When electric current is passed through acidi | vedclass

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Question

(C) The reduction reaction at the cathode is: $2H^+ + 2e^- \rightarrow H_2$. Number of moles of $H_2$ gas collected = $\frac{1120 \,mL}{22400 \,mL/mol

Vedclass · Accepted Answer

$5.0$

Vedclass · Answer

(C) The reduction reaction at the cathode is: $2H^+ + 2e^- ightarrow H_2$. Number of moles of $H_2$ gas collected = $\frac{1120 \,mL}{22400 \,mL/mol} = 0.05 \,mol$. From the stoichiometry of the reaction, $1 \,mol$ of $H_2$ requires $2 \,mol$ of electrons. Therefore, $0.05 \,mol$ of $H_2$ requires $0.05 	imes 2 = 0.1 \,mol$ of electrons. Total charge $Q = n 	imes F = 0.1 \,mol 	imes 96500 \,C/mol = 9650 \,C$. Using the relation $Q = I 	imes t$, where $t = 1930 \,s$: $I = \frac{Q}{t} = \frac{9650 \,C}{1930 \,s} = 5.0 \,A$.

When electric current is passed through acidified water for $1930 \,s$, $1120 \,mL$ of $H_2$ gas is collected (at $STP$) at the cathode. What is the current passed in amperes?

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