Balance the redox reaction where $Fe^{+2}$ in an acidic medium reduces $Cr_2O_7^{-2}$ ion to $Cr^{+3}$ ion,while $Fe^{+2}$ itself is oxidized to $Fe^{+3}$.

Step-$1$: Write the unbalanced ionic equation:
$Fe_{(aq)}^{+2} + Cr_2O_{7(aq)}^{-2} \rightarrow Fe_{(aq)}^{+3} + Cr_{(aq)}^{+3}$
Step-$2$: Separate into half-reactions:
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3}$
Reduction half: $Cr_2O_{7(aq)}^{-2} \rightarrow Cr_{(aq)}^{+3}$
Step-$3$: Balance atoms other than $O$ and $H$:
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3}$
Reduction half: $Cr_2O_{7(aq)}^{-2} \rightarrow 2Cr_{(aq)}^{+3}$
Step-$4$: Balance $O$ atoms using $H_2O$ and $H$ atoms using $H^+$:
Reduction half: $Cr_2O_{7(aq)}^{-2} + 14H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{+3} + 7H_2O_{(l)}$
Step-$5$: Balance charges by adding electrons:
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3} + e^-$
Reduction half: $Cr_2O_{7(aq)}^{-2} + 14H_{(aq)}^{+} + 6e^- \rightarrow 2Cr_{(aq)}^{+3} + 7H_2O_{(l)}$
Step-$6$: Equalize electrons and add the half-reactions:
Multiply oxidation half by $6$: $6Fe_{(aq)}^{+2} \rightarrow 6Fe_{(aq)}^{+3} + 6e^-$
Add both: $6Fe_{(aq)}^{+2} + Cr_2O_{7(aq)}^{-2} + 14H_{(aq)}^{+} \rightarrow 6Fe_{(aq)}^{+3} + 2Cr_{(aq)}^{+3} + 7H_2O_{(l)}$