When an electric current is passed through ac | vedclass

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Question

(A) The reduction reaction at the cathode is: $2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)}$.At $N.T.P.$,$1 \ mole$ of $H_{2}$ gas occupies $22400 \ mL

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$1.0$

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(A) The reduction reaction at the cathode is: $2H^{+}_{(aq)} + 2e^{-} ightarrow H_{2(g)}$.At $N.T.P.$,$1 \ mole$ of $H_{2}$ gas occupies $22400 \ mL$.Therefore,the number of moles of $H_{2}$ produced is: $n = \frac{112 \ mL}{22400 \ mL/mol} = 0.005 \ mol$.From the stoichiometry of the reaction,$1 \ mole$ of $H_{2}$ requires $2 \ moles$ of electrons. Thus,$0.005 \ mol$ of $H_{2}$ requires $0.005 	imes 2 = 0.01 \ moles$ of electrons.Using Faraday's law: $Q = n 	imes F = I 	imes t$,where $F = 96500 \ C/mol$.$0.01 \ mol 	imes 96500 \ C/mol = I 	imes 965 \ s$.$I = \frac{0.01 	imes 96500}{965} = 1.0 \ A$.

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

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