Consider the reaction: \(Cr_{2}O_{7}^{2-}\) + 14H++ 6e−→2Cr3+ + 7H2O
The quantity of electricity in coulombs needed to reduce 1 mol of \(Cr_{2}O_{7}^{2-}\) is-
1. 5F
2. 3F
3. 6F
4. 2F
Subtopic: Faraday’s Law of Electrolysis |
74%
From NCERT
Other Reason
Add Note
Highlight in NCERT
Please attempt this question first.
Hints
Please attempt this question first.
Upgrade Your Plan
Upgrade now and unlock your full question bank experience with daily practice.