Q.Balance the following ionic equation in basic (alkaline) medium by the ion-electron method: $\text{MnO}_4^{-} + \text{I}^{-} \to \text{MnO}_2 + \text{I}_2$.

[!TLDR] Balance the acidic skeleton (Mn gains 3e⁻, I loses 1e⁻ each) then convert H⁺ to OH⁻/H2O for basic medium. [!ANSWER] $2\text{MnO}_4^{-} + \text{H}_2\text{O} + 6\text{I}^{-} \to 2\text{MnO}_2 + 2\text{OH}^{-} + 3\text{I}_2$ (equivalently written as $2\text{MnO}_4^{-} + 4 …

Reduction half (basic form): $\text{MnO}_4^{-}\to\text{MnO}_2$; balance O by adding $2\text{H}_2\text{O}$ to the right (product has 2 fewer O than reactant): $\text{MnO}_4^{-}\to\text{MnO}_2+2\text{H}_2\text{O}$; this introduces 4 H on the right, balanced in basic medium by adding $4\text{OH}^{-}$ to the left AND $4\text{H}_2\text{O}$... the cleanest route is to balance in acidic form first, then convert. Acidic form: $\text{MnO}_4^{-}+4\text{H}^{+}\to\text{MnO}_2+2\text{H}_2\text{O}$; balance charge ($+3$ left vs $0$ right) with $3e^{-}$: $\text{MnO}_4^{-}+4\text{H}^{+}+3e^{-}\to\text{MnO}_2+2\text{H}_2\text{O}$. Convert to basic medium by adding $4\text{OH}^{-}$ to both sides: $\text{MnO}_4^{-}+4\text{H}_2\text{O}+3e^{-}\to\text{MnO}_2+2\text{H}_2\text{O}+4\text{OH}^{-}$, which simplifies (cancel $2\text{H}_2\text{O}$ from both sides) to $\text{MnO}_4^{-}+2\text{H}_2\text{O}+3e^{-}\to\text{MnO}_2+4\text{OH}^{-}$.\n\nOxidation half: $2\text{I}^{-}\to\text{I}_2+2e^{-}$ (no H or O involved, so acidic/basic makes no difference here).\n\nCombine: LCM of 3 and 2 is 6; multiply the reduction half by 2 and the oxidation half by 3: $2\text{MnO}_4^{-}+4\text{H}_2\text{O}+6e^{-}\to2\text{MnO} …

Balance each half-reaction in acidic form first, then convert any $\text{H}^+$ to $\text{H}_2\text{O}$/$\text{OH}^-$ by adding equal $\text{OH}^-$ …

Leaving $\text{H}^+$ in the final 'basic medium' equation (forgetting the conversion step entirely); mis-c …