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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: - The law of equilibrium was first given by
- Which of the following has highest chlorine content
- Which of the following compounds has the highest boiling point
- The molecular weight of a gas is 45. Its density at STP is
- What is the approximate percentage of H₂O₂ in a sample labelled as 10 V?
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The law of equilibrium was first given by
- Which of the following has highest chlorine content
- Which of the following compounds has the highest boiling point
- The molecular weight of a gas is 45. Its density at STP is
- What is the approximate percentage of H₂O₂ in a sample labelled as 10 V?
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”