Electrochemical Potential – Non Standard Conditions Variations in Concentration

1. Electrochemical Potential – Non Standard ConditionsVariations in Concentration Standard Conditions for Concentration are 1 M Solutions If the Concentrations are changed, then an adjustment must be made in calculating the electrochemical potential, E Equation for adjusting electrochemical equation for non-standard concentrations – Nernst equation E = E - RT ln Q = E - 0.0592 log Q nF n

2. Equation for adjusting electrochemical equation for non-standard concentrations – Nernst equation E = E - RT ln Q = E - 0.0592 log Q nF n R = gas constant for joules = 8.31 J mol –1 K-1 T = temperature in K = 298 K F = Faraday’s constant = 96500 C mol -1 n = mole electrons transferred

3. Equation for adjusting electrochemical equation for non-standard concentrations – Nernst equation E = E - RT ln Q = E - 0.0592 log Q nF n Q = reaction quotient The ratio of the product concentrations to the reactants concentrations. For the reaction: 2 A + B  3 C + D Reaction Quotient Q = [ C ]3 [ D ] [ A ]2 [ B ] The coefficients in the balanced equation are included in the reaction quotient 1Pure Solids - The quotient focuses on concentration. Since concentration is constant for a solid, any solid in the reaction is left out of the reaction quotient

4. Example Problem: A voltaic cell is set up at 25 C with one half cell consisting of an aluminum electrode in 0.00100 M aluminum nitrate and the other cell consisting of a nickel electrode in 0.500 M nickel II nitrate. Identify the oxidation and reduction half reactions. The relevant reduction half reactions are: Al 3+ + 3 e - Al -1.66 V Ni2+ + 2 e - Ni -0.25 V The Al reaction must be reversed to yield a positive potential. Balance the two half reactions so that the total electrons transferred are equivalent Oxidation: 2 Al  2 Al 3+ + 6 e - 1.66 V Reduction: 3 Ni2+ + 6 e - 3 Ni -0.25 V

5. Example Problem: A voltaic cell is set up at 25 C with one half cell consisting of an aluminum electrode in 0.00100 M aluminum nitrate and the other cell consisting of a nickel electrode in 0.500 M nickel II nitrate. Write a balanced net ionic equation. Oxidation: 2 Al  2 Al 3+ + 6 e - 1.66 V Reduction: 3 Ni2+ + 6 e - 3 Ni -0.25 V Balanced net ionic equation: 2 Al + 3 Ni2+ 2 Al 3+ + 3 Ni

6. Example Problem: A voltaic cell is set up at 25 C with one half cell consisting of an aluminum electrode in 0.00100 M aluminum nitrate and the other cell consisting of a nickel electrode in 0.500 M nickel II nitrate. Determine the standard electrochemical potential, E Oxidation: 2 Al  2 Al 3+ + 6 e - 1.66 V Reduction: 3 Ni2+ + 6 e - 3 Ni -0.25 V 1.41 V

7. Example Problem: A voltaic cell is set up at 25 C with one half cell consisting of an aluminum electrode in 0.00100 M aluminum nitrate and the other cell consisting of a nickel electrode in 0.500 M nickel II nitrate. Determine the reaction quotient Q for the reaction as described. 2 Al + 3 Ni2+ 2 Al 3+ + 3 Ni Q = [Al 3+] 2 = [0.00100 M] 2 = 8.00 x 10 -6 [Ni 2+] 3 [0.500] 3

8. Example Problem: A voltaic cell is set up at 25 C with one half cell consisting of an aluminum electrode in 0.00100 M aluminum nitrate and the other cell consisting of a nickel electrode in 0.500 M nickel II nitrate. Determine the electrochemical potential, E, for the reaction as described. Ecell = E- 0.0592 log Q = 1.41 – 0.0592 log (8.00 x 10 –6) = 1.46 V n 6

9. Example Problem: A voltaic cell is set up at 25 C with one half cell consisting of an aluminum electrode in 0.00100 M aluminum nitrate and the other cell consisting of a nickel electrode in 0.500 M nickel II nitrate. Determine the change in free energy,  G, for the reaction as described Since Ecell = 1.46 V  Gcell = - n F Ecell = - (6) (96500 C) (1.46 V) = - 845,000 J