# Ordering the reduction potentials for Cr3+, Mn3+ and Fe3+?

I have been asked to order, with explanation, the standard reduction potentials for the following:

1. $\ce{Cr^3+ + e- -> Cr^2+}$

2. $\ce{Mn^3+ + e- -> Mn^2+}$

3. $\ce{Fe^3+ + e- -> Fe^2+}$

(all ions in aqueous conditions)

I am really uncertain what factors to consider; I am even struggling to make this a general born-haber process. I know that $\mathrm{3^{rd}}$ ionization enthalpies will come into play, and that $\ce{Fe}$ will be expected to have the highest one. Will hydration enthalpies also be relevant? Will there be a ligand exchange component (with the solvent, water) that I have to consider also?

Firstly, I'd like to say the explanations we propose here are merely our efforts trying to explain what we observed experimentally in the first place. Let's get some data:

$${\begin{array}{|c|c|} \hline \ce{Cr^3+/ Cr^2+} & \ce{Mn^3+ / Mn^2+} & \ce{Fe^3+ / Fe^2+} \\ \hline \pu{-0.41 V} & \pu{+1.57 V} & \pu{+1.97 V} \\ \hline \end{array}}$$

And while we're at it, let's ponder on electronic configurations as well:

$${\begin{array}{|c|c|} \hline \ce{Mn^2+} & \pu{[Ar] 3d5 4s0} \\ \hline \ce{Fe^3+} & \pu{[Ar] 3d5 4s0} \\ \hline \end{array}}$$

You were correct in thinking that $\ce{Fe^3+}$ will have the highest reduction potential due to the third ionization enthalpy coming into play, all we've got to do now is to think for manganese and chromium. Why don't we put it differently:

Why is $\ce{Cr^2+}$ reducing and $\ce{Mn^3+}$ oxidizing when they both have the same $\pu{d4}$ configuration?

It's because if manganese acts as an oxidizer, it'll get converted into $\ce{Mn^2+}$, which is stable owing to the $\pu{d5}$ configuration. Chromium, on the other hand is notorious for showing the $+3$ oxidation state in its compounds. So, that's a win-win kind of situation we're dealing with.

The subtle role played by the solvent molecules when they're acting as ligands comes into limelight on asking yet another relevant question:

Which is a stronger reducing agent: $\ce{Cr^2+}$ or $\ce{Fe^2+}$?

Answer: $\ce{Cr^2+}$. This is because of more CFSE stabilization of $\ce{Cr^3+}$ as compared to $\ce{Fe^3+}$ when water molecules behave as a ligand for our complex.

• this has really helped with pointing me in the right direction-thank you so much!!! Apr 30 '17 at 22:35