Use oxidation numbers to find the oxidant and the reductant!
$MnO_4^-$ $(o.n.(Mn)=7)$ is the oxidant which is reduced to $Mn^{2+}$ $(o.n.(Mn)=2)$. $Fe^{2+}$ $(o.n.(Fe)=2)$ is the reductant which is oxidised to $Fe^{3+}$ $(o.n.(Fe)=3)$.
How many electrons are exchanged by the oxidant and the reductant?
$MnO_4^-$ catches 5 $e^-$. $Fe^{2+}$ looses 1 $e^-$.
Write the partial oxidation and reduction reactions in an acid solution.
$MnO_4^- $ $+$ $5e^-$ $+$ $8H^+ $ $\longrightarrow$ $ Mn^{2+}$ $+$ $7H_2O$ $Fe^{2+}$ $-$ $e^-$ $\longrightarrow$ $Fe^{3+}$
Balance by the number of exchanged electrons and write the redox reaction!
$MnO_4^- $ $+$ $5e^-$ $+$ $8H^+ $ $\longrightarrow$ $ Mn^{2+}$ $+$ $7H_2O$
$5Fe^{2+}$ $-$ $5e^- $ $\longrightarrow$ $ 5Fe^{3+}$
$MnO_4^-$ $+$ $5Fe^{2+}$ $+$ $8H^+ $ $\longrightarrow$ $ Mn^{2+}$ $+$ $5Fe^{3+}$ $+$ $4H_2O$
Complete for the ions which are not involved et annotate!
$2MnO_4^- $ $+$ $10Fe^{2+}$ $+$ $16H^+$ $\longrightarrow$ $ 2Mn^{2+}$ $+$ $10Fe^{3+}$ $+$ $8H_2O$ $2K^+$ $+$ $2MnO_4^- $ $+$ $10Fe^{2+}$ $+$ $10SO_4^{2-}$ $+$ $16H^+$ $+$ $8SO_4^{2-} $ $\longrightarrow$ $2Mn^{2+}$ $+$ $2SO_4^{2-} $ $+$ $10Fe^{3+}$ $+$ $15SO_4^{2-}$ $+$ $2K^+$ $+$ $SO_4^{2-}$ $+$ $8H_2O$ Potassium permanganate, iron(II) sulfate and sulfuric acid produce manganese(II) sulfate, iron(III) sulfate, potassium sulfate and water.