When an electric current is passes through acidified water, $112 \mathrm{~mL}$ of hydrogen gas at N.T.P. was…
When an electric current is passes through acidified water, $112 \mathrm{~mL}$ of hydrogen gas at N.T.P. was collected at the cathode in $965$ seconds. The current passed, in ampere, is :
$2.0$
$0.1$
$0.5$
$1.0$
Solution
Reduction at cathode:
$2 \mathrm{e}^{-}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}$
(valence factor) $\mathrm{H}_2=2$
At NTP $22400 \mathrm{~mL}$ of $\mathrm{H}_2=1$ mole of $\mathrm{H}_2 ~112 \mathrm{~mL}$ of $\mathrm{H}_2=\frac{1}{22400} \times 112=0.005 \text { mole of } \mathrm{H}_2$
Moles of $\mathrm{H}_2$ produced
$\begin{aligned}
&=\frac{\mathrm{I} \times \mathrm{t}}{96500} \times \text { mole ratio } \\
&0.005=\frac{\mathrm{I} \times 965}{96500} \times \frac{1 \text { mole of } \mathrm{H}_2}{2 \text { mole of }e^{-}} \\
&\mathrm{I}=1.0 \mathrm{~A}
\end{aligned}$