Chemistry - Electrochemistry Question with Solution | TestHub
ChemistryElectrochemistryFaraday's Law of Electrolysis/Electrolytic CellsHard2 minQB
ChemistryHardmultiple choice
Given the half-reactions (a) H⁺ + e⁻ → ½H₂, E° = 0.00 V; (b) 2H₂O → O₂ + 4H⁺ + 4e⁻, E° = 1.23 V; (c) 2SO₄²⁻ → S₂O₈²⁻ + 2e⁻, E° = 1.96 V: select the correct statements.
Options:(select one or more)
Answer:
A, C
Solution:
H⁺ is reduced at the cathode in dilute acid, and water is oxidised at the anode; peroxodisulfate formation needs a higher potential.
Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Faraday's Law of Electrolysis/Electrolytic Cells
⏱ 2mℹ️ Source: QB
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