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Chapter 18 Electrochemistry The batteries we use to power portable computers and other electronic devices all rely on redox reactions to generate an electric current. Redox reactions are central to the development of small, light, long-lasting power sources. To a large extent, the future development of technology depends on the capabilities of these power sources. In this chapter, we see what is involved in using chemical reactions to generate electricity. Assignment for Chapter 18 18.2; 18.6; 18.17; 18.27 Electrochemistry in Biological Systems Ion transport Muscle contraction Neuron excitation Psychological activities Electrochemistry • Electrochemistry is a branch of chemistry that deals with the use of spontaneous chemical reactions to produce electricity and the use of electricity to bring about spontaneous chemical change. Why electricity and chemical reaction may be connected? Because both involve electrons! The motion or transfer of electrons! Proton transfer: acid-base reactions Electron transfer: redox reactions Half-Reactions • Chemical equations showing the changes involved only in oxidation or reduction. Oxidation half-reaction: 2 Mg(s) Mg (aq) 2e Reduction half-reaction: Fe (aq)+3e Fe(s) 2+ - Example The oxidation of iron(II) to iron(III): Fe (aq) Fe (aq)+e 2+ 3+ - The reduction of copper(II) to copper metal: Cu (aq)+2e Cu(s) 2+ - Exercise: Aluminum metal is oxidized to Al3+ in aqueous solution: Al(s) Al (aq)+3e 3+ - Balancing Redox Reactions MnO4-(aq) + H2C2O4 (aq) Mn2+ (aq) + CO2 (g) ! • The chemical equation of a reduction halfreaction is added to that of an oxidation half-reaction to form the chemical equation for the overall redox reaction. Use H2O and H+ and OH- as “extra” chemicals Balancing Redox Reactions in acidic solutions MnO4-(aq) + H2C2O4 (aq) Mn2+ (aq) + CO2 (g) ! Reduction: MnO4- Mn2+ (aq) Oxidation: H2C2O4 (aq) CO2 (g) Balance all elements except H and O: Reduction: MnO4- Mn2+ (aq) Oxidation: H2C2O4 (aq) 2CO2 (g) Balance O using water and H using H+: Reduction: MnO4- + 8H+ Mn2+ (aq)+4H2O Oxidation: H2C2O4 (aq) 2CO2 (g) + 2H+ Balancing Redox Reactions Balance Charge: Reduction: MnO4- + 8H+ +5e- Mn2+ (aq)+4H2O Oxidation: H2C2O4 (aq) 2CO2 (g) + 2H+ + 2eReduction x 2 + Oxidation x 5 2MnO4- + 16H+ +10e- 2Mn2+ (aq)+8H2O 5H2C2O4 (aq) 10CO2 (g) + 10H+ + 10e2MnO4- (aq)+ 5H2C2O4 (aq) +6H+ (aq) 2Mn2+ (aq)+8H2O(l) + 10CO2 (g) Figure 18.1 A schematic diagram of how to balance a redox equation by balancing the half-reactions separately and then combining them. Example • Write down the balanced net ionic reaction for the reaction: Cu+HNO3Cu2++NO. Cu+H++NO3-Cu2++NO CuCu2+ 2NO3-2NO + 4H2O CuCu2+ 8H++2NO3-2NO + 4H2O CuCu2++2e- 8H++2NO3-+6e-2NO + 4H2O x3 3Cu(s)+2NO3-(aq)+8H+(aq)3Cu2+(aq)+2NO(g)+4H2O(l) Exercise MnO4++H2SO3 HSO4- + Mn2+. Write down the balanced net ionic reaction. Balancing Redox Reactions in basic solutions Br- (aq)+ MnO4-(aq) MnO2(s)+BrO3-(aq) Br- BrO3- MnO4-(aq) MnO2(s) Br- + 3H2OBrO3- MnO4-(aq) MnO2(s) +2H2O Br- + 3H2O+6OH-BrO3-+6H2O MnO4-(aq)+4H2O MnO2(s) +2H2O+4OHBr- + 6OH-BrO3-+3H2O +3eMnO4-(aq)+2H2O MnO2(s) +4OH-+6e2MnO4-(aq)+ Br-(aq) +H2O(l) MnO2(s)+BrO3-(aq)+2OH-(aq) Electrochemical Cell • A device in which an electric current is either produced by a spontaneous chemical reaction or is used to bring about a nonspontaneous reaction. A galvanic cell is an electrochemical cell in which a spontaneous chemical reaction is used to generate an electric current. Figure 18.2 In an electrochemical cell, a reaction takes place