Question

96500 coulomb of electricity is passed through CuSo4 solution it will liberate

Answer 1

At the cathode
Cu2+ (aq) + 2e-  ---> Cu(s)

At the anode4OH-(aq) ----> 2H2O (l) + O2(g) + 4e-

Faraday’s constant = 96500C/mol
To deposit 1 mole of copper, we need 2x96500CSo, 96500C will deposit 0.5 moles of copper = 0.5x63.5 = 31.75g
To liberate one mole of oxygen molecules,we need  4x96500CSo, 96500C will liberate = ¼  = 0.25 moles = 0.25x32 = 8g

Answer 2

Answer:

96500 coulomb of electricity passed through CuSO₄ solution will liberate 31.75g of Copper.

Explanation:

The reaction at the cathode will be,

Cu²⁺(aq)+2e⁻→Cu(s)

Whereas the reaction at anode will be,

4OH⁻(aq)→2H₂O(l)+O₂(g)+4e⁻

We know that,

Faraday's constant=96500 Cmol⁻¹

For depositing one mole of Copper, two units of Faraday's constant is needed.

Therefore, 96500C will only deposit half a mole of copper.

∴0.5*63.5=31.75g

Thus, 96500 coulomb of electricity passed through CuSO₄ solution will liberate 31.75g of Copper.