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19/04/2015 Oxidation number Red-ox reactions Electrochemistry Lecture 9 • The oxidation number of an atom (oxidation state) represents the number of electrons lost, gained or unequally shared by an atom. • It could be zero, positive or negative. • If the oxidation number is zero it means that the atom has the same number of electrons as in the free neutral atom. Rules for assigning oxidation number • Positive oxidation number means the atom has fewer electrons assigned to it than in neutral atom. • Negative oxidation number means the atom has more electrons assigned to it than in the neutral atom. • All elements in their free state have an oxidation number of zero • H is +1 (except metal hydrides where is -1, e.g. NaH) • O is -2 (except in peroxides where is -1 and OF2, where is +2) • The metallic element in an ionic compound has a positive oxidation number 1 19/04/2015 Rules …. • In covalent compounds the negative oxidation number is assigned to the most electronegative atom. • The algebraic sum of the oxidation numbers of the elements in a compound is zero. • The algebraic sum of the oxidation numbers of the elements in a poliatomic ion is equal to the charge of the ion. Ion/ atom H+ Ca2+ Fe3+ P O2Cl- Oxidation Molecule number/s +1 +2 +3 0 -2 -1 Oxidation numbers SO4-2 CH4 NH3 H2O H3PO4 Na2S2O3 Multiple oxidation numbers: • A) N2; N2O; NO; N2O3; NO2; N2O5 0 +1 +3 +5 • B) P; P2O3; P2O5 0 +3 +5 • C) Cl ; Cl2; Cl2O; Cl2O3; HClO3; HClO4 -1; 0; +1; +3 +5 +7 2 19/04/2015 Quiz • Find the oxidation number of: • Br (a) or S (b) in the following compounds; a) HBr; HBrO2; HBrO3; HBrO4 b) H2S; H2SO4; Na2S2O3 Oxidation –reduction reaction(redox) • Redox is a chemical process in which the oxidation number of an element (elements) is changed. • This process may involve the complete transfer of electrons to form ionic bond or only a partial transfer or shift of electrons to form covalent bonds. Oxidation Reduction • This process occurs whenever the oxidation number of an element increases as a result of losing electrons. Br0 – 5e →Br+5 Cu0 - 2e → Cu2+ S-2 – 8e → S+6 • Reduction occurs every time the oxidation number of an element decreases as a result of gaining electrons: Fe3+ + e→ Fe2+ Mn+7 + 5e → Mn+2 Cl+5 + 6e → Cl-1 3 19/04/2015 • Losing of electrons from an atom or molecule is called oxidation, and gaining of electrons is reduction. • Oxidation and reduction occur simultaneously in a chemical reaction; one cannot take place without the other. This can be easily remembered through the use of mnemonic devices. Two of the most popular are: "OIL RIG" (Oxidation Is Loss, Reduction Is Gain) Oxidizing agent vs. reducing agent • The substance that causes an increase in the oxidation state of another substance is called an oxidizing agent. • The substance that causes a decrease in the oxidation state of another substance is called a reducing agent. http://www.landfood.ubc.ca/soil200/interaction/orgmatter_air.htm 4 19/04/2015 Zn + 2HCl(aq)→ ZnCl2 + H2 O.n. 0 +1;-1 +2; -1; 0 Zn0 is oxidized, so it is the reducing agent. H+1 is reduced, so it is the oxidizing agent. http://www.emc.maricopa.edu/faculty/farabee/biobk/biobookenzym.html Quiz • In the following reactions, identify what is oxidized, and what is reduced, oxidizing and Ballancing red-ox reactions • The total increase in oxidation numbers must equal the total decrease in oxidation numbers. reducing agent : • A) Cl2(g)+ 2NaBr →2NaCl + Br2 • B) 2PbO(s)→ 2Pb(s) + O2(g) 5 19/04/2015 All balanced equations must satisfy two criteria. • 1. There must be mass balance. That is, the same number of atoms of each kind must appear in reactants and products. • 2. There must be charge balance. The sums of actual charges on the left and right sides of the equation must equal to • In a balanced formula unit equation, the total charge on each side will be equal to zero. • In a balanced net ionic equation, the total charge on each side might not be zero, but it still must be equal on the two sides of the equation. each other. Check if the reaction is well ballanced • 1. H2O2 + H+ + MnO4-→ O2 + Mn2+ + H2O 2. Mg + 2H+ →Mg2+ + H2 3. ClO2+ OH- → ClO2- + ClO3- + H2O THE HALF-REACTION METHOD 1. Write as much of the overall unbalanced equation as possible, omitting spectator ions. 2. Construct unbalanced oxidation and reduction half-reactions (these are usually incomplete as well as unbalanced). Show complete formulas for polyatomic ions and molecules. 