When electric current is passed through acidi | vedclass

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(C) The reduction reaction at the cathode is: $2H^{+} + 2e^{-} \rightarrow H_2$. Given volume of $H_2$ gas $= 1120 \ mL = 1.12 \ L$. Number of moles o

Vedclass · Accepted Answer

$5.0$

Vedclass · Answer

(C) The reduction reaction at the cathode is: $2H^{+} + 2e^{-} ightarrow H_2$. Given volume of $H_2$ gas $= 1120 \ mL = 1.12 \ L$. Number of moles of $H_2$ gas $= \frac{1.12 \ L}{22.4 \ L/mol} = 0.05 \ mol$. According to 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 $Q = n 	imes F = 0.1 	imes 96500 \ C = 9650 \ C$. Using the formula $Q = I 	imes t$,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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