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 $=96500C$ mol ${}^{-1}$)

• 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 $19.3\times {10}^4\sec$

$2H_{2}O+2e^{-}\to H_2+2O{H}^{-}$

$1mol{H}_2$ requires $2$ faraday current

$\therefore 0.01$ mol will require $=0.02$ faraday current.

$Q=0.02\times 96500C$

$Q=i\times t$

$i=10$ mili ampere $=10\times {10}^{-3}$ ampere

$Q=10\times {10}^{-3}\times t$

$\Rightarrow 0.02\times 96500={10}^{-2}\times t$

$\Rightarrow t=19.30\times {10}^{4}sec$

Option B is correct.