Question

Chromium metal can be plated out from an acidic solution containing $CrO _{3}$ according to the following equation.
$CrO _{3}(a q)+6 H ^{+}(a q)+6 e^{-} \longrightarrow Cr (s)+3 H _{2} O$
Calculate (i) How many grams of chromium will be plated out by $24,000 C$ and (ii) How long will it take to plate out $1.5 \,g$ of chromium by using $12.5\, A$ current?

Answer 1

Molar mass of $Cr =52\, g$
Equivalent mass of $Cr =\frac{52}{6} g$
(i) Mass of Cr deposited on passing $24000$ Coulombs
$=\frac{24000}{96500} \times \frac{52}{6} g =2.15\, g$
(ii) Number of gram equivalent of $Cr =\frac{1.5}{52} \times 6=\frac{9}{5 \text {. }}$
$\Rightarrow$ Coulombs required for $1.5 \,g \,Cr =\frac{9}{52} \times 96500=I t$
$\Rightarrow t=\frac{9 \times 96500}{52 \times 12.5} s =22.27 \min$