Permanganate $(VII)$ ion,$MnO_4^-$,in basic solution oxidises iodide ion,$I^-$,to produce molecular iodine $(I_2)$ and manganese $(IV)$ oxide $(MnO_2)$. Write a balanced ionic equation to represent this redox reaction.

(N/A) Step $1:$ Write the skeletal ionic equation:
$MnO_4^-(aq) + I^-(aq) \rightarrow MnO_2(s) + I_2(s)$
Step $2:$ Identify the two half-reactions:
Oxidation half: $I^-(aq) \rightarrow I_2(s)$
Reduction half: $MnO_4^-(aq) \rightarrow MnO_2(s)$
Step $3:$ Balance $I$ atoms in the oxidation half-reaction:
$2I^-(aq) \rightarrow I_2(s) + 2e^-$
Step $4:$ Balance $O$ and $H$ atoms in the reduction half-reaction in basic medium:
$MnO_4^-(aq) + 2H_2O(l) + 3e^- \rightarrow MnO_2(s) + 4OH^-(aq)$
Step $5:$ Equalize the number of electrons by multiplying the oxidation half-reaction by $3$ and the reduction half-reaction by $2$:
$6I^-(aq) \rightarrow 3I_2(s) + 6e^-$
$2MnO_4^-(aq) + 4H_2O(l) + 6e^- \rightarrow 2MnO_2(s) + 8OH^-(aq)$
Step $6:$ Add the two half-reactions to obtain the balanced net ionic equation:
$2MnO_4^-(aq) + 6I^-(aq) + 4H_2O(l) \rightarrow 2MnO_2(s) + 3I_2(s) + 8OH^-(aq)$