Question

$2.5F$ of electricity are passed through a $CuS{O}_4$ solution. The number of gm equivalent of Cu deposited on anode is:

• A. Zero
• B. $2.5$
• C. $1.25$
• D. $63.5$

Answer 1

The correct option is D $63.5$

We know that

$W=ZIt$

Where,

W = amount of substance deposited

Z = electrochemical equivalent

I = amount of current

t = time

$It=Q=2F\left(given\right)$

$Z=\frac{atomicmass}{nF}=\frac{63.5}{2F}$

Hence,

$W=ZIt$

$W=\frac{63.5}{2F}\times 2F=63.5g$