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I need someone to help me with a problem:

We have a solution which is $\pu{0.1 M}$ of $\ce{KMnO4}$ and $\pu{0.1 M}$ $\ce{Fe^2+}$ and has $\mathrm{pH}=0$. Given the electrical reduction potential of $\ce{MnO4-/Mn^2+}$ and $\ce{Fe^3+/Fe^2+}$, we are asked to find the electric potential of the reaction.

This is my work on the question:

The electrical reduction potential of $\ce{KMnO4/Mn^2+}$ is bigger, so the $\ce{Fe^2+}$ becomes $\ce{Fe^3+}$ and $\ce{MnO4-}$ becomes $\ce{Mn^2+}$. I've come to this reaction:

$\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}$

I tried to find a potential using Nernst equation, but the concentration of $\ce{Mn^2+}$ and $\ce{Fe^3+}$ is initially zero. So the logarithm tends to minus infinity. On the other hand, I tried to look at equilibrium but got a zero potential (as expected because at equilibrium Gibbs free energy equals 0)

How can I solve it?

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