3. Balance by inspection all elements in each halfreaction, except H and O. 6 19/04/2015 4. Balance the charge in each half-reaction by adding electrons as “products” or “reactants.” 5. Balance the electron transfer by multiplying the balanced half-reactions by appropriate Example • I2 + S2O3-2→ 2I- + S4O6-2 integers. 6. Add the resulting half-reactions and eliminate any common terms. I2→ I(reduction half reaction) I2→ 2 II2 + 2e→ 2 I-(balanced reduction half reaction) S2O3-2→ S4O6-2 (oxidation half reaction) 2S2O3-2→ S4O6-2 2S2O3-2→ S4O6-2 – 2e (balanced oxidation half reaction) Adding H+; OH- or H2O In acidic solution: we add only H + or H2O (not OH- in acidic solution) In basic solution: we add only OH - or H2O (not H+ in alkalic solution) I2 + 2S2O3-2 +2e→ 2I- + S4O6-2 – 2e Full red-ox reaction 7 19/04/2015 The following chart shows how to balance hydrogen and oxygen In acidic solution: To balance O add H2O and/ then to balance H add H+ In basic solution To balance O: For each O needed : add two OH- to side needing O and add one H2O to other side To balance H: Add one H2O to side needing H and add one OH- to other side Change in oxidation method 1. Write as much of the overall unbalanced equation as possible. 2. Assign oxidation numbers to find the elements that undergo changes in oxidation numbers. 3. a) Draw a bracket to connect atoms of the element that is oxidised. Show the increase in oxidation number per atom. Draw a bracket to connect atoms of the element that is reduced. Show the decrease in oxidation number per atom. b) determined the factors that will make the total increase and decrease in oxidation numbers equal. • 4. Insert coefficients into the equation to make the total increase and decrease in oxidation number equal. • 5. Balance the other atoms by inspection. 8 19/04/2015 Quiz • Balance the following red-ox reaction with both methods: S+ HNO3→ H2SO4 + NO S0 → S+6 reduction half reaction S0→ S+6 + 6e N+5 +3e → N+2 oxidation half reaction N+5 + 3e → N+2 • Number of electrons lost: 6 Number of electrons gained: 3 = The second mode 0 - 6e S0 + 2N+5→ S+6 + 2N+3 equation +2 S+ HNO3→ H2SO4 + NO this part should doubled S0 + 2HNO3 →H2SO4 + 2NO – +6 +5 3e balanced Oxidation Reduction S0 → S+6 + 6e N+5 + 3e→N+2 The numbers of electrons from oxidation is 6. The numbers of electrons from reduction is 3. So the numbers of electrons from reduction should be twiced. 9 19/04/2015 Quiz 1 S+ 2HNO3→ 1 H2SO4 + 2NO • Ag + HNO3→ AgNO3 + NO + H2O • Balance the following equations. For each equation tell what is oxidized, what is reduced, what is the oxidizing • agent, and what is the reducing agent. • (a) Zn(s) + HClO4(aq) → Zn(ClO4)2(aq) H2(g) • (b) K(s) + H2O(l) → KOH(aq) + H2(g) heat • (c) NaClO3(s) NaCl(s) + O2(g) Redox titration • KI + K2CrO4 + H2SO4 → I2 + Cr2(SO4)3 + K2SO4 + H2O • H2O2 + KMnO4 + H2SO4→ O2 + K2SO4 + MnSO4 + H2O 10 19/04/2015 • In reaction of some amount Fe2+ cation with permanganate ions (KMnO4; M=0.02 mole/L) in acidic solution 20 ml of the titrant was used. Solution • Start from chemical equation: Fe2+ + MnO4- + H+ → Fe3+ + Mn2+ + H2O Balance it. • How many moles (grams) of iron ions were present in the sample? 5Fe2+ + MnO4- + 8 H+ → 5Fe3+ + +Mn2+ + 4H2O 5 moles of Fe2+ will react with 1 mole of MnO4- • Step 3. calculate number of KMnO4 moles used: n = M . V = 0.02 mol/l . 