# 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 : $$\ce{MnO4- +5e- + 8H+ ->Mn^{2+}}$$ $$\ce{MnO2 + 4H+ +2e- -> Mn^{2+}}$$

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?

## 1 Answer

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-} &\ce{-> Mn^2+ + 4H2O} & \Delta G_1^\circ &= -728.46~\mathrm{kJ~mol^{-1}} \tag{1} \\ \ce{MnO2 + 4H+ + 2e-} &\ce{-> Mn^2+ + 2H2O} & \Delta G_2^\circ &= -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:

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

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 &= \Delta G_1^\circ - \Delta G_2^\circ \\ &= -728.46~\mathrm{kJ~mol^{-1}} + 237.35~\mathrm{kJ~mol^{-1}} \\ &= 491.11~\mathrm{kJ~mol^{-1}} \\ E_3^\circ &= \frac{-\Delta G_3^\circ}{3F} \\ &= +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 $+1.70~\mathrm{V}$.

• Well I still have a question - The balanced cell reaction would $2$ times $(1) -$ $5$ times of $(2)$ and thus in total 10 $e^{-1}$ are used, So shouldn't we multiply the gibb's energy by them and after subtracting divide by 10? Why would that be wrong? – user127 Dec 16 '15 at 7:31
• Can you calculate what the resultant chemical equation is if you take $2\times (1) - 5 \times (2)$? Is that what you are trying to find $\Delta G^\circ$ for? – orthocresol Dec 16 '15 at 8:00