Question

When an electric current is passed through acidified water, $112\mathrm{mL}$ of Hydrogen gas at N.T.P. was collected at the Cathode in $965\mathrm{seconds}$. The current passed, in Ampere, is

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

Answer 1

The correct option is B

$1.0$

Explanation for correct option:

(B) $1.0$

At cathode reduction will happen i.e gaining of electrons

$$\begin{gathered} 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\left(\mathrm{g}\right) \\ \mathrm{At}\mathrm{N}.\mathrm{T}.\mathrm{P},22.4\mathrm{L}(22400\mathrm{mL})\mathrm{of}{\mathrm{H}}_2=1\mathrm{mole}\mathrm{of}{\mathrm{H}}_2 \\ \mathrm{Then},112\mathrm{mL}\mathrm{of}{\mathrm{H}}_2=\frac{112}{22400}\mathrm{x}1 \\ =0.005\mathrm{mole}\mathrm{of}{\mathrm{H}}_2 \\ \mathrm{Moles}\mathrm{of}{\mathrm{H}}_2\mathrm{produced}=\frac{\mathrm{I}\left(\mathrm{A}\right)\mathrm{x}\mathrm{t}\left(\mathrm{s}\right)}{96500(\mathrm{C}/\mathrm{mol}{\mathrm{e}}^{-})}\mathrm{x}\mathrm{mole}\mathrm{ratio} \\ 0.005=\frac{\mathrm{I}\mathrm{x}965}{96500(\mathrm{C}/\mathrm{mol}{\mathrm{e}}^{-})}\mathrm{x}\frac{1\mathrm{mol}\mathrm{of}{\mathrm{H}}_2}{2\mathrm{mol}\mathrm{of}{\mathrm{e}}^{-}} \\ \mathrm{I}=1\mathrm{A} \end{gathered}$$

Explanation for incorrect options:

The current passed is $1.0\mathrm{A}$. Thus, options A, C, and D are incorrect.

Hence, the correct option is B i.e. $1.0\mathrm{A}$.