Question

By passing 0.1 Faraday of electricity through fused sodium chloride, the amount of chlorine liberated is

• A. 35.45 g
• B. 70.9 g
• C. 3.55 g
• D. 17.77 g

Answer 1

Answer: (C)

3.55 g

Answer 2

$Cl^{-} \rightarrow \frac{1}{2}Cl_{2} + e^{-}$ i.e., 1Faraday liberate $Cl_{2} = \frac{1}{2}$mol $= \frac{1}{2} \times 71 = 35.5$ $(\because wt.\text{of}Cl = 35.5)$

$\therefore$ 0.1Faraday will liberate $Cl_{2} = 35.5 \times 0.1 = 3.55g.$