Question

Potassium chlorate is prepared by the electrolysis of $KCl$ in basic solution $6O{H}^{-}+C{l}^{-}\to Cl{O}_3^{-}+3H_{2}O+6e^{-}$

If only $60\%$ of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce $10\text{g}$ of $KCl{O}_3$ using a current of $2\text{A}$ is

$(\text{Given}:F=96500{\text{C mol}}^{-1};\text{molar mass of}KCl{O}_3=122{\text{g mol}}^{-1})$

• A. 11.0
• B. 11.00
• C. 11

Answer 1

Answer: (A), (B), (C)

$F=96500{\text{C mol}}^{-1}$

Total Current passed = $2\text{A}$

Current used is $60\%$ of $2\text{A}=1.2\text{A}$

Amount of $KCl{O}_3$ Produced = $10\text{g}$

Molar mass of $KCl{O}_3=122{\text{g mol}}^{-1}$

$6O{H}^{-}+C{l}^{-}\to CI{O}_3^{-}+3H_{2}O+6e^{-}$

$\text{Chemical equivalent of}KCl{O}_3=\frac{\text{Molar mass of}KCl{O}_3}{6}$
$\text{Chemical equivalent of}KCl{O}_3=\frac{122}{6}$

$\text{Electrochmical equivalent,}Z=\frac{\text{Chemical equivalent}}{96500}$

$\text{Electrochemical equivalent of}KCl{O}_3=\frac{\text{Chemical equivalent of}KCl{O}_3}{96500}$

$\text{Electrochemical equivalent of}KCl{O}_3=\frac{\left(\frac{122}{6}\right)}{96500}$

$W=Z\times I\times t$

$\Rightarrow 10=\frac{\left(\frac{122}{6}\right)}{96500}\times 1.2\times t\text{sec}$

$\Rightarrow t=\frac{10\times 96500\times 6}{122\times 1.2}\text{sec}$

$\Rightarrow t=39549\text{sec}$.
$\Rightarrow t=659\text{min}$
$\Rightarrow t=10.9858\text{hr}$
The time (rounded to the nearest hour) required to produce $10\text{g}$ of $KCl{O}_3$ is $11\text{hr}$.