Question

When an electric current is passed through acidulated water, $112\text{mL}$ of hydrogen gas at NTP collects at the cathode in $965\text{sec}$. The current passes, in ampere is

• A. $0.1$
• B. $0.5$
• C. $1.0$
• D. $2.0$

Answer 1

The correct option is C $1.0$

As $22400\text{mL}$ of hydrogen at STP (or NTP) = $2\text{g}$
So,$112\text{mL}$ of hydrogen at STP
$=\frac{2\text{g}\times 112\text{mL}}{22400\text{mL}}={10}^{-2}\text{g}$
$\underset{2\text{F}}{2H^{+}}+2e^{-}\to \underset{1\text{mol}}{H_2}$
$=2\times 96500\text{C}=2\text{g}$
$2\text{g}$ hydrogen is deposited by = $2\times 96500\text{C}\times {10}^{-2}\text{g}$
hydrogen will be deposited by
$\Rightarrow \frac{2\times 96500\times {10}^{-2}\text{g}}{2\text{g}}=965\text{C}$
$Charge\left(Q\right)=I.t$
$\Rightarrow I=\frac{Q}{t}=\frac{965\text{C}}{965\text{s}}=1{\text{C s}}^{-1}=1\text{A}$