| ⇦ | 
| ⇨ |
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: - The ionic radii of isoelectronic species N³⁻, O²⁻ and F⁻ are in the order
- ΔG = ΔH – TΔS was given by
- The best coagulant for the precipitation of Fe(OH)₃ is
- Green chemistry means such reaction which
- In which of the following compounds, nitrogen exhibits highest oxidation state
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The ionic radii of isoelectronic species N³⁻, O²⁻ and F⁻ are in the order
- ΔG = ΔH – TΔS was given by
- The best coagulant for the precipitation of Fe(OH)₃ is
- Green chemistry means such reaction which
- In which of the following compounds, nitrogen exhibits highest oxidation state
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”