Question

Electrolysis of dilute aqueous $NaCl$ solution was carried out by passing $10$ milli ampere current. The time required to liberate $0.01$ mol of $H_2$ gas at the cathode is: ($1$ Faraday $=96500Cmo{l}^{-1}$)

• (A) $9.65\times {10}^4$ sec
• (B) $19.3\times {10}^4$ sec
• (C) $28.95\times {10}^4$ sec
• (D) $38.6\times {10}^4$ sec

Answer 1

The Correct Option is B

Explanation:
$Q=i\times t$$Q=10\times {10}^{-3}\times t$$2H_{2}O+2e^{-}\to H_2+2O{H}^{-}$To liberate $0.01$ mole of $H_2$, $0.02$ Faraday charge is required$Q=0.02\times 96500C$$\therefore 0.02\times 96500={10}^{-2}\times t$$\Rightarrow t=19.3\times {10}^4$ secHence, the correct option is (B).