Electrochemistry. York University CHEM Electrochemistry - 1

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1 Electrochemistry Reading: from Petrucci, Harwood and Herring (8th edition): Required for Part 1: Sections 21-1 through Examples for Part 1: 21-1 through Problem Set for Part 1: Review: Chapter 5 questions Chapter 21 questions: 15-17, 32, 34, 43, 53 Additional problems from Chapter 21: York University CHEM Electrochemistry - 1

2 Applications of Electrochemistry Spontaneous chemical reactions can be used to produce an electric current and do work. (batteries, fuel-cells) An electric current can be used to force non-spontaneous chemical reactions to occur. (electrolysis) Reactions can be made to occur in a specific place. (electroplating, electropolishing) The voltage produced by a reaction can be used as an analytical tool. (ph electrodes) The current produced by a reaction can be used as an analytical tool. York University CHEM Electrochemistry - 2

3 G and non-pv Work For a reversible process at constant T and P: H = q P + w E (definition of enthalpy) w E = non-pv work done on the system q P = T S (reversible, constant T) Let w max = maximum non-pv work done by the system w max = -w E (reversible) So w max = - H + T S = - G Conclusion: The maximum non-pv work that can be obtained from a process is equal to - G. York University CHEM Electrochemistry - 3

4 Work from Chemical Reactions Spontaneous chemical reactions can be used to do work. How? One possibility: burn fuel to release heat boil water use the expanding steam to do work Disadvantages: inefficient (only part of heat can be turned into work) can not readily carry out the reverse process York University CHEM Electrochemistry - 4

5 Work from Redox Reactions Another method: Use redox reactions. Cu(s) + 2Ag + Cu Ag(s) r G = kj mol -1 The Cu(s) is oxidized (gives up electrons). Half-reaction: Cu(s) Cu e - The Ag + (aq) is reduced (receives electrons). Half-reaction: Ag + + e - 2Ag(s) This reaction is spontaneous. If we can transfer the electrons through an external circuit, we can use it do electrical work. York University CHEM Electrochemistry - 5

6 Cu is oxidized at one electrode: Cu(s) Cu e - Ag + is reduced at the other electrode: e - + Ag + Ag(s) Electrons travel through the wire. Ions travel through the salt bridge. Electrochemical Cells York University CHEM Electrochemistry - 6

7 Atomic View of an Electrochemical Cell York University CHEM Electrochemistry - 7

8 Electrochemical Cells - Terminology The anode is the electrode at which oxidation occurs. The cathode is the electrode at which reduction occurs. The cell potential is the voltage difference when no current flows between the electrodes. Cell potential is also called cell voltage or EMF (electromotive force). Cells in which spontaneous reactions produce a current are called voltaic cells or galvanic cells. In electrolytic cells electricity is used to force a nonspontaneous reaction to occur. York University CHEM Electrochemistry - 8

9 Cell Diagrams Anode (oxidation) is placed on left side of diagram. Cathode (reduction) is placed on right side of diagram. Single vertical line,, indicates a boundary between different phases (i.e., solution solid). Double vertical line,, indicates a boundary (salt bridge) between different half-cell compartments. Example: Cu(s) Cu 2+ Ag + Ag(s) At the anode: Cu(s) Cu e - At the cathode: Overall: Ag + + e - Ag(s) Cu(s) + 2Ag + Cu Ag(s) York University CHEM Electrochemistry - 9

10 Cell Diagrams - examples Diagram the cell in which the following overall reaction occurs: Pb(s) + 2AgCl(s) PbCl 2 (s) + 2Ag(s) Answer: Pb(s) Cl - (aq) PbCl 2 (s)) AgCl(s) Ag(s) Cl - (aq) Write the half-cell reactions for the following cell: Ag(s) Ag + (aq) Cl - (aq) AgCl(s) Ag(s) Answer: Anode (oxidation): Ag(s) Ag + (aq) + e - Cathode (reduction): AgCl(s) + e - Ag(s) + Cl - (aq) York University CHEM Electrochemistry - 10

11 Balancing Redox Reactions - Review Example: SO MnO 4- SO Mn 2+ (unbalanced) S from +4 to +6 (oxidized). Mn from +7 to +2 (reduced). (1) Write balanced half-reactions for oxidation and reduction. Oxidation half-reaction Skeleton reaction: SO 3 2- SO e - Balanced: SO H 2 O(l) SO H + + 2e - Reduction half-reaction Skeleton reaction: MnO e - Mn 2+ Balanced: MnO H + + 5e - Mn H 2 O(l) York University CHEM Electrochemistry - 11

