I am working the following USNCO problem (#41 from 2002). Based on this question and answer Deriving a reduction potential from two other reduction potentials, it seems that $\Delta G$ must be calculated and then added. However, when I do this, I get an answer that is not one of the given choices.

The problem is:

Use the given standard reduction potentials to determine the reduction potential for this half-reaction: $$\ce{MnO4- + 3e- +4H+ -> MnO2 + 2H2O}$$

The given reactions are:

$$\begin{align} \ce{MnO4- + e-} &\rightarrow \ce{MnO4^2-} & E &= +0.564~\mathrm{V} \\ \ce{MnO4^2- + 2e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= +2.261~\mathrm{V} \end{align}$$

The possible answers are: $1.695~\mathrm{V}$, $2.825~\mathrm{V}$, $3.389~\mathrm{V}$, and $5.086~\mathrm{V}$.

The correct answer is A. Using $\Delta G$, I got $E = 1.928$, but this is not a choice. Am I doing the math wrong, or is my method incorrect? Thanks!


It's somewhat unclear what you mean by "using $\Delta G$", but here's the full working. The information you are given is:

$$\begin{align} \ce{MnO4- + e-} &\rightarrow \ce{MnO4^2-} & E &= +0.564~\mathrm{V} \tag{1} \\ \ce{MnO4^2- + 2e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= +2.261~\mathrm{V} \tag{2} \end{align}$$

Converting these to Gibbs free energy changes is fundamentally the correct step. We have $\Delta G = -nFE$. (Standard state symbols omitted for clarity, they are implied throughout.) For half-equation (1), $n = 1$, and for half-equation (2), $n = 2$. So:

$$\begin{align} \Delta G_1 &= -(1)(96485~\mathrm{C~mol^{-1}})(0.564~\mathrm{V}) = -54417.7~\mathrm{J~mol^{-1}} \\ \Delta G_2 &= -(2)(96485~\mathrm{C~mol^{-1}})(2.261~\mathrm{V}) = -436306.7~\mathrm{J~mol^{-1}} \\ \end{align}$$

You need to now sum up these two equations, which also means summing up the changes in Gibbs free energy. Let's define a new reaction $(3) = (1) + (2)$:

$$\begin{align} \ce{MnO4- + 3e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= E_3 \tag{3} \end{align}$$

As you can see, all I have done is to add the two half-equations together and cancel out any terms that appear on both sides. We have $n = 3$ for this half-reaction.

$$\begin{align} \Delta G_3 &= \Delta G_1 + \Delta G_2 \\ &= -490724.4~\mathrm{J~mol^{-1}} \\ E_3 &= -\frac{\Delta G_3}{nF} \\ &= 1.695~\mathrm{V} \end{align}$$

Last thing, please always quote your units in your answer! You cannot have an electrode potential that is unitless.


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