In the reaction between acidified ${\mathrm{KMnO}}_4$ and ${\mathrm{FeSO}}_4$ solution, the number of electrons involved in the process is

The reaction of acidified ${\mathrm{KMnO}}_4$ with ${\mathrm{FeSO}}_4$ produces ${\mathrm{MnSO}}_4,{ \mathrm{K}}_2{\mathrm{SO}}_4$ and ${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_2$. In this reaction manganese undergo reduction whereas iron undergo oxidation

$2{\mathrm{KMnO}}_4+10{\mathrm{FeSO}}_4+8{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+5{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3+8{\mathrm{H}}_2\mathrm{O}$

Manganese oxidation state changes from $+7$ (reactant) to $+2$ (product). The total number of electrons involved in the reaction are $+5$