Question

On passing 0.1 Faraday of electricity through aluminium chloride, the amount of aluminium metal deposited on cathode is $\rightarrow$

• A. 0.9 gm
• B. 0.3 gm
• C. 0.27 gm
• D. 2.7 gm

Answer 1

Answer: (A)

0.9 gm

Answer 2

At cathode; $Al^{3 +} + 3e^{-} \rightarrow Al$

$E_{Al} = \frac{27}{3} = 9$

$W_{Al} = E_{Al} \times \text{No.offaradays} = 9 \times 0.1 = 0.9gm$.