JEE MAIN - Chemistry (2020 - 6th September Morning Slot - No. 3)
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)
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
11
Explanation
For synthesis of 1 mole of ClO3- , 6F of charge
is required.
$$ \therefore $$ To synthesise $${{10} \over {122}}$$ moles of KClO3,
Charge required = $${{10} \over {122}} \times 6$$ F
$${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \times {{60} \over {100}}$$
$$ \Rightarrow $$ t(hr) = $${{965 \times 100} \over {122 \times 2 \times 36}}$$ = 10.98 hr $$ \simeq $$ 11 Hr
$$ \therefore $$ To synthesise $${{10} \over {122}}$$ moles of KClO3,
Charge required = $${{10} \over {122}} \times 6$$ F
$${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \times {{60} \over {100}}$$
$$ \Rightarrow $$ t(hr) = $${{965 \times 100} \over {122 \times 2 \times 36}}$$ = 10.98 hr $$ \simeq $$ 11 Hr
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