One Faraday of electricity liberates x times | vedclass

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(B) The electrode reaction for the reduction of copper ions is: $Cu^{2+} + 2e^{-} \rightarrow Cu$.According to the stoichiometry of the reaction,$2 \

Vedclass · Accepted Answer

$5$

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(B) The electrode reaction for the reduction of copper ions is: $Cu^{2+} + 2e^{-} ightarrow Cu$.According to the stoichiometry of the reaction,$2 \ Faraday$ of electricity is required to deposit $1 \ mol$ (or $1 \ 	ext{gram atom}$) of copper.Therefore,$1 \ Faraday$ of electricity will deposit $1/2 = 0.5 \ 	ext{mol}$ of copper.Since $0.5 \ 	ext{mol} = 0.5 \ 	ext{gram atom}$,we can express this as $5 	imes 10^{-1} \ 	ext{gram atom}$.Comparing this with $x 	imes 10^{-1}$,we get $x = 5$.

One Faraday of electricity liberates $x \times 10^{-1}$ gram atom of copper from copper sulphate,$x$ is . . . . . . .

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