| ⇦ | 
| ⇨ |
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: - Treatment of acetaldehyde with ethyl magnesium bromide and subsequent hydrolysi
- Water is oxidised to oxygen by
- The method of zone refining of metals is based on the principle of
- n-Butane and isobutane are
- Which one of the following is an example of heterogenous catalysis
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Treatment of acetaldehyde with ethyl magnesium bromide and subsequent hydrolysi
- Water is oxidised to oxygen by
- The method of zone refining of metals is based on the principle of
- n-Butane and isobutane are
- Which one of the following is an example of heterogenous catalysis
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”