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 :

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) 1.0 (B) 0.5 (C) 0.1 (D) 2.0

Tardigrade Admin · Accepted Answer

At cathode: 2 H +( aq )+2 e- arrow H 2( g ) At NTP, 22.4 L or 22400 mL of H 2=1 mol of H 2 112 mL of H 2=(1/22400) × 112=0.005 mol of H 2 We know n =(I × t/F) 0.005 =(I × 965/2 × 96500) ⇒ I=1 A

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

1.0

0.5

0.1

2.0

At cathode: $2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$

At NTP, $22.4 L$ or $22400 m L$ of $H_{2} = 1 m o l$ of $H_{2}$

$112 m L$ of $H_{2} = \frac{1}{22400} \times 112 = 0.005 m o l$ of $H_{2}$

We know $n = \frac{I \times t}{F}$

$0.005 = \frac{I \times 965}{2 \times 96500} \Rightarrow I = 1 A$

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

1.0

0.5

0.1

2.0

At cathode: 2H+(aq)+2e−→H2​(g) At NTP, 22.4L or 22400mL of H2​=1mol of H2​ 112mL of H2​=224001​×112=0.005mol of H2​ We know n=FI×t​ 0.005=2×96500I×965​⇒I=1A

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

At cathode: $2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$

At NTP, $22.4 L$ or $22400 m L$ of $H_{2} = 1 m o l$ of $H_{2}$

$112 m L$ of $H_{2} = \frac{1}{22400} \times 112 = 0.005 m o l$ of $H_{2}$

We know $n = \frac{I \times t}{F}$

$0.005 = \frac{I \times 965}{2 \times 96500} \Rightarrow I = 1 A$