in two separate negative regions. Oxidation occurs at one electrode, and the electrons released travel through the external circuit to the other electrode, where they cause reduction. The site of oxidation is called the anode, and the site of reduction is called the cathode. positive Any two objects that have different (first) ionization energies may function as a cell. - 1.234 + - 0.02 + Figure 18.3(a) When a bar of zinc is placed in a beaker of copper(II) sulfate solution, copper is deposited on the zinc and the blue copper (II) ions are gradually replaced by colorless zinc ions. (b) The residue in the beaker is copper metal. No more copper ions can be seen in solution. Zn(s)+Cu (aq) Zn (aq)+Cu(s) 2+ 2+ Figure 18.4 The reaction shown in Fig. 18.3 takes place all over the surface of the zinc as electrons are transferred to the Cu2 ions in solution. Cu 2+ (aq)+2e- Cu(s) Zn(s) Zn (aq)+2e 2+ - Figure 18.5 The Daniell cell consists of copper and zinc electrodes dipping into solutions of copper(II) sulfate and zinc sulfate, respectively. The two solutions make contact through the porous pot, which allows ions to pass through to complete the electrical circuit. Zn(s)|Zn 2+ Cu 2+ |Cu(s) Zn(s) Zn 2+ (aq)+2e- Cu (aq)+2e Cu(s) 2+ - Electrodes and Cell Diagram Zn(s)|Zn 2+ Cu 2+ |Cu(s) + H (aq)|H 2 (g)|Pt(s) 3+ 2+ Fe (aq),Fe (aq) |Pt(s) 2+ 2+ Zn(s)|Zn (aq)|Cu (aq) |Cu(s) Figure 18.6 This cell is typical of galvanic cells used in the laboratory. The two electrodes are connected by an external circuit and a salt bridge. The latter completes the electrical circuit within the cell. Pt(s)|Fe3+ (aq),Fe2+ (aq) ||Cu 2+ (aq) |Cu(s) Figure 18.7 The cell potential is measured with an electronic voltmeter, a device that draws negligible current so that the composition of the cell does not change during the measurement. The display shows a positive value when the terminal of the meter is connected to the cathode of the galvanic cell. 2+ 2+ Zn(s)|Zn (aq)||Cu (aq) |Cu(s) Cell Potential E = 1.1 V The cell potential • An indication of the electron-pulling and – pushing power of the cell reactions; cell reactions at equilibrium generate zero potential. Figure 18.8 Electrons produced by oxidation leave a galvanic cell at the anode (), travel through the external circuit, and reenter the cell at the cathode (), where they cause reduction. The circuit is completed inside the cell by migration of ions through the salt bridge. A salt bridge is unnecessary when the two electrodes share a common electrolyte. positive negative Figure 18.9 This schematic picture of a galvanic cell indicates the identities of the anode and cathode, displays the oxidation and reduction half-reactions, and shows the direction of electron flow. Describing a galvanic cell and identifying the cell reaction Hg2+ (aq)+2e- 2Hg(l) 2Hg(l)+2Cl- (aq) Hg 2Cl2 (s)+2e- Hg (aq)+2Cl (aq) Hg 2Cl2 (s) 2+ - Hg(l)|Hg 2 Cl2 (s)|HCl(aq)||Hg (NO ) (aq) |Hg(l) 2 3 2 (KCl gel) Exercise: Describing a galvanic cell and identifying the cell reaction (Assume platinum electrode is used) H2 (g) 2H+ (aq)+2e- Co3+ (aq)+e- Co2+ (aq) H2 (g)+2Co (aq) 2H (aq)+ 2Co (aq) 3+ + + 3+ 2+ 2+ Pt(s)|H 2 (g)||H (aq)||Co (aq),Co (aq) |Pt(s) Figure 18.12 The cell potential can be thought of as being the difference of the two reduction potentials produced by the two electrodes. The cell potential is positive if the cathode has a higher potential than the anode. Cell potential and electrode potential standard cell potential: E 0 E 0 (cathode) E 0 (anode) Fe(s)|Fe2+ (aq)||Ag + (aq) | Ag(s): E 0 1.24 V Hydrogen