How would we represent the reduction of $M n O_{2}$ $MnO_2$ to give $M n^{2 +}$ $Mn^(2+)$ in acidic conditions?

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Answer 1 · 2016-09-01T15:36:12

Manganese is reduced from $M n^{I V +}$ $Mn^(IV+)$ to $M n^{I I +}$ $Mn^(II+)$ ,....

And of course, for every reduction there is a corresponding oxidation: here, chloride ion is oxidized to zerovalent chlorine gas.

The overall (balanced) redox equation is:

$M n O_{2} (s) + 4 H C l (a q) \to M n C l_{2} (s) + C l_{2} (g) + 2 H_{2} O (l)$ $MnO_2(s) + 4HCl(aq) rarr MnCl_2(s) + Cl_2(g) +2H_2O(l)$

Is that balanced? Do not trust my arithmetic!

How would we represent the reduction of MnO2MnO_2 to give Mn2+Mn^(2+) in acidic conditions?