0.02 l (conversion of 20 ml into l) n= 4 . 10-4 moles of KMnO4 used n for Fe = 5 . 4. 10-4 = 2 .10-3 moles of Fe2+ 11 19/04/2015 An activity series place metals in order from most active to least active • Na<Mg<Al<Zn<Fe<Cd<Co<Ni<Pb< H active A more active metal will displace a less active metal from solution. A more active metal: (i) loses electrons more easily (ii) is the stronger reductant • H<Cu<Hg<Ag<Au • noble (iii) is the weaker oxidant (iv) has a lower standard reduction potential The less active metal is less likely lose electrons. It (ii) is the weaker reductant (iii) is the stronger oxidant (iv) has a higher standard reduction potential Electrochemistry • Electrochemistry deals with the chemical changes produced by electric current and with the production of electricity by chemical reactions. http://www.ausetute.com.au/redox.html 12 19/04/2015 Many metals are purified or are plated onto jewelry by electrochemical methods. Digital watches, automobile starters, • Batteries serve as power sources for all types of gadgets • The energy in a battery comes from calculators, and pacemakers are just a few a spontaneous redox reaction where devices that depend on electrochemically the electron transfer is forced to produced power. take place through a wire Corrosion of metals is an electrochemical process. • The apparatus that provides electricity in this way is called a galvanic or voltaic cell • Cell reactions are obtained by adding the half-reactions. A galvanic cell. The cell consists of two half-cells where the oxidation and reduction half-reactions take place. The salt bridge is required for electrical neutrality. The overall cell reaction is: 2Ag+(aq)+Cu(s) 2Ag(s)+Cu2+ (aq) • Half-reactions are balanced using the ion-electron method. 13 19/04/2015 The electrodes are assigned the name anode or cathode. Reduction (electron gaining) occurs at the cathode. Electrons appear as reactants in this halfreaction. Oxidation (electron lossing) occurs at the anode. Electrons appear as products in this halfreaction. • There are two types of electrical conduction in a galvanic cell. The movement of ions through the salt bridge and in solution is required for charge neutrality. • Metallic conduction occurs when electrons move through the wires. • Electrolytic conduction occurs through the liquid by movement of ions, not electrons. 14 19/04/2015 • The anode has negative polarity Quiz because the electrons left behind by the Cu2+ ions give it a slightly negative charge. Among the following cations find • The cathode has positive polarity those from cathode and anode part because of the Ag+ ions “joining” the (remember about activity series): electrode give it a slightly positive Cu2+; Ag+, Zn2+, Ca2+, charge. For convenience, a standard cell notation has been developed by chemists. Cathode Cu2+ Ag+ Cu2+ Ag+ Anode Ca2+ Cu2+ Zn2+ Ca2+ Anode half-cell is specified on the left. Cathode half-cell is specified on the right. Phase boundaries are indicated using “|”. The salt bridge separates the anode and cathode and is indicated using “||” 15 19/04/2015 • The cell diagram for the coppersilver galvanic cell is Galvanic cells can push electrons through a wire. The magnitude of this ability Cu2+ Cu(s) | (anode) (aq) || Ag+ (aq) |Ag(s) (cathode) is expressed as a potential. The maximum potential a given cell can generate is called the cell potential, Ecell • The tendency for a species to • The standard cell potential, E0cell, is the cell potential measured at 298 K (25oC) with all ion concentration 1.00 M. gain electrons and be reduced is its reduction potential. • When measured at standard condition, it is called the standard reduction potential, E0. 