Zn\(^{2+}\) + 2e\(^-\) → Zn 2F will be liberated 65g of Zn, if 0.65gm of Zn is liberated, the quantity of electricity used X = (0.65/65) X 2 = 0.02F ∴ Hg\(^{2+}\) + 2e\(^-\) → Hg i.e 2F → 201gm of Hg ∴ 0.02 will liberated Y = \(\frac{0.02}{2}\) X 201 = 2.01g