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.)
standard electrode emf E^{o}=0.34 v
Cu2+ + 2e- --> Cu
E = E^{o} - (0.0591/n)log{[Cu]/[Cu^{2+}]}
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 10^{48})
E = 1.51 - 0.01182 X47.346
E 0.95 v
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