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During electrolysis of conc. H2SO4, perdisulphuric acid (H2S2O8), and O2
form in equimolar amount. The amount of H2 that will form simultaneously will be :
1. Thrice that of O2 in moles.
2. Twice that of O2 in moles.
3. Equal to that of O2 in moles.
4. Half of that of O2 in moles.
Subtopic: Faraday’s Law of Electrolysis |
Level 3: 35%-60%
Hints
When \(CuSO_4\) solution using copper electrodes is electrolysed, pH of the solution
1. Increases
2. Decreases
3. Remains the same
4. Firstly increases and then decreases
1. Increases
2. Decreases
3. Remains the same
4. Firstly increases and then decreases
Subtopic: Faraday’s Law of Electrolysis |
Level 3: 35%-60%
Please attempt this question first.
Hints
Please attempt this question first.
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