Potassium chlorate is prepared by the electrolysis of KCl in basic solution 6 OH ^ - + Cl ^ - → ClO _ 3 ^ - + 3 H _ 2 O + 6 e ^ - . If only 60 % of the current is utilized in the reaction, the time (rounded to the

Correct answer: 0 Number of faradays = (I × t)/(96500) = (2 × t × 60 × 60)/(96500)F Mass of KClO3 produce = 10Gm 2 × t × 60 × 60 / 96500 × 0 . 60 × 122 × ( 1 / 6 ) = 10 So t = 10 . 98 hours

JEE Main 20206 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 6 September
Shift: Morning Shift
Subject: Chemistry
Chapter: Electrochemistry

JEE Main 2020 · 6 September · Morning Shift · Chemistry · Electrochemistry

Potassium chlorate is prepared by the electrolysis of KCl in basic...

Potassium chlorate is prepared by the electrolysis of

KCl

in basic solution

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

. If only

60 %

of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce

10g

{KClO}_{3}

using a current of

2A

\ldots \ldots . .

(Given :

F = 96, 500C

mol

; molar mass of

{KClO}_{3} = 122g{mol}^{- 1}

)

Correct answer

Step-by-step solution

Number of faradays = $\dfrac{I \times t}{96500}$

= $\dfrac{2 \times t \times 60 \times 60}{96500}F$

Mass of $KClO3$ produce = $10Gm$

${ 2 \times t \times 60 \times 60 / 96500 } \times 0 . 60 \times 122 \times ( 1 / 6 ) = 10$
So $t = 10 . 98$ hours

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