An aqueous solution of copper sulphate is subjected to electrolysis using platinum electrodes. Calculate the number of electrons flowing through the circuit when 0.3175 g of copper is deposited. Also calculate the number of molecules present in anodic product (atomic mass of copper = 63.5)

Explanation

`CuSO_(4) + Cu^(2+) + So_(4)^(-2)`
`H_(2)O rarr H^(+) + OH^(-)`
Among the cations `Cu^(+2) and H^(+), Cu^(+2)` requires lesser discharge potential compared to `H^(+)` so Cu is discharged at cathode.
At cathode : `Cu^(+2) + 2e^(-) rarr Cu`
For the deposition of 63.5 g of Cu, `2 xx 6.023 xx 10^(23) e^(-)` have to flow through the circuit, therefore, 0.3175 g of Cu requires ? electrons
`= (0.3175 xx 2 xx 6.023 xx 10^(23))/(63.5)`
`= 6.023 xx 10^(21)` electrons are required
AT anode : `4OH^(-) rarr 2H_(2)O + 4e^(-) + O_(2)`
For the formation of 1 mole of oxygen 4 moles of electrons are required
`96,500 C = (N)/(4)` moles of oxygen
`= (6.023 xx 10^(23))/(4)` molecule
(`6 xx 10^(21)` electrons = 965 C)
No. of molecules of oxygen liberated for 965 C
`= (965 xx 6.023 xx 10^(23))/(4 xx 96,500) = 1.5 xx 10^(21)` molecules