Volume of $\mathrm{M} / 8 \mathrm{KMnO}_4$ solution required to react completely with $25.0 \mathrm{~cm}^3$…
Volume of $\mathrm{M} / 8 \mathrm{KMnO}_4$ solution required to react completely with $25.0 \mathrm{~cm}^3$ of $M / 4 \mathrm{FeSO}_4$ in acidic medium is.
8.0 mL
5.0 mL
15.0 mL
10.0 mL
Solution
The balance ionic equation for the reaction is
$\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}$
From the above equation it is clear that 1 mole of $\mathrm{KMnO}_4=5$ moles of $\mathrm{FeSO}_4$.
Apply molarity equation to balance the redox reaction
$\begin{aligned}
\frac{M_1 V_1}{n_1}\left(\mathrm{KMnO}_4\right) & =\frac{M_2 V_2}{n_2}\left(\mathrm{FeSO}_4\right) \\
\frac{1 \times V_1}{8 \times 1} & =\frac{1}{4} \times \frac{25}{5} \\
V_1 & =\frac{1 \times 25 \times 8}{4 \times 5} \\
V_1 & =10 \mathrm{~cm}^3 \text { or } V_1=10.0 \mathrm{~mL}
\end{aligned}$