Chemistry - Electrochemistry Question with Solution | TestHub

Number of moles of$\mathrm{C}{\mathrm{r}}^{3+}$before electrolysis$=\frac{\mathrm{M}\mathrm{V} \left(\mathrm{mol}\right)}{1000}$
$=\frac{0.2\times 250}{1000}=0.05$
Number of moles of$\mathrm{C}{\mathrm{r}}^{3+}$after electrolysed$=\frac{\mathrm{M}\mathrm{V}}{1000}$
$=\frac{0.1\times 250}{1000}=0.025$
Number of moles of$\mathrm{C}{\mathrm{r}}^{3+}$electrolysed$=0.05-0.025=0.025$
Mass of chromium electrolysis$=0.025\times 52=1.3$g
Equivalent mass of chromium$=\frac{\text{Atomic mass}}{\text{Valency of ions}}$
$=\frac{52}{3}=17.33$
$\mathrm{W}=\frac{\mathrm{ItE}}{96500}\Rightarrow \frac{\mathrm{W}\times 96500}{\mathrm{I}\times \mathrm{E}}=\frac{1.3\times 96500}{96.5\times 17.33}$
$\mathrm{t}=75.14$sec
$\approx 75$sec.