Q.Consider the following half-cell reactions: $\text{Mn}^{2+} + 2e^{-} \rightarrow \text{Mn}$, $E^{o} = -1.18$ V $\text{Mn}^{2+} \rightarrow \text{Mn}^{3+} + e^{-}$, $E^{o} = -1.51$ V The $E^{o}$ for the reaction $3\text{Mn}^{2+} \rightarrow \text{Mn} + 2\text{Mn}^{3+}$, and the possibility of the forward reaction, are respectively:

ΔG°(total) = 2.36F + 3.02F = 5.38F for n=2 electrons overall, giving $E^{o} = -5.38F/2F = -2.69$ V.

Step 1. The target reaction, $3\text{Mn}^{2+} \rightarrow \text{Mn} + 2\text{Mn}^{3+}$, is a disproportionation: one Mn²⁺ is reduced to Mn(0) while two Mn²⁺ are oxidised to Mn³⁺. Because E° values cannot simply be added when the number of electrons differs between half-reactions, work through Gibbs free energies instead, using $\Delta G = -nFE^{o}$ for each half-reaction as written.

Step 2. Reduction half (n=2): $\text{Mn}^{2+}+2e^- \rightarrow \text{Mn}$, $E^o=-1.18$ V, so $\Delta G_1 = -(2)(F)(-1.18) = +2.36F$.

Step 3. Oxidation half, doubled for 2 mol Mn²⁺ (total n=2 electrons released): $2\text{Mn}^{2+} \rightarrow 2\text{Mn}^{3+}+2e^-$, using the given $E^o=-1.51$ V (this value already IS the potential of the reaction as written, so for 2 electrons $\Delta G_2 = -(2)(F)(-1.51) = +3.02F$.

Step 4. Total $\Delta G^{o} = \Delta G_1 + \Delta G_2 = 2.36F + 3.02F = 5.38F$. The overall reaction transfers n = 2 electrons (the 2 electrons released by oxidation are exactly consumed by reduction), so $E^{o}_{overall} = -\dfrac{\Delta G^{o}}{nF} = -\dfrac{5.38F}{2F} = -2.69$ V.

Step 5. Since $E^{o}$ is negative, $\Delta G^{o}$ is positive, so the forward reaction is non-spontaneous — consistent with the chemical fact that Mn²⁺ is the most stable oxidation state of manganese in solution and does not spontaneously disproportionate.

Convert each half-reaction's E° to ΔG using $\Delta G=-nFE^{o}$ (scaling n for the actual electrons transferred in that half), sum the ΔG values, then convert the total ΔG back to an overall E° using the overall n. Never add E° values directly when the two half-reactions involve different numbers of electrons.

The most common error is adding or averaging the two given E° values directly (e.g. -1.18 + -1.51, or their average) — this is invalid whenever the electron counts differ between the half-reactions; always route through ΔG = -nFE° instead.