In neutral medium- KM nO 4 acts as a powerful oxidizing agent according to the reaction-2 KM nO 4- H 2 O - 2 MnO 2-2 KOH -3 O - The change involvesA- Increase of 3 e -B- Loss of 3 e -C- Loss of 5 e -D- Gain of 5 e -

The correct option is A Increase of 3e^ MnO_4^ + 2H_2O + 3e^ → MnO_2 + 4OH^ ...

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Potassium Permanganate

In neutral me...

Question

In neutral medium, $K M n O_{4}$ acts as a powerful oxidizing agent according to the reaction:
$2 K M n O_{4} + H_{2} O \to 2 M n O_{2} + 2 K O H + 3 (O)$ . The change involves

Loss of

3 e^{-}

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Increase of

3 e^{-}

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Loss of

5 e^{-}

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Gain of

5 e^{-}

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Solution

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Similar questions

K M n O_{4}

2 K M n O_{4} + H_{2} O \to 2 M n O_{2} + 2 K O H + 3 (O)

\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = a x + b ln (4 e^{x} + 5 e^{- x}) + C

K M n O_{4}

2 K M n O_{4} + 2 K O H \to 2 K_{2} M n O_{4} + H_{2} O + O

K M n O_{4}

K M n O_{4}

K = 39.09

M n = 54.94

O = 16.0

2 K M n O_{4} + 2 K O H \to 2 K_{2} {M n O}_{4} + H_{2} O + [O]

K M n O_{4}

K M n O_{4}

K M n O_{4}

\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = A x + B ln | 4 e^{x} + 5 e^{- x} | + C,

A, B

C

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