Question

Chromium metal can be plated out from an
acidic solution containing $Cr{O}_3$ according to the following reaction:
$Cr{O}_3+6H^{+}+6e^{-}\to Cr+3H_{2}O$
Calculate the mass (m) of chromium plated out by $24000coulomb$.
Also, find the time (t) it will take to plate out $1.5g$ of chromium using $12.5A$ current.
Current efficiency is $40\%$
Molecular weight of $Cr$ is $52gmo{l}^{-1}$

• A. $m=8.44gt=26min$
• B. $m=2.15gt=55.6min$
• C. $m=8.44gt=12.41min$
• D. $m=2.15gt=88.32min$

Answer 1

The correct option is B $m=2.15gt=55.6min$

From the following reaction:
$Cr{O}_3+6H^{+}+6e^{-}\to Cr+3H_{2}O$
(a)
$\text{From 6 mol of electron or 6F charge , 1 mol of Cr is obtanied}$
$\Rightarrow 6F⟶1molCr=52g24000C⟶\frac{52\times 24000}{6\times 96500}g=2.15gCr\text{is obtained}$

(b)
$\text{For plating 1 mol or 52 g Cr, charge required is 6F}$
$\text{Charge required for plating 1.5 g Cr}=\frac{6\times 96500\times 1.5}{52}\text{Charge required for plating 1.5 g Cr}=16701.92C$
$\text{Current supplied}=12.5A\text{Efficiency}=40\%\text{Actual current}=12.5\times 0.4=5A$

$\text{Time}=\frac{\text{Charge}}{\text{Actual Current}}t=\frac{16701.92}{5}=3340.38s=55.67min$