Question

When 0.04 faraday of electricity is passed through a solution of $\rightarrow$, then the weight of $\rightarrow$ metal deposited at the cathode is.

• A. 0.2 gm
• B. 0.4 gm
• C. 0.6 gm
• D. 0.8 gm

Answer 1

Answer: (D)

0.8 gm

Answer 2

$Ca^{+ +} + 2e^{-} \rightarrow Ca$

$E_{Ca} = \frac{40}{2} = 20$

$W_{Ca} = E_{Ca}$× No. of faradays $= 20 \times 0.04 = 0.8gm$.