron ……….…1.8…..Fe 15. Cobalt …….....1.9…..Co 16. Nickel ………1.8,…..Ni 17. Tin ………….1.8… ,.Sn 18. Lead ………...1.8.….Pb 19. Hydrogen .….2.1…...H 20. Copper …..….1.9…..Cu 21. Silver …….…1.9…. Ag 22. Mercury ……2.0….. Hg 23. Platinum …...2.3 Pt 24. Gold ……..…2.5.…..Au (an important refining process) Metal – increasing ease of oxidation from bottom to top Li K Ba Na Sr Ca Mg Al Ti Mn Zn Cr Fe Cd Co Ni Sn Pb Cu Ag Hg Pt Au SRA Metal Ion – increasing ease of reduction from top to bottom St. Reduction Potentials Metals that react with water - 3.04 Ca 2 + - 2.93 - 2.91 - 2.89 - 2.87 Na + - 2.71 Ba2+ Sr2+ Metals that react with acids nonoxidizing acids: oxidizing acids: Mg 2+ - 2.37 3+ Cr 3+ - 1.66 - 0.76 - 0.45 - 0.74 Ni 2 + - 0.26 Sn 2 + - 0.14 2+ - 0.13 Al Zn 2 + Fe 2 + Pb 0.00 Cu 2+ 0.52 Ag + Metals that are highly unreactive Hg Au 0.80 2+ 0.80 3+ 1.50 SOA Notes on Combustion Reactions (reacting with Oxygen): 1. An organic compound reacting with oxygen often results in the production of a large amount of heat, 2. Some metals particularly alkali and alkaline earth metals readily react with oxygen. 3. Magnesium burns with a bright whilte flame in and continue to burn even in an atmosphere of : . 4. When finely divided into a powder, many metals burns in . Steel wool burns when placed in a flame. Powdered metals such as aluminum are classified as highly combustible, or even explosive. 5. Iron & Steel react poorly with gaseous . 6. Platinum, gold, silver, and copper do not react with . 7. Electrochemistry 12 Most nonmetals combine as readily with oxygen as do the metals. Usually, these reactions occur rapidly enough to be described as combustion. The most important reaction of this type is the combustion of carbon as a source of heat. If plenty of oxygen is available, the combustion product is but when the supply of oxygen is limited, can be formed as well. Common Oxidizing Agents: (1) Reactions of the Permanganate Ion MnO4− is a very versatile oxidizing agent. It reacts differently in acidic, neutral, and basic solutions. (a) In acidic solutions the half-reaction is: MnO 4− + 8 H + + 5e − → Mn 2 + + 4 H 2 O 5 electrons are used in this reduction, and the soluble Mn 2 + ion is colorless. In addition, this reaction is slow and the presence of Mn 2+ ions catalyzes the process. (b) In neutral or slightly acidic solutions the permanganate half-reaction is: MnO 4− + 4 H + + 3e − → MnO 2 + 2 H 2 O The hydrogen ions come from the dissolution of water molecules. The product MnO 2 is an insoluble black precipitate. (c) In basic solutions the reaction involves a 1 electron transfer: MnO4− + e − → MnO42− The reaction mixture changes color from the deep violet of the MnO4− ion to green for the MnO 42 − ion. (2) Reactions of Chromium (VI) When chromium is in the +6 oxidation state, it forms different compounds depending on the pH of the solution. (a) In neutral or basic solutions the chromate ion, CrO42 − , predominates. (b) In acidic solutions the dichromate ion predominates: 2CrO 42− + 2 H + → Cr2 O72− + H 2 O (c) In very acid solutions the dichromate ion becomes protonated to form chromic acid: Cr2 O 72 − + 2 H + + H 2 O → 2 H 2 CrO 4 (d) As the acidity of a solution increases, the strength of chromium (VI) as an oxidizing agent increases. Chromate salts are dissolved in concentrated sulfuric acid to produce chromic acid, H 2 CrO 4 , which is very effective cleaning agent because it oxidize most organic materials. Or example, since the dichromate ion is orange and the Cr 3 + ion is green, the reaction of dichromate with ethyl alcohol I producing Cr 3 + ion is used to test the sobriety of drivers: 8 H + + Cr2 O 72 − + 3CH 3 CH 2 OH → 2Cr 3+ + 3CH 3 CHO + 7 H 2 O (3) Iodine Iodine – is a weak oxidizing agent. A dilute solution of iodine in alcohol, known as tincture of iodine, is an effective antiseptic for minor wounds. Because it is a weak oxidizing agent, iodine can be used in reaction mixtures when chemists want to oxidize only the more active reducing agents that are present. (4) Hydrogen Peroxide, - is another weak oxidizing agent. Its mode of action is to decompose into water and atomic oxygen (O). Atomic oxygen normally combines to form molecular . However, in the brief time that is available, atomic oxygen acts as a very good oxidizing agent if it encounters a suitable reactant. Hydrogen peroxide is used, in 3% solutions, as a household disinfectant and hair bleach. In higher concentrations, 30%, it is a very powerful oxidizing agent that must be handled with care. In fact, higher concentrations of have been used as rocket fuel because of the oxygen this compound supplies. (5) If a positively charged metal ion reacts, then it is usually converted to a metal atom by gaining back an electron(s). hence the metal ion is behaving as an oxidizing agent. Common Reducing Agents: (1) The sulfite ion ( or the bisulfate ion ( oxidized to sulfate ion: is a very potent reducing agent, and when it reacts with an oxidizing agent it is (2) Thiosulfate ion, - is a reducing agent, one of very few that is stable in air because it reacts slowly with . Thiosulfate solutions are used as reducing agents in chemical analysis, often with iodine as the oxidizing agent. Chlorine is itself a powerful oxidizing agent. It is a gas having an irritating odor and in even relatively small concentrations can cause lung damage if inhaled. Therefore it is wise to prevent it from escaping into the air. Carrying out the reaction in an apparatus that directs any gases into a solution containg thiosulfite ion will remove chlorine from them by the reaction: . Electrochemistry 13 (3) Another weak acid that serves as a common reducing agent for reactions in aqueous solutions is oxalic acid, . Oxidation of oxalic acid in an acidic solution produces carbon dioxide: of its salt, or in the form (4) Tin is often used as a reducing agent, it is oxidized to tin (IV): . Active metals such as zinc and magnesium can also serve as reducing agents in chemical reactions. In introductory chemistry, the reactivity series or activity series is an empirical series of metals, in order of "reactivity" from highest to lowest. It is used to summarize information about the reactions of metals with acids and water, single displacement reactions and the extraction of metals from their ores. Going from bottom to top, the metals: • increase in reactivity; • lose electrons more readily to form positive ions; • corrode or tarnish more readily; • require more energy (and different methods) to be separated from their ores; • become stronger reducing agents. K K+ Na Na+ Li Li+ Sr Sr Ca Ca electrolysis 2+ Mg Mg2+ Al Al3+ C react with water 2+ react with acids included for comparison Zn Zn2+ Cr Cr2+ There is no unique and fully consistent way to define the reactivity series, but it is common to use the three types of reaction listed below, many of which can be performed in a high-school laboratory (at least as demonstrations). Reaction with water and acids The most reactive metals (for example, sodium) will react with cold water to produce hydrogen and the metal hydroxide: 2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g) Fe Fe2+ Cd Cd2+ Co Co2+ Sn Sn2+ Pb Pb2+ There is some ambiguity at the borderlines between the groups. Magnesium, aluminium and zinc can react with water, but the reaction is usually very slow unless the metal samples are specially prepared to remove the surface layer of oxide which protects the rest of the metal. Copper and silver will react with nitric acid, but not by the simple equation shown for iron. 2+ An iron nail placed in a solution of copper sulfate will quickly change colour as metallic copper is deposited. The iron is converted into iron(II) sulfate: Fe (s) + CuSO4 (aq) → Cu (s) + FeSO4 (aq) In general, a metal can displace any of the metals which are lower in the reactivity series: the higher metal reduces the ions of the lower metal. This is used in the thermite reaction for preparing small quantities of metallic iron, and in the Kroll process for preparing titanium (Ti comes at about the same level as Al in the reactivity series). Al (s) + Fe2O3 (s) → Fe (s) + Al2O3 (s) 2Mg (s) + TiCl4 (l) → Ti (s) + 2MgCl2 (s) However, other factors can come into play, as in the preparation of metallic potassium by the reduction of potassium chloride with sodium at 850 ºC: although sodium is lower than potassium in the reactivity series, the reaction can proceed because potassium is more volatile, and is preferentially distilled off from the mixture. smelting with coke Ni Ni2+ Metals in the middle of the reactivity series (for example, iron) will react with acids, but not water, to give hydrogen and a metal salt: Fe (s) + H2SO4 (aq) → FeSO4 (aq) + H2 (g) Single displacement reactions react with acids H2 H+ included for comparison Cu Cu Ag Ag+ Hg Hg2+ Au Au3+ Pt Pt2+ heat or highly unreactive physical extraction Electrochemistry 14 Use the C11-Redox Table to identify the strongest oxidizing and reducing agents in a mixture and then predict which reactions will occur: 1. Choose the strongest oxidizing agent present in the mixture by starting at the top left corner of the redox table and going down the list until the oxidizing agent in the mixture is found. 2. Choose the strongest reducing agent in the mixture by starting at the bottom right corner of the redox table and going up the list until the reducing agent in the mixture is found. Electrochemistry 15 3. Reduction half –reaction equations are read from left to right (following the forward arrow). 4. Oxidation half-reaction equations are read from right to left (following the reverse arrow). 5. Any substances not present in the redox table will be assumed to be spectator ions and need not be considered. Electrochemistry 16 Electrochemistry 17 Redox Titration Redox titrations have the special advantage that many involve intensely colored species that show a striking color change at the endpoint. For example, MnO4− is deep purple in color, while Mn 2 + is colorless. Thus when MnO4− has been added to Fe 2 + is very slight excess, the color of the solution changes permanently to purple. ← Molecular Equation: 10 FeSO 4 + 2 KMnO 4 + 8 H 2 SO 4 → 5 Fe 2 ( SO 4 ) 3 + 2 MnSO 4 + K 2 SO 4 + 8 H 2 O Net Ionic Equation: ← 5Fe 2 + + MnO4− + 8H + → 5 Fe 3+ + Mn 2+ + 4 H 2 O Many redox titration procedures are indirect and involve additional preliminary reactions of the sample before the titration can be carried out. For example, a soluble calcium salt will not take part in a redox reaction with potassium permanganate. (i) However, adding ammonium oxalate to the solution containing Ca 2 + causes the quantitative precipitation of calcium oxalate: Ca(2aq) + C 2 O42(−aq) → CaC2 O4( s ) (ii) After the precipitate is filtered and washed, it is dissolved in sulfuric acid to form oxalic acid: CaC 2 O 4 ( s ) + 2 H (+aq ) → Ca (2aq+ ) + H 2 C 2 O( aq ) (iii) Finally, the oxalic acid is titrated with permanganate solution of accurately known concentration, taking advantage of the redox reaction: + 2 MnO4−( aq ) + 5 H 2 C 2 O4( aq ) + 6 H aq → 2 Mn(2aq+ ) + 10CO2 ( g ) + 8 H 2 O( l ) ) In this way, the quantity of calcium can be determined indirectly by reactions that involve precipitation, acid-base, and redox steps. Electrochemistry 18 IV) (A).Reactions Producing Covalent Compound from Ionic Reactants – many of these compounds are organic acids (acetic, formic, benzoic acids) or gases such as H 2 S , NH 3 , SO 2 , and CO2 : → 2 NaCl ( aq ) + H 2 S ( g ) e. g. 1) Na 2 S ( aq ) + 2 HCl ( aq ) → 2 KNO3( aq.) + H 2 O (l ) + SO2( g ) 2) K 2 SO3 + 2 HNO3 3) NaHCO 3 + HCl → NaCl + H 2CO 3 ⇒ H 2 O + CO 2 ( g ) unstable 4) NaHSO 3 + HCl → NaCl + H 2SO 3 ⇒ H 2 O + SO 2 ( g ) unstable 5) ( NH 4 ) 2 SO 4 + NaOH → NaSO 4 + NH 4+ + OH − ⇒ H 2 O + NH 3( g ) unstable 6) KC 2 H 3 O2 + HCl → HC 2 H 3O 2 + KCl --------------- a covalent compound formed (B). Other Types of Reactions Producing Gas: 1) When heated, some solids decompose to give gaseous products (Decomposition-metathetical): (i) 2 HgO( s ) heat → 2 Hg (l ) + O2( g ) Electrochemistry 19 → CaO( s ) + CO2( g ) - Marble consists primarily of calcium carbonate which decomposes to give (ii) CaCO3( s ) heat quicklime. (iii) NH 4 Cl ( s ) → NH 3 ( g ) + HCl ( g ) heat 2) Some gas reactions proceed explosively (Decomposition): (i) 4C 3 H 5 ( NO3 ) 3 (l ) → 2 N 2 (l ) + 12CO2( g ) + O2( g ) + 10H 2 O( g ) 3) Several elements react with oxygen to form gaseous oxides (Synthesis): (i) S ( s ) + O 2 ( g ) → SO2 ( g ) (ii) 2 SO2 ( g ) + O2( g ) → 2SO3( g ) (iii) N 2( g ) + O2 ( g ) → 2 NO( g ) (iv) 2 NO ( g ) + O2 ( g ) → 2 NO 2 ( g ) 4) Many chemical reactions that involve oxygen produce energy (heat) so rapidly that a flame results (Combustion): → CO2( g ) + 2 H 2 O( g ) (i) CH 4( g ) + 2O2( g ) 5) The actions of acids on certain ionic solids are another important class of reactions that form gases (Double Displacement): (i) CaCO 3 ( s ) + 2 HCl ( g ) → CaCl 2 ( s ) + H 2 CO 3 ⇒ CO 2 ( g ) + H 2 O ( l ) unstable Electrochemistry 20 A) Application of Redox Reactions - Oxidation-Reduction reactions are the basis for processes that are an absolute necessity for life itself. Some important examples: i) fossil fuel + O2 → CO2 + H 2 O + energy . ii) Redox reactions, with oxygen as the oxidizing agent, are also the source of energy for a living organism. For humans, the substances oxidized are our foods: carbohydrates, fats, and proteins. These oxidations, which are multistep in nature, occur at a very slow, controlled rate, allowing cells to trap and use the released energy. The blood supplies the required oxygen. The blood returns the waste product carbon dioxide to the lungs to be exhaled. a ) carbohydrate + O2 → CO2 + H2 O + energy b) fat + O2 → CO2 + H2 O + energy c) protein + O2 → CO2 + H2 O + energy + nitrogen − containing products iii) Photosynthesis is another redox process that is of prime importance to humans: glucose − a building + 6O 2 6CO 2 + 6 H 2 O + solar energy → C 6 H 12 O6 block of carbohydra tes B) Other Applications of Redox Reactions: iv) Corrosion: The combination of iron (II) ions and hydroxide ions forms a low-solubility precipitate of iron (II) hydroxide, which is further oxidized by oxygen and water to form iron (III) hydroxide, a yellow-brown solid. The familiar red-brown rust is formed by the dehydration of iron (III) hydroxide to form a mixture of iron (III) hydroxide and hydrated iron (III) oxide. The amount of the hydroxide and the oxide . varies, so rust is referred to as a hydrated oxide of indeterminate formula, Preventing Corrosion – Corrosion such as rusting of iron caused by the oxidation into iron oxide ( Fe2 O3 ) can be prevented by coating the metal with oil, paint, plastic, or another metal (tin or zinc). Tin adheres well to the iron and the outer surface of the tin coating has a thin, strongly adhering layer of tin oxide that protects the tin, which in turn protect the iron. Iron is more easily oxidized than tin , but zinc is more easily oxidized than iron and therefore zinc plating of steel or iron can provide double protection – a protective layer and preferential corrosion of the zinc. Corrosion is also prevented by electro-chemical processes (cathodic protection) that are based on the fact that when two metals are in contact, the metal that is more readily oxidized will corrode in preference to the other metal. For example, magnesium or zinc attached to iron can prevent the iron from corroding. When oxygen and moisture attack the iron, electrons flow from the magnesium to the iron, causing the magnesium to corrode in preference to the iron. v) Bleaching – Bleaches are chemical substances that are used to eliminate unwanted colour from fabrics and other materials. Colour is caused by the movements of electrons between different energy levels of the atoms of the materials. Consequently, the colour of a material can often be eliminated by removing these electrons by the process of oxidation. Thus bleaches are oxidizing agents. Three of the most common commercial and household bleaches are chlorine ( Cl2 ), hypochlorite ion ( ClO− ), and hydrogen peroxide ( H2 O2 ). NaClO is the active compound in common liquid household bleaches. Solid chlorine bleaches contain Ca (ClO) 2 . In both compounds, the hypochlorite ion is the oxidizing agent that causes bleaching. Hydrogen peroxide is used as a mild antiseptic. It kills bacteria by oxidizing them. vi) Fuels and Explosives – Fuels release energy as they are oxidized. Explosives contain both oxidizing and reducing agents, often within the same