JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

Potassium chlorate is prepared by electrolysis of $\mathrm{KCl}$ in basic solution as shown by following equation.

$6{\mathrm{OH}}^{-}+{\mathrm{Cl}}^{-}\rightarrow {\mathrm{ClO}}_{3}^{-}+3{H}_{2}O+6{e}^{-}$

A current of $\mathrm{xA}$ has to be passed for $10h$ to produce $10.0g$ of potassium chlorate. the value of $x$ is __________. (Nearest integer)

(Molar mass of ${\mathrm{KClO}}_{3}=122.6g{\mathrm{mol}}^{-1}F=96500C$ )

Official previous-year question

Held on 20 Jul 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

Given balanced equation is

$\overset{\circleddash }{60H}+\overset{\circleddash }{C}l\longrightarrow {\mathrm{ClO}}_{3}^{-}+3{H}_{2}O+6{e}^{-}$

$\rightarrow 10{\mathrm{gKClO}}_{3}\Rightarrow \frac{10}{122.6}{\mathrm{molKCO}}_{3}$ in obtained

$\rightarrow$ from the above reaction, it is concluded that by

$6F$ charge $1{\mathrm{molKClO}}_{3}$ is obtained.

$\rightarrow$ By the passage of $6F$ charge $=1{\mathrm{molKClO}}_{3}$

$\therefore$ By the passage of $\frac{x\times 10\times 60\times 60}{96500}F$ charge

$=\frac{1}{6}\times \frac{x\times 10\times 60\times 60}{96500}$

Now $\frac{x\times 10\times 60\times 60}{6\times 96500}=\frac{10}{122.6}$

$\Rightarrow x=\frac{10\times 965}{60\times 122.6}=\frac{965}{735.6}=1.311\simeq 1$

$W=\frac{E}{F}\times I\times t$

$10=\frac{122.6}{96500\times 6}\times x\times 10\times 3600$

$X=1.311$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

Solution

Given balanced equation is

$\overset{\circleddash }{60H}+\overset{\circleddash }{C}l\longrightarrow {\mathrm{ClO}}_{3}^{-}+3{H}_{2}O+6{e}^{-}$

$\rightarrow 10{\mathrm{gKClO}}_{3}\Rightarrow \frac{10}{122.6}{\mathrm{molKCO}}_{3}$ in obtained

$\rightarrow$ from the above reaction, it is concluded that by

$6F$ charge $1{\mathrm{molKClO}}_{3}$ is obtained.

$\rightarrow$ By the passage of $6F$ charge $=1{\mathrm{molKClO}}_{3}$

$\therefore$ By the passage of $\frac{x\times 10\times 60\times 60}{96500}F$ charge

$=\frac{1}{6}\times \frac{x\times 10\times 60\times 60}{96500}$

Now $\frac{x\times 10\times 60\times 60}{6\times 96500}=\frac{10}{122.6}$

$\Rightarrow x=\frac{10\times 965}{60\times 122.6}=\frac{965}{735.6}=1.311\simeq 1$

$W=\frac{E}{F}\times I\times t$

$10=\frac{122.6}{96500\times 6}\times x\times 10\times 3600$

$X=1.311$

JEE Main 2021 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2021 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry