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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: - Which one of the following ions will be the smallest in size?
- Picric acid is
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Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which one of the following ions will be the smallest in size?
- Picric acid is
- 1-bromopropane on treatment with ethanolic potassium hydroxide produces
- According to Raoult’s law, the relative lowering of vapour pressure
- The s-block element used as a catalyst in the manufacture of Buna-S rubber is
Topics: Electrochemistry and Chemical Kinetics (87)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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“Mn⁺⁶ O₄²⁻→Mn⁺⁷O₄⁻ +e⁻
0.1 mole
Quantity of electricity required
= 0.1F = 0.1×96500 = 9650 C”