Question

Two Faraday of electricity is passed through a solution of $CuS{O}_4$. The mass of copper deposited at the cathode is:

(At. mass of $Cu$ = 63.5 amu)

• A. $0g$
• B. $63.5g$
• C. $2g$
• D. $127g$

Answer 1

The correct option is B $63.5g$

$C{u}^{2+}+2e^{-}\to Cu$.
From the above-given half-reaction, two moles of electrons react with one mole of $Cu\left(II\right)$ to deposit one mole of $Cu$. This corresponds to 63.5 g (molecular weight) of copper.

Two Faraday's of electricity corresponds to 2 moles of electrons. Hence, they will deposit 63.5 g of copper.