Question

A 250 mL sample of a $0.2MCr^{3 +}$ is electrolysed with a current of 96.5 amp; if the remaining concentration of $Cr^{3 +}$ ion is 0.1 M, the duration of electrolysis is given : atomic mass of Cr = 52

• A. 75 sec
• B. 150 sec
• C. 225 sec
• D. 25 sec

Answer 1

Answer: (A)

Number of moles of $Cr^{3 +}$ before electrolysis $=\frac{M V \left(\right. mol \left.\right)}{1000}$
$=\frac{0.2 \times 250}{1000}=0.05$
Number of moles of $Cr^{3 +}$ after electrolysed $=\frac{M V}{1000}$
$=\frac{0.1 \times 250}{1000}=0.025$
Number of moles of $Cr^{3 +}$ electrolysed $=0.05-0.025=0.025$
Mass of chromium electrolysis $=0.025\times 52=1.3$ g
Equivalent mass of chromium $=\frac{\text{Atomic mass}}{\text{Valency of ions}}$
$=\frac{52}{3}=17.33$
$W=\frac{ItE}{96500}\Rightarrow \frac{W \times 96500}{I \times E}=\frac{1 . 3 \times 96500}{96 . 5 \times 17 . 33}$
$t=75.14$ sec
$\approx75$ sec.