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(D) The reduction reaction at the cathode is: $2H_2O + 2e^- 	o H_2 + 2OH^-$. From the stoichiometry,$2 \ moles$ of electrons are required to produce

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$1.0$

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(D) The reduction reaction at the cathode is: $2H_2O + 2e^- 	o H_2 + 2OH^-$. From the stoichiometry,$2 \ moles$ of electrons are required to produce $1 \ mole$ of $H_2$ gas. At $N.T.P.$,$1 \ mole$ of gas occupies $22400 \ mL$. Therefore,the number of moles of $H_2$ produced is: $n(H_2) = \frac{112 \ mL}{22400 \ mL/mol} = 0.005 \ mol$. Using Faraday's law,the moles of electrons required is: $n(e^-) = 2 	imes n(H_2) = 2 	imes 0.005 = 0.01 \ mol$. Since $Q = I 	imes t$ and $Q = n(e^-) 	imes F$,where $F \approx 96500 \ C/mol$: $I = \frac{n(e^-) 	imes 96500}{t} = \frac{0.01 	imes 96500}{965} = 1.0 \ A$.

When an electric current is passed through acidified water,$112 \ mL$ of hydrogen gas at $N.T.P.$ was collected at the cathode in $965 \ seconds$. The current passed,in ampere,is

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