Question

On passing $C$ ampere of current for time $t$ sec through $1L$ of $2MCuS{O}_4$ solution (atomic weight of $Cu=63.5$), the amount $m$ of $Cu$ (in gram) deposited on cathode will be:

• A. $m=\frac{Ct}{(63.5\times 96500)}$ g
• B. $m=\frac{Ct}{(31.75\times 96500)}$ g
• C. $m=\frac{(C\times 96500)}{(31.75\times t)}$ g
• D. $m=\frac{(31.75\times C\times t)}{96500}$ g

Answer 1

The correct option is D $m=\frac{(31.75\times C\times t)}{96500}$ g

$m=\frac{EIt}{F}$ $=\frac{\frac{63.5}{2}\times C\times t}{96500}=\frac{31.75\times C\times t}{96500}g$

Option D is correct.