The volume of oxygen gas liberated at NTP by passing a current of 9650 coulombs through acidified water is

Question

The volume of oxygen gas liberated at NTP by passing a current of 9650 coulombs through acidified water is (A) 1.12 litre (B) 2.24 litre (C) 11.2 litre (D) 0.56 litre

Tardigrade Admin · Accepted Answer

Oxidation reaction at anode, upon electrolysis of water:

2 H_{2} O (l) ⟶ O_{2} (g) + 4 H^{+} (a q) + 4 e^{-}; E_{ce ll}^{\circ} = + 1.23 V

Thus, 1 mole of oxygen is liberated by 4 moles of electrons.

4 \times 96500

coulombs electricity liberates

= 22.4 L_{2} O_{2}

gas

9650

coulombs electricity liberates

= \frac{22.4}{4 \times 96500} \times 9650

= 0.56 L . O_{2}

gas

Q. The volume of oxygen gas liberated at NTP by passing a current of $9650$ coulombs through acidified water is

$1.12$ litre

$2.24$ litre

$11.2$ litre

$0.56$ litre

Q. The volume of oxygen gas liberated at NTP by passing a current of 9650 coulombs through acidified water is

1.12 litre

2.24 litre

11.2 litre

0.56 litre

Q. The volume of oxygen gas liberated at NTP by passing a current of $9650$ coulombs through acidified water is

$1.12$ litre

$2.24$ litre

$11.2$ litre

$0.56$ litre