| ⇦ | 
| ⇨ |
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: - When CO₂ is passed through solution of calcium hydroxide,Which one of the
- If 18g of water is electrolysed then volume of oxygen obtained at STP is
- Among the metals Cr,Fe,Mn,Ti,Ba and Mg, the one that cannot be obtained by
- If pH of a saturated solution of Ba(OH)₂ is 12, the value of its Ksp is
- In the gas equation ,PV = nRT
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When CO₂ is passed through solution of calcium hydroxide,Which one of the
- If 18g of water is electrolysed then volume of oxygen obtained at STP is
- Among the metals Cr,Fe,Mn,Ti,Ba and Mg, the one that cannot be obtained by
- If pH of a saturated solution of Ba(OH)₂ is 12, the value of its Ksp is
- In the gas equation ,PV = nRT
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”