Question

When an electric current is passed 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 _______.

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

Answer 1

Answer: (C)

$1.0$

Reduction at cathode: $2H^{+}\left(aq\right)+2e^{-}\to H_2\left(g\right)$

At N.T.P, 22.4 L (or 22400 mL) of $H_2=$ 1 mole of $H_2$

$112$ mL of $H_2=\frac{112}{22400}\times 1=0.005$ mole of $H_2$

Moles of $H_2$ produced $=\frac{I\left(A\right)\times t\left(s\right)}{96500\left(C/mol{e}^{-}\right)}\times$ mole ratio

$0.005mol=\frac{I\left(A\right)\times 965s}{96500\left(C/mol{e}^{-}\right)}\times \frac{\text{1 mol}{H}_2}{\text{2 mol}{e}^{-}}$

$I=1A$