A current of 0.75 A is passed through an acidic CuSO4 solution for 5 min. The volume of oxygen at STP liberated at Pt anode is

A
$13.05 mL$
B
$26.1 mL$
C
$39.15 mL$
D
$52.2 mL$

Solution:

Quantity of electricity passed , $Q = I t = (0.75 \times 5 \times 60) C$

Amount o electricity passed $n = \frac{Q}{F} = \frac{0.75 \times 5 \times 60}{96500} mol$

The reaction at anode is $2 H_{2} O (l) \to 4 H^{+} (aq) + O_{2} (g) + 4 e$

Thus, $4 mol$ of electrons causes the release of $1 mol$ of $O_{2} (= 22.414 Lat STP) .$ Hence, for the amount n of electricity, volume of $O_{2}$ released will be

$V = \frac{1}{4} (\frac{0.75 \times 5 \times 60}{96500} mol) (22.414 L {mol}^{- 1}) = 0.01306 L = 13.06 mL$

A current of $0.75 A$ is passed through an acidic ${CuSO}_{4}$ solution for 5 min. The volume of oxygen at STP liberated at Pt anode is

Solution:

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