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
B
19.3×104sec
C
28.95×104sec
D
38.6×104sec

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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