compound. For example, nitroglycerin ( C3 H5 ( NO3 ) 3 ) contains C and H atoms that are oxidized to form CO2 and H2 O , as well as N atoms that are reduced to form N 2 . The C and H atoms are the reducing agents and the NO3 groups are the oxidizing agents. When suitably activated, a highly exothermic redox reaction occurs in which large amounts of gaseous products form. The rapid formation and expansion of the gas creates the shock wave that accompanies the detonation of an explosive. Nitroglycerin, which can be activated merely by shaking, decomposes as follows: 4C3 H5 ( NO3 ) 3( l ) → 6 N 2( g ) + 12O2 ( g ) + 10 H2O( g ) . vii) Photography – Photography involves the capturing of a light image on a light-sensitive medium and the processing of the image to make a permanent record. These processes are based on the redox reactions of silver halides. The films is an emulsion of silver bromide ( AgBr ) deposited onto a plastic backing. The silver bromide grains in the emulsion are highly light sensitive. A single light photon can activate a grain by initially oxidizing a bromide ion and subsequently reducing a silver ion: (a) oxidation: , (b) reduction : The activated grains represent the captured image, although at this stage you would not be able to see the image. As the film is being +1 processed, the remaining Ag ions in the activated grains are first converted to free silver (Ag) by a reducing agent. Next, the un-reacted Electrochemistry 21 AgBr in the non-activated grains is removed by an appropriate solution process. At this stage you would recognize the film as a negative. The final step involves the printing of the negative onto photographic paper. This process also involves redox reactions. Electrochemistry 22 C) Another Major Application - Electrochemistry Devices that use redox reactions to either produce or use electricity are called electrochemical cells. Electrons enter and exit electrochemical cells with the help of electrodes. Electrodes are made of electrical conductors such as metals or graphite, and they provide the surfaces on which oxidation and reduction occur with sites of oxidation and reduction separated. There are two types of electrochemical cells: (I) Voltaic Cells (or simply called electric cell) – those that produce electricity as the result of spontaneous redox reactions. A battery consists of one or more voltaic cells. For example a common flashlight battery consists of single voltaic cells that generate 1.5 volts, whereas a 12-volts automobile battery consists of six voltaic cells that each generates 2 volts. Electric cells adapted for scientific study are often called galvanic cells. (II) Electrolytic Cells – those in which electrical energy drives non-spontaneous redox reactions. Electrolytic cells are used to plate precious metals onto jewelry, to purify metals, and to obtain active metals from their compounds. For example, both aluminum and magnesium metals are obtained from their compounds using electrolytic cells. Cell Potential (cell voltage) – The ability of a cell reaction to move electrons through a wire from one electrode of a voltaic cell to another is described by a quantity called the electrical potential of the cell or merely the cell potential. The cell potential represents the difference in electrical potential energy between the two electrodes. It is measured in volts. Standard Cell Potential – The magnitude of a voltaic cell’s potential depends largely on the nature of the cell reaction. Each electrode of a voltaic cell makes a characteristic contribution to the cell potential: E cell = E oxidation + E reduction . Standard electrode potentials and standard cell potentials refer to those potentials measured under the standard conditions of 1 atmosphere, 25o C , and the concentration of solutions of 1M. The electrode potentials due to isolated oxidation and reduction half reactions cannot be directly measured. Only the difference between the two electrode potentials can be measured. So relative electrode potentials are obtained by measuring the different half-cell electrode potentials using the standard hydrogen electrode as the same reference half-cell electrode and then combining them into cell potentials. The standard hydrogen electrode reaction is the reduction of H + to form H2 . This reduction reaction is assigned a standard reduction potential of exactly zero: 2 H + ( aq , 1 M ) + 2e − → H2 ( g , 1atm) E 0 = 0.00V The standard reduction and oxidation potentials for zinc are measured as follows: Oxidation − anode ( − ): + − Zn( s) → Zn(2aq ) + 2e Re duction − cathode ( + ): 2 H + ( aq ,1 M ) + 2e − → H2( g ,1atm) overall reaction: + Zn( s) + 2 H + ( aq ,1 M ) → Zn(2aq ) + H2 ( g ,1atm) Reduction potentials for other substances are determined in a similar way. Once the standard reduction potentials are determined, then oxidation potentials are obtained by reversing the sign. Standard electrode potentials are useful in a variety of ways, two of which are: (i) Determining ease of oxidation and reduction of a substance - The more negative the standard reduction potential, the harder the species is to reduce. (ii) Calculating cell potentials using the equation: Another way of defining the standard cell potential by using the standard reduction potential ability of a standard half-cell to attract electrons, thus undergoing reduction: which represents the where Rules for Analyzing Standard Cells given the contents of the Cell: 1. The cathode is the electrode where the strongest oxidizing agent present in the cell reacts, i.e., the oxidizing agent in the cell that is closest to the top on the left side of the redox table C11. If required, copy the reduction half-reaction for the strongest oxidizing agent and its reduction potential from the table. 2. The anode is the electrode where the strongest reducing agent present in the cell reacts, i.e., the reducing agent in the cell that is closest to the bottom on the right side of the redox table C11. If required, copy the oxidation half-reaction (reverse the half-reaction by reading from right to left) for the strongest reducing agent and the reduction potential listed in the table. 3. Balance the electrons for the two half-reaction equations (but do not change the ) and add the half-reaction equations to obtain the overall or net cell reaction. Electrochemistry 23 4. Calculate the standard cell potential, . From thermodynamics, = (maximum work). In an electrical system, work is supplied by the electric current that is pushed along by the potential of the cell. It can be calculated from the equation: maximum work = (joules). So, . where is the Faraday constant ( , so the quantity of charge of 1 mol of electrons is )= From page 105, the Faraday constant. . So, This relationship allows equilibrium constants to be calculated from standard cell voltages, as illustrated by the following example 12.11: When all of the ion concentrations in a cell are one molar, the cell potential is equal to the standard potential. When the concentrations change, however, so does the potential. For example, in an operating cell or battery, the potential gradually drops as the reactants are consumed. The cell approaches equilibrium, and when it gets there the potential has dropped to zero – the battery is dead.the effect of concentration on the cell potential can be calculated from the Nernst equation: where is the cell potential at is the cell potentiall at and non-standard concentrations and standard concentrations (1 mol/L) n is the amount, in moles, of electrons transferred according to the cell reaction Q is the reaction quotient. Standard cell potentials can be determined readily using standard reduction potentials. If the concentrations are not standard, the cell potential can be calculated using the Nernst equation as illustrated in above example 12.10. Electrochemistry 24 Measuring Standard Reduction Potential: In order to assign values for standard reduction potentials, the “reducing” strength of all possible half-cells is measured against an accepted, standard half-cell. The standard half-cell used for this purpose is the standard hydrogen half-cell, whose electrode potential is defined to be exactly zero volts. The standard hydrogen half-cell consists of an inert platinum electrode immersed in a 1.00 mol/L solution of hydrogen ion, with hydrogen gas at a pressure of 100 kPa bubbling over the electrode. Examples of Galvanic Cells Set Up: Electrochemistry 25 Example 1: Example 2: illustrated by figure on next page Example 3: An acidic dichromate solution is a strong oxidizing agent that reacts spontaneously with copper metal. So carbon is used as the inert cathode electrode that will not interfere with the desired cell reaction. The carbon electrode (cathode) remains unchanged, but the orange color of the dichromate solution becomes less intense and more yellow, evidence that reduction is occurring in this half cell. Electrochemistry 26 (I) Voltaic Cell Setup Voltaic Cell Setup Oxidation − anode ( −): + − Zn( s) → Zn(2aq ) + 2e + − Re duction − cathode ( +): Cu(2aq ) + 2e → Cu( s) Overall reaction: Zn( s) + CuSO4( aq ) → ZnSO4( aq ) + Cu( s) The zinc strip in ZnSO4( aq ) , and the copper strip in CuSO4( aq ) are placed in separate compartments to permit only indirect electron transfers through the wire from the anode to the cathode. In this setup, the voltaic cell consists of two half-cells. .A tube, called the salt bridge, containing an electrolyte such as Na 2 SO4( aq ) allows ions to move from one compartment to another but prevents the solutions from mixing totally. This movement of ions keeps the number of positive and negative ions equal in each half-cell, thus completing the electrical circuit. Otherwise, the charge will start to accumulate in each half-cell and the flow of electrons will cease. Other devices such as a porous clay barrier can be used instead of the salt bridge. Electrochemistry 27 Electrochemistry 28 Common Batteries: Non-Rechargeable (Primary Cell) Batteries (1) Common Dry Cell (2) Alkaline Dry Cell + − Zn( s) → Zn(2aq ) + 2e Oxidation − anode ( − ): Zn( s) + 2OH − → ZnO( s) + H2 O( l ) + 2e − Oxidation − anode ( − ): Re duction − cathode ( + ): 2 MnO2( s) + 2 NH4+( aq ) + 2e − 2 MnO2 ( s) + H2 O( l ) + 2e − Re duction − cathode ( + ): → Mn2 O3( s) + 2OH − → Mn2 O3( s) + 2 NH3( aq ) + H2 O( l ) overall reaction: Zn( s) + 2 MnO2( s) + 2 NH4 Cl( aq ) → + Zn(2aq ) Zn( s) + 2 MnO2 ( s) → ZnO( s) + Mn2 O3( s) overall reaction: − + 2Cl + Mn2 O3( s) + 2 NH3( aq ) Advantages: (i) low cost Disadvantages: (i) not re-chargeable; (ii) voltage drops if current runs rapidly; (iii) zinc reacts with ammonium ions slightly, causing cell to run down. Advantage: (i) Longer shelf life because zinc does react with NaOH as readily as with acidic NH4 Cl ; (ii) maintains a steady voltage even under high current loads; (iii) generates 50% more total energy for same size. Disadvantages: (i) higher cost due to more complicated design. Re-chargeable (Secondary Cell) Batteries (1) Lead Storage Battery Oxidation − anode ( − ): (2) Nickel-Cadmium Battery Pb ( s) + SO42(−aq ) → PbSO4 ( s) + 2 e − Oxidation − anode ( − ): Cd ( s) + 2OH − → Cd (OH ) 2( s) + 2e − Re duction − cathode ( + ): PbO2 ( s) + 4 H (+aq ) + SO42(−aq ) + 2 e − → PbSO4 ( s) + 2 H 2 O( l ) Re duction − cathode ( + ): NiO2( s) + 2 H2 O( l ) + 2e − overall reaction: Pb ( s) + PbO2 ( s) + 2 H 2 SO4 ( s) → 2 PbSO4 ( s) + 2 H 2 O( l ) overall reaction: Cd ( s) + NiO2 ( s) + 2 H2O( l ) → Cd ( OH ) 2( s) + Ni ( OH ) 2 ( s) Re-charge: → Ni (OH ) 2 ( s) + 2OH − ( aq ) Pb ( s) + PbO2( s) + 4 H(+aq ) + 2 SO42(−aq ) Like the lead storage battery, the products formed in the electrode reactions are insoluble and adhere to the electrode surfaces, thus permitting recharging discharge → ← 2 PbSO4 ( s) + 2 H2 O( l ) charge The lead sulfate formed at anode and cathode during discharge is insoluble and adheres to the electrode surfaces. This makes it possible to reverse the reactions when the battery is re-charged. Advantages: (i) simple, cheap, and reliable Disadvantages: (i) heavy with 15 to 20 kg mass; (ii) lead is toxic Advantages: (i) lightweight; (ii) produces a constant voltage during discharge. Disadvantages: (i) cadmium is toxic; (ii) If it is discharged only partially and then re-charged, it develops the tendency to need recharging after only a short use. Fuel Cells A Fuel Cell is a Voltaic Cell in which a fuel is continuously supplied from an external reservoir to the cell. The most common one is the hydrogen-oxygen fuel cell, which is used in space shuttle. Oxidation − anode ( − ): 2H 2 ( g ) + 4OH − ( aq ) → 2 H2O( l ) + 4e − Re duction − cathode ( + ): O2( g ) + 2 H2O( l ) + 4e − → 4OH − ( aq ) overall reaction: 2 H2 ( g ) + O2( g ) → 2 H2O( l ) − The cell contains concentrated KOH, which supplies OH ions that participate in the electrode reactions. Advantages: (i) Because the fuel is supplied continuously, fuel cells do not run down like ordinary batteries. (ii) Very efficient – about 90% - at converting chemical energy into electrical energy. Electrochemistry 29 Electrolysis for non-spontaneous redox reaction in electrolytic cells: If a cell does not contain all oxidized and reduced species shown in the half-reaction equation, it is possible that the reactants (electrodes and electrolyte) present will not react spontaneously. For example, if lead electrodes are placed in a solution of zinc sulfate and the electrodes are connected with a wire, nothing happens, --------------------------------------- Since the , it is concluded that the lead will not be oxidized spontaneously in the zinc sulfate solution. Electrochemistry 30 (II) Electrolytic Cells – The process by which electricity in an electrolytic cell is used to bring about a non-spontaneous chemical change is called electrolysis. An electrolytic cell consists of a source of direct electrical current that is connected to the electrodes, which are immersed in the reactants. Inert electrodes are often used because they do not participate in the redox reactions themselves, but merely serve as surfaces on which the redox reactions take place. (1) Down’s Cell - Electrolysis of Molten NaCl (In the molten state, the Na + and Cl − ions are mobile. Oxidation − anode ( − ): 2Cl − ( l ) → Cl2 ( g ) + 2e − Re duction − cathode ( + ): 2 Na + ( l ) + 2e − → 2 Na( l ) overall reaction: 2 NaCl ( l ) → 2Na (l) + Cl2 ( g ) The electrolysis of molten NaCl is the most practical means for obtaining metallic Na. The use of chemical reducing agents is (2) Electrolysis of Brine (Aqueous NaCl ) – the primary way to produce hydrogen, chlorine and sodium hydroxide. Oxidation − anode ( − ): 2Cl −( l ) → Cl2( g ) + 2e− Re duction − cathode ( + ): 2 H2O( l ) + 2e− → 2 H2( g ) + 2OH(−aq ) overall reaction: 2Cl − ( aq ) + 2 H2O( l ) → Cl2( g ) + 2 H2( g ) + 2OH(−aq ) Na + is the spectator ion: 2 NaCl( aq ) + 2 H2O( l ) → H2 ( g ) + 2 NaOH Electrochemistry 31 impractical because of sodium’s extremely high reactivity. The Na + ions are not reduced because Na is extremely reactive. (3) Electrolysis of (4) Electroplating – the use of electrolysis to deposit a thin coating of metal on an object, which could be decorative or to protect the metal underneath from corrosion. Oxidation − anode ( −): H 2O 2 H2O(l ) → 4 H + (aq) + O2( g) + 4e− Re duction − cathode ( +): 4 H2O(l ) + 4e− → 2 H2( g ) + 4OH(−aq ) overall reaction: 6 H2O(l ) → 2 H2( g) + O2( g) + 4 H + (aq ) + 4OH(−aq) OH − ions and H + ions form around the cathode and anode respectively. They migrate through the solution and neutralize each other, forming water. So the net reaction becomes: 2 H2 O( l ) → 2 H2 ( g ) + O2 ( g ) Adding a small amount of an electrolyte such as Na2 SO4 or H2 SO4 to the water increases the current flow through the solution and increases the rate of electrolysis reaction. For example to put a silver coating on a spoon, the spoon is made the cathode (+), and a silver bar is made the anode (-): When a power source is turned on, silver is oxidized at the anode: Oxidation − anode ( −): Ag ( s ) → Ag + + e − Silver ions the migrate through the solution to the cathode where they are reduced, forming a silver plate on the spoon: Re duction − cathode ( + ): Ag + ( aq ) + e − → Ag ( s) Voltage applied to the cell is controlled at a low value so to prevent oxidation and reduction of water at the electrodes. Procedure for Analysing Electrolytic Cells: Step 1 – Use the cell notation as a list or make a list of all substances present. Label all possible oxidizing and reducing agents present, including