JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Redox Reactions and Electrochemistry

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

6 HO^{-} + 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 10 g of

KCIO_{3}

using a current of 2 A is ____.
(Given :

F = 96, 500 C mol^{- 1}

; molar mass of

KClO_{3} = 122 g mol^{- 1}

)

Answer: 11

Solution

The amount of potassium chlorate wanted is

\frac{10}{122} mol

, and the half-reaction shows that

6

moles of electrons are needed per mole of

ClO X_{3} X^{-}

, so the electrons required are

6 \times \frac{10}{122} mol

. Only

60%

of the current does useful work, so the electrons actually supplied must be scaled up by

\frac{100}{60}

, giving

\frac{10}{122} \times 6 \times \frac{100}{60} = \frac{100}{122} mol

. The charge is

\frac{100}{122} \times 96500 C

, and with

Q = I t

I = 2 A

the time is

t = \frac{100 \times 96500}{122 \times 2} \approx 39500 s \approx 11 hours

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Integer answer