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(C) The reduction reaction at the cathode is: $Cu^{2+} (aq) + 2e^- 	o Cu (s)$.To deposit $1 \ mole$ of $Cu$ $(63.5 \ g)$,$2 \ moles$ of electrons are

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$\frac{N_A}{5}$

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(C) The reduction reaction at the cathode is: $Cu^{2+} (aq) + 2e^- 	o Cu (s)$.To deposit $1 \ mole$ of $Cu$ $(63.5 \ g)$,$2 \ moles$ of electrons are required.Number of moles of $Cu$ to be deposited $= \frac{	ext{Given mass}}{	ext{Atomic mass}} = \frac{6.35 \ g}{63.5 \ g/mol} = 0.1 \ mol$.Since $1 \ mole$ of $Cu$ requires $2 \ moles$ of electrons,$0.1 \ mole$ of $Cu$ requires $0.1 	imes 2 = 0.2 \ moles$ of electrons.Total number of electrons $= 0.2 	imes N_A = \frac{2}{10} 	imes N_A = \frac{N_A}{5}$.

How many electrons would be required to deposit $6.35 \ g$ of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate? (Atomic mass of copper $= 63.5 \ u$,$N_A =$ Avogadro's constant)

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