water for aqueous electrolytes. Step 2 – Use the Redox Table C11 to identify the strongest oxiding agent present in the cell. Write the equation for its reduction half reaction, including the reduction potential (this is the reaction at the cathode). Electrochemical and Electrolytic Cells—Summary In BOTH Cells: Losing Electrons Oxidation at the Anode Electrochemical Cells Gaining Electrons Reduction at the Cathode Electrolytic Cells Electrochemistry 32 Higher Half-Rx. on Table Is the Cathode Anode is + Cathode is – SPONTANEOUS Eo is Positive + NON-SPONTANEOUS Eo is Negative At the Cathode: At the Cathode: Reduction of the Cation in the beaker. Reduction of Cation in the Electrolyte Eg. Cu2+ + 2e- Cu Eg. Cu2+ + 2e- Cu or Reduction of Water: H2O +2e- H2 + 2OHAt the Anode: Oxidation of the Metal Anode Electrode Eg. Pb Pb2+ + 2e- At the Anode: Oxidation of the Anion in Solution: Eg. 2Cl- Cl2 + 2e- or Oxidation of Water: + H2O ½ O2 + 2H + 2e or Oxidation of the Metal Anode Electrode: Eg. Cu Cu2+ + 2e(whichever has the highest oxidation potential ie. lowest on the right side of table) Electrons go from A C in the wire External power supply pushes e-‘s on to the Cathode, making it – and takes e-‘s from the Anode making it + Cations move toward Cathode Anions move toward Anode Cations(+) attracted to the – Cathode where they (or water) are reduced. Anions(-) attracted to the + Anode, where they, water or the anode is oxidized In the Salt Bridge Applications include: Automobile (Pb/Acid) Applications include: Electrolysis to Battery, Zn/C (Dry) Cells, Alkaline Cells, decompose compounds into elements, Fuel Cells Downs Cell, Electrorefining, Electroplating, Electrowinning (eg. Production of Al) Electrochemistry 33 Electrolytic Cells - Voltaic cells are driven by a spontaneous chemical reaction that produces an electric current through an outside circuit. These cells are important because they are the basis for the batteries that fuel modern society. But they aren't the only kind of electrochemical cell. The reverse reaction in each case is non-spontaneous and requires electrical energy to occur. The general form of the reaction can be written as: Spontaneous ----------> Reactants ⟶ Products + Electrical Energy <----------- Non spontaneous It is possible to construct a cell that does work on a chemical system by driving an electric current through the system. These cells are called electrolytic cells, and operate through electrolysis. Electrolysis is used to drive an oxidation-reduction reaction in a direction in which it does not occur spontaneously by driving an electric current through the system while doing work on the chemical system itself, and therefore is non-spontaneous. Electrolytic cells, like galvanic cells, are composed of two half-cells--one is a reduction half-cell, the other is an oxidation half-cell. The direction of electron flow in electrolytic cells, however, may be reversed from the direction of spontaneous electron flow in galvanic cells, but the definition of both cathode and anode remain the same, where reduction takes place at the cathode and oxidation occurs at the anode. Because the directions of both half-reactions have been reversed, the sign, but not the magnitude, of the cell potential has been reversed. Similarities and Differences Galvanic and Electrolytic Cell: Electrolytic cells are very similar to voltaic (galvanic) cells in the sense that both require a salt bridge, both have a cathode and anode side, and both have a consistent flow of electrons from the anode to the cathode. However, there are also striking differences between the two cells. The main differences are outlined below: Differences between a Galvanic cell and an Electrolytic cell Electrochemical cell (Galvanic Cell) A Galvanic cell converts chemical energy into electrical energy. Electrolytic cell An electrolytic cell converts electrical energy into chemical energy. Here, the redox reaction is spontaneous and is responsible for the The redox reaction is not spontaneous and electrical energy has production of electrical energy. to be supplied to initiate the reaction. The two half-cells are set up in different containers, being connected Both the electrodes are placed in a same container in the solution through the salt bridge or porous partition. of molten electrolyte. Here, the anode is positive and cathode is the negative electrode. Here the anode is negative and cathode is the positive electrode. The The reaction at the anode is oxidation and that at the cathode is reaction at the anode is oxidation and that at the cathode is reduction. reduction. The electrons are supplied by the species getting oxidized. They The external battery supplies the electrons. They enter through Electrochemistry 34 Electrochemical cell (Galvanic Cell) Electrolytic cell move from anode to the cathode in the external circuit. the cathode and come out through the anode. What are the differences between an electrochemical cell and an electrolytic cell? electrochemical cell electrolytic cell Produces electricity Consumes electricity Spontaneous Nonspontaneous Salt bridge Two beakers No salt bridge One beaker Metal Electrodes Inert C or Pt Electrodes What are the similarities between an electrochemical cell and an electrolytic cell? electrochemical cell or electrolytic cell The Anode is the site of oxidation and the cathode is the site of reduction. Anions migrate to the anion and cations migrate to the cathode. Both have redox reactions. Electrons flow through a wire from the anode to the cathode Electrolytic Cell - To explain what happens in an electrolytic cell let us examine the decomposition of molten sodium chloride into sodium metal and chlorine gas. The reaction is written below. ---------> Non spontaneous ( electrolytic cell ) 2 Na Cl (l) 2 Na (s) ⟶ + Cl2 (g) <--------- Spontaneous ( electrochemical cell ) Water is capable of undergoing both oxidation: H2O → O2(g) + 4 H+ + 2 e– E° = -1.23 v and reduction: 2 H2O + 2 e– → H2(g) + 2 OH– E° = -0.83 v Thus if an aqueous solution is subjected to electrolysis, one or both of the above reactions may be able to compete with the electrolysis of the solute. For example, if we try to electrolyze a solution of sodium chloride, hydrogen is produced at the cathode instead of sodium: cathode: H2O + 2 e– → H2(g) + 2 OH– anode: Cl– → ½ Cl2(g) + e– net: – Cl + H2O → 2 H2(g) + ½ Cl2(g) + 2 OH E =+0.41 v ([OH–] = 10-7 M) E° = –1.36 v – E = –0.95 v If molten NaCl (l) is placed into the container and inert electrodes of C(s) are inserted , attached to the + and - terminals of a battery, an electrolytic reaction will occur. Electrochemistry 35 || 1. Electrons from the negative terminal travel to the cathode and are used to reduce sodium ions into sodium atoms. The sodium will plate onto the cathode as it forms. The sodium ion are migrating towards the cathode. Na + Na 0 (s) + e- ⟶ 2. The negative Chlorine ions migrate towards the anode and release electrons as they oxidize to form chlorine atoms. The chlorine atoms will combine together to form chlorine gas which will bubble away. 2 Cl - ⟶ Cl2 (g) + e- 3. Note that the site of oxidation is still the anode and the site of reduction is still the cathode, but the charge on these two electrodes are reversed. Anode is now + charged and the cathode has a - charged. 4. The conditions under which the electrolyte cell operates are very important. The substance that is the strongest reducing agent (the substance with the highest standard cell potential value in the table) will undergo oxidation. The substance that is the strongest oxidizing agent will be reduced. If a solution of sodium chloride (containing water) was used in the above system, hydrogen would undergo reduction instead of sodium, because it is a stronger reducing agent than sodium. Electrochemistry 36 Predicting Electrolysis Reaction There are four factors that determine whether or not electrolysis will take place even if the external voltage exceeds the calculated amount: 1) An overpotential or voltage excess is sometimes needed to overcome interactions at the electrode surface. This case happens more frequently with gases. Ex1) H2 (g) : 1.5 V overpotential Pt (s) : 0 V overpotential 2)There might be more than one electrode reaction that occurs meaning that there may be more than one half-reaction leaving two or more possibilities for the cell reaction. 3) The reactants may be in nonstandard conditions which means that the voltage for the half cells may be less or more than the standard condition amount Ex1) Concentration of chloride ion = 5.5M not the unit activity of 1M. This means that the reduction of chloride = 1.31V not 1.36V Ex2) The standard condition is to have a pH of 4 in the anode half cell but sometimes during nonstandard states, the pH may be higher or lower changing the voltage. 4) An inert electrode’s ability to electrolysis depend on the reactants in the electrolyte solution while an active electrode can run on its own to perform the oxidation or reduction half reaction. If all four of these factors are accounted for, we can successfully predict electrode half reactions and overall reactions in electrolysis. Example) Predict the electrode reactions and the overall reaction when the anode is made of (a) copper and (b) platinum. Quantitative Aspects of Electrolysis Michael Faraday discovered in 1833 that there is always a simple relationship between the amount of substance produced or consumed at an electrode during electrolysis and the quantity of electrical charge Q which passes through the cell. For example, the half-equation Ag+ + e– → Ag tells us that when 1 mol Ag+ is plated out as 1 mol Ag, 1 mol e– must be supplied from the cathode. Since the negative charge on a single electron is known to be 1.6022 × 10–19 C, we can multiply by the Avogadro constant to obtain the charge per mole of electrons. This quantity is called the Faraday Constant, symbol F: F = 1.6022 × 10–19 C × 6.0221 × 1023 mol–1 = 9.649 × 104 C mol–1 Thus in the case of Eq. (1), 96 490 C would have to pass through the cathode in order to deposit 1 mol Ag. For any electrolysis the electrical charge Q passing through an electrode is related to the amount of electrons ne– by Thus F serves as a conversion factor between ne– and Q. Often the electrical current rather than the quantity of electrical charge is measured in an electrolysis experiment. Since a coulomb is defined as the quantity of charge which passes a fixed point in an electrical circuit when a current of one ampere flows for one second, the charge in coulombs can be calculated by multiplying the measured current (in amperes) by the time (in seconds) during which it flows: Q = It In this equation I represents current and t represents time. If you remember that coulomb = 1 ampere × 1 second 1C=1As you can adjust the time units to obtain the correct result. Now that we can predict the electrode half-reactions and overall reactions in electrolysis, it is also important to be able to calculate the quantities of reactants consumed and the products produced. For these calculations we will be using the Faraday constant: 1 mol of electron = 96,485 C charge (C) = current (C/s) x time (s) (C/s) = 1 coulomb of charge per second = 1 ampere (A) Simple conversion for any type of problem: 1. Convert any given time to seconds 2. Take the current given (A) over the seconds, [1 c = (A)/(s)] 3. Finally use the stoichiometry conversion of 1 mol of electron = 96,485 C (Faraday's Constant) Example) The electrolysis of dissolved Bromine sample can be used to determine the amount of Bromine content in sample. At the cathode, the reduction half reaction is Br2+(aq) + 2 e- -> 2 Br-. What mass of Bromine can be deposited in 3.00 hours by a current of 1.18 A? Electrochemistry 37 Sol) 3.00 hours x 60 min/hour x 60 sec/1 min x 1.18 C(A) / 1 sec x 1 mol e-/96,485 C = 0.132 mol e- Electrochemistry 38 Problems 1) Predict the products of electrolysis by filling in the graph: Cl-, Br-, I-, H+, OH-, Cu2+, Pb2+, Ag+, K+, Na+, 2) Calculate the quantity of electrical charge needed to plate 1.386 mol Cr from an acidic solution of K2Cr2O7according to halfequation: H2Cr2O7(aq) + 12H+(aq) + 12e– → 2Cr(s) + 7 H2O(l) 3) Hydrogen peroxide, H2O2, can be manufactured by electrolysis of cold concentrated sulfuric acid. The reaction at the anode is 2H2SO4 → H2S2O8 + 2H+ + 2e– When the resultant peroxydisulfuric acid, H2S2O8, is boiled at reduced pressure, it decomposes: 2H2O + H2S2O8 → 2H2SO4 + H2O2 Calculate the mass of hydrogen peroxide produced if a current of 0.893 flows for 1 hour. 4) The electrolysis of dissolved Cholride sample can be used to determine the amount of Chloride content in sample. At the cathode, the reduction half reaction is Cl2+(aq) + 2 e- -> 2 Cl-. What mass of Chloride can be deposited in 6.25 hours by a current of 1.11 A? 5) In an electrolytic cell the electrode at which the electrons enter the solution is called the ______ ; the chemical change that occurs at this electrode is called _______. 1. anode, oxidation 2. anode, reduction 3. cathode, oxidation 4. cathode, reduction 5. cannot tell unless we know the species being oxidized and reduced. 6)How long (in hours) must a current of 5.0 amperes be maintained to electroplate 60 g of calcium from molten CaCl2? 1. 27 hours 2. 8.3 hours 3. 11 hours 4. 16 hours 5. 5.9 hours 7) How long, in hours, would be required for the electroplating of 78 g of platinum from a solution of [PtCl6]2-, using an average current of 10 amperes at an 80% electrode efficiency? 1. 8.4 2. 5.4 3. 16.8 4. 11.2 5. 12.4 8) How many faradays are required to reduce 1.00 g of aluminum(III) to the aluminum metal? 1. 1.00 2. 1.50 3. 3.00 4. 0.111 Electrochemistry 39 5. 0.250 9) Find the standard cell potential for an electrochemical cell with the following cell reaction. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Answers: 1). Cl- chlorine H+ hydrogen Cl- chlorine Cu2+ copper I- iodine H+ Hhydrogen 2) 12 mol e– is required to plate 2 mol Cr, giving us a stoichiometric ratio S(e–/Cr). Then the Faraday constant can be used to find the quantity of charge. nCr n e– Q × = 8.024 × 105 C Q = 1.386 mol Cr × 3) The product of current and time gives us the quantity of electricity, Q. Knowing this we easily calculate the amount of electrons, ne–. From the first half-equation we can then find the amount of peroxydisulfuric acid, and the second leads to nH2O2 and finally to mH2O2. × g H2O2 = 0.5666 g H2O2 = 05666 4) 0.259 mol e5) d 6) d 7) b 8) d 9) Write the half-reactions for each process. Zn(s) → Zn2+(aq) + 2 eCu2+(aq) + 2 e- → Cu(s) Look up the standard potentials for the reduction half-reaction. Eoreduction of Cu2+ = + 0.339 V Eoreduction of Zn2+ = - 0.762 V Determine the overall standard cell potential. Eocell = + 1.101 V Electrochemistry 40 Application of Electrolytic Cells (A) Production of Elements Most elements occur naturally combined with other elements in compounds because the reduction potentials for many metal elements are very negative and they are easily oxidized. Many metals can be produced by electrolysis of solutions of their ionic compounds, but two difficulties arise. First, many naturally occurring ionic compounds have low solubility in water, and second, water is a stronger oxidizing agent than active metal cations. To overcome these difficulties, instead of using water, ionic compounds are melted at high temperature. In the electrolysis of molten binary ionic compounds, only one oxidizing agent and one reducing agent are present. Production of Sodium: Fig. 3 shows the set-up for the production of sodium from molten sodium chloride in a Down’s Cell. Mixing with calcium chloride, the melting point of sodium chloride is reduced from to . Production of Aluminum: Fig. 5 shows the set-up for the production of aluminum from molten aluminum oxide ( ) in a HallHeroult Cell. The coomon ore has a high melting point of , but it was found that it dissolves in a molten mineral called . The molten conducting mixture has a melting point around .. cryolite, Electrochemistry 41 The Chlor-Alkali Process – Instead of eliminating or replacing water as a solvent in the electrolytic production of elements, the ChlorAlkali Process overcomes the difficulty of producing active metals by simply “overpower” the reduction of water. A high voltage leads to the reduction of metal ions rather than water because the reduction of water I a relatively slow reaction. Aqueous sodium chloride can be electrolyzed in this manner to produce chlorine, hydrogen, and sodium hydroxide. As shown