Question

How many c.c. of oxygen will be liberated by 2 Ampere current flowing for 193 seconds through acidified water

• A. 11.2 c.c.
• B. 33.6 c.c.
• C. 44.8 c.c.
• D. 22.4 c.c.

Answer 1

Answer: (D)

22.4 c.c.

Answer 2

$Q = I \times t = 2 \times 193 = 386C$

$\mathbf{H}_{\mathbf{2}}\mathbf{O \rightarrow}\mathbf{H}_{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{O}_{\mathbf{2}}$ i.e., $O^{2 -} \rightarrow \frac{1}{2}O_{2} + 2e^{-}$

i.e., $2F = 2 \times 96500$ gives

$O_{2} = \frac{1}{2}$mole =11200 c.c.

$\therefore$ 386 c will give $O_{2} = \frac{11200}{2 \times 96500} \times 386 = 22.4c.c.$