Question

A 250 mL sample of a $0.2MC{r}^{3+}$ is electrolysed with a current of 96.5 amp; if the remaining concentration of $C{r}^{3+}$ 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 $C{r}^{3+}$ before electrolysis $=\frac{MV(mol)}{1000}$
$=\frac{0.2\times 250}{1000}=0.05$
Number of moles of $C{r}^{3+}$ after electrolysed $=\frac{MV}{1000}$
$=\frac{0.1\times 250}{1000}=0.025$
Number of moles of $C{r}^{3+}$ 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
$\approx 75$ sec.