in Fig. 6, the high voltage forces the reduction of aqueous sodium ions to sodium metal. The sodium metal is then reacted with water to produce hydrogen gas and sodium hydroxide. Chlorine is prefrerentially produced at the anode instead of oxygen, in spite of its more favourable position in the redox table. Newer chlor-alkali plants now use a process that relies on an ion-exchange membrane to separate the sodium and chloride ions during electrolysis, thus eliminating mercury and making the process safer and cheaper. (B) Refining of Metals Electrochemistry 42 (C) Electroplating Plating of a metal at the cathode of an electrolytic cell is called electroplating and is a common technology that is used to cover the surface of an object with a thin layer of metal. Today, we still have examples of technological processes that are not fully understand, such as chromium plating and silver plating. For example, there is no satisfactory explanation for why silver deposited in an electrolysis of a silver nitrate solution does not adhere well to any surface, whereas silver plated from silver cyanide solution does. Chromium is best plated from a solution of chromic acid. A thin layer of chromium metal is very shiny and, like aluminum, protects itself from corrosion by forming a tough oxide layer. Stoichiometry of Cell Reactions In the production of elements, the refining of metals, and electroplating, the quantity of electricity that passes through a cell determines the masses (or volumes for gases) of substances that react or are produced at the electrodes. One coulomb C is the quantity of charge transferred by a current of one ampere (A) during a time of one second: . Electrochemistry 43 Electrochemistry 44 Electrochemistry 45 To calculate the volume of a gas that is produced. The following formula is used: For 1 mol of an ideal gas at 0o C (273 K ) and 1 atm, the volume of the gas given by the ideal gas law is: V= nRT (100 . mol )(0.08206 L atm / K mol )(273 K ) = = 22.4 L . P 100 . atm The condition 0o C (273 K ) and 1 atm are called standard temperature and pressure (STP) Electrochemistry 46 Electrochemistry 47 WRITING IONIC EQUATIONS FOR REDOX REACTIONS This page explains how to work out electron-half-reactions for oxidation and reduction processes, and then how to combine them to give the overall ionic equation for a redox reaction. This is an important skill in inorganic chemistry. Don't worry if it seems to take you a long time in the early stages. It is a fairly slow process even with experience. Take your time and practise as much as you can. Electron-half-equations - What is an electron-half-equation? When magnesium reduces hot copper(II) oxide to copper, the ionic equation for the reaction is: You can split the ionic equation into two parts, and look at it from the point of view of the magnesium and of the copper(II) ions separately. This shows clearly that the magnesium has lost two electrons, and the copper(II) ions have gained them. These two equations are described as "electron-half-equations" or "half-equations" or "ionic-half-equations" or "half-reactions" - lots of variations all meaning exactly the same thing! Any redox reaction is made up of two half-reactions: in one of them electrons are being lost (an oxidation process) and in the other one those electrons are being gained (a reduction process). Working out electron-half-equations and using them to build ionic equations In the example above, we've got at the electron-half-equations by starting from the ionic equation and extracting the individual halfreactions from it. That's doing everything entirely the wrong way round! In reality, you almost always start from the electron-halfequations and use them to build the ionic equation. Example 1: The reaction between chlorine and iron(II) ions Chlorine gas oxidises iron(II) ions to iron(III) ions. In the process, the chlorine is reduced to chloride ions. You would have to know this, or be told it by an examiner. In building equations, there is quite a lot that you can work out as you go along, but you have to have somewhere to start from! You start by writing down what you know for each of the half-reactions. In the chlorine case, you know that chlorine (as molecules) turns into chloride ions: The first thing to do is to balance the atoms that you have got as far as you possibly can: ALWAYS check that you have the existing atoms balanced before you do anything else. If you forget to do this, everything else that you do afterwards is a complete waste of time! Now you have to add things to the half-equation in order to make it balance completely. All you are allowed to add are: • electrons • water • hydrogen ions (unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead) In the chlorine case, all that is wrong with the existing equation that we've produced so far is that the charges don't balance. The lefthand side of the equation has no charge, but the right-hand side carries 2 negative charges. That's easily put right by adding two electrons to the left-hand side. The final version of the half-reaction is: Now you repeat this for the iron(II) ions. You know (or are told) that they are oxidised to iron(III) ions. Write this down: The atoms balance, but the charges don't. There are 3 positive charges on the right-hand side, but only 2 on the left. You need to reduce the number of positive charges on the right-hand side. That's easily done by adding an electron to that side: Combining the half-reactions to make the ionic equation for the reaction What we've got at the moment is this: Electrochemistry 48 It is obvious that the iron reaction will have to happen twice for every chlorine molecule that reacts. Allow for that, and then add the two half-equations together. But don't stop there!! Check that everything balances - atoms and charges. It is very easy to make small mistakes, especially if you are trying to multiply and add up more complicated equations. You will notice that I haven't bothered to include the electrons in the addedup version. If you think about it, there are bound to be the same number on each side of the final equation, and so they will cancel out. If you aren't happy with this, write them down and then cross them out afterwards! Example 2: The reaction between hydrogen peroxide and manganate(VII) ions The first example was a simple bit of chemistry which you may well have come across. The technique works just as well for more complicated (and perhaps unfamiliar) chemistry. Manganate(VII) ions, MnO4-, oxidise hydrogen peroxide, H2O2, to oxygen gas. The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulphuric acid. During the reaction, the manganate(VII) ions are reduced to manganese(II) ions. Let's start with the hydrogen peroxide half-equation. What we know is: The oxygen is already balanced. What about the hydrogen? All you are allowed to add to this equation are water, hydrogen ions and electrons. If you add water to supply the extra hydrogen atoms needed on the right-hand side, you will mess up the oxygens again - that's obviously wrong! Add two hydrogen ions to the right-hand side. Now all you need to do is balance the charges. You would have to add 2 electrons to the right-hand side to make the overall charge on both sides zero. Now for the manganate(VII) half-equation: You know (or are told) that the manganate(VII) ions turn into manganese(II) ions. Write that down. The manganese balances, but you need four oxygens on the right-hand side. These can only come from water - that's the only oxygencontaining thing you are allowed to write into one of these equations in acid conditions. By doing this, we've introduced some hydrogens. To balance these, you will need 8 hydrogen ions on the left-hand side. Now that all the atoms are balanced, all you need to do is balance the charges. At the moment there are a net 7+ charges on the lefthand side (1- and 8+), but only 2+ on the right. Add 5 electrons to the left-hand side to reduce the 7+ to Electrochemistry 49 2+. This is the typical sort of half-equation which you will have to be able to work out. The sequence is usually: • Balance the atoms apart from oxygen and hydrogen. • Balance the oxygens by adding water molecules. • Balance the hydrogens by adding hydrogen ions. • Balance the charges by adding electrons. Combining the half-reactions to make the ionic equation for the reaction The two half-equations we've produced are: You have to multiply the equations so that the same number of electrons are involved in both. In this case, everything would work out well if you transferred 10 electrons. But this time, you haven't quite finished. During the checking of the balancing, you should notice that there are hydrogen ions on both sides of the equation: You can simplify this down by subtracting 10 hydrogen ions from both sides to leave the final version of the ionic equation - but don't forget to check the balancing of the atoms and charges! You will often find that hydrogen ions or water molecules appear on both sides of the ionic equation in complicated cases built up in this way. Always check, and then simplify where possible. Example 3: The oxidation of ethanol by acidified potassium dichromate(VI) This technique can be used just as well in examples involving organic chemicals. Potassium dichromate(VI) solution acidified with dilute sulphuric acid is used to oxidise ethanol, CH3CH2OH, to ethanoic acid, CH3COOH. The oxidising agent is the dichromate(VI) ion, Cr2O72-. This is reduced to chromium(III) ions, Cr3+. We'll do the ethanol to ethanoic acid half-equation first. Using the same stages as before, start by writing down what you know: Balance the oxygens by adding a water molecule to the left-hand side: Add hydrogen ions to the right-hand side to balance the hydrogens: And finally balance the charges by adding 4 electrons to the right-hand side to give an overall zero charge on each side: The dichromate(VI) half-equation contains a trap which lots of people fall into! Electrochemistry 50 Start by writing down what you know: What people often forget to do at this stage is to balance the chromiums. If you don't do that, you are doomed to getting the wrong answer at the end of the process! When you come to balance the charges you will have to write in the wrong number of electrons which means that your multiplying factors will be wrong when you come to add the half-equations . . . A complete waste of time! Now balance the oxygens by adding water molecules . . . . . . and the hydrogens by adding hydrogen ions: Now all that needs balancing is the charges. Add 6 electrons to the left-hand side to give a net 6+ on each side. Combining the half-reactions to make the ionic equation for the reaction What we have so far is: What are the multiplying factors for the equations this time? The simplest way of working this out is to find the smallest number of electrons which both 4 and 6 will divide into - in this case, 12. That means that you can multiply one equation by 3 and the other by 2. Note: Don't worry too much if you get this wrong and choose to transfer 24 electrons instead. All that will happen is that your final equation will end up with everything multiplied by 2. The multiplication and addition looks like this: Now you will find that there are water molecules and hydrogen ions occurring on both sides of the ionic equation. You can simplify this to give the final equation: Note: You have now seen a cross-section of the sort of equations which you could be asked to work out. Now you need to practice so that you can do this reasonably quickly and very accurately! Aim to get an averagely complicated example done in about 3 minutes. WRITING IONIC EQUATIONS FOR REDOX REACTIONS DONE UNDER ALKALINE CONDITIONS This page explains how to work out electron-half-reactions for oxidation and reduction processes which are carried out under alkaline conditions, and then how to combine them to give the overall ionic equation for the redox reaction. Combining them is easy; working them out may be more difficult than under acidic conditions. Why is it more difficult to write electron-half-equations for these reactions? What you already know: When you are trying to balance electron-half-equations, you are only allowed to add: • electrons • water • hydrogen ions (unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead) If you are working under acidic or neutral conditions, the sequence is usually: • Balance the atoms apart from oxygen and hydrogen. • Balance the oxygens by adding water molecules. • Balance the hydrogens by adding hydrogen ions. Electrochemistry 51 • Balance the charges by adding electrons. The whole process is fairly automatic, and provided you take care, there isn't much to go wrong. How is this different under alkaline conditions? The problem is that the water and the hydroxide ions that you add to balance the equations under alkaline conditions contain both hydrogen and oxygen. - To balance the oxygens, you could in principle add either H2O or OH to the equation. The same thing is true for balancing the hydrogens. How do you know what to start with? How to tackle the problem? In some cases, it is obvious how to build up the half-equation using hydroxide ions. Always check this before you get involved in anything more difficult. You will see what I mean shortly. If it isn't immediately obvious, work out the electron-half equation as if it were being done under acidic conditions just as you have learnt to do on the previous page - in other words by writing in water molecules, hydrogen ions and electrons. Once you have got a balanced half-equation, you then convert it to alkaline conditions. You will see how to do that in the following examples. Study the Four examples below (Don't worry if the chemistry in these examples is unfamiliar to you. It doesn't matter in the slightest. All that matters is how you work out the equations): Example 1: The oxidation of cobalt(II) to cobalt(III) by hydrogen peroxide If you add an excess of ammonia solution to a solution containing cobalt(II) ions, you get a complex ion formed called the 2+ hexaamminecobalt(II) ion, Co(NH3)6 . This is oxidised rapidly by hydrogen peroxide solution to the hexaamminecobalt(III) ion, 3+ Co(NH3)6 . Ammonia solution is, of course, alkaline. The half-equation for the cobalt reaction is easy. Start by writing down what you know (or are told): Everything balances apart from the charges. Add an electron to the right-hand side to give both sides an overall charge of 2+. The hydrogen peroxide half-equation isn't very difficult either, except that you aren't told what is formed and so have to make a guess. It would balance very nicely if you ended up with 2 hydroxide ions on the right-hand side. This is a good example of a case where it is fairly obvious where to put hydroxide ions. You would then just have to add 2 electrons to the left-hand side to balance the charges. Combining the half-reactions to make the ionic equation for the reaction What we have so far is: The multiplication and addition looks like this: And that's it - an easy example! Example 2: The oxidation of iron(II) hydroxide by the air Electrochemistry 52 If you add sodium hydroxide solution to a solution of an iron(II) compound you get a green precipitate of iron(II) hydroxide. This is quite quickly oxidised by oxygen in the air to an orange-brown precipitate of iron(III) hydroxide. The half-equation for the iron(II) hydroxide is straightforward. Start with what you know: You obviously need another hydroxide ion on the left-hand side. This is even more straightforward than the previous example. To balance the charges, add an electron to the right-hand side. The half-equation for the oxygen isn't so easy. You don't know what is being formed. It isn't at all obvious whether you need to balance the oxygens by adding water molecules or hydroxide ions to the right-hand side. OK - treat it as if it were being done under acidic conditions, and the problem disappears! In this case, you can only balance the oxygens by adding water molecules to the right-hand side. Balance the hydrogens by adding hydrogen ions to the left-hand side. And then balance the charges by adding 4 electrons: Now you have got a perfectly balanced half-equation. The problem is, of course, that it only applies under acidic conditions. We should have alkaline conditions - with hydroxide ions present not hydrogen ions. So . . . get rid of the hydrogen ions! Add enough hydroxide ions to both sides of the equation so that you can neutralise all the hydrogen