standard potential 2H +2e H2 (g): E 0 V + - 0 Zn(s)|Zn (aq)||H (aq) | H 2 (g)|Pt(s): E 0.76 V 2+ + 0 E 0 E 0 (H + / H 2 ) E 0 (Zn 2+ / Zn) 0.76 V E 0 (Zn 2+ / Zn) 0.76 V Figure 18.13 The variation of standard potentials in the main groups of the periodic table. Note that the most negative values occur in the s block and the most positive values occur close to fluorine. Table 18-1 p856 Example: deducing the standard potential of an electrode The standard potential of a Zn2+/Zn electrode is -0.76 V and the standard potential of the cell Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) is 1.10 V. What is the standard potential of Cu2+/Cu electrode? E 0 (Cu 2+ /Cu) E 0 (Zn 2+ /Zn) 1.10 V E (Cu / Cu ) 1.10 V E (Zn / Zn) 1.10 V - 0.76 V=+0.34 V 0 2+ 0 2+ Exercise: deducing the standard potential of an electrode The standard potential of a Fe2+/Fe electrode is -0.44 V and the standard potential of the cell Fe(s)|Fe2+(aq)||Pb2+(aq)|Pb(s) is 0.31 V. What is the standard potential of Pb2+/Pb electrode? E 0 (Pb2+ /Pb) E 0 (Fe2+ /Fe) 0.31 V E ( Pb / Pb) 0.31 V+E ( Fe / Fe) 0.31 V 0.44 V =-0.13 V 0 2+ 0 2+ Figure 18.14 The significance of standard potentials. Only couples with negative standard potentials (and hence lying below hydrogen) can reduce hydrogen ions to hydrogen gas. The reducing power increases as the standard potential becomes more negative. E 0 (Cu 2+ / Cu)=+0.34 V E (Cu /Cu) E (H /H) 0.34 V 0 2+ 0 + Cu (aq)+H 2 (g) Cu(s)+H (aq) 2+ + E 0 (Zn2+ / Zn)=-0.76 V E (Zn /Zn) E (H /H) 0.76 V 0 2+ 0 + Zn(s)+H (aq) Zn (aq)+H2 (g) + 2+ Figure 18.15 Although aluminum has a negative standard potential, signifying that it can be oxidized by hydrogen ions (as in the hydrochloric acid, left), nitric acid (right) stops reacting with it as soon as an impenetrable layer of aluminum oxide has formed on its surface. This resistance to further reaction is termed passivation of the metal. E 0 (Al3+ /Al)=-1.66 V E (Al /Al) E (H /H) 1.66 V 0 3+ 0 + Al(s)+H (aq) Al (aq)+H 2 (g) + 3+ 18.4 Cell Potential, Electrical Work, and Free Energy Work Work is never the maximum possible if any current is flowing. In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum. Basic Concepts 7.) Electric Potential (E) Combining definition of electrical charge and potential G work E q Relation between free energy difference and electric potential difference: q nF G nFE Describes the voltage that can be generated by a chemical reaction Figure 18.16 The relation between the standard potential of a reaction (reactants, purple; products, yellow) and the equilibrium constant. For a (half) reaction (Mn+ + ne- M), the higher the reduction potential (the more negative ΔGo), the more easily it proceeds. Galvanic Cells 1.) Galvanic or Voltaic cell Example: Calculate the voltage for the following chemical reaction G = -150kJ/mol of Cd n – number of moles of electrons Solution: G nFE E G nF 150 10 3 J E 0.777 J 0.777 V C C ( 2 mol ) 9.649 10 4 mol 18.4 Cell Potential, Electrical Work, and Free Energy Maximum Cell Potential Directly related to the free energy difference between the reactants and the products in the cell. ΔG° = –nFE° F = 96,485 C/mol e– Example 18.5 Calculating G for a Cell Reaction Using the data in Table 18.1, calculate G for the reaction Cu2+(aq) + Fe(s) Cu(s) + Fe2+(aq) Is this reaction spontaneous? Example 18.5 Calculating G for a Cell Reaction The half-reactions are We can calculate G from the equation G G o RTlnQ G -nFE E Eol - RT nF lnQ Example 18.5 Calculating G for a Cell Reaction Since two electrons are transferred per atom