JEE Advanced · Chemistry · 17. Electrochemistry
Electrolysis of dilute aq. \(\mathrm{NaCl}\) solution was carried out by passing \(10 \mathrm{~mA}\) current. The time required to liberate \(0.01\) mole of \(\mathrm{H}_2\) gas at the cathode is (1 Faraday \(=96500 \mathrm{C} \mathrm{mol}^{-1}\) )
- A \(9.65 \times 10^4 \mathrm{~s}\)
- B \(19.3 \times 10^4 \mathrm{~s}\)
- C \(28.95 \times 10^4 \mathrm{~s}\)
- D \(38.6 \times 10^4 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(19.3 \times 10^4 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(2 \mathrm{H}_2 \mathrm{O}+2 e^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}\)
For \(0.01\) mole of \(\mathrm{H}_2, 0.02\) mole of electrons are consumed Charge required \(=0.02 \times 96500 \mathrm{C}=i \times t\)
\(
\text { Time required }=\frac{0.02 \times 96500}{10 \times 10^{-3}}=19.3 \times 10^4 \mathrm{~s}
\)
For \(0.01\) mole of \(\mathrm{H}_2, 0.02\) mole of electrons are consumed Charge required \(=0.02 \times 96500 \mathrm{C}=i \times t\)
\(
\text { Time required }=\frac{0.02 \times 96500}{10 \times 10^{-3}}=19.3 \times 10^4 \mathrm{~s}
\)
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