Deriving a reduction potential from two other reduction potentials
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:

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

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