12 Balancing Redox Reactions - continued (2) Adjust coefficients so the two half-reactions have the same numbers of electrons. 5SO H 2 O(l) 5SO H e - 2MnO H e - 2Mn H 2 O(l) (3) Add the two half-reactions. 5SO MnO H + 5SO Mn H 2 O(l) 10 electrons are transferred (important for later). (4) Check that the reaction is balanced for both atoms and charge. York University CHEM Electrochemistry - 12

13 Redox Reactions in Basic Solution 5SO MnO H + 5SO Mn H 2 O(l) Since this balanced reaction involves H +, it is appropriate for acidic solution For the reaction in basic solution, add This gives: 6 {H 2 O(l) H + + OH - } 5SO MnO H 2 O(l) 5SO Mn OH - This method can also be used to get half-reactions in basic solution. York University CHEM Electrochemistry - 13

14 Current and Charge Current is the amount of charge transferred per unit time. The amount of charge transferred is determined by the stoichiometry of the cell reaction. The charge on one mole of electrons is 96,485 coulombs. This is called the Faraday constant, F. F = 96,485 C mol -1 = 96,485 J V -1 mol -1 Example: Cu(s) + 2Ag + Cu Ag(s) C are transferred per mole of Cu oxidized. The current produced by a cell is determined by the reaction kinetics and the resistance of the circuit. York University CHEM Electrochemistry - 14

15 Electrical Work The work done in an electrical circuit is Units: w elec = charge (potential difference) coulombs volts = joules The maximum possible non-pv work equals - G. Let n moles of electrons transferred per mole of reaction nf = total charge (coulombs) transferred E cell maximum possible cell potential (requires zero current) Then: r G = -nfe cell York University CHEM Electrochemistry - 15

16 Electrical Work - continued r G = -nfe cell There is a fundamental connection between r G and cell potential. Because of this: Cell potentials depend on concentrations. Electrochemical cells can be used to measure concentrations. (ph electrodes, for example) Electrochemical cells can be used to measure r G (and to determine r G ). Tabulated thermodynamic data can be used to determine cell potentials. York University CHEM Electrochemistry - 16

17 Cell Potential - Example A cell is constructed in which the half-cell reactions are: Anode: H 2 (g) 2H + (aq) + 2e - Cathode: Cl 2 (g) + 2e - 2Cl - (aq) When P H2 = P Cl2 = atm and [H + ] = [Cl - ] = M, the cell potential is found to be volts. Find r G under these conditions for H 2 (g) + Cl 2 (g) 2H + (aq) + 2Cl - (aq) Solution: r G = -nfe cell and n = 2 for the overall reaction. r G = -2(96,485 J V -1 mol -1 )( V) = kj mol -1 York University CHEM Electrochemistry - 17

18 Cell Potentials and Spontaneity r G = -nfe cell If r G < 0, the reaction is spontaneous as written. If r G > 0, the reaction is non-spontaneous as written (reverse reaction is spontaneous). If the reaction proceeds as written, then n > 0. Therefore: If E cell > 0, the reaction is spontaneous as written. If E cell < 0, the reaction is non-spontaneous as written. York University CHEM Electrochemistry - 18

19 Standard Cell Potentials Definition: The standard cell potential, E cell, is the cell potential that would obtain if all reactants and products were in their standard states. Therefore: r G = -nfe cell Standard states may be hypothetical. The standard cell potential is used in calculations of actual cell potentials. The standard cell potential is the sum of standard potentials for the individual half-cells. York University CHEM Electrochemistry - 19

20 Standard Cell Potential - example The cell with the overall reaction H 2 (g) + Cl 2 (g) 2H + (aq) + 2Cl - (aq) has a standard cell potential of V. Determine f G for Cl - (aq). Solution: r G = -nfe cell r G = -2(96,485 C mol -1 )( V) = kj mol -1 For H 2 (g), Cl 2 (g), and H + (aq); f G = 0. r G = 2 f G (Cl - (aq)) f G (Cl - (aq)) = kj mol -1 York University CHEM Electrochemistry - 20

