Question

How would we represent the reduction of `MnO_2` to give `Mn^(2+)` in acidic conditions?

Answer 1

Manganese is reduced from `Mn^(IV+)` to `Mn^(II+)`,....

Explanation:

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:

`MnO_2(s) + 4HCl(aq) rarr MnCl_2(s) + Cl_2(g) +2H_2O(l)`

Is that balanced? Do not trust my arithmetic!