Potassium chlorate is prepared by the
electrolysis of KCl in basic solution
6OH- + Cl- $\to$ ClO3- + 3H2O + 6e-
If only 60% of the current is utilized in the
reaction, the time (rounded to the nearest hour)
required to produce 10 g of KClO3 using a
current of 2 A is_________.
(Given : F = 96,500 C mol–1; molar mass of
KCIO3 = 122 g mol–1)
Answer (integer)
11
Solution
For synthesis of 1 mole of ClO<sub>3</sub><sup>-</sup> , 6F of charge
is required.
<br><br>$\therefore$ To synthesise
${{10} \over {122}}$ moles of KClO<sub>3</sub>,
<br><br>Charge required = ${{10} \over {122}} \times 6$ F
<br><br>$${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \times {{60} \over {100}}$$
<br><br>$\Rightarrow$ t(hr) = ${{965 \times 100} \over {122 \times 2 \times 36}}$ = 10.98 hr $\simeq$ 11 Hr
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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