21 Standard Hydrogen Electrode To create a voltage, we need two half-cells. So we can't measure individual half-cell potentials. Convention: The standard hydrogen electrode is assigned a half-cell potential of zero. 2H + (aq) + 2e - H 2 (g) E = 0 volts Standard states (a=1): [H + ] 1 M, P H2 = 1 bar 1 atm The half-cell potential will differ from zero if H + and/or H 2 are not in their standard states. York University CHEM Electrochemistry - 21

22 Standard Hydrogen Electrode - continued H 2 (g) at one bar bubbled over a platinum electrode. Pt acts a catalyst for the reaction. 2H + + 2e - H 2 (g) Used as a basis for calculations. Not really very practical. York University CHEM Electrochemistry - 22

23 Standard Electrode Potentials The standard electrode potential for a half-cell is the potential when all species are in their standard states. refers to reduction at the electrode (these days) measured relative to a standard hydrogen electrode as the anode Example: Cell for measuring E for Cu 2+ /Cu. Anode: H 2 (g) 2H + + 2e - (oxidation) Cathode: Cu 2+ (1 M) + 2e - Cu(s) (reduction) Cell Diagram: Pt H 2 (1 bar) H + (1 M) Cu 2+ (1 M) Cu(s) Cell potential is V. So E = V for Cu 2+ /Cu. York University CHEM Electrochemistry - 23

24 Standard Electrode Potentials - continued Reduction Half-Reaction E (volts) F 2 (g) + 2e - 2F - (aq) O 2 (g) + 4H + (aq) + 4e - 2H 2 O(l) H + (aq) + 2e - H 2 (g) Zn 2+ (aq) + 2e - Zn(s) Li + (aq) + e - Li(s) F 2 is easiest to reduce (largest E ). F - is hardest to oxidize. Li + is hardest to reduce. Li is easiest to oxidize. F 2 is best oxidizing agent; Li is best reducing agent. York University CHEM Electrochemistry - 24

25 Using Standard Electrode Potentials A standard cell potential, E cell, may be calculated from the standard electrode potentials for the cathode, E cathode, and anode, E anode : E cell = E cathode - E anode The anode potential is subtracted since the potential is for reduction and the anode reaction is oxidation. Standard electrode potentials are listed in tables. Standard electrode potentials do not depend on how a reaction is written since they are related to r G per mole of electrons. ( r G = -nfe cell ) York University CHEM Electrochemistry - 25

26 Standard Electrode Potentials - example Find the standard cell potential for the reaction: Zn(s) + Cl 2 (g) Zn 2+ (aq) + 2Cl - (aq) Solution: Write half-cell reactions and find E values. Oxidation: Zn(s) Zn 2+ (aq) + 2e - Reduction: Cl 2 (g) + 2e - 2Cl - (aq) From Table 21.1: Zn 2+ (aq) + 2e - Zn(s) E = V Cl 2 (g) + 2e - 2Cl - (aq) E = V E cell = E cathode - E anode = (-0.763) = V York University CHEM Electrochemistry - 26

27 Cell Potential and Equilibrium Constant We have derived the following two equations: r G = -nfe cell and r G = -RTln K eq Combining these gives Uses of this equation: E cell = lnk eq Calculating K eq from standard half-cell potentials (see example 21-7 in text). Relating E cell for different reactions. York University CHEM Electrochemistry - 27

28 Cell Potential and Equilibrium - Example At K, the standard reduction potential for O 2 (g) in acidic solution is V: O 2 (g) + 4H + (aq) + 4e - 2H 2 O(l) E 1 = V Find the standard reduction potential for O 2 (g) in basic solution: O 2 (g) + 2H 2 O(l) + 4e - 4OH - (aq) E 2 =? Solution: The second reaction is equal to the first plus 4H 2 O(l) 4H + (aq) + 4OH - (aq) K eq = K W 4 So K 2 = K 1 K W 4, E 2 = (RT/nF)ln(K 1 K W 4 ), n = 4 E 2 = E 1 + (RT/F)lnK W = ln( ) E 2 = V York University CHEM Electrochemistry - 28

29 Dissolving Metals with Acids Many metals are dissolved by acids with the evolution of H 2 (g). Oxidation: M(s) M n+ (aq) + ne - E = E M Reduction: 2H + (aq) + 2e - H 2 (g) E = 0 Overall: M(s) + nh + (aq) M n+ (aq) + (n/2)h 2 (g) E = -E M Conclusions: The more negative the standard reduction potential of the metal ion, the easier the metal is to dissolve. Lowering the ph promotes the dissolution of metals. York University CHEM Electrochemistry - 29

