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Question

Electrolysis of dilute aqueous $$\mathrm{N}\mathrm{a}\mathrm{C}l$$ solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of $$\mathrm{H}_{2}$$ gas at the cathode is (1 Faraday $$=96500\mathrm{C}$$ mol $$^{-1}$$)


A
9.65×104sec
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B
19.3×104sec
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C
28.95×104sec
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D
38.6×104sec
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Solution

The correct option is B $$ 19.3\times 10^{4}\sec$$
$$2H_2O+2e^−\rightarrow H_2+2OH^−$$

$$1 mol \  H_2$$ requires $$2$$ faraday current 

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

$$Q= 0.02 \times 96500 \ C$$

$$Q=i\times t$$ 

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

$$Q=10×10^{−3} \times t$$

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

$$\Rightarrow t=19.30×10^4 \ sec$$

Option B is correct.

Chemistry

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