Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryFaraday's Law of Electrolysis/Electrolytic CellsHard2 minPYQ_2008

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Electrolysis of dilute aq. $NaCl$ solution was carried out by passing $10 mA$ current. The time required to liberate $0.01$ mole of $H_{2}$ gas at the cathode is (1 Faraday $= 96500 C mol^{- 1}$ )

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Answer:

Solution:

$2 H_{2} O + 2 e^{-} ⟶ H_{2} + 2 OH^{-}$ For $0.01$ mole of $H_{2}, 0.02$ mole of electrons are consumed Charge required $= 0.02 \times 96500 C = i \times t$ $Time required = \frac{0.02 \times 96500}{10 \times 1 0 ^{- 3}} = 19.3 \times 1 0^{4} s$

Stream:JEE_ADVSubject:ChemistryTopic:ElectrochemistrySubtopic:Faraday's Law of Electrolysis/Electrolytic Cells

⏱ 2mℹ️ Source: PYQ_2008

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