in the reaction, 2 moles of electrons are required per mole of reactants and products. Thus n = 2 mol e-, F = 96,485 C/mol e-, and = 0.78 V = 0.78 J/C. Therefore, C J G (2 mol e ) 96, 485 0.78 mol e C 1.5 105 J The process is spontaneous, as indicated by both the negative sign of G and the positive sign of Example 18.5 Calculating G for a Cell Reaction This reaction is used industrially to deposit copper metal from solutions resulting from the dissolving of copper ores. 18.5 Dependence of Cell Potential on Concentration A Concentration Cell 18.5 Dependence of Cell Potential on Concentration Nernst Equation E cell Eocell - RT nF lnQ The relationship between cell potential and concentrations of cell components At 25°C: 0.0591 E = E log Q n At equilibrium (E=0, Q=K): 0.0591 E= log K n aA + ne- » bB Eo b A RT E Eo ln B a nF AA b 0.05916 V [ B ] E Eo log n [ A]a Eo and the Equilibrium Constant 1.) A Galvanic Cell Produces Electricity because the Cell Reaction is NOT at Equilibrium Concentration in two cells change with current Concentration will continue to change until Equilibrium is reached E = 0V at equilibrium Battery is “dead” Consider the following ½ cell reactions: aA + ne- » cC dD + ne- » bB E+o E-o Cell potential in terms of Nernst Equation is: E cell E E E o 0.05916 [ C ] c o 0.05916 [ B ]b log E log a d n n [ A] [ D ] Simplify: E cell ( E o E o 0.05916 [C ]c [ D ]d ) log n [ A]a [ B ]b 18.5 Dependence of Cell Potential on Concentration CONCEPT CHECK! Explain the difference between E and E°. When is E equal to zero? When the cell is in equilibrium ("dead" battery). When is E° equal to zero? E is equal to zero for a concentration cell. 18.5 Dependence of Cell Potential on Concentration EXERCISE! A concentration cell is constructed using two nickel electrodes with Ni2+ concentrations of 1.0 M and 1.00 × 10-4 M in the two half-cells. Calculate the potential of this cell at 25°C. 0.118 V 1 E 0.0591 log 0.118 2 110-4 o 18.5 Dependence of Cell Potential on Concentration CONCEPT CHECK! You make a galvanic cell at 25°C containing: A nickel electrode in 1.0 M Ni2+(aq) A silver electrode in 1.0 M Ag+(aq) Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential. 1.03 V 1.03 V NO3- K+ Ni Ag Ni2 + E Ag/Ag 0.78 o NO3Ag+ E Ni/Ni2 -0.25 o Example 18.7 The Effects of Concentration on For the cell reaction predict whether is larger or smaller than for the following cases. a. [Al3+] = 2.0 M, [Mn2+] = 1.0 M b. [Al3+] = 1.0 M, [Mn2+] = 3.0 M E cell E cell o 0.0591 n logQ E cell o 0.0591 n [Al3 ]2 log [Mn 2 ]3 Example 18.7 The Effects of Concentration on Solution a. A product concentration has been raised above 1.0 M. This will oppose the cell reaction and will cause to be less than ( < 0.48 V). E cell E cell o 0.0591 n log 22 13 b. A reactant concentration has been increased above 1.0 M, and will be greater than ( > 0.48 V). E cell E cell o 0.0591 n log 12 33 Investigating Matter 18.1 (a) A glass electrode in a protective plastic sleeve (left) is used to measure pH. It is sometimes used in conjunction with a calomel electrode (right) in pH meters such as this one. Table 18-2 p869 Calomel reference electrodes The potential of the calomel electrode is known vs the SHE. This is used as the reference electrode in the measurement of pH Hg 2Cl2 (s) 2e 2Hg (l ) 2Cl The other electrode in a pH probe is a glass electrode which has a Ag wire coated with AgCl dipped in HCl(aq). A thin membrane separates the HCl from the test solution Eo and the Equilibrium Constant 1.) A Galvanic Cell Produces Electricity because the Cell Reaction