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

6 OH^- + Cl^- $C l O_{3}^{-}$ -> + 3H₂O + 6e^-

A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is__________. (Nearest Integer)

(Molar mass of KClO₃ = 122.6g mol^-1, F = 96500C)

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Vishal Baghel | Contributor-Level 10

2 months ago

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

Mole of $C l O_{3}^{-} o r K C l O_{3} = \frac{10}{122.6} = 0.8456 m o l$

1 mole of $C l O_{3}^{-}$ or KClO₃ produced by 6 faraday (F).

$∴ 0.08456$ mole of $C l O_{3}^{-}$ or KClO₃ produced by (6 × 0.08156) F

$\frac{l \times t}{F} = 6 \times 0.8156$

$l = \frac{6 \times 0.08156 \times 965000}{10 \times 60 \times 60} A = 1.311 A \approx 1 A$

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