ions. Because it is now a balanced equation, you must add the same number to both sides to maintain the balance. The hydrogen ions and hydroxide ions on the left-hand side would turn into 4 water molecules: Finally, there are water molecules on both sides of the equation. Cancel out any which aren't changed. This has all been a bit tedious - although you haven't actually had to think very much! Don't forget to re-check that everything balances now that you have finished. Combining the half-reactions to make the ionic equation for the reaction From here on it's all back to the usual routine. We've worked out the two half-equations: The iron equation will have to happen 4 times to supply enough electrons to the oxygen. Notice that the hydroxide ions on each side cancel out. (Perhaps to your surprise - certainly to mine when I worked this out!) Example 3: The reduction of manganate(VII) ions to manganate(VI) ions by hydroxide ions Electrochemistry 53 This is a fairly obscure reaction but it is not too difficult to work out and balance. It is unusual in that hydroxide ions are acting as reducing agents. Dark purple potassium manganate(VII) solution is slowly reduced to dark green potassium manganate(VI) solution by concentrated potassium hydroxide solution. Bubbles of oxygen gas are also given off. Note: Unless the potassium manganate(VII) solution is very dilute, its strong purple colour tends to mask the green of the potassium manganate(VI) in the short term. The half-equation for the conversion of manganate(VII) ions to manganate(VI) ions is easy (provided, of course, that you know their formulae!). So what is happening to the hydroxide ions to turn them into oxygen gas? It isn't actually very difficult to work out the half-equation directly, but suppose you want to avoid thinking and go through a standard routine: Write down what you know, balancing the oxygens in the process: Balance the hydrogens by adding hydrogen ions: . . . and now balance the charges: Now get rid of the hydrogen ions by adding enough hydroxide ions to both sides of the equation: . . . and tidy it all up! Combining the half-reactions to make the ionic equation for the reaction What we have so far is: The manganese reaction will have to happen four times in order to use up the four electrons produced by the hydroxide equation. Putting this together, you get: The chemistry may be unfamiliar, but working out the equation isn't too hard! Example 4: The oxidation of chromium(III) to chromium(VI) If you add an excess of sodium hydroxide solution to a solution containing chromium(III) ions, you get a dark green solution 3containing the complex ion hexahydroxochromate(III), Cr(OH)6 . 2- This can be oxidised to bright yellow chromate(VI) ions, CrO4 by warming it with hydrogen peroxide solution. We've already worked out the hydrogen peroxide half-equation where it is acting as an oxidising agent under alkaline conditions: So this time we just need to put together the half-equation for the chromium ions. What we know is: It isn't the least bit obvious where to put hydroxide ions or water molecules, so treat it as if it were being done under acidic Electrochemistry 54 conditions. That way, you start by balancing the oxygens by adding water molecules. To get 6 oxygens on each side, you need two waters on the right-hand side: Now balance the hydrogens by adding hydrogen ions: . . . and then balance the charges by adding electrons: Finally, convert from acidic to alkaline conditions by adding enough hydroxide ions to both sides to turn the hydrogen ions into water: . . . and tidy it all up: Combining the half-reactions to make the ionic equation for the reaction The two half-equations are: If you multiply one equation by 3 and the other by 2, that transfers a total of 6 electrons. Finally, tidy up the hydroxide ions that occur on both sides to leave the overall ionic equation: (I) How to Calculate Cell EMF for an Electrochemical Cell The cell electromotive force, or cell EMF is the net voltage between the oxidation and reduction half-reactions taking place between two redox half-reactions. Cell EMF is used to determine whether or not the cell is galvanic. This example problem shows how to calculate the cell EMF using standard reduction potentials. The Table of Standard Reduction Potentials is needed for this example. Problem 1: Consider the redox reaction: Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g) a) Calculate the cell EMF for the reaction. b) Identify if the reaction is galvanic. Solution: Step 1: Break the redox reaction into reduction and oxidation half-reactions. Hydrogen ions, H+ gain electrons when forming hydrogen gas, H2. The hydrogen atoms are reduced by the half-reaction: 2 H+ + 2 e- → H2 Magnesium loses two electrons and is oxidized by the half-reaction: Mg → Mg2+ + 2 eStep 2: Find the standard reduction potentials for the half-reactions. Reduction: E0 = 0.0000 V The table shows reduction half-reactions and standard reduction potentials. To find E0 for an oxidation reaction, reverse the reaction. Reversed reaction: Mg2+ + 2 e- → Mg This reaction has a E0 = -2.372 V. E0Oxidation = - E0Reduction E0Oxidation = - (-2.372 V) = + 2.372 V Step 3: Add the two E0 together to find the total cell EMF, E0cell E0cell = E0reduction + E0oxidation E0cell = 0.0000 V + 2.372 V = +2.372 V Electrochemistry 55 Step 4: Determine if reaction is galvanic. Redox reactions with a positive E0cell value are galvanic. This reaction's E0cell is positive and therefore galvanic. Answer: The cell EMF of the reaction is +2.372 Volts and is galvanic. * Notice that we multiply the half-reactions by factors to make the electrons cancel, but we do not multiply the reduction potentials by these factors. To obtain the cell potentials, we simply subtract one reduction potential from the other. (II) Cell Potential and Free Energy Example Problem - Calculating Maximum Theoretical Energy of an Electrochemical Cell Cell potentials are measured in volts or energy per unit charge. This energy can be related to the theoretical maximum free energy or Gibbs free energy of the total redox reaction driving the cell. Problem 1: For the following reaction: Cu(s) + Zn2+(aq) ↔ Cu2+(aq) + Zn(s) a. Calculate ∆G°. b. Will zinc ions plate out onto the solid copper in the reaction? Solution: Free energy is related to cell EMF by the formula: ∆G° = -nFE0cell , where ∆G° is the free energy of the reaction n is the number of moles of electrons exchanged in the reaction F is Faraday's constant (9.648456 x 104 C/mol) E0cell is the cell potential. Step 1: Break the redox reaction into oxidation and reduction half-reactions. Cu → Cu2+ + 2 e- (oxidation) Zn2+ + 2 e- → Zn (reduction) Step 2: Find the E0cell of the cell. Cu → Cu2+ + 2 e- E0 = -0.3419 V Zn2+ + 2 e- → Zn E0 = -0.7618 V E0cell = E0reduction + E0oxidation E0cell = -0.4319 V + -0.7618 V E0cell = -1.1937 V Step 3: Find ∆G°. There are 2 moles of electrons transferred in the reaction for every mole of reactant, therefore n=2. Another important conversion is 1 volt = 1 Joule/Coulomb ∆G° = -nFE0cell ∆G° = -(2 mol)(9.648456 x 104 C/mol)(-1.1937 J/C) ∆G° = 230347 J or 230.35 kJ The zinc ions will plate out if the reaction is spontaneous. Since ∆G° > 0, the reaction is not spontaneous and the zinc ions will not plate out onto the copper at standard conditions. Answer: a. ∆G° = 230347 J or 230.35 kJ b. Zinc ions will not plate out onto the solid copper. (III) Equilibrium Constant of an Electrochemical Cell Reaction Example Problem: Using the Nernst Equation to Determine the Equilibrium Constant The equilibrium constant of an electrochemical cell's redox reaction can be calculated using the Nernst equation and the relationship between standard cell potential and free energy. This example problem shows how to find the equilibrium constant of a cell's redox reaction. Problem 1: The following two half-reactions are used to form an electrochemical cell: Oxidation: SO2(g) + 2 H20(ℓ) → SO4-(aq) + 4 H+(aq) + 2 e- E°ox = -0.20 V Reduction: Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(ℓ) E°red = +1.33 V What is the equilibrium constant of the combined cell reaction at 25 °C? Solution: Step 1: Combine and balance the two half-reactions. The oxidation half-reaction produces 2 electrons and the reduction half-reaction needs 6 electrons. To balance the charge, the oxidation Electrochemistry 56 reaction must be multiplied by a factor of 3. 3 SO2(g) + 6 H20(ℓ) → 3 SO4-(aq) + 12 H+(aq) + 6 e+ Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(ℓ) 3 SO2(g) + Cr2O72-(aq) + 2 H+(aq) → 3 SO4-(aq) + 2 Cr3+(aq) + H2O(ℓ) By balancing the equation, we now know the total number electrons exchanged in the reaction. This reaction exchanged six electrons. Step 2: Calculate the cell potential. E°cell = E°ox + E°red E°cell = -0.20 V + 1.33 V E°cell = +1.13 V Step 3: Find the equilibrium constant, K. When a reaction is at equilibrium, the change in free energy is equal to zero. The change in free energy of an electrochemical cell is related to the cell potential by the equation: ∆G = -nFEcell where ∆G is the free energy of the reaction n is the number of moles of electrons exchanged in the reaction F is Faraday's constant (96484.56 C/mol) E is the cell potential. If ∆G = 0:, solve for Ecell 0 = -nFEcell Ecell = 0 V This means, at equilibrium, the potential of the cell is zero. The reaction progresses forward and backwards at the same rate meaning there is no net electron flow. With no electron flow, there is no current and the potential is equal to zero. Now there is enough information known to use the Nernst equation to find the equilibrium constant. The Nernst equation is: Ecell = E°cell - (RT/nF) x log10Q where Ecell is the cell potential E°cell refers to standard cell potential R is the gas constant (8.3145 J/mol·K) T is the absolute temperature n is the number of moles of electrons transferred by the cell's reaction F is Faraday's constant (96484.56 C/mol) Q is the reaction quotient At equilibrium, the reaction quotient Q is the equilibrium constant, K. This makes the equation: Ecell = E°cell - (RT/nF) x log10K From above, we know the following: Ecell = 0 V E°cell = +1.13 V R = 8.3145 J/mol·K T = 25 °C = 298.15 K F = 96484.56 C/mol n = 6 (six electrons are transferred in the reaction) Solve for K: 0 = 1.13 V - [(8.3145 J/mol·K x 298.15 K)/(6 x 96484.56 C/mol)]log10K -1.13 V = - (0.004 V)log10K log10K = 282.5 K = 10282.5 K = 10282.5 = 100.5 x 10282 K = 3.16 x 10282 Electrochemistry 57 Answer: The equilibrium constant of the cell's redox reaction is 3.16 x 10282. Electrochemistry 58 (IV) Nernst Equation Example Problem: Calculate Cell Potential in Nonstandard Conditions Standard cell potentials are calculated in standard conditions. The temperature and pressure are at standard temperature and pressure and the concentrations are all 1 M aqueous solutions. In non-standard conditions, the Nernst equation is used to calculate cell potentials. It modifies the standard cell potential to account for temperature and concentrations of the reaction participants. This example problem shows how to use the Nernst equation to calculate a cell potential. Problem: Find the cell potential of a galvanic cell based on the following reduction half-reactions at 25 °C Cd2+ + 2 e- → Cd E0 = -0.403 V Pb2+ + 2 e- → Pb E0 = -0.126 V where [Cd2+] = 0.020 M and [Pb2+] = 0.200 M. Solution: The first step is to determine the cell reaction and total cell potential. In order for the cell to be galvanic, E0cell > 0. For this reaction to be galvanic, the cadmium reaction must be the oxidation reaction. Cd → Cd2++ 2 e- E0 = +0.403 V Pb2+ + 2 e- → Pb E0 = -0.126 V The total cell reaction is: Pb2+(aq) + Cd(s) → Cd2+(aq) + Pb(s) and E0cell = 0.403 V + -0.126 V = 0.277 V The Nernst equation is: Ecell = E0cell - (RT/nF) x lnQ where Ecell is the cell potential E0cell refers to standard cell potential R is the gas constant (8.3145 J/mol·K) T is the absolute temperature n is the number of moles of electrons transferred by the cell's reaction F is Faraday's constant 96485.337 C/mol ) Q is the reaction quotient, where Q = [C]c·[D]d / [A]a·[B]b where A, B, C, and D are chemical species; and a, b, c, and d are coefficients in the balanced equation: aA+bB→cC+dD In this example, the temperature is 25 °C or 300 K and 2 moles of electrons were transferred in the reaction. RT/nF = (8.3145 J/mol·K)(300 K)/(2)(96485.337 C/mol) RT/nF = 0.013 J/C = 0.013 V The only thing remaining is to find the reaction quotient, Q. Q = [products]/[reactants] **For reaction quotient calculations, pure liquid and pure solid reactants or products are omitted.** Q = [Cd2+]/[Pb2+] Q = 0.020 M / 0.200 M Q = 0.100 Combine into the Nernst equation: Ecell = E0cell - (RT/nF) x lnQ Ecell = 0.277 V - 0.013 V x ln(0.100) Ecell = 0.277 V - 0.013 V x -2.303 Ecell = 0.277 V + 0.023 V Ecell = 0.300 V Answer: The cell potential for the two reactions at 25 °C and [Cd2+] = 0.020 M and [Pb2+] = 0.200 M is 0.300 volts. Electrochemistry 59 Electrochemistry 60 (V) Electrochemistry Calculations 1. For the half reaction, M2+(aq) +2e- → M(s), what range of Eo is possible if: a. M displaces MnO4-(aq), but not AgCl(s). M is unable to react with HCl, but reacts with HNO3. Solution: Since the reaction can displace MnO4-(aq), it would be able to have a standard reaction up to +0.56. However, since it cannot displace AgCl(s), its standard reaction potential would be above +0.2223. Therefore: +0.2223V<E0<+0.560V 2. A Lithium-ion battery has a theoretical Eo cell of 1.5V. Find the Eo for the reduction half-reaction: Li+(aq) + e- -> Li(s) Solution: The E0 for the reaction Li+ (aq) + e- -> Li(s) equals -3.040 according to the Standard Electrode Potential Table 3. Write the voltaic cell for the reaction between I2(s) and Sn4+(aq). Determine which half reaction is the cathode and the anode and calculate the reaction standard potential. Solution: I2(s) + 2e -> 2I-(aq) Sn4+(aq) + 2 e- -> Sn2+(aq) E0: +0.535V Reduction => Cathode E0: +0.154V Oxidation => Anode => Pt(s)|Sn2+(aq), Sn4+(aq)|| I2(s)| 2I- (aq) Note: Pt(s) serves as an electrode conductor and does not affect the overall reaction. Since both the Sn reactant and product are both aqueous solutions, Pt or other metal conductors can be used in place. E0 = E0(cathode) – E0(anode) E0= (+0.535V)-( +0.154V)= E0= +0.0381V 4. Given the reaction: 2MnO4-(aq) + PbO(s) + 6H+(aq) -> 2MnO2 (s) + 5PbO2 (s) + 3H2O (l) E0 cell: +1.23V What does E0 PbO2(s) equal? Solution: MnO4- -> MnO2 (s) PbO (s) + 6H+(aq) -> 2MnO2 (s) + 3H2O (l) Mn oxidation number: Reactant: +7 Product: +4 =>Mn is the reduction reaction: Mn is the cathode =>E0 = 1.51 E0 cell= E0 cathode - E0 anode 1.23= 1.51- E0 anode E0anode= +0.28V 5. Determine whether a forward reaction is spontaneous or non-spontaneous: a. Ag2+(aq) + Cr2+(aq) -> Ag+ (aq) E0cell= E0 cathode – E0 anode E0 cell= (E0Ag2+/Ag-) - (E0Cr3+/Cr2+) E0 cell= 1.98V – (-0.424V) E0 cell= +2.404V => Spontaneous b. Sn4+(aq) + 2I-(aq) → Sn2+ (aq) + I2 (aq) ^Reduced ^Oxidized E0 cell = (+0.154V) – (+0.535V) Electrochemistry 61 E0 cell = -0.381V => Non-spontaneous 6. Given the following voltaic cell diagram and using the standard gold electrode as the reduction half reaction, and metal M as an oxidizing half reaction, find the E0 reduction half reaction: Au3+(1M) +3e- -> Au(s) E0= +1.52 3+ M (1M) + 3eE0= ? a. In E0= 1.858 V b. Al E0= 3.20 V c. La E0= 3.90 V d. U E0= 3.18 V Solution: Au is the reduction half reaction E0= +1.52= E0 cathode E0 cell= E0 cathode - E0 anode In 1.858 = +1.52- E0 anode 1.858 + E0 anode = 1.52 E0 anode= -0.338V Al 3.20=1.52 - E0 anode 3.20 + E0 anode = 1.52 E0anode= -1.68V La 3.90 = 1.52 - E0 anode 3.90 + E0 anode = 1.52 E0 anode= -2.38V U 3.18 = 1.52 - E0 anode 3.18 + E0 anode=1.52 E0anode= -1.66V 7. Calculate ∆Go for the following reactions and determine if spontaneous: a. NO3-(aq) + Al(s) + 4H+(aq) → NO(g) + Al3+(aq) + 2H2O (l) Solution: ∆G0= -nFE0cell E0cell = (E0NO3-/NO-) - (E0Al3+/Al) E0 cell = (0.956) – (-1.676) E0 cell = 2.63V N=3 since: N oxidation: reactant = +5 Al oxidation: reactant = 0 product= +2 product= +3 F=96,485 C/mol e- ∆G0=-[(3)(96,485 c/mol e-)(2.63V)] ∆G0= -761kJ Spontaneous b. F2(g) + 2Li(s) -> 2F- (aq) + 2Li+ (aq) Solution: E0cell = (E0F2/F-)- (E0Li+/Li) E0cell = 2.866 + 3.040 = 5.906V N=2 since: F2 + 2e- -> 2F2Li+ + 2e- 2Li(s) Electrochemistry 62 ∆G0= -[(2 mol e-)(96,485 C/mol e-)(5.906V)] ∆G0=-1,140 kJ Spontaneous 8. Find ∆G0 by combining the following half reactions to calculate the E0 cell at: Ag2+ + 2e- -> Ag(s) Ag2+(aq) + e- -> Ag+(aq) E0= +1.98V Ag+(aq) + e- -> Ag(s) E0= +0.800V Solution: Ag2+(aq) + e- -> Ag+(aq) ∆G0= -(1 x F x (1.98V)) Ag+ (aq) + e- -> Ag (s) ∆G0= -(1 x F x (0.800V)) Ag2+ (aq) + 2e- -> Ag(s) ∆G0= -(1.98V)F – (0.800V)F= ∆G0= (-2.78V)F ∆G0= -nFE0cell (-2.78V)F = -2FE0cell E0 cell= (-2.78V)F/(-2F) E0 cell= +1.39V 9. A voltaic cell with Ecell=1.500V has what [Ag+] in the cell? Zn(s)|Zn2+(1.50M)||Ag+(?M)|Ag(S) Solution: Solve for Ecell: Oxidation reaction: Zn(s) --> Zn2+(aq) + 2e E°=-0.763 Reduction reaction: (Ag+(aq)+e --> Ag(s))Ÿ2 E°=0.800V Net Ionic Equation: Zn(s)+2Ag+(aq) --> Zn2+(aq) +2Ag(s) E°cell=1.563V Use Nernst equation to solve for [Ag+]: E=E°cell-(0.0592/2)log([Zn2+]/[Ag+]2) 1.500V = 1.563-(0.0592/2)log((1.5M)/(x2)) log((1.5M)/(x2)) = -2(1.500-1.563)/0.0592 log((1.5M)/(x2)) = 2.13 x = 0.4222 [Ag+]= 0.4222 M 10. [Zn2+]= 1.2M What [Cu2+] will allow a forward spontaneous reaction? Net Reaction: Zn(S) + Cu2+ --> Cu(s) + Zn 2+ E°cell = 1.100V Solution: Nernst equation setting E=0.000V & [Zn2+] =1.0M E=E°cell-(0.0592/2)log([Zn2+]/[Cu2+]) 0.00V = 1.100-(0.0592/2)log((1.2M)/[Cu2+]) (0.00-1.100)/(-0.0296) = log((1.2M)/[Cu2+]) 37.2=log((1.2M)/[Cu2+]) [Cu2+]= 8.4×10-17M 11. A voltaic cell: Ag(s)|Ag+(saturated Ag2CrO4)||Ag+(0.125M)|Ag(s) What is Ecell if Ksp=1.2×10-11 for Ag2CrO4? Solution: First must solve for [Ag+] in saturated solution Ksp = [Ag+]2[CrO42-] = (2s)2(s) = 4s3 S = Cube root of (1.2x10-11/4) = 1.44×10-4 M [Ag+]anode = 2s = 2.88×10-4M Ag(s)+Ag+(0.125M)-->Ag(s)+Ag+(2.88×10-4M) Ecell=E°cell—(0.0592/2)log((2.88×10-4M)/(0.125M)) Electrochemistry 63 Ecell=0.000+0.078V =0.078V 12. Describe what would happen to a nail that is in a mixture which demonstrates a pink color if there is a base present and a blue color in the presence of Turbull's blue KFe[Fe(CN)6] in each situation: a) Copper wire is wrapped around the tip of the nail b) The center of the nail has the outer layer filed off c) The nail is coated with Zn (galvination) Solution: a) Iron is a stronger reducing agent than Copper and so blue precipitate will form in the areas not covered by Copper (oxidation will occur). Hydroxide would be produced at the cathode (copper wire) and pink will form. b) Newly exposed metal will oxidize producing a blue precipitate c) The Zinc will oxidize rather than the Iron producing hydroxide ions (pink color). 