is NOT at Equilibrium Since Eo=E+o- E-o: 0.05916 [C ]c [ D ]d E cell E log n [ A]a [ B ]b o At equilibrium Ecell =0: Definition of equilibrium constant Eo 0.05916 log K n K 10 nE o 0.05916 at 25oC at 25oC Investigating Matter 18.1 (b) This durable portable pH meter can be used for quick measurements of pH in the field. Its accuracy is not as high as that of a laboratory pH meter. The pH meter is an electrochemical cell Overall cell potential is proportional to pH Ecell 0.06V pH Eref pH Ecell Eref 0.06V In practice, a hydrogen electrode is impractical 18.6 Batteries One of the Six Cells in a 12–V Lead Storage Battery Anode (negative plate) reaction: Pb(s) + HSO−4(aq) → PbSO4(s) + H+(aq) + 2e− Cathode (positive plate) reaction: PbO2(s) + HSO−4(aq) + 3H+ (aq) + 2e− → PbSO4(s) + 2H2O(l) The total reaction: Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) Figure 18.18 One cell of a leadacid battery like those used in automobiles. A lead-acid battery is an example of a secondary cell. It needs to be charged before it can produce a current. The electrolyte is dilute sulfuric acid. 18.6 Batteries A Common Dry Cell Battery Zn(s) + 2OH−(aq) → ZnO(s) + H2O(l) + 2e− [E° = -1.28 V] 2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH−(aq) [E° = +0.15 V] The total reaction: Zn(s) + 2MnO2(s) ⇌ ZnO(s) + Mn2O3(s) [E°cell = +1.43 V] Figure 18.17 A commercial dry cell consists of a graphite cathode in a zinc container; the latter acts as the anode. The container is filled with a moist paste of NH4Cl, MnO2, finely divided carbon, and an inert filler such as starch. In the cell reaction, manganese(IV) is reduced to manganese(III) and zinc metal is oxidized to Zn2 ions. The Rechargeable Lithium Ion Battery The cathode (marked +) half-reaction is: The anode (marked -) half reaction is: The overall reaction has its limits. Overdischarge supersaturates lithium cobalt oxide, leading to the production of lithium oxide, possibly by the following irreversible reaction: The Disposable Lithium Battery (1.5 – 3.7 V) 3 2Li(s) → 2Li+ + 2e− [E° = 3.0 V] 2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH−(aq) [E° = +0.15 V] https://en.wikipedia.org/wiki/Lithium_battery 18.6 Batteries Mercury batteries use either pure mercury(II) oxide (HgO)—also called mercuric oxide—or a mixture of HgO with manganese dioxide (MnO2) as the cathode. Mercuric oxide is a nonconductor, so some graphite is mixed with it; the graphite also helps prevent collection of mercury into large droplets. The half-reaction at the cathode is: HgO + H2O + 2e− → Hg + 2OH− with a standard potential of +0.0977 V. The anode is made of zinc (Zn) and separated from the cathode with a layer of paper or other porous material soaked with electrolyte; this is known as a salt bridge. Two half-reactions occur at the anode. The first consists of an electrochemical reaction step: Zn + 4OH− → Zn(OH)4−2 + 2e− followed by the chemical reaction step: Zn(OH)4−2 → ZnO + 2OH− + H2O yielding an overall anode half-reaction of: Zn + 2OH− → ZnO + H2O + 2e− The overall reaction for the battery is: Zn + HgO → ZnO + Hg In other words, during discharge, zinc is oxidized (loses electrons) to become zinc oxide (ZnO) while the mercuric oxide gets reduced (gains electrons) to form elemental mercury. A little extra mercuric oxide is put into the cell to prevent evolution of hydrogen gas at the end of life. Fuel cells – a battery with a difference Reactants are not contained within a sealed container but are supplied from outside sources anode : 2H 2 ( g ) 4OH (aq) 4H 2O(l ) 4e cathode : O2 ( g ) 2H 2O(l ) 4e 4OH (aq) overall : 2H 2 ( g ) O2 ( g ) 2H 2O(l ) 18.6 Batteries (New Generation) Schematic of the Hydrogen-Oxygen Fuel Cell Anode Reaction: 2H2 + 2O2− → 2H2O + 4e− Cathode Reaction: O2 + 4e− → 2O2− Overall Cell Reaction: 2H2 + O2 → 2H2O 18.7 Corrosion Process of returning metals to their natural state – the ores from which they were originally obtained. Involves oxidation of the metal. Figure 18.22 The mechanism of rust formation. (a) Oxidation of the iron occurs at a point out of contact with the oxygen of the air, and the surface of the metal acts as an anode in a tiny galvanic cell. (b) Further oxidation of Fe2 to Fe3 results in the deposition of rust on the surface. 18.7 Corrosion The Electrochemical Corrosion of Iron 18.7 Corrosion Corrosion Prevention Application of a coating (like paint or metal plating) Galvanizing Alloying Cathodic Protection Protects steel in buried fuel tanks and pipelines. 18.7 Corrosion Cathodic Protection Fig. 18.24 In the cathodic protection of a buried pipeline, or other large metal construction, the artifact is connected to a number of buried blocks of metal, such as magnesium or zinc. The sacrificial anodes (the magnesium block in this illustration) supply electrons to the pipeline (the cathode of the cell), thereby preserving it from oxidation. Figure 18.21 Iron nails stored in oxygen-free water (left) do not rust because the oxidizing power of water itself is weak. When oxygen is present (as a result of air dissolving in the water, right), oxidation is thermodynamically spontaneous and rust soon forms. 18.8 Electrolysis Forcing a current through a cell to produce a chemical change for which the cell potential is negative. Galvanic cell Electrolytic cell Electrolysis Electrolysis of a molten salt using inert electrodes Signs of electrodes: In electrolysis, anode is positive because electrons are removed from it by the battery In a galvanic cell, the anode is negative because is supplies electrons to the external circuit Anode : 2Cl (l ) Cl2 ( g ) 2e Cathode : 2 Na (l ) 2e 2 Na(l ) Overall : 2 Na (l ) 2Cl (l ) 2 Na(l ) Cl2 ( g ) Figure 18-26 p886 Electrolysis in aqueous solutions – a choice of process There are (potentially) competing processes in the electrolysis of an aqueous solution Cathode Cathode:2 Na (l) 2e 2 Na(l)...E 2.71V Cathode : 2H 2O(l ) 2e H 2 ( g ) 2OH (aq)...E 0.83V Anode Anode : 2Cl (l ) Cl2 ( g ) 2e...E 1.36V Anode : 2H 2O(l ) O2 ( g ) 4H 4e...E 1.23V (See later, overvoltage) Thermodynamics or kinetics? On the basis of thermodynamics we choose the processes which are favoured energetically (processes with lowest potential) Anode : 2H 2O(l ) O2 ( g ) 4H 4e...E 1.23V Cathode : 2H 2O(l ) 2e H 2 ( g ) 2OH (aq)...E 0.83V But…chlorine is evolved at the anode Electrolysis of water In aqueous solutions of most salts or acids or bases the products will be O2 and H2 Cathode : 2H 2O(l ) 2e H 2 ( g ) 2OH (aq)...E 0.83V Anode : 2H 2O(l ) O2 ( g ) 4H 4e...E 1.23V Other applications of electrolysis Figure 18.28 A schematic picture showing the electrolytic process for refining copper. The anode is impure copper. It undergoes oxidation, and the Cu2 ions so produced migrate to the cathode, where they are reduced to pure copper metal. A similar arrangement is used for electroplating objects. Plating Order The higher the reduction potential of the half reaction, the easier the plating occurs. Cathode : M i Mj mj mi (aq ) m i e - M i (s) E i (aq ) m je - M j (s) E j o o If E i E j , then M i forms before M j as power volt age increases. o o Electrolysis