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!


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.