Balance the following redox reactions by the ion-electron method:
$(a)$ $MnO_4^-(aq) + I^{-}(aq) \rightarrow MnO_2(s) + I_2(s)$ (in basic medium)
$(b)$ $MnO_4^-(aq) + SO_2(g) \rightarrow Mn^{2+}(aq) + HSO_4^-(aq)$ (in acidic solution)
$(c)$ $H_2O_2(aq) + Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + H_2O(l)$ (in acidic solution)
$(d)$ $Cr_2O_7^{2-} + SO_2(g) \rightarrow Cr^{3+}(aq) + SO_4^{2-}(aq)$ (in acidic solution)

$(a)$ Oxidation half: $2I^{-} \rightarrow I_2 + 2e^{-}$. Reduction half: $MnO_4^{-} + 2H_2O + 3e^{-} \rightarrow MnO_2 + 4OH^{-}$. Multiplying by $3$ and $2$ respectively: $6I^{-} + 2MnO_4^{-} + 4H_2O \rightarrow 3I_2 + 2MnO_2 + 8OH^{-}$.
$(b)$ Oxidation half: $SO_2 + 2H_2O \rightarrow HSO_4^{-} + 3H^{+} + 2e^{-}$. Reduction half: $MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O$. Multiplying by $5$ and $2$ respectively: $2MnO_4^{-} + 5SO_2 + 2H_2O + H^{+} \rightarrow 2Mn^{2+} + 5HSO_4^{-}$.
$(c)$ Oxidation half: $Fe^{2+} \rightarrow Fe^{3+} + e^{-}$. Reduction half: $H_2O_2 + 2H^{+} + 2e^{-} \rightarrow 2H_2O$. Multiplying oxidation by $2$: $2Fe^{2+} + H_2O_2 + 2H^{+} \rightarrow 2Fe^{3+} + 2H_2O$.
$(d)$ Oxidation half: $SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^{+} + 2e^{-}$. Reduction half: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O$. Multiplying oxidation by $3$: $Cr_2O_7^{2-} + 3SO_2 + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O$.