Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (At. mass of Cu = 63.5 u)

Question

Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (At. mass of Cu = 63.5 u) (A) 0 g (B) 63.5 g (C) 2 g (D) 127 g

Tardigrade Admin · Accepted Answer

Equivalent mass of copper =( text Atomic mass / V text alency ) =(63.5/2)=31.75 Amount of copper deposited by 2 faraday =2 × 31.75 =63.5 g.

Q. Two Faraday of electricity is passed through a solution of $C u S O_{4}$ . The mass of copper deposited at the cathode is
(At. mass of $C u = 63.5 u$ )

$0 g$

$63.5 g$

$2 g$

$127 g$

Equivalent mass of copper $= \frac{Atomic mass}{V alency}$
$= \frac{63.5}{2} = 31.75$
Amount of copper deposited by $2$ faraday $= 2 \times 31.75$
$= 63.5 g$ .

Q. Two Faraday of electricity is passed through a solution of CuSO4​. The mass of copper deposited at the cathode is (At. mass of Cu=63.5u)

0g

63.5g

2g

127g