QJEEAdv08Cha84d: JEE Advanced 2008 Chemistry - Class 12 JEEAdv08Cha84d Solution: Electrolysis of dilute aqueous NaCl solution was carried out by …

Given,

Current passed a 10 mA.

Moles of Hydrogen released are 0.01 moles.

In the electrolysis of aqueous solution sodium chloride, as the reduction potential of sodium ion is less than the hydrogen ion, the reaction that takes place at the cathode could be given as,

2H$${}^{+}$$ + 2e$${}^{-}$$ $$\to$$ H$${}_2$$

0.01 moles of hydrogen would require 2 $$\times$$ 0.01 = 0.02 moles of electrons (or) 0.02 Faradays.

We know the formula to calculate the amount of Faradays passed.

Amount of Faraday's passed = $$\frac{I(A)\times t(s)}{96500\mathrm{C}{\mathrm{mol}}^{-1}}$$

On rearranging the equation to calculate the time taken we get,

t = Amount of Faraday's passed $$\times$$ 96500 $$\times$$ I

$$0.02=\frac{\frac{10mA}{1000mA/A}\times t(s)}{96500Cmo{l}^{-1}}$$

$$\Rightarrow t=0.02\times 96500\times 0.01$$

On multiplying we get,

$$\Rightarrow t=19.3\times {10}^4$$ s

The time required to liberate 0.01 moles of H$${}_2$$ gas at the cathode is 19.3 $$\times$$ 10$${}^4$$ s.