Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation. \(6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}\) A current of xA has to be passed for 10 h to produce 10.0 g of potassium chlorate. The value of \(x\) is . (Nearest Integer) (Molar mass of \(\mathrm{KCIO}_{3}=122.6 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~F}=96500 \mathrm{C}\) )

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Redox Reactions and Electrochemistry

Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.

6 OH^{-} + Cl^{-} \to ClO_{3}^{-} + 3 H_{2} O + 6 e^{-}

A current of xA has to be passed for 10 h to produce 10.0 g of potassium chlorate. The value of

x

is . (Nearest Integer)
(Molar mass of

KCIO_{3} = 122.6 g mol^{- 1}, F = 96500 C

)

Answer: 1

Solution

The half reaction shows six electrons released for every chlorate ion formed, so six faradays are needed per mole of potassium chlorate. The amount formed is

n = \frac{10.0}{122.6} = 0.0816 mol

so the charge required is

Q = 6 \times 0.0816 \times 96500 = 4.72 \times 1 0^{4} C

This charge passes in

10 \times 3600 = 3.6 \times 1 0^{4}

s, so

I = \frac{4.72 \times 1 0 ^{4}}{3.6 \times 1 0 ^{4}} = 1.31 A

To the nearest integer,

x = 1

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Redox Reactions and Electrochemistry

Solution

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