Question

$$96500$$ coulombs of electric current liberates from $$CuSO_{4}$$ solution

Answer 1

Correct option is B. $$31.75\ gm\ Cu$$

At the cathode

$$Cu^{2+} (aq) + 2e- \rightarrow Cu(s)$$

At the anode $$4OH-(aq) \rightarrow2H_2O (l) + O_2(g) + 4e^-$$

Faraday’s constant = $$96500C/mol$$

To deposit 1 mole of copper, we need $$2*96500C$$ So, $$96500 C$$ will deposit 0.5 moles of copper = $$0.5*63.5 = 31.75g$$