16 19/04/2015 • When two half-cells are connected: • The one with the larger reduction • The difference in the two standard reduction potentials gives the standard cell potential potential will acquire electrons and undergo reduction. • The half-cell with the lower reduction potential will give up electrons and standard reduction standard reduction o Ecell potential of the potential of the substance reduced substance oxidized undergo oxidation. Calculate the standard potential of cells formed from: Cu│Cu2+(aq) and Zn2+ (aq)│Zn E0 = +0.34; E0= -0.76 V Cu│Cu2+ - substance reduced (oxidizing agent; cathode); The galvanic cell formula is as follows: Zn│Zn2+(aq) ││ Cu2+(aq) │Cu Solution E0cell = (+0.34V) – (-0.76V) = 1.12V Zn2+│Zn – substance oxidized (reducing agent; anode) 17 19/04/2015 Determine oxidized and reducing agent for half cells presented below. Write the formula for galvanic cell. Calculate the standard potential of cells formed from: 1. Ag│Ag+ and Na+│Na E0 = +0.80; E0= -2.71 V 2. Fe│Fe2+ and Mg2+│Mg E0 = -0.44; E0= -2.37V It is not possible to measure the reduction potential of an isolated half-cell The standard hydrogen electrode (SHE). • A reference electrode, called the standard hydrogen electrode (SHE), has been assigned the potential of exactly 0 V 18 19/04/2015 • Using a hydrogen half-cell, other reduction potentials can be measured A galvanic cell comprised of copper and hydrogen half-cells. The reaction is Cu2+(aq)+H2(g) Cu(s)+2H+(aq) Cell notation: Pt(s), H2(g)|H+(aq)||Cu2+(aq)|Cu(s) Galvanic series • The galvanic series (or electropotential series) determines the nobility of metals and semi-metals. • It is the order of metal and its cation pairs according to increase of their • For any pair of reduction potentials ordered most reactive to least reactive: • The higher (more positive) reduction potential will occur as a reduction • The lower (more negative) reduction potential will be reversed and occur as an oxidation standard reduction potentials, E0. 19 19/04/2015 Cathode (Reduction) Half-Reaction Standard Potential E0 (volts) Li+(aq) + e- -> Li(s) -3.04 K+(aq) + e- -> K(s) -2.92 Ca2+(aq) -2.76 + 2e- -> Ca(s) 2H+(aq) + 2e- -> H2(g) 0.00 Cu2+(aq) + e- -> Cu+(aq) 0.16 AgCl(s) + e- -> Ag(s) + Cl-(aq) 0.22 Cu2+(aq) + 2e- -> Cu(s) 0.34 • In a galvanic cell, the calculated cell potential for the spontaneous reaction is always positive. • If the calculated cell potential is negative, the cell is spontaneous in the reverse direction. The free energy change (ΔG) for a system can also be used to predict if a reaction is spontaneous. Free energy changes and cell potentials are related. 20 19/04/2015 •The maximum useful work that can be obtained from a reaction is: G maximum work ΔG0 = - n F E0 cell or ΔG = - n F Ecell n = number of moles of electrons transferred F = Faraday constant = 96,485 C/mol e E0cell = standard cell potential in volts Calculate the ΔG0 for the following reaction: ΔG Ecell Reaction <0 >0 Spontaneous 0 0 At equilibrium >0 <0 Non-spontaneous Zn(s) + Cu2+(aq) ↔ Cu(S) + Zn2+(aq) Step 1: Break the redox reaction into oxidation and reduction half-reactions. Cu2+ +2 e- → Cu (reduction) Zn→ Zn2++ 2 e- (oxidation) 21 19/04/2015 Step 2: Find the E0cell of the cell. From the Table of standard reduction potentials Cu2+ + 2e → Cu; E0 = 0.342V Zn2+ + 2 e- → Zn; E0 = -0.762 V The two equilibria which are set up in the half cells are: E0cell = E0reduction - E0oxidation E0cell = 0.342 V – (- 0.762 V) E0cell = 1.104V The negative sign of the zinc E° value shows that it releases electrons more readily than hydrogen does. The positive sign of the copper E° shows that it releases electrons less readily than hydrogen. Step 3: Find ΔG0. 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 Self-checking quiz Calculate the ΔG0 for the following reaction: 2Hg+ + Cd0→2Hg0 + Cd+2 E0 for Hg+│Hg cell is + 0.615V Joule/Coulomb E0 for Cd2+│Cd cell is -0.403V ΔG0 = -nFE0cell The cell consists of CdSO4, Cd, Hg2SO4, Hg is named Weston Cell. ΔG0 = -(2 mol)(9.65 x 104 C/mol)(1.104J/C) ΔG0 = - 213072 J or - 213.07 kJ 22 19/04/2015 Self-checking quiz Calculate the ΔG0 for the following reaction: Ni2+ + Cd0→Ni0 + Cd+2 E0 for Ni2+│Ni cell is - 0.25 V E0 for Cd2+│Cd cell is -0.403V The cell which consists from Cd (anode) and NiO(OH) (catode) forms gumstick batteries Lithium – ion battery, LIB Disassembled Ni–Cd battery from cordless drill. 