of Mixtures of Ions Ag (l ) e- Ag ...E 0.8 V Cu 2 (l ) 2e- Cu ...E 0.34 V Zn 2 (l ) 2e- Zn...E -0.76 V The larger the reduction potential, the more negative the free energy change for the half reaction, the more easily it proceeds. Therefore, as the electrolysis voltage increases gradually from very low, the metals plate out in this order (silver being the first to deposit): Ag Cu 2 Zn 2 Electrolysis of Mixtures of Ions Ce4 4e - Ce ...E 1.7 V 2 VO 2 2H e - VO 2 H 2O ...E 1.0 V Fe3 e- Fe2 ...E 0.77 V Oxidizing order (plating-out order): 4 Ce VO 2 Fe3 Be aware of overvoltage In the electrolysis of aqueous solution of NaCl, the half reactions at the anode are: Anode : 2Cl (l ) Cl2 ( g ) 2e...E 1.36V Anode : 2H 2O(l ) O2 ( g ) 4H 4e...E 1.23V It seems oxygen will be produced before chlorine. In reality, however, this is no so – chlorine is produced first. Why so? That is because of overvoltage – caused by the electrode —solution interface that blocks the transfer of electrons from solution to the atoms on electrode. Figure 18.29 A schematic representation of the stoichiometric relations that are used to calculate the amount of product formed by electrolysis or the amount of time current must flow to produce a given product. Anode : aA pR n qS ne Cathode : bC me rP sQ - 18.8 Electrolysis Stoichiometry of Electrolysis How much chemical change occurs with the flow of a given current for a specified time? current and time quantity of charge moles of electrons moles of analyte grams of analyte 18.8 Electrolysis Stoichiometry of Electrolysis current and time quantity of charge Coulombs of charge = amps (C/s) × seconds (s) quantity of charge moles of electrons mol e = Coulombs of charge 1 mol e 96, 485 C 18.8 Electrolysis CONCEPT CHECK! An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate 0.0719 g of the metal from a solution containing M(NO3)3. What is the metal? gold (Au) Solution in detail The charge: 2 A x 52.8 s = 105.6 C which is 105.6 C / F = 105.6/96485 mol for a monovalent metal. From the formula M(NO3)3, M is a trivalent metal. Therefore, the number of moles for M is 105.6/96485/3 mol The atomic weight therefore is 0.0719/(105.6/96485/3) g/mol = 197.1 g/mol. 18.8 Electrolysis CONCEPT CHECK! Consider a solution containing 0.10 M of each of the following: Pb2+, Cu2+, Sn2+, Ni2+, and Zn2+. Predict the order in which the metals plate out as the voltage is turned up from zero. Cu2+, Pb2+, Sn2+, Ni2+, Zn2+ Do the metals form on the cathode or the anode? Explain. The plating order is the order of reduction potential of the electrode (Table 18.1) Example 18.11 Electroplating How long must a current of 5.00 A be applied to a solution of Ag+ to produce 10.5 g silver metal? Example 18.11 Electroplating Solution In this case, we must use the steps given earlier in reverse: 1 mol Ag 10.5 g Ag 9.73 102 mol Ag 107.868 g Ag Example 18.11 Electroplating Each Ag+ ion requires one electron to become a silver atom: Ag+ + e Ag Thus 9.73 102 mole of electrons is required, and we can calculate the quantity of charge carried by these electrons: 9.73 10 2 96, 485 C 3 mol e 9.39 10 C mol e Example 18.11 Electroplating The 5.00 A (5.00 C/s) of current must produce 9.39 103 C of charge. Thus C 3 5.00 (time, in s) 9.39 10 C s 9.39 103 s Time s 1.88 103 s 31.3 min 5.00 Example 18.11 describes only the half-cell of interest. There also must be an anode at which oxidation is occurring. 