Q13 / 160 · AP EAMCET 2008 · Chemistry
When electric current is passed through acidified water for $1930 \mathrm{~s}, 1120 \mathrm{~mL}$ of $\mathrm{H}_2$ gas is collected (at STP) at the cathode. What is the current passed in amperes?
- A$0.05$
- B$0.50$
- Correct answerC$5.0$
- D$50$
Solution
Electrolysis of water takes place as follows
$$
\mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { cathode }}{\mathrm{H}^{+}}+\underset{\text { anode }}{\mathrm{OH}^{-}}
$$
At anode
$$
\begin{aligned}
& \mathrm{OH}^{-} \stackrel{\text { Oxidation }}{\longrightarrow} \mathrm{OH}+e^{-} \\
& 4 \mathrm{OH} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2
\end{aligned}
$$
At cathode
$$
2 \mathrm{H}^{+}+2 e^{-} \stackrel{\text { Reduction }}{\longrightarrow} \mathrm{H}_2
$$
Given, time, $\quad t=1930 \mathrm{~s}$
Number of moles of hydrogen collected
$$
\begin{aligned}
& =\frac{1120 \times 10^{-3}}{22.4} \text { moles } \\
& =0.05 \mathrm{moles}
\end{aligned}
$$
$\because 1$ mole of hydrogen is deposited by $=2$
moles of electrons
$\therefore 0.05$ moles of hydrogen will be deposited by
$$
\begin{aligned}
& =2 \times 0.05 \\
& =0.10 \text { mole of electrons }
\end{aligned}
$$
Charge,
$$
\begin{aligned}
Q & =n F \\
& =0.1 \times 96500
\end{aligned}
$$
Charge, $Q=i t$
$$
\begin{aligned}
0.1 \times 96500 & =i \times 1930 \\
i & =\frac{0.1 \times 96500}{1930} \\
& =5.0 \mathrm{~A}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.