Balance the half-reaction $\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+}$ in acidic medium. How many electrons are transferred per $\mathrm{MnO}_4^-$ ion?

1. Start with the skeleton equation in acidic medium: $\mathrm{MnO}_4^- + \,\mathrm{H}^+ + n\,\mathrm{e}^- \rightarrow \mathrm{Mn}^{2+} + \mathrm{H_2O}$. 2. Balance oxygen: 4 O atoms on LHS, so add 4 $\mathrm{H_2O}$ on RHS. 3. Balance hydrogen: 8 H atoms on RHS, so add 8 $\mathrm{H}^+$ on LHS. 4. Balance charge: LHS so far is $-1 + 8(+1) = +7$; RHS is $+2$. To make LHS match $+2$, add 5 electrons on LHS. 5. Final equation: $\mathrm{MnO}_4^- + 8\,\mathrm{H}^+ + 5\,\mathrm{e}^- \rightarrow \mathrm{Mn}^{2+} + 4\,\mathrm{H_2O}$. Five electrons per $\mathrm{MnO}_4^-$. 6. Confirm with oxidation states: Mn goes $+7 \rightarrow +2$, a 5-electron gain per Mn atom. Matches. _Source: syllabus Chemistry Part 2, Ch 7, §7.4 (half-reaction method, $\mathrm{MnO}_4^-$ example), p. 12–13._