\begin{align} \ce{Co^{3+}(aq) + e- &-> Co^{2+}(aq)} & E° &= \pu{+1.82 V} \\ \ce{Co^{2+}(aq) + 2e- &-> Co(s)} & E° &= \pu{-0.28 V} \end{align}

what is the standard reduction potential for $\ce{Co^{3+}(aq) +3e- -> Co(s)}$?

If I add up these two half-equations to get the desired half-equation, I get $E^\circ = \pu{+1.54 V}$. But the correct answer is $\pu{+ 0.42 V}$!

Am I missing something?


1 Answer 1


That is because you cannot simply add the electrode potentials algebraically. In both cases, the electrode potentials have to be multiplied by $n$. What you can do instead is use Gibbs free energy change for each reaction which then can be added algebraically.

$$\Delta G^\circ = -nFE^\circ$$

Using this you can get the $\Delta G^\circ$ for each reaction which can be then added algebraically in the same way you add the reactions. The resultant $\Delta G^\circ_\mathrm{net}$, on equating with $-nFE^\circ_\mathrm{net}$, we can get the value of $E^\circ_\mathrm{net}$, which should be your answer.

In this approach, since the value of $n$ for both the reactions are different, the relative weightage of the electrode potentials of each reaction in determining the resultant $E^\circ$ will differ and therefore the final $E^\circ$ will be different than what you get by just adding up the electrode potentials.

For your case:

$$\begin{align} \ce{Co^3+ + e- &-> Co^2+} & E^\circ_1 &= \pu{+1.82 V} & \Delta G_1^\circ &= \pu{-175.6 kJ mol-1} \tag{1} \\ \ce{Co^2+ + 2e- &-> Co} & E^\circ_2 &= \pu{-0.28 V} & \Delta G_2^\circ &= \pu{+54.0 kJ mol-1} \tag{2} \\ \end{align}$$

$$\begin{align} \Delta G^\circ_\mathrm{net} &= \Delta G_1^\circ + \Delta G_2^\circ \\ &= \pu{-121.6 kJ mol-1} \\[3pt] E^\circ_\mathrm{net} &= -\frac{\pu{-121.6 kJ mol-1}}{3F} \\ &= \pu{+0.42 V} \end{align}$$


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