30 Dissolving Metals with Acids - example Determine the concentrations of each of the following metals that will dissolve at ph = 7.00 and ph = Cu 2+ (aq) + 2e - Cu(s) Pb 2+ (aq) + 2e - Pb(s) Zn 2+ (aq) + 2e - Zn(s) Solution: M(s) + 2H + (aq) M 2+ (aq) + H 2 (g) E = V E = V E = V n = 2 E cell = -E T = K P H2 1 bar E cell = (RT/nF)lnK eq K eq = exp(-(77.85 V -1 )E ) [M 2+ ] [H + ] 2 K eq = [H + ] 2 exp(-(77.85 V -1 )E ) York University CHEM Electrochemistry - 30

31 Dissolving Metals with Acids - continued [M 2+ ] [H + ] 2 exp(-(77.85 V -1 )E ) ph = 7 ph = 0 E (V) [Cu 2+ ] eq = M M [Pb 2+ ] eq = M M [Zn 2+ ] eq = M M Metals with E 0 are difficult to dissolve even in strong acids. Metals with E 0 will dissolve in strong acids. Metals with E 0 will dissolve in water. York University CHEM Electrochemistry - 31

32 Enhancing Dissolution of Metals Concentrated HNO 3 will dissolve Cu(s): Cu 2+ (aq) + 2e - Cu(s) NO H + + 4e - NO(g) + H 2 O(l) E = V E = V For the overall reaction, E = ( ) V = V. This is very favorable. Gold can be dissolved using aqua regia (1 part HNO 3 to 3 parts HCl): Au 3+ (aq) + 3e - Au(s) E = 1.52 V Au 3+ (aq) + 4Cl - (aq) [AuCl 4 ] - (aq) York University CHEM Electrochemistry - 32

33 We have shown that The Nernst Equation r G = -nfe cell But r G depends on the concentrations of reactants and products: r G = r G + RTlnQ Therefore, E cell also depends on concentrations. Combining these gives nfe cell = nfe cell - RTlnQ. So This is known as the Nernst Equation. York University CHEM Electrochemistry - 33

34 The Nernst Equation - continued Here: R = J mol -1 K -1 F = 96,485 C mol -1 = 96,485 J V -1 mol -1 lnq = ln(10) logq = logq So E cell = E cell - ( V K -1 )(T/n)logQ If T = K (25 C), then E cell = E cell - (1/n)( V)logQ York University CHEM Electrochemistry - 34

35 Using the Nernst Equation Find E cell at 298 K for the cell Pt Fe 2+ (0.10 M),Fe 3+ (0.20 M) Ag + (1.0 M) Ag(s) Solution: First find E cell, then use the Nernst Equation. Anode: Fe 2+ Fe 3+ + e - E = V Cathode: Ag + + e - Ag(s) E = V Cell: Fe 2+ + Ag + Fe 3+ + Ag(s) E cell = E cathode - E anode = V n = 1 Q = [Fe 3+ ] / [Fe 2+ ][Ag + ] = (0.20) / (0.10)(1.0) = 2.0 E cell = E cell - (1/n)( V)logQ = V York University CHEM Electrochemistry - 35

36 We can make a cell with the same reaction occuring at both electrodes. E cell = 0 The cell voltage is due to the difference in concentration. Concentration Cells York University CHEM Electrochemistry - 36

37 Determining K sp With saturated AgI(aq) at the anode and [Ag + ] = M at the cathode, E cell = V. Use this to find K sp. Anode: Ag(s) Ag + (aq, sat. AgI) + e - Cathode: Cell: Ag + (aq, 0.1M) + e - Ag(s) Ag + (aq, 0.1M) Ag + (aq, sat. AgI) E cell = 0. Q = [Ag + ] sat,ki / [0.1 M]. n = 1. The Nernst equation becomes (at K): V = E cell = - ( V) log([ag + ] sat,ki / [0.1 M]) [Ag + ] sat,ki = [0.1 M] = M K sp = [I - ][Ag + ] = ( ) 2 = York University CHEM Electrochemistry - 37

38 Electrochemistry Basics - Summary Electrochemical cells permit us to couple electrical work to chemical reactions. Cell potential and r G are directly related: r G = -nfe cell. Standard reduction potentials are tabulated and may be used to compute E cell and r G. The effect of concentration on cell potential is given by the Nernst equation: York University CHEM Electrochemistry - 38