Write the balanced chemical equation for the following reaction in an acidic medium using the ion-electron method:
Permanganate ion $(MnO_4^-)$ reacts with sulphur dioxide gas to produce $Mn^{+2}$ and hydrogen sulphate ion $(HSO_4^-)$.

(N/A) The unbalanced equation is:
$MnO_{4_{(aq)}}^{-} + SO_{2_{(g)}} \rightarrow Mn_{(aq)}^{+2} + HSO_{4_{(aq)}}^{-}$
Step $1$: Separate into two half-reactions:
Reduction Half-Reaction $(R.H.R.)$ : $MnO_{4_{(aq)}}^{-} \rightarrow Mn_{(aq)}^{+2}$
Oxidation Half-Reaction $(O.H.R.)$ : $SO_{2_{(g)}} \rightarrow HSO_{4_{(aq)}}^{-}$
Step $2$: Balance atoms and charges in each half-reaction in acidic medium:
$R.H.R.$ : $MnO_{4_{(aq)}}^{-} + 8H_{(aq)}^{+} + 5e^{-} \rightarrow Mn_{(aq)}^{+2} + 4H_{2}O_{(l)}$
$O.H.R.$ : $SO_{2_{(g)}} + 2H_{2}O_{(l)} \rightarrow HSO_{4_{(aq)}}^{-} + 3H_{(aq)}^{+} + 2e^{-}$
Step $3$: Equalize the number of electrons by multiplying $R.H.R.$ by $2$ and $O.H.R.$ by $5$:
$2MnO_{4_{(aq)}}^{-} + 16H_{(aq)}^{+} + 10e^{-} \rightarrow 2Mn_{(aq)}^{+2} + 8H_{2}O_{(l)}$
$5SO_{2_{(g)}} + 10H_{2}O_{(l)} \rightarrow 5HSO_{4_{(aq)}}^{-} + 15H_{(aq)}^{+} + 10e^{-}$
Step $4$: Add the two half-reactions:
$2MnO_{4_{(aq)}}^{-} + 5SO_{2_{(g)}} + 16H_{(aq)}^{+} + 10H_{2}O_{(l)}$ $\rightarrow 2Mn_{(aq)}^{+2} + 5HSO_{4_{(aq)}}^{-} + 15H_{(aq)}^{+} + 8H_{2}O_{(l)}$
Simplifying the equation:
$2MnO_{4_{(aq)}}^{-} + 5SO_{2_{(g)}} + 2H_{2}O_{(l)} + H_{(aq)}^{+} \rightarrow 2Mn_{(aq)}^{+2} + 5HSO_{4_{(aq)}}^{-}$