Equivalent mass of potassium permanganate in acid medium is

Question

Equivalent mass of potassium permanganate in acid medium is (A) 1/2 molecular mass (B) 1/3 molecular mass (C) 1/5 molecular mass (D) 1/6 molecular mass.

Tardigrade Admin · Accepted Answer

In acidic medium, M+7 nO-4 + 8H+ + 5e- arrow Mn2+ + 4H2O Total number of electrons gained = 5 Total change in the oxidation number of Mn = 7 - 2 = 5 Eq. mass of KMnO4 = ( Mol.mass of KMnO4/5)

Q. Equivalent mass of potassium permanganate in acid medium is

1/2 molecular mass

1/3 molecular mass

1/5 molecular mass

1/6 molecular mass.

In acidic medium,
$M^{+ 7} n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Total number of electrons gained $= 5$
Total change in the oxidation number of $M n = 7 - 2 = 5$
Eq. mass of $K M n O_{4} = \frac{M o l . ma ss o f K M n O _{4}}{5}$

Q. Equivalent mass of potassium permanganate in acid medium is

1/2 molecular mass

1/3 molecular mass

1/5 molecular mass

1/6 molecular mass.

In acidic medium, M+7nO4−​+8H++5e−→Mn2++4H2​O Total number of electrons gained =5 Total change in the oxidation number of Mn=7−2=5 Eq. mass of KMnO4​=5Mol.massofKMnO4​​

In acidic medium,
$M^{+ 7} n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Total number of electrons gained $= 5$
Total change in the oxidation number of $M n = 7 - 2 = 5$
Eq. mass of $K M n O_{4} = \frac{M o l . ma ss o f K M n O _{4}}{5}$