13. An aqueous solution containing [ a) Zn2+ b) Al3+ c) Ag+ d) Ni2+] has 2.15A of current passed through for 60 min, how much metal (in g) is formed at the cathode? Solution: First calculate charge passed through and number of moles of electrons transferred: Mol e- = 60 min ×(60 s/1 min)×(2.15 C/1 s)×(1 mol e-/96485 C) = 0.08 mol ea) Zn2+ 0.08 mol e- × (1 mol Zn2+/2 mol e-) × (1 mol Zn/1 mole Zn2+) × (65.39g Zn/1 mol Zn) = 2.62g Zn b) Al3+ 0.08 mol e- × (1 mol Al3+/3 mol e-) ×(1 mol Al/1 mole Al3+) × (26.98g Al/1 mol Al) = 0.72g Al c) Ag+ 0.08 mol e- × (1 mol Ag+/1 mol e-) ×(1 mol Ag/1 mole Ag+) × (107.9g Al/1 mol Ag)= 8.63g Ag d) Ni2+ 0.08 mol e- × (1 mol Ni2+/2 mol e-) ×(1 mol Ni/1 mole Ni2+) × (58.69g Ni/1 mol Ni)= 2.35g Ni 14. What voltage is required for the electrolysis of the following reactions? All reactants are in standards states. a) Zn(s) + Sn2+(aq) --> Zn2+(aq) + Sn(s) b) 2Fe2+(aq) + Hg2+(aq) --> 2Fe3+(aq) + Hg(l) c) Cu(s) + Sr2+(aq) --> Cu2+(aq) + Sr(s) Solution: a) oxidation reaction: Zn(s) --> Zn2+(aq) + 2e- E○ cell= -0.763V reduction reaction: Sn2+(aq) + 2e- --> Sn(s) E○ cell= -0.137V net reaction: Zn(s) + Sn2+(aq) --> Zn2+(aq) + Sn(s) E○ cell= 0.626V This reaction is spontaneous b) oxidation reaction: 2Fe2+ --> 2Fe3+ + 2ereduction reaction: Hg2+(aq) + 2e- --> Hg(l) E○ cell= 0.771V E○ cell= 0.854V net reaction: 2Fe2+(aq) + Hg2+(aq) --> 2Fe3+(aq) + Hg(l) E○ cell= 0.083 V This reaction is spontaneous c) oxidation reaction: Cu(s) --> Cu2+(aq) + 2ereduction reaction: Sr2+(aq) + 2e-(aq) --> Sr(s) E○ cell= 0.337V E○ cell= -2.89V net reaction: Cu(s) + Sn4+(aq) --> Cu2+(aq) + Sn2+(aq) E○ cell= -3.227V reaction requires voltages >3.227V 15. N2H4(aq) +O2(g) --> N2(g) + 2H2O(l) E°cell=1.559V Calculate ∆G°f for [N2H4(aq)] Solution: Calculate ∆G°f = -nFE°cell = -4 mol e- × (96485 C/1 mol e-) × 1.559V = -6.017×105 J = -601.7 kJ Find ∆G°f for (N2H4): -601.7 kJ = ∆G°f[N2(g)] + 2 ∆G°f[H2O(l)] - ∆G°f[N2H4(aq)] - ∆G°f[02(g)] -601.7kJ = 0.00kJ + 2(-237.2) + 601.7 = 127.3 kJ Electrochemistry 64 Electrochemistry 65 Galvanic cell A galvanic cell, or voltaic cell, named after Luigi Galvani, or Alessandro Volta respectively, is an electrochemical cell that derives electrical energy from spontaneous redox reaction taking place within the cell. It generally consists of two different metals connected by a salt bridge, or individual halfcells separated by a porous membrane. Volta was the inventor of the voltaic pile, the first electrical battery. In common usage, the word "battery" has come to include a single galvanic cell, but a battery properly consists of multiple cells. True or False 1. Hydrogen has oxidation potentials of 0. 2. The standard oxidation potential is not much like the standard reduction potential. 3. The standard reduction cell potential and the standard oxidation cell potential can never be combined. Solutions 1. True 2. False: the standard oxidation potential is much like the standard reduction potential 3. False: The standard reduction cell potential and the standard oxidation cell potential can be combined to determine the overall cell potential Practice Problems 1. What does the standard reduction potential measure? 2. What are the differences between the standard reduction potential and standard oxidation potential, and how are the two related? 3. What conditions must be met for a potential to be standard? 4. When standard reduction potentials are measured, what are the potentials relative to? 5. How is a standard reduction potential measured? 6. Explain how the activity series is used. 7. Based on the activity series, which species will be oxidized and reduced: Zn2+ or H+. 8. Explain how standard reduction potentials or standard oxidation potentials are applied. 9. Draw and label a SHE. 10. The standard reduction potential of Fe3+ is +0.77V. What is its standard oxidation potential. Solutions 1. Standard reduction potential measures the tendency for a given chemical species to be reduced. 2. The standard oxidation potential measures the tendency for a given chemical species to be oxidized as opposed to be reduced. For the same chemical species the standard reduction potential and standard oxidation potential are opposite in sign. 3. The cell must be at 298K, 1atm, and all solutions must be at 1M. 4. Standard reduction potentials are measured with relativity to hydrogen which has be universally set to have a potential of zero. 5. A standard reduction potential is measured using a galvanic cell which contains a SHE on one side and an unknown chemical half cell on the other side. The amount of charge that passes between the cells is measured using a voltmeter. 6. The activity series is a list of standard reduction potentials in descending order of the tendency for chemical species to be reduced. Species at the top are more likely to be reduced while species at the bottom are more likely to be oxidized. 7. H+ is farther up on the activity series then Zn2+ so H+ is reduced while Zn2+ is oxidized. 8. Standard reduction and oxidation potentials can be applied to solve for the standard cell potential of two different non hydrogen species. Examples can be seen in Cell Potentials. 9. See Figure (2). 10. The standard oxidation potential and standard reduction potential are always opposite in sign for the same species. The oxidation potential is -0.77V. Electrochemistry 66 Standard Electrode Potentials In an electrochemical cell, an electric potential is created between two dissimilar metals. This potential is a measure of the energy per unit charge which is available from the oxidation/reduction reactions to drive the reaction. It is customary to visualize the cell reaction in terms of two half-reactions, an oxidation half-reaction and a reduction half-reaction. Reduced species -> oxidized species + ne- Oxidation at anode - Oxidized species + ne -> reduced species Reduction at cathode The cell potential (often called the electromotive force or emf) has a contribution from the anode which is a measure of its ability to lose electrons - it will be called its "oxidation potential". The cathode has a contribution based on its ability to gain electrons, its "reduction potential". The cell potential can then be written Ecell = oxidation potential + reduction potential If we could tabulate the oxidation and reduction potentials of all available electrodes, then we could predict the cell potentials of voltaic cells created from any pair of electrodes. Actually, tabulating one or the other is sufficient, since the oxidation potential of a half-reaction is the negative of the reduction potential for the reverse of that reaction. Two main hurdles must be overcome to establish such a tabulation 1. 2. The electrode potential cannot be determined in isolation, but in a reaction with some other electrode. The electrode potential depends upon the concentrations of the substances, the temperature, and the pressure in the case of a gas electrode. In practice, the first of these hurdles is overcome by measuring the potentials with respect to a standard hydrogen electrode. It is the nature of electric potential that the zero of potential is arbitrary; it is the difference in potential which has practical consequence. Tabulating all electrode potentials with respect to the same standard electrode provides a practical working framework for a wide range of calculations and predictions. The standard hydrogen electrode is assigned a potential of zero volts. The second hurdle is overcome by choosing standard thermodynamic conditions for the measurement of the potentials. The standard electrode potentials are customarily determined at solute concentrations of 1 Molar, gas pressures of 1 atmosphere, and a standard temperature which is usually 25°C. The standard cell potential is denoted by a degree sign as a superscript. E°Cell 1. Measured against standard hydroden electrode. 2. Concentration 1 Molar 3. Pressure 1 atmosphere 4. Temperature 25°C The example below shows some of the extreme values for standard cell potentials. Cathode (Reduction) Standard Potential Half-Reaction E° (volts) Li+(aq) + e- -> Li(s) -3.04 K+(aq) + e- -> K(s) -2.92 Ca2+(aq) + 2e- -> Ca(s) -2.76 Na+(aq) + e- -> Na(s) -2.71 Zn2+(aq) + 2e- -> Zn(s) -0.76 Cu2+(aq) + 2e- -> Cu(s) 0.34 O3(g) + 2H+(aq) + 2e- -> O2(g) + H2O(l) 2.07 F2(g) + 2e- -> 2F-(aq) 2.87 The values for the table entries are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential. The link below takes you to a more extensive table. Useful applications of the standard electrode potentials include the following. Strengths of oxidizing and reducing agents Electrode potentials under non-standard conditions Potentials for voltaic cells Relationship to Gibbs free energy Relationship to equilibrium constants Electrochemistry 67 Calculation of Voltaic Cell Potentials When an electrochemical cell is arranged with the two half-reactions separated but connected by an electrically conducting path, a voltaic cell is created. The maximum voltage which can be produced between the poles of the cell is determined by the standard electrode potentials under the standard conditions under which those potentials are defined. Consider the historic Daniell cell in which zinc and copper were used as electrodes. The data from the table of standard electrode potentials is Cathode (Reduction) Standard Potential Half-Reaction E° (volts) 2+ Zn (aq) + 2e -> Zn(s) -0.76 Cu2+(aq) + 2e- -> Cu(s) 0.34 The cell potential can be written Ecell = oxidation potential + reduction potential Since the tabulated standard electrode potentials are reduction potentials, the one which is most negative will need to be reversed in sign to get its oxidation potential. When that is done, it is clear that the theoretical standard cell potential for the zinc-copper cell is 1.10 volts. In general, a real voltaic cell will differ from the standard conditions, so we need to be able to adjust the calculated cell potential to account for the differences. This can be done with the application of the Nernst equation. Presuming that there are two metal electrodes with their ions, the standard potential for a cell with anode potential and cathode potential volts volts is E°cell = If the ion concentration at the anode is [A] = volts M and the concentration at the cathode is [C] = M, then the thermodynamic reaction quotient Q = [A]/[C] = For temperature = °C = K and number of electrons exchanged n = Ecell = , the calculated cell potential is volts By numerically exploring the above calculation, you can confirm that large concentration differences between the anions and the cations can cause the voltage to differ significantly from the standard cell potential, and that such large concentration differences also make the cell potential more temperature dependent. Electrochemistry 68 The Nernst Equation The cell potential for a voltaic cell under standard conditions can be calculated from the standard electrode potentials. But real voltaic cells will typically differ from the standard conditions. The Nernst equation relates the cell potential to its standard cell potential. R= gas constant T = temperature in Kelvins Q = thermodynamic reaction quotient F = Faraday's constant n = number of electrons transferred The quantity Q, the thermodynamic reaction constant, is like a dynamic version of the equilibrium constant in which the concentrations and gas pressures are the instantaneous values in the reaction mixture. For a reaction the reaction quotient has the form where [C] is understood to be the molar concentration of product C, or the partial pressure in atmospheres if it is a gas. Applied to the Daniell cell where zinc and copper form the electrodes, the reaction is Zn(s) + Cu2+(aq) <-> Zn2+(aq) + Cu(s) , the form of Q is since the concentrations of the pure metal solids are assigned the value 1. This implies that the departure of the cell potential from its standard value of 1.10 volts will be influenced by the temperature and the ion concentrations. One implication is that the cell potential will be reduced from the standard value if the concentration of Zn2+(aq) is greater than that of Cu2+(aq) at the standard temperature. An excess concentration of Cu2+(aq) will give a higher voltage. The graph at right shows the increase in cell voltage with increasing concentration of the cation. Note that the horizontal axis is logarithmic, and that the straight line variation of the voltage represents an logarithmic variation with Q. Note that the cell potential is equal to the standard value if the concentrations are equal even if they are not equal to the standard value of 1M, since the logarithm gives the value zero. Consider a concentration of 10-5Molar for Zn2+(aq) and 0.1 Molar for Cu2+(aq) as a test case for temperature dependance. We can see that the cell potential tends to increase with temperature, or that a colder cell prodices less voltage - a commonly observed phenomenon with dry cell batteries. The variation with temperature is linear with temperature, but quite small for this cell. The large variations of practical output voltage with temperature for dry cells does not arise from the Nernst equation alone. Electrochemistry 69 Background for the Nernst Equation The Nernst equation allows us to predict the cell potential for voltaic cellsunder conditions other than standard conditions of 1M, 1 atmosphere, 25°C. The effects of different temperatures and concentrations may be tracked in terms of the Gibbs free energy change ∆G. This free energy change depends upon the temperature and concentrations according to where ∆G° is the free energy change under standard conditions and Q is the thermodynamic reaction quotient. The free energy change is related to the cell potential Ecell by so for non-standard conditions or which is called the Nernst equation. Lithium-ion battery Lithium-ion battery Varta Lithium-ion battery, Museum Autovision, Altlußheim, Germany Energy/weight 100-160 Wh/kg Energy/size 250-360 Wh/L Power/weight ~250-~340 W/kg Charge/discharge efficiency 80-90% Energy/consumer-price 2.8-5 Wh/US$ Self-discharge rate 8(21°C), 15(40°C) and 31%(60°C) per month Time durability (24-36) months Cycle durability ~1200 cycles Nominal cell voltage 3.6 / 3.7 V Cylindrical 18650-cell before closing Electrochemistry 