Question

$10$g fairly concentrated solution of $CuS{O}_4$ is electrolysed using $0.01$ faraday of electricity. The number of equivalents of acid or alkali in solution.

[Atomic mass of $Cu=63.5$]

• A. $0.01eq.of{H}^{+}$
• B. $0.21eq.of{H}^{+}$
• C. $0.11eq.of{H}^{+}$
• D. None of these

Answer 1

Answer: (A)

$0.01eq.of{H}^{+}$

Half cell reaction :

$2H_{2}O\to O_2+4H^{+}+4e^{-}$

From the reaction,

4 faraday of electriity will produce 4 equivalent of acid or $H^{+}$ ions

Thuss, 1 faraday of electriity will produce 1 equivalent of acid or $H^{+}$ ions.

$\therefore$ $0.01$ faraday will produce $=$ 0.01eq. of $H^{+}$

Thus, when $10$ g fairly concentrated solution of $CuS{O}_4$ is electrolysed using $0.01$ faraday of electricity, the number of equivalents of hydrogen ions formed is $0.01eq.$