During a redox titration involving a solution containing Fe ^ 2 +...

For redox titration of a solution containing& ${Fe}^{2 +}$~against~${MnO}_{4}^{-}$,~$H^{+}$~ions are present in excess i.e., the solution~is acidic.

$\overset{+ 7}{Mn}{O}_{4}^{-} + {Fe}^{2 +} \to {Fe}^{3 +} + {Mn}^{2 +} ( acidicmedium )$

Here, oxidation number of~$Mn$~is reduced from~$+ 7to + 2$~(reduction)~and oxidation number of~$Fe$~is increased from~$+ 2to + 3$(oxidation).

Reduction half cell,

${MnO}_{4}^{-} + 8H^{+} + 5e^{-} ⟶ {Mn}^{2 +} + 4{H}_{2}O$....(equation 1)

Oxidation half cell,

[$Fe^{2 +} ⟶ Fe^{3 +} + e^{-} ] \times 5$.....(equation 2)

By adding equation 1 and 2, final balanced reaction will be,

${MnO}_{4}^{-} + 8H^{+} + 5{Fe}^{2 +} ⟶ {Mn}^{2 +} + 5{Fe}^{3 +} + 4{H}_{2}O$
Therefore, five electrons gets transferred.