70 A lithium-ion battery (sometimes abbreviated Li-ion battery) is a type of rechargeable battery in which the cathode (positive electrode) contains lithium. The anode (negative electrode) is generally made of a type of porous carbon. During discharging, the current flows within the battery (when the external circuit is connected) from the anode to the cathode, as in any type of battery: the internal process is the movement of Li+ ions from the anode to the cathode, through the non-aqueous electrolyte and separator diaphragm. During charging, an external electrical power source (the charging circuit) forces the current to pass in the reverse direction; the positive terminal from the charging circuit has to be connected to the cathode of the battery, and the anode has to be connected to the negative terminal of the external circuit. The lithium ions then migrate from the cathode to the anode, where they become embedded in the porous electrode material in a process known as intercalation. The electrolyte is of such nature that it complexes with the lithium ions. These non-aqueous electrolytes are generally based on patented formulations containing manganese or cobalt salts. Pure lithium, like sodium, is very reactive. It will vigorously react with water to form lithium hydroxide and hydrogen gas is liberated. Thus a nonaqueous electrolyte is used, and water is rigidly excluded from the battery pack by using a sealed container. Lithium-ion batteries are common in portable consumer electronics because of their high energy-to-weight ratios, lack of memory effect, and slow self-discharge when not in use. In addition to consumer electronics, lithium-ion batteries are increasingly used in defense, automotive, and aerospace applications due to their high energy density. However, certain kinds of mistreatment may cause conventional Li-ion batteries to explode. The three primary functional components of a lithium-ion battery are the anode, cathode, and electrolyte, for which a variety of materials may be used. Commercially, the most popular material for the anode is graphite. The cathode is generally one of three materials: a layered oxide (such as lithium cobalt oxide), one based on a polyanion (such as lithium iron phosphate), or a spinel (such as lithium manganese oxide), although materials such as TiS2 (titanium disulfide) originally were also used. Depending on the choice of material for the anode, cathode, and electrolyte, the voltage, capacity, life, and safety of a lithium-ion battery can change dramatically. Recently, novel architectures have been employed to improve the performance of these batteries. Lithium-ion batteries are not to be confused with lithium batteries, the key difference being that lithium batteries are primary batteries, containing metallic lithium, while lithium-ion batteries are secondary batteries, containing an intercalation anode material. History Lithium-ion batteries were first proposed by M.S. Whittingham (Binghamton University), then at Exxon, in the 1970s. Whittingham used titanium(II) sulfide as the cathode and lithium metal as the anode. The electrochemical properties of the lithium intercalation in graphite were first discovered in 1980 by Rachid Yazami et al. at the Grenoble Institute of Technology (INPG) and French National Centre for Scientific Research (CNRS) in France. They showed the reversible intercalation of lithium into graphite in a lithium/polymer electrolyte/graphite half cell. Their work was published in 1982 and 1983. It covered both the thermodynamics (staging) and the kinetics (diffusion) aspects of the lithium intercalation into graphite together with reversibility. Lithium batteries in which the anode is made from metallic lithium pose severe safety issues. As a result, lithium-ion batteries were developed in which the anode, like the cathode, is made of a material containing lithium ions. In 1981, Bell Labs developed a workable graphite anode to provide an alternative to the lithium battery. Following groundbreaking cathode research by a team led by John Goodenough,the first commercial lithium-ion battery was released by Sony in 1991. The cells used layered oxide chemistry, specifically lithium cobalt oxide. These batteries revolutionized consumer electronics. In 1983, Michael Thackeray, John Goodenough, and coworkers identified manganese spinel as a cathode material. Spinel showed great promise, since it is a low-cost material, has good electronic and lithium ion conductivity, and possesses a three-dimensional structure which gives it good structural stability. Although pure manganese spinel fades with cycling, this can be overcome with additional chemical modification of the material. Manganese spinel is currently used in commercial cells. In 1989, Arumugam Manthiram and John Goodenough of the University of Texas at Austin showed that cathodes containing polyanions, eg. sulfates, produce higher voltage than oxides due to the inductive effect of the polyanion. In 1996, Akshaya Padhi, John Goodenough and coworkers identified the lithium iron phosphate (LiFePO4) and other phospho-olivines (lithium metal phosphates with olivine structure) as cathode materials for lithium-ion batteries. LiFePO4 is superior over other cathode materials in terms of cost, safety, stability and performance, and is most suitable for large batteries for electric automobiles and other energy storage applications such as load saving, where safety is of utmost importance. It is currently being used for most lithium-ion batteries powering portable devices such as laptop computers and power tools. In 2002, Yet-Ming Chiang and his group at MIT published a paper in which they showed a dramatic improvement in the performance of lithium batteries by boosting the material's conductivity by doping it with aluminium, niobium and zirconium, though at the time, the exact mechanism causing the increase became the subject of a heated debate. Electrochemistry 71 In 2004, Chiang again increased performance by utilizing iron-phosphate particles of less than 100 nm in diameter. This miniaturized the particle density by almost a hundredfold, increased the surface area of the electrode and improved the battery's capacity and performance. Commercialization of the iron-phosphate technology led to a competitive market and a patent infringement battle between Chiang and Goodenough. Electrochemistry The three participants in the electrochemical reactions in a lithium-ion battery are the anode, cathode, and electrolyte. Both the anode and cathode are materials into which and from which lithium can migrate. The process of lithium moving into the anode or cathode is referred to as insertion (or intercalation ), and the reverse process, in which lithium moves out of the anode or cathode is referred to as extraction (or deintercalation). When a lithium-based cell is discharging, the lithium is extracted from the anode and inserted into the cathode. When the cell is charging, the reverse process occurs: lithium is extracted from the cathode and inserted into the anode. During discharge, the anode of a conventional Li-ion cell is made from carbon, the cathode is a metal oxide, and the electrolyte is a lithium salt in an organic solvent. Useful work can only be extracted if electrons flow through a (closed) external circuit. The following equations are written in units of moles, making it possible to use the coefficient x. The cathode half-reaction (with charging being forwards) is: The anode half reaction is: The overall reaction has its limits. Overdischarge will supersaturate lithium cobalt oxide, leading to the production of lithium oxide, possibly by the following irreversible reaction: Overcharge up to 5.2V leads to the synthesis of cobalt(IV) oxide, as evidenced by x-ray diffraction In a lithium-ion battery the lithium ions are transported to and from the cathode or anode, with the transition metal, Co, in LixCoO2 being oxidized from Co3+ to Co4+ during charging, and reduced from Co4+ to Co3+ during discharge. Cathodes Cathode Material Average Voltage Gravimetric Capacity Gravimetric Energy LiCoO2 3.7 V 140 mAh/g 0.518 kW·h/kg LiMn2O4 4.0 V 100 mAh/g 0.400 kW·h/kg LiNiO2 3.5 V 180 mAh/g ? kW·h/kg LiFePO4 3.3 V 150 mAh/g 0.495 kW·h/kg Li2FePO4F 3.6 V 115 mAh/g 0.414 kW·h/kg LiCo1/3Ni1/3Mn1/3O2 3.6 V 160 mAh/g ? kW·h/kg Li(NixMnyCoz)O2 ? mAh/g ? kW·h/kg ?V Anodes Anode Material Average Voltage Gravimetric Capacity Gravimetric Energy Graphite (LiC6) 0.1-0.2 V 372 mAh/g 0.0372-0.0744 kW·h/kg Hard Carbon (LiC6) ? V ? mAh/g ? kW·h/kg Titanate (Li4Ti5O12) 1-2 V 160 mAh/g 0.16-0.32 kW·h/kg Silicium (Li22Si6) ?V ? mAh/g ? kW·h/kg Si (Li4.4Si) 0.5-1 V 4212 mAh/g 2.106-4.212 kW·h/kg Ge (Li4.4Ge) 0.7-1.2 V 1624 mAh/g 1.137-1.949 kW·h/kg Electrochemistry 72 See uranium trioxide for some details of how the cathode works. While uranium oxides are not used in commercially-made batteries, intercalation and deintercalation function in the same way as with lithium-based cells. Electrolytes The cell voltages given in the section above are larger than the potential at which aqueous solutions would electrolyze. Therefore, nonaqueous solutions are used. Liquid electrolytes in lithium-ion batteries consist of lithium salts, such as LiPF6, LiBF4 or LiClO4 in an organic solvent, such as ethylene carbonate. A liquid electrolyte conducts lithium ions, acting as a carrier between the cathode and the anode when a battery passes an electric current through an external circuit. Typical conductivities of liquid electrolyte at room temperature (20 oC) are in the range of 10 mS/cm, increasing by approximately 3040% at 40 oC and decreasing by a slightly smaller amount at 0 oC. Unfortunately, organic solvents are easily decomposed on anodes during charging. However, when appropriate organic solvents are used as the electrolyte, the solvent is decomposed on initial charging and forms a solid layer called the solid electrolyte interphase (SEI), which is electrically insulating yet sufficiently conductive to lithium ions. The interphase prevents decomposition of the electrolyte after the second charge. For example, ethylene carbonate is decomposed at a relatively high voltage, 0.7 V vs. Li, and forms a dense and stable interface. Advantages and disadvantages Advantages • Lithium-ion batteries can be formed into a wide variety of shapes and sizes so as to efficiently fill available space in the devices they power. • Lithium-ion batteries are lighter than other energy-equivalent secondary batteries—often much lighter. A key advantage of using lithium-ion chemistry is the high open circuit voltage that can be obtained in comparison to aqueous batteries (such as lead acid, nickel-metal hydride and nickel-cadmium). Lithium-ion batteries do not suffer from the memory effect. They also have a self-discharge rate of approximately 5-10% per month, compared with over 30% per month in common nickel metal hydride batteries, approx. 1.25% per month for Low Self-Discharge NiMH batteries and 10% per month in nickel-cadmium batteries. According to one manufacturer, Li-ion cells (and, accordingly, "dumb" Li-ion batteries) do not have any self-discharge in the usual meaning of this word. What looks like a self-discharge in these batteries is a permanent loss of capacity (see below). On the other hand, "smart" Li-ion batteries do self-discharge, mainly due to the small constant drain of the built-in voltage monitoring circuit. • Disadvantages of traditional Li-ion technology Shelf life • • • A disadvantage of lithium-ion cells lies in their relatively poor cycle life: upon every (re)charge, deposits form inside the electrolyte that inhibit lithium ion transport, resulting in the capacity of the cell to diminish. The increase in internal resistance affects the cell's ability to deliver current, thus the problem is more pronounced in high-current than low-current applications. The increasing capacity hit means that a full charge in an older battery will not last as long as one in a new battery (although the charging time required decreases proportionally, as well). Also, high charge levels and elevated temperatures (whether resulting from charging or being ambient) hasten permanent capacity loss for lithium-ion batteries. The heat generated during a charge cycle is caused by the traditional carbon anode, which has been replaced with good results by lithium titanate. Lithium titanate has been experimentally shown to drastically reduce the degenerative effects associated with charging, including expansion and other factors. See "Improvements of lithium-ion technology" below. At a 100% charge level, a typical Li-ion laptop battery that is full most of the time at 25 °C or 77 °F will irreversibly lose approximately 20% capacity per year. However, a battery in a poorly ventilated laptop may be subject to a prolonged exposure to much higher temperatures, which will significantly shorten its life. Different storage temperatures produce different loss results: 6% loss at 0 °C (32 °F), 20% at 25 °C (77 °F), and 35% at 40 °C (104 °F). When stored at 40%–60% charge level, the capacity loss is reduced to 2%, 4%, 15% at 0, 25 and 40 degrees Celsius respectively. Internal resistance The internal resistance of lithium-ion batteries is high compared to other rechargeable chemistries such as nickel-metal hydride and nickel-cadmium. It increases with both cycling and chronological age. Rising internal resistance causes the voltage at the terminals to drop under load, reducing the maximum current that can be drawn from them. Eventually they reach a point at which the battery can no longer operate the equipment it is installed in for an adequate period. High drain applications such as power tools may require the battery to be able to supply a current that would drain the battery in 1/15 hour if sustained; e.g. 22.5 A for a battery with a capacity of 1.5 A·h). Lower-power devices such as MP3 players, on the other hand, may draw low enough current to run for 10 hours on a charge (e.g. 150 mA for a battery with a capacity of 1500 mA·h). With similar battery technology, the MP3 player's battery will effectively last much longer, since it can tolerate a much higher internal resistance. To power larger devices, such as electric cars, it is much more efficient to connect many smaller batteries in a parallel circuit rather than using a single large battery. Safety requirements Li-ion batteries are not as durable as nickel metal hydride or nickel-cadmium designs and can be extremely dangerous if mistreated. They may explode if overheated or if charged to an excessively high voltage. Furthermore, they may be irreversibly damaged if discharged below a certain Electrochemistry 73 voltage. To reduce these risks, lithium-ion batteries generally contain a small circuit that shuts down the battery when it is discharged below about 3 V or charged above about 4.2 V. In normal use, the battery is therefore prevented from being deeply discharged. When stored for long periods, however, the small current drawn by the protection circuitry may drain the battery below the protection circuit's lower limit, in which case normal chargers are unable to recharge the battery. More sophisticated battery analyzers can recharge deeply discharged cells by slow-charging them to reactivate the safety circuit and allow the battery to accept charge again. Other safety features are also required for commercial lithium-ion batteries: • • • • shut-down separator (for overtemperature), tear-away tab (for internal pressure), vent (pressure relief), and thermal interrupt (overcurrent/overcharging). These devices occupy useful space inside the cells, and reduce their reliability; typically, they permanently and irreversibly disable the cell when activated. They are required because the anode produces heat during use, while the cathode may produce oxygen. Safety devices and recent and improved electrode designs greatly reduce or eliminate the risk of fire or explosion. These safety features increase the cost of lithium-ion batteries compared to nickel metal hydride cells, which only require a hydrogen/oxygen recombination device (preventing damage due to mild overcharging) and a back-up pressure valve. Many types of lithium-ion cell cannot be charged safely below 0 °C. Product recalls About 1% of lithium-ion batteries are recalled. Specifications and design A lithium-ion battery from a mobile phone. • • Specific energy density: 150 to 220 Wh/kg (540 to 720 kJ/kg) Volumetric energy density: 250 to 530 Wh/l (900 to 1900 J/cm³) Specific power density: 