Why is $\ce{MnO4-}$ more stable in basic than acidic conditions? I know that electrode potential values suggest this, but what is the underlying chemical explanation for the phenomenon?

  • 2
    $\begingroup$ $\ce{{MnO_4}^{-}}$ reduction starts as an electron transfer, which happens easier if the ion is protonated first.. $\endgroup$
    – permeakra
    Jan 31, 2016 at 21:42

1 Answer 1


The obviously feasible and spontaneous disproportionation reaction can be explained by considering the standard electrode potentials (standard reduction potential) involved (quoted as half–cell reductions, as is the convention).

  1. $\displaystyle \ce{MnO4- (aq) + e- -> MnO4^2- (aq)} \quad\quad (E^0 = +0.56~\mathrm{V})$

  2. $\displaystyle \ce{MnO4^2- (aq) + 4H+(aq) + 2e- -> MnO2(s) + 2H2O(l)} \quad\quad (E^0 = +1.70~\mathrm{V}, \text{in acid solution})$

2. has the more positive potential, so this will be the reduction half–cell reaction.

1. has the less positive potential, so this will be (reversed) the oxidation half–cell reaction.

$$E^0_\text{reaction} = E^0_\text{reduction} – E^0_\text{oxidation} = (+1.70~\mathrm{V}) - (+0.56~\mathrm{V}) = +1.14~\mathrm{V} \gg 0~\mathrm{V}$$

Therefore, the reaction is very feasible.

Incidentally: Given the two half–cell reactions, you get the complete balanced equation by adding 2. plus 2 x 1. reversed.

The greater stability of the manganate(VI) ion in alkali can also be explained by considering the electrode potential for 2. in an alkaline media.

  1. $\displaystyle \ce{MnO4^2- (aq) + 2H2O(l) + 2e- -> MnO2(s) + 4OH- (aq)}\quad \quad (E^0 = +0.59~\mathrm{V}, \text{in alkaline solution})$

so, re–calculating gives

$$E^0_\text{reaction} = E^0_\text{reduction} - E^0_\text{oxidation} = (+0.59~\mathrm{V}) - (+0.56~\mathrm{V}) = +0.03~\mathrm{V} > 0~\mathrm{V}$$

Therefore just feasible! but on the basis of an equilibrium argument, here, the far lower $E^0_\text{reaction}$, suggests the $\ce{MnO4^2-}$ ion is far more likely to exist, i.e. more stable, in a very high pH solution and in practice it is stable for a few hours in alkali.

  • 1
    $\begingroup$ Please try to improve the formatting of your post using $\LaTeX$. For more information on how to do this, see here and here. Additionally, you can visit this chatroom for more assistance. $\endgroup$
    – bon
    Jan 31, 2016 at 14:10
  • $\begingroup$ That's an empirical argument based on comparing electrode potential values. I would like to know, in terms of coordination complex formation, why is the MnO4- complex less stable in acidic than in alkaline conditions? $\endgroup$
    – kane9530
    Jan 31, 2016 at 14:43
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    $\begingroup$ @kane9530 It would be bad to think of $\ce{MnO4-}$’s stability in terms of complex stability. Ligand exchange is not an option, the oxygens are bound much too tightly to the manganese. $\endgroup$
    – Jan
    Jan 31, 2016 at 21:16

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