Which expression correctly represents the Nernst equation for a redox couple at 25°C?

Which expression correctly represents the Nernst equation for a redox couple at 25°C?

• A. Ecell = E° + (0.059/n) log([Ox]/[Red])
• B. Ecell = E° - (0.059/n) log([Red]/[Ox])
• C. Ecell = E° + (0.059/n) log([Red]/[Ox])
• D. Ecell = E° - (0.059/n) log([Ox]/[Red])

Answer: (D)

The Nernst equation shows how the actual electrode potential deviates from the standard potential based on the ratio of the oxidized and reduced species. For a redox pair written as Ox + ne− ⇌ Red, the reaction quotient is Q = [Red]/[Ox], and at 25°C the equation becomes E = E° − (0.059/n) log([Red]/[Ox]). Because log([Ox]/[Red]) = −log([Red]/[Ox]), this same relationship can be written in equivalent forms by choosing which ratio appears inside the log and adjusting the sign accordingly. Writing it as E = E° − (0.059/n) log([Ox]/[Red]) is simply another valid form, since it corresponds to the same dependence on the ratio when the logarithm is taken of the reciprocal. The essential idea is that the potential depends on the ratio of oxidized to reduced species, with the 0.059/n factor fixed at 25°C.

The Nernst equation shows how the actual electrode potential deviates from the standard potential based on the ratio of the oxidized and reduced species. For a redox pair written as Ox + ne− ⇌ Red, the reaction quotient is Q = [Red]/[Ox], and at 25°C the equation becomes E = E° − (0.059/n) log([Red]/[Ox]). Because log([Ox]/[Red]) = −log([Red]/[Ox]), this same relationship can be written in equivalent forms by choosing which ratio appears inside the log and adjusting the sign accordingly. Writing it as E = E° − (0.059/n) log([Ox]/[Red]) is simply another valid form, since it corresponds to the same dependence on the ratio when the logarithm is taken of the reciprocal. The essential idea is that the potential depends on the ratio of oxidized to reduced species, with the 0.059/n factor fixed at 25°C.