On passing C ampere of current for time t sec through 1 L of 2 (M) CuSO4 solution (atomic weight of Cu = 63.5), the amount m of Cu (in gram) deposited on cathode will be

Question

On passing C ampere of current for time t sec through 1 L of 2 (M) CuSO4 solution (atomic weight of Cu = 63.5), the amount m of Cu (in gram) deposited on cathode will be (A) m=(C t/(63.5×96500)) (B) m=(C t/(31.25×96500)) (C) m=(C×96500/(31.25× t)) (D) m=(31.25× C× t/96500)

Tardigrade Admin · Accepted Answer

m= (E C t/F) =((63.5/2) × C × t/96500)=(31.75 × C × t/96500)

Q. On passing $C$ ampere of current for time $t$ sec through $1 L$ of $2 (M) C u S O_{4}$ solution (atomic weight of $C u = 63.5$ ), the amount $m$ of $C u$ (in gram) deposited on cathode will be

$m = \frac{Ct}{( 63.5 \times 96500 )}$

$m = \frac{Ct}{( 31.25 \times 96500 )}$

$m = \frac{C \times 96500}{( 31.25 \times t )}$

$m = \frac{31.25 \times C \times t}{96500}$

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

Q. On passing C ampere of current for time t sec through 1L of 2(M)CuSO4​ solution (atomic weight of Cu=63.5), the amount m of Cu (in gram) deposited on cathode will be

m=(63.5×96500)Ct​

m=(31.25×96500)Ct​

m=(31.25×t)C×96500​

m=9650031.25×C×t​