| ⇦ | 
| ⇨ |
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: - In the following reaction, the product(A)
- Which one of the following elements has the maximum electron affinity?
- When sodium is added to ethanol
- Which of the following can be termed a phenol formaldehyde resin
- Formula of Plaster of Paris is
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the following reaction, the product(A)
- Which one of the following elements has the maximum electron affinity?
- When sodium is added to ethanol
- Which of the following can be termed a phenol formaldehyde resin
- Formula of Plaster of Paris is
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”