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(B) The reaction at the cathode for the reduction of copper ions is: $Cu^{2+} (aq) + 2e^- \rightarrow Cu (s)$.According to Faraday's first law of elec

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(B) The reaction at the cathode for the reduction of copper ions is: $Cu^{2+} (aq) + 2e^- ightarrow Cu (s)$.According to Faraday's first law of electrolysis,the passage of $1 \ F$ $(96500 \ C)$ of electricity deposits $1 \ 	ext{gram equivalent}$ of the metal.For copper,the equivalent mass is $\frac{	ext{molar mass}}{2} = \frac{63.5}{2} = 31.75 \ g$.Since $1 \ 	ext{mole}$ of $Cu$ weighs $63.5 \ g$,$31.75 \ g$ corresponds to $\frac{31.75}{63.5} = 0.5 \ mol$ of $Cu$.Thus,passing $96500 \ C$ of electricity deposits $0.5 \ mol$ of copper.

When $96500 \ C$ of electricity is passed through a copper sulphate solution,the amount of copper deposited will be ............... $mol$.

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