Question

Nernst Equation Applied to Half-Reactions The Nernst equation can be applied to half-reactions.

1) Calculate the reduction potential (at 25°C) of the half-cell Cu/Cu2+ (1.5×10-2 M).
(The half-reaction is Cu2+ + 2e- --> Cu.)

2) Calculate the reduction potential (at 25°C) of the half-cell MnO4- (1.80×10-1 M)/ Mn2+ (4.00×10-2 M) at pH = 6.00.
(The half-reaction is MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O.)

Answer 1

standard electrode emf E^o=0.34 v
Cu2+ + 2e- --> Cu

E = E^o - (0.0591/n)log{[Cu]/[Cu²⁺]}
E = 0.34 - (0.0591/2) log {1/1.5 x 10^-2}

E = 0.34 - 0.0539

E = 0.286 v

b)

The half-reaction is MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O...... E° = 1.51 V

pH = 6

-log [H+] = 6

log [H+] = -6

[H+] = 10^-6 M

[MnO4-] = 1.80×10-1M or 0.18 M

[Mn2+] =4.00×10-2 M or 0.04 M

so E = E° - (0.0591/n) log { [Mn2+] . [H2O]⁴ } / { [MnO4-] .[H+]⁸ }

for pure solids and pure liquids ...the molar concentration is taken as unity...so [H2O] = 1

E = 1.51 - (0.0591/5)log {0.04/(0.18 X 10^-48)}

E = 1.51 - 0.01182 log (0.222 X 10⁴⁸)

E = 1.51 - 0.01182 X47.346

E 0.95 v