[Big-Daddy]

I know that the redox couple MnO₄^-/Mn²⁺ is typically written MnO₄^- + 8H⁺ + 5e^-  Mn²⁺ + 4H₂O in acidic conditions and MnO₄^- + 4H₂O + 5e^-  Mn²⁺ + 8OH^- in basic conditions. The second reaction is simply the first with the addition of 8K_w.

But then how is it admissible to add K_w so freely? I tried to analyse the redox potential of this couple as it varies based on pH, but I cannot come out with a decently consistent equation to express it. Based on the acidic-conditions reaction the Q term from the Nernst equation should be [Mn²⁺]/[MnO₄^-] * 1/[H⁺]⁸. But based on the basic-conditions reaction the Q term should be [Mn²⁺]/[MnO₄^-] * K_w⁸/[H⁺]⁸. This implies that a different method of calculating E (cell potential) would be needed based on conditions but this surely cannot be true as pH is continuous?

(Also, I am aware that in alkaline conditions Mn(OH)₂ precipitates. I will factor that into my calculations later using the K_sp, in a method I am accustomed to. But please do not worry about it for the sake of this problem.)

[Borek]

MnO₄^- doesn't get reduced to Mn²⁺ in high pH, but to MnO₄^2-.

[Big-Daddy]

Sure, that is the more stable reaction. But can't we work out E for the couple MnO₄^-/Mn²⁺ in the same way as before, regardless of pH? In fact it should be (barring for now the precipitation of the hydroxide) exactly the same reaction as before, so why am I getting the wrong expression for it in Q?

I had wanted to plot E against [H⁺] for this couple ...

[Borek]

Check your math, you are doing a kindergarten mistake in your Q.

[Big-Daddy]

What is it? I can't see any? Looks like under acidic conditions Q = [Mn²⁺]/([MnO₄^-] * [H⁺]⁸) but under basic conditions Q = [Mn²⁺] * [OH^-]⁸/[MnO₄^-] which is a fundamental contradiction because of the difference of the factor K_w⁸.