Question

96500 C electricity is passed through $CuSO_4$. The amount of copper precipitated is

• A. 0.25 mole
• B. 0.5 mole
• C. 10 mole
• D. 2 .00 mole

Answer 1

Answer: (B)

0.5 mole

Answer 2

$\underset{1\,mole}{Cu^{2+}} + \underset{2 Faraday = 2\times 96500 \,C}{2e^-} {->} Cu$ $2\times 96500 \,C$ deposits $1$ mole of copper $96500 \,C$ deposits $0.5$ mole of copper.