The coulombs of electricity required for reduction of 1 mole of MnO 4-to Mn 2- areA- 96500 CB- 4-83 - 105 C- 9365 - 106 CD- 1-93 - 105 C

The correct option is B 4.83 × 10^5CReduction of MnO_4^ to Mn^2f is MnO_4^ +8H^+ +5e^ → Mn^2++4H_2O ∴ Reduction of d mole of MnO_4^ Requires = 5 × 96500 C ...

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Standard XII

Chemistry

Law of Radioactivity

The coulombs ...

Question

The coulombs of electricity required for reduction of 1 mole of $M n O_{4}^{-}$ to $M n^{2 +}$ are

96500 C

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1.93 \times 10^{5} C

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4.83 \times 10^{5} C

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9365 \times 10^{6} C

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Solution

The correct option is C $4.83 \times 10^{5} C$
Reduction of $M n O_{4}^{-}$ to $M n^{2} f$ is
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
$∴$ Reduction of d mole of $M n O_{4}^{-}$
Requires = 5 $\times$ 96500 C = 4.825 $\times 10^{5} C$

Suggest Corrections

The coulombs of electricity required for reduction of 1 mole of MnO−4 to Mn2+ are

The correct option is C 4.83×105CReduction of MnO−4 to Mn2f is MnO−4+8H++5e−→Mn2++4H2O ∴ Reduction of d mole of MnO−4 Requires = 5 × 96500 C = 4.825 ×105C

The coulombs of electricity required for reduction of 1 mole of $M n O_{4}^{-}$ to $M n^{2 +}$ are

The correct option is C $4.83 \times 10^{5} C$
Reduction of $M n O_{4}^{-}$ to $M n^{2} f$ is
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
$∴$ Reduction of d mole of $M n O_{4}^{-}$
Requires = 5 $\times$ 96500 C = 4.825 $\times 10^{5} C$