When 0.1 mol MnO₄²⁻ is oxidised the quantity of electricity required to completely

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.

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View Comments (1)

  • "Mn⁺⁶ O₄²⁻→Mn⁺⁷O₄⁻ +e⁻
    0.1 mole
    Quantity of electricity required
    = 0.1F = 0.1×96500 = 9650 C"

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