Explain the balancing of a redox reaction by the half-reaction method with a suitable example.

Step-$1$: Write the unbalanced equation for the reaction in ionic form:
$Fe_{(aq)}^{+2} + Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow Fe_{(aq)}^{+3} + Cr_{(aq)}^{+3}$
Step-$2$: Separate the equation into two half-reactions.
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3}$
Reduction half: $Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow Cr_{(aq)}^{+3}$
Step-$3$: Balance the atoms other than $O$ and $H$ in each half-reaction. For the reduction half-reaction,multiply $Cr^{+3}$ by $2$ to balance $Cr$ atoms:
$Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow 2Cr_{(aq)}^{+3}$
Step-$4$: For reactions in acidic medium,add $H_{2}O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms:
$Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Step-$5$: Add electrons to balance the charges. Multiply the oxidation half-reaction by $6$ to equalize the electrons:
Oxidation half: $6Fe_{(aq)}^{+2} \rightarrow 6Fe_{(aq)}^{+3} + 6e^{-}$
Reduction half: $Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} + 6e^{-} \rightarrow 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Step-$6$: Add the two half-reactions to obtain the balanced overall reaction:
$6Fe_{(aq)}^{+2} + Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} \rightarrow 6Fe_{(aq)}^{+3} + 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$