When an electric current is passes through acidified water, 112 ~mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is :

Correct answer: D. 1.0 Reduction at cathode: 2 e^ - +2 H_2 O arrow H_2+2 OH^ - (valence factor) H_2=2 At NTP 22400 ~mL of H_2=1 mole of H_2 ~112 ~mL of H_2=(1)/(22400) × 112=0.005 mole of H_2 Moles of H_2 produced aligned &= I × t 96500 × mole ratio &0.005= I × 965 96500 × 1 mole of H_2 2 mole of e^ - &I=1.0 ~A aligned

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Chemistry
Chapter: Electrochemistry

JEE Main 2018 · 15 April · Morning Shift · Online · Chemistry · Electrochemistry

When an electric current is passes through acidified water, 112 ~mL...

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 :

Options

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

Correct answer

D. $1.0$

Step-by-step 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}

\mathrm{H}_2=1

mole of

\mathrm{H}_2 ~112 \mathrm{~mL}

\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\text{ of } } \mathrm{H}_2}{2 \text{ mole of }e^{-}} \\ &\mathrm{I}=1.0 \mathrm{~A} \end{aligned}

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