1: outer metal casing (also negative terminal) 2: separator (between electrodes) 3: positive electrode 4: negative electrode with current collector (metal grid, connected to metal casing). Everything is rolled. Li-ion batteries provide light weight, high energy density power sources. They are used in : Portable devices: mobile phones, smartphones, The positive electrode half reaction (with charging being forwards) is: laptops and tablets, digital cameras and many The negative electrode half-reaction is: Electric vehicles: Because of their light weight others. Li-ion batteries are used for energy storage for many electric vehicles, electric wheelchairs, They are based on lithium cobalt oxide (LiCoO2), which offers high energy density, but have well-known safety concerns, especially when damaged from radio-controlled models and model aircrafts to the Mars Curiosity rover. 23 19/04/2015 Cell potentials depend on concentrations • Equating: G nFEcell (nonstanda rd conditions) G nFE (standard conditions) o o cell • The cell potential can be related to the equilibrium constant K o G o RT ln K nFEcell thus o Ecell o RT ln K or K e nFEcell / RT nF G G 0 RT ln Q thus o nFEcell nFEcell RT ln Q or o Ecell Ecell RT ln Q nF • The last expression is a form of the Nernst equation which relates ion concentrations to the cell potential • Use molar concentrations (M) for ions and partial pressures of gases in atmospheres when calculating Q Self-check quiz Example: In a certain zinc-copper cell: Zn(s)+Cu2+(aq)Zn2+(aq)+Cu(s) the ion concentrations are: [Cu2+]=0.0100 M and [Zn2+]=1.0 M. What is the cell potential at 298 K? o o o Ecell ECu 0.34 V (0.76) V 1.10 V 2 E Zn2 o Ecell Ecell 2 RT [ Zn ] ln 1.04 V 2 F [Cu 2 ] Find the cell potential of a galvanic cell based on the following reduction half-reactions at 25°C Cd2+ + 2 e- → Cd Pb2+ + 2 e- → Pb E0 = -0.403 V E0 = -0.126 V where [Cd2+] = 0.020 M and [Pb2+] = 0.200 M. 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 For this two electron change at 298 K 24 19/04/2015 The Nernst equation could be also expressed as: • Galvanic cells, commonly called batteries, can be classified as either primary or secondary cells • Primary cells are not designed to be recharged In any concentration cell, the spontaneous • Secondary cells are able to be recharged reaction is always from the more concentrated solution to the more diluted solution. 25 19/04/2015 Secondary cells Primary cells A rechargeable battery, storage battery, or accumulator is a type of electrical battery. It comprises one or more electrical cells, and is a type of energy accumulator. It is known as a secondary cell because its electrochemical reactions are electrically reversible. • A battery is usually a collection of cells connected in series • When connected in series, the voltage of each cell is added to provide the total voltage of the battery (−) Zn | NH4Cl | MnO2,C (+). 26 19/04/2015 Battery in medicine Electricity in nature Pacemaker Tool for hearing improvement Electrochemistry in medicine. Electrocardiography (ECG or EKG from Electroencephalography (EEG) is the recording of electrical activity along the scalp transthoracic (across the thorax or chest) Greeek: kardia, meaning heart) is a interpretation of the electrical activity of the heart over a period of time, as detected by electrodes attached to the surface of the skin and recorded by a device external to the body Epileptic spike and wave discharges monitored with EEG 27 19/04/2015 Voltage gained ion channels Voltage-gated ion channels for cations (sodium, calcium, and potassium channels) and anions (chloride channels) are described together with electrophysiology, pharmacology and other medical sciences. 28