Deriving a reduction potential from two other reduction potentials


> 
> Given:
> 
> 
> \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? 




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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