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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: - If 2 g of aluminium is treated, first with excess of dilute H₂SO₄ and then
- Highest electron affinity is shown by
- For the reversible reaction,The equilibrium shifts in forward direction
- M⁺³ ion loses 3e⁻. Its oxidation number will be
- Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85g
Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If 2 g of aluminium is treated, first with excess of dilute H₂SO₄ and then
- Highest electron affinity is shown by
- For the reversible reaction,The equilibrium shifts in forward direction
- M⁺³ ion loses 3e⁻. Its oxidation number will be
- Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85g
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”