⇦ | ⇨ |
When 0.1 mol MnO₄²⁻ is oxidised the quantity of electricity required to completely oxidise MnO₄²⁻ to MnO₄⁻ is
Options
(a) 96500 C
(b) 2 ˣ 96500 C
(c) 9650 C
(d) 96.50 C
Correct Answer:
9650 C
Explanation:
Mn⁺⁶ O₄²⁻→Mn⁺⁷O₄⁻ +e⁻
0.1 mole
Quantity of electricity required
= 0.1F = 0.1×96500 = 9650 C.
Related Questions: - Toluene can be oxidised to benzoic acid by
- H₂S acts only as a reducing agent while SO₂, can act both as a reducing
- The intermediate formed in aldol condensation is
- In acidic medium, the equivalent weight of KMnO₄ is
- Which of the following is diamagnetic
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Toluene can be oxidised to benzoic acid by
- H₂S acts only as a reducing agent while SO₂, can act both as a reducing
- The intermediate formed in aldol condensation is
- In acidic medium, the equivalent weight of KMnO₄ is
- Which of the following is diamagnetic
Topics: Electrochemistry and Chemical Kinetics (87)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
“Mn⁺⁶ O₄²⁻→Mn⁺⁷O₄⁻ +e⁻
0.1 mole
Quantity of electricity required
= 0.1F = 0.1×96500 = 9650 C”