1. 1 4-5-2011

2. 2 Calculation of the standard emf of an electrochemical cell The procedure is simple: 1.Arrange the two half reactions placing the one with the larger E o up (the cathode). 2.The half reaction with the lower E o is placed down (the anode). 3.E o cell = E o cathode - E o anode

3. 3 A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO 3 ) 2 and a Ag electrode in a 1.0 M AgNO 3 solution. Find the overall reaction and the standard cell emf. From table of standard reduction potentials, we have: Ag + + e  Ag(s)E o = 0.80V Mg 2+ + 2e  Mg(s)E o = -2.37V From diagonal rule Ag + will react with Mg(s), therefore the reaction will be as follows:

4. 4 Multiply the first half reaction by 2 to account for the number of electrons in the second one and reverse the second half reaction: 2Ag + + 2e  2Ag(s)E o = 0.80V Mg(s)  Mg 2+ + 2e E o = +2.37V 2Ag + + Mg(s)  Mg 2+ + 2Ag(s)E o = 3.17V E o rxn = E o cathode - E o anode E o rxn = 0.80 – (- 2.37) = 3.17V Remember to use E o values with sign as it appears in the table (reduction potential).

5. 5 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 V 2e - + Cd 2+ (1 M) Cd (s) Cr (s) Cr 3+ (1 M) + 3e - Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) x 2 x 3

6. 6 Remember that multiplying a half reaction with any coefficient will not change the value of E o. E 0 = E cathode - E anode cell 00 E 0 = -0.40 – (-0.74) cell E 0 = 0.34 V cell

7. 7 Spontaneity of Redox Reactions  G = -nFE cell  G 0 = -nFE cell 0 n = number of moles of electrons in reaction F = 96,500 J V mol = 96,500 C/mol  G 0 = -RT ln K = -nFE cell 0 E cell 0 = RT nF ln K (8.314 J/K mol)(298 K) n (96,500 J/V mol) ln K = = 0.0257 V n ln K E cell 0 = 0.0592 V n log K E cell 0

8. 8 What is the equilibrium constant for the following reaction at 25 0 C? Fe 2+ (aq) + 2Ag (s)  Fe (s) + 2Ag + (aq) = 0.0257 V n ln K E cell 0 E 0 = -0.44 – (0.80) E 0 = -1.24 V e 0.0257 V x nE0E0 cell K = n = 2 = e 0.0257 V x 2x 2-1.24 V K = 1.23 x 10 -42 E 0 = E Fe /Fe – E Ag /Ag 00 2+ + From the reaction as written, Fe half cell is the cathode, and Ag half cell is the anode: 2Ag + + 2e  Ag(s)E o = 0.80V Fe 2+ + 2e  Fe(s)E o = -0.44V From diagonal rule, the rxn is non spontaneous

9. 9 Calculate the equilibrium constant for the following reaction at 25 o C: Sn(s) + 2Cu 2+  Sn 2+ + 2Cu + From the reaction as written, Cu half cell is the cathode, and Sn half cell is the anode: 2Cu 2+ + 2e  2Cu + E o = 0.15V Sn 2+ + 2e  Sn(s)E o = -0.14V According to the diagonal rule the reaction occurs spontaneously as written: E o cell = E o cathode (Cu2+, Cu+) – E o anode(Sn2+, Sn(s)) E o cell = 0.15 – (-0.14) = 0.29 V

10. 10 e 0.0257 V x nE0E0 cell K = K = 6.3*10 9 K = e 0.0257 x 20.29 K = 6.3*10 9

11. 11 Calculate the standard free energy for the following reaction at 25 o C: 2Au(s) + 3Ca 2+  2Au 3+ + 3Ca(s) First, we write the two half cells and calculate the cell potential as they appear in the reaction. From the reaction as written, Ca half cell is the cathode, and Au half cell is the anode: 2Au 3+ + 6e  2Au(s) E o = +1.50Vanode 3Ca 2+ + 6e  3Ca(s)E o = -2.87Vcathode According to the diagonal rule the reaction will not occur spontaneously as written: E o cell = E o cathode – E o anode E o cell = -2.87 – 1.50 = -4.37 V  G o = - nFE o cell  G o = - (6)(96500)(-4.37) = 2.53*10 6 J  G o = 2.53*10 3 kJ The large +ve  G o suggests a non-spontaneous reaction at the standard states.

12. 12 The Effect of Concentration on Cell Emf  G =  G 0 + RT ln Q  G = -nFE  G 0 = -nFE 0 -nFE = -nFE 0 + RT ln Q E = E 0 - ln Q RT nF Nernst equation At 298 K - 0.0257 V n ln Q E 0 E = - 0.0592 V n log Q E 0 E =

13. 13 Will the following reaction occur spontaneously at 25 0 C if [Fe 2+ ] = 0.60 M and [Cd 2+ ] = 0.010 M? Fe 2+ (aq) + Cd (s) Fe (s) + Cd 2+ (aq) E 0 = -0.44 – (-0.40) E 0 = -0.04 V E 0 = E Fe /Fe – E Cd /Cd 00 2+ - 0.0257 V n ln Q E 0 E = - 0.0257 V 2 ln -0.04 VE = 0.010 0.60 E = 0.013 E > 0 Spontaneous From the reaction as written, Fe half cell is the cathode, and Cd half cell is the anode, use these to calculate the standard electrode potential: We cannot use E o to predict whether the rxn is spontaneous or not since we do not have standard conditions

14. 14 Will the following reaction occur spontaneously at 25 0 C if [Fe 2+ ] = 0.68 M and [Co 2+ ] = 0.15 M? Fe 2+ (aq) + Co (s) Fe (s) + Co 2+ (aq) E 0 = -0.44 – (-0.28) E 0 = -0.16 V E 0 = E Fe /Fe – E Co /Co 00 2+ - 0.0257 V n ln ([Co 2+ ]/[Fe 2+ ]) E 0 E = From the reaction as written, Fe half cell is the cathode, and Co half cell is the anode, use these to calculate the standard electrode potential:

15. 15 - 0.0257 V 2 ln -0.16 VE = 0.15 0.68 E = - 0.14 E < 0  Non-spontaneous Now think at what ratio of [Co 2+ ]/[Fe 2+ ] will the reaction be spontaneous? To find such a ratio we must pass by the point where E o cell = 0 - 0.0257 V n ln ([Co 2+ ]/[Fe 2+ ]) - 0.160 = [Co 2+ ]/[Fe 2+ ] = 4*10 -6, the ratio must be less than 4*10 -6

16. 16 Concentration Cells Galvanic cell from two half-cells composed of the same material but differing in ion concentrations.

17. 17 Selected Problems 2, 3, 6, 11-13, 15, 16, 17, 18, 21, 22, 24, 25, 29, 32, 34.