For the oxidation of $0.2 \mathrm{M} \mathrm{FeSO}_4$ solution 0.965 amperes current is passed through it…
For the oxidation of $0.2 \mathrm{M} \mathrm{FeSO}_4$ solution 0.965 amperes current is passed through it for 1 hour. The volume of the solution that is oxidised in $\mathrm{mL}$ is
70
80
60
90
Solution
The equation is
$
\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}
$
1 mole is deposited by $2 \times 96500^{\circ} \mathrm{C}=193000^{\circ} \mathrm{C}$
The charged supplied $=0.965 \times 60 \times 60$
$
=3474 \mathrm{C} \quad(\because Q=i \times t)
$
Number of moles deposited by this charge
$
\begin{aligned}
& =\frac{3474}{193000}=0.018 \mathrm{~mol} \\
& \text { Molarity }=\frac{\text { Number of moles } \times 1000}{\text { Volume in } \mathrm{mL}} \\
& \text { Volume }(\text { in } \mathrm{mL})=\frac{\text { Number of moles } \times 1000}{M(\text { molarity })} \\
& =\frac{0.018 \times 1000}{0.2}=90 \mathrm{~mL} \\
&
\end{aligned}
$