# Electric potential of a redox reaction (without solids)

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?