300 to 1500 W/kg (@ 20 seconds and 285 Wh/l) • Because lithium-ion batteries can have a variety of cathode and anode materials, the energy density and voltage vary accordingly. Lithium-ion batteries with a lithium iron phosphate cathode and graphite anode have a nominal open-circuit voltage of 3.2 V and a typical charging voltage of 3.6 V. Lithium nickel manganese cobalt (NMC) oxide cathode with graphite anodes have a 3.7 V nominal voltage with a 4.2 V max charge. The charging procedure is performed at constant voltage with current-limiting circuitry (i.e., charging with constant current until a voltage of 4.2 V is reached in the cell and continuing with a constant voltage applied until the current drops close to zero). Typically, the charge is terminated at 7% of the initial charge current. In the past, lithium-ion batteries could not be fast-charged and typically needed at least two hours to fully charge. Currentgeneration cells can be fully charged in 45 minutes or less; some lithium-ion varieties can reach 90% in as little as 10 minutes. Charging procedure Stage 1: Apply charging current limit until the voltage limit per cell is reached. Stage 2: Apply maximum voltage per cell limit until the current declines below 3% of rated charge current. Stage 3: Periodically apply a top-off charge about once per 500 hours. The charge time is about three to five hours, depending upon the charger used. Generally, cell phone batteries can be charged at 1C and laptop-types at 0.8C, where C is the current that would discharge the battery in one hour. Charging is usually stopped when the current goes below 0.03C but it can be left indefinitely depending on desired charging time. Some fast chargers skip stage 2 and claim the battery is ready at 70% charge. Laptop battery chargers sometimes gamble, and try to charge up to 4.35v then disconnect battery. This helps to compensate internal resistance and charge up to 100% in short time. Top-off charging is recommended to be initiated when voltage goes below 4.05 V/cell. Electrochemistry 74 Lithium-ion cells are charged with 4.2 ± 0.05 V/cell,except for military long-life cells where 3.92 V is used to extend battery life. Most protection circuits cut off if either 4.3 V or 90 °C is reached. If the voltage drops below 2.50 V per cell, the battery protection circuit may also render it unchargeable with regular charging equipment. Most battery protection circuits stop at 2.7–3.0 V per cell. For safety reasons it is recommended to stay within the manufacturer's stated voltage and current ratings during both charge and discharge cycles. Improvements focus on several areas, and often involve advances in nanotechnology and microstructures. • • • Increasing cycle life and performance (decreases internal resistance and increases output power) by changing the composition of the material used in the anode and cathode, along with increasing the effective surface area of the electrodes (related developments have helped ultracapacitors) and changing materials used in the electrolyte and/or combinations thereof (e.g., Li-VOx-based cells with polymer electrolyte). Improving capacity by improving the structure to incorporate more active materials. Improving the safety of lithium-ion batteries. Manganese spinel cathodes LG (Lucky Goldstar Chemical), which is the third-largest producer of lithium-ion batteries, uses the lithium manganese spinel for its cathode. It is working with its subsidiary CPI to commercialize lithium-ion batteries containing manganese spinel for HEV applications. Several other companies are also working with manganese spinel, including NEC and Samsung. Lithium iron phosphate cathode with traditional anode The University of Texas first licensed its patent for lithium iron phosphate cathodes to the Canadian utility Hydro-Québec. Phostech Lithium inc. was later spun-off from Hydro-Québec for the sole development of lithium iron phosphate. Valence Technology, located in Austin, Texas, is also working on lithium iron magnesium phosphate cells. Since March 2005, the Segway Personal Transporter has been shipping with extended-range lithium-ion batteries made by Valence Technology using iron magnesium phosphate cathode materials. Segway, Inc. chose to build their large-format battery with this cathode material because of its improved safety over metal-oxide materials. To date Valence has shipped 100,000 batteries to Segway. In November 2005, A123Systems announced the development of lithium iron phosphate cells based on research licensed from MIT. While the battery has slightly lower energy density than other competing lithium-ion technologies, a 2 Ah cell can provide a peak of 70 Amps without damage and operate at temperatures above 60 degrees C. Their first cell has been in production since 2006 and is being used in consumer products including DeWalt power tools, aviation products, automotive hybrid systems and PHEV conversions. High power cathode using lithium nickel manganese cobalt (NMC) Imara Corporation, based in Menlo Park, CA is commercializing a new materials-agnostic technology first applied on an NMC material which has the effect of lowering impedance and extending cycle life. These high power-capable cells have high energy density relative to other high power cells in the market.[53][unreliable source?] The batteries are being deployed in power tools, outdoor power equipment and hybrid vehicles; Sony and Sanyo use NMC and NCA blended with LMO (spinel) for high-powered applications. NMC has a significant safety advantage over cobalt oxide and 50% greater energy density than FePO4, but suffers from a poor cycle life. Nissan Motor Co. has nearly completed development of a lithium-ion battery using a lithium nickel manganese cobalt oxide cathode (NMC). The new system, which will reportedly offer almost double the capacity of Nissan/AESC’s current manganese spinel cell. Traditional cathode with lithium titanate anode Altairnano, a small firm based in Reno, Nevada, has announced a nano-sized titanate electrode material for lithium-ion batteries. It is claimed the prototype battery has three times the power output of existing batteries and can be fully charged in six minutes. However, total energy capacity per cell is about half that of normal lithium-ion cells. The company also says the battery cells have now achieved a life of over 9,000 charge cycles while still retaining up to 85% charge capacity. Durability and battery life are therefore much longer, estimated to be around 20 years, or four times longer than regular lithium-ion batteries. The batteries can operate from -50 °C to over 75 °C and will not explode or experience thermal runaway, even under severe conditions, because they do not contain graphite-coated-metal anode electrode material. The batteries are currently being tested in a new production car made by Phoenix Motorcars which was on display at the 2006 SEMA motorshow. They're also being tested, on a one MW grid scale, in the PJM Interconnection Regional Transmission Organization control area in Norristown, Pennsylvania as well as by several branches of the United States Department of Defense. In addition, the batteries are being demonstrated by Proterra in their all-electric EcoRide BE35 vehicle, a lightweight 35-foot bus. Altairnano is currently working with three different cell chemistries for various energy and power storage applications, with another new cell chemistry expected in the fall of 2009. The nature of their latest cathode materials is currently proprietary. Combined anode and cathode developments EnerDel, which is jointly owned by Ener1 and Delphi, is working to commercialize cells containing a titanate anode and manganese spinel cathode. Although the cells show excellent thermal properties and cyclability, their low voltage may hamper commercial success. Research claims In April 2006, a group of scientists at MIT announced a process which uses viruses to form nano-sized wires. These can be used to build ultrathin lithium-ion batteries with three times the normal energy density. Electrochemistry 75 As of June 2006, researchers in France have created nanostructured battery electrodes with several times the energy capacity, by weight and volume, of conventional electrodes. In the September 2007 issue of Nature, researchers from the University of Waterloo, Canada, reported a new cathode chemistry, in which the hydroxyl group in the iron phosphate cathode was replaced by fluorine. The advantages seem to be two-fold. First, there is less volume change in the cathode over a charge cycle which may improve battery life. Secondly, the chemistry allows the substitution of the lithium in the battery with either sodium or a sodium/lithium mixture (hence their reference to it as an Alkali-Ion battery). In November 2007, Subaru unveiled their concept G4e electric vehicle with a lithium vanadium oxide-based lithium-ion battery, promising double the energy density of a conventional lithium-ion battery (lithium cobalt oxide and graphite). In the lab, lithium vanadium oxide anodes, paired with lithium cobalt oxide cathodes, have achieved 745Wh/l, nearly three times the volumetric energy density of conventional lithium-ion batteries. In December 2007, researchers at Stanford University reported creating a lithium-ion nanowire battery with ten times the energy density (amount of energy available by weight) through using silicon nanowires deposited on stainless steel as the anode. The battery takes advantage of the fact that silicon can hold large amounts of lithium, and helps alleviate the longstanding problem of cracking by the small size of the wires. To gain a tenfold improvement in energy density, the cathode would need to be improved as well; however, even just improving the anode could provide "several" times the energy density, according to the team. The team leader, Yi Cui, expects to be able to commercialize the technology in about five years. Having a large capacitive anode will not increase the capacity of the battery as predicted by the author when the cathode material is far less capacitive than the anode. However, current lithium-ion capacity is mainly limited by the low theoretical capacity (372 mAh g−1) of the graphite in use as the anode material, so improvement could be significant and would then be limited by the cathode material instead. There are trials with metal hydrides as anode material for lithium-ion batteries. A practical electrode capacity as high as 1480 mAh g−1 has been reported. In April 2009 a report in New Scientist claimed that Angela Belcher's team at MIT had succeeded in producing the first full virus-based 3-volt lithiumion battery. In November 2009, engineers at the University of Dayton Research Institute developed the world's first solid-state, rechargeable lithium air battery which was designed to address the fire and explosion risk of other lithium rechargeable batteries and make way for development of large-size lithium rechargeables for a number of industry applications, including hybrid and electric cars. Recent studies performed at Binghamton University by M. S. Whittingham et al. determined that vanadium ions can be incorporated into the ironcontaining olivine structure of LiFePO4; a small amount of vanadium (around 5%) enhancing the rate capability of the LiFePO4 olivine cathode material. The resulting compound material had higher electronic and ionic conductivities, and they were of comparable magnitude. The doping reaction kinetics were optimal under reducing atmosphere during the synthesis of the LiFe0.95V0.05PO4 material. Guidelines for prolonging lithium-ion battery life • Lithium-ion batteries should never be depleted below their minimum voltage (2.4 to 2.8 V/cell, depending on chemistry). If a lithium-ion battery is stored with too low a charge, there is a risk that the charge will drop below the low-voltage threshold, resulting in an unrecoverable dead battery. Usually this does not instantly damage the battery itself but a charger or device which uses that battery will refuse to charge a dead battery. The battery appears to be dead or not existent because the protection circuit disables further discharging and there is zero voltage on the battery terminals. • Lithium-ion batteries should be kept cool. Ideally they are stored in a refrigerator. • Aging will take its toll much faster at high temperatures Prolonging life in multiple cells through cell balancing Analog front ends that balance cells and eliminate mismatches of cells in series or parallel significantly improve battery efficiency and increase the overall pack capacity. As the number of cells and load currents increase, the potential for mismatch also increases. There are two kinds of mismatch in the pack: state-of-charge (SOC) and capacity/energy (C/E) mismatch. Though the SOC mismatch is more common, each problem limits the pack capacity (mAh) to the capacity of the weakest cell. Safety Lithium-ion batteries can rupture, ignite, or explode when exposed to high-temperature environments, e.g. in an area that is prone to prolonged direct sunlight. Short-circuiting a lithium-ion battery can cause it to ignite or explode and any attempt to open or modify the casing or circuitry is dangerous. For this reason they normally contain safety devices that protect the cells from abuse. Contaminants inside the cells can defeat these safety devices. For example, the mid-2006 recall of approximately 10 million Sony batteries used in Dell, Sony, Apple, Lenovo/IBM, Panasonic, Toshiba, Hitachi, Fujitsu and Sharp laptops was stated to be as a consequence of internal contamination with metal particles. Under some circumstances, these can pierce the separator, causing the cell to short, rapidly converting all of the energy in the cell to heat resulting in an exothermic oxidizing reaction, increasing the temperature to a few hundred degrees Celsius in a fraction of a second. This causes the neighboring cells to heat up, causing a chain reaction. The mid-2006 Sony laptop battery recall was not the first of its kind; it was, however, the largest to date. During the past decade, there have been numerous recalls of lithium-ion batteries in cellular phones and laptops owing to overheating problems. In October 2004, Kyocera Wireless recalled approximately 1 million batteries used in cellular phones due to counterfeit batteries produced in Kyocera's name. In December 2006, Dell recalled approximately 22,000 batteries from the U.S. market. In March 2007, Lenovo recalled approximately 205,000 9-cell lithium-ion batteries due to an Electrochemistry 76 explosion risk. In August 2007, Nokia recalled over 46 million lithium-ion batteries, warning that some of them might overheat and possibly explode. One such incident occurred in the Philippines involving an Nokia N91, which uses the BL-5C battery. Replacing the lithium cobalt oxide cathode material in lithium-ion batteries with lithiated metal phosphate leads to longer cycle and shelf life, improves safety, but lowers capacity. Currently these 'safer' lithium-ion batteries are mainly used in electric cars and other large-capacity battery applications, where safety issues are critical. Another option is to use a manganese oxide or iron phosphate cathode. A new class of high power cathode materials, lithium nickel manganese cobalt (NMC) oxide has recently been introduced that have a significantly higher temperature tolerance compared to lithium cobalt oxide (see above). In the event of a lithium-ion battery explosion, dense white smoke which can cause severe irritation to the respiratory tract, eyes and skin will be generated. All precautions must be taken to limit exposure to these fumes. Restrictions on transportation As of January 2008, the United States Department of Transportation issued a new rule that permits passengers on board commercial aircraft to carry lithium batteries in their checked baggage IF the batteries are installed in a device. Types of batteries affected by this rule are those containing lithium, including Li-ion, lithium polymer, and lithium cobalt oxide chemistries. Lithium-ion batteries containing more than 25 grams Equivalent Lithium Content (ELC) are exempt from the rule and are forbidden in air travel. The purpose of this restriction is that it greatly reduces the chances of the batteries becoming short-circuited and causing a fire. A limited number of replacement batteries can be carried in hand luggage providing they are kept in their original protective packaging or in individual containers or plastic bags.
