Question

In $CuS{O}_4$ solution when electric current equal to $2.5$ faraday is passed, the gm equivalent deposited on the cathode is:

• A. $1$
• B. $1.25$
• C. $2$
• D. $2.5$

Answer 1

Answer: (B)

$1.25$

$W=ZIt$

$It=Q$

$2$ faraday electricity deposites $Cu=1g$ equivalent

so $2.5F$ electticity deposites $Cu=\frac{1g\times 2.5}{2}=1.25$gm equivalent.