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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: - Choose the correct statements
- IUPAC name of Pt(NH₃)₃Br(NO₂)Cl]Cl is
- The oxidation number of As in H₂AsO₄⁻ is
- Which of the following statements about amorphous solids is incorrect
- The first fractional product of petroleum from top to bottom is
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Choose the correct statements
- IUPAC name of Pt(NH₃)₃Br(NO₂)Cl]Cl is
- The oxidation number of As in H₂AsO₄⁻ is
- Which of the following statements about amorphous solids is incorrect
- The first fractional product of petroleum from top to bottom 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”