What amount of mercury would be liberated if the same quantity of electricity that liberates 0.65g of zinc is applied?
-
2.01g
- 1.00g
- 4.02g
- 8.04g
The correct answer is: A
Explanation
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