Standard reduction potential of MnO4-/MnO2 couple This is the data given:

$$\ce{MnO4- /Mn^{2+}} = 1.5~\mathrm{V}$$

$$\ce{MnO2 /Mn^{2+}}= 1.23~\mathrm{V}$$

I know that :
$$\ce{MnO4- +5e- + 8H+ -&gt;Mn^{2+}}$$
$$\ce{MnO2 + 4H+ +2e- -&gt; Mn^{2+}}$$ 

After that here's what's done which I absolutely don't get. After using $\Delta G^\circ = -nFE^\circ$, won't we have to multiply these energy values by the coefficients of the balanced equations before we can get $E_{\mathrm{cell}}$?

In particular, shouldn't the $\Delta G^\circ$ of the first equation be multiplied by two and that of the second by three?
The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.



Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations

$$\begin{align}
\ce{MnO4- + 8H+ + 5e-} &amp;\ce{-&gt; Mn^2+ + 4H2O} &amp; \Delta G_1^\circ &amp;= -728.46~\mathrm{kJ~mol^{-1}} \tag{1} \\
\ce{MnO2 + 4H+ + 2e-} &amp;\ce{-&gt; Mn^2+ + 2H2O} &amp; \Delta G_2^\circ &amp;= -237.35~\mathrm{kJ~mol^{-1}} \tag{2}
\end{align}$$

The trick is to find the appropriate combination of equations $(1)$ and $(2)$ that give the equation $(3)$, which you are interested in:

$$\ce{MnO4- + 4H+ + 3e- -&gt; MnO2 + 2H2O} \qquad \qquad \Delta G_3^\circ = \,\,? \qquad \qquad \tag{3}$$

If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:

$$\begin{align}
\Delta G_3^\circ &amp;= \Delta G_1^\circ - \Delta G_2^\circ \\
&amp;= -728.46~\mathrm{kJ~mol^{-1}} + 237.35~\mathrm{kJ~mol^{-1}} \\
&amp;= 491.11~\mathrm{kJ~mol^{-1}} \\
E_3^\circ &amp;= \frac{-\Delta G_3^\circ}{3F} \\
&amp;= +1.697~\mathrm{V}
\end{align}$$

If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.