18.9 Commercial Electrolytic Processes Production of aluminum Purification of metals Metal plating Electrolysis of sodium chloride Production of chlorine and sodium hydroxide Electrolysis for Producing Aluminum Al 3 3e - Al ...E -1.66 V 2H 2O(l ) 2e- H 2 ( g ) 2OH ...E 0.83V Aluminum metal would not be plated out of an aqueous solution of Al3+. Alumina (Al2O3) has high melting point (2050 oC), too hot for electrolysis. Hall and Heroult (1886) found that if alumina is mixed with Na3AlF6, the melting point drops to 1000 oC, low enough for electrolysis. 3 Cathode :AlF6 3e- Al 6FAnode :2Al 2OF6 12F- C 4AlF6 CO2 4e2- 3- Overall :2Al 2O3 3C 4Al 3CO2 Charles Martin Hall (1863-1914) 18.9 Commercial Electrolytic Processes Producing Aluminum by the Hall-Heroult Process Miscellaneous applications Figure 18-24 p884 18.9 Commercial Electrolytic Processes The Downs Cell for the Electrolysis of Molten Sodium Chloride Figure 18.30 In the Downs process, molten sodium chloride is electrolyzed with a graphite anode (at which the Cl ions are oxidized to chlorine) and a steel cathode (at which the Na ions are reduced to sodium). The sodium and chlorine are kept apart by the hoods surrounding the electrodes. Calcium chloride is present to lower the melting point of sodium chloride to a more economical value. Figure 18.25 A schematic picture of the electrolytic cell used in the Dow process for magnesium. The electrolyte is molten magnesium chloride. As the current generated by the external source passes through the cell, magnesium metal is produced at the cathode and chlorine gas is produced at the anode. Figure 18.26 In this schematic picture of an electrolysis experiment, the electrons emerge from a galvanic cell at its anode () and enter the electrolytic cell at its cathode (), where they bring about reduction. Electrons are drawn out of the electrolytic cell through its anode () and into the galvanic cell at its cathode (). If the cell reaction in the galvanic cell is more strongly spontaneous than the reaction in the electrolytic cell is nonspontaneous, then the overall process is spontaneous. This experiment is an example of one reaction driving another to which it is coupled. Figure 18.27 Michael Faraday (1791–1867) giving a public lecture on chemistry at the Royal Institution in London. Figure 18.20 The electric eel (Electrophorus electricus) lives in the Amazon. The average potential difference it produces along its length (1 m) is about 700 V. Figure 18.23 Metal girders are galvanized by immersion in a bath of molten zinc. Figure 18.31 Chromium plating lends decorative flair as well as protection to the steel of this motorcycle. Large quantities of electricity are needed to plate chromium because six electrons are required to produce each atom of chromium. Case Study 18 (a) One of the three alkali fuel cells used on the space shuttle. Although only one cell is needed to provide lifesupport, electricity, and drinking water, shuttle flight rules require that all three be functioning. Case Study 18 (b) High-pressure hydrogen tanks run across the top of this bus provided by Ballard Power Systems for testing hydrogen-oxygen fuel cells in Chicago. The bus is pollution free and can go 250 miles before needing to be refueled. History of Electrochemistry https://en.wikipedia.org/wiki/History_of_electrochemistry History of Battery https://en.wikipedia.org/wiki/History_of_the_battery Introduction to Fuel Cells https://www.youtube.com/watch?v=cQ8dFM4GFxk Nobel Prize in Chemistry 2007 https://www.youtube.com/wath?v=j3FecuuZ_kg http://www.nobelprize.org/nobel_prizes/chemistry/laureates/2007/p opular-chemistryprize2007.pdf Assignment for Chapter 18 18.2; 18.6; 18.17; 18.27