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Potassium permanganate gives the followi...

Potassium permanganate gives the following reactions in neutral medium `MnO_4^(-) + 2H_2O + 3e^(-) to MnO_2 + 4OH^-` . The equivalent weight of `KMnO_4` is (atomic mass of Mn= 55u )

A

158

B

79

C

52.66

D

31.6

Text Solution

Verified by Experts

The correct Answer is:
C
Equivalent weight of `KMnO_2` in neutral medium
`MnO_4^(-) + 2H_2O+ 3e^(-) to MnO_2 + 4OH^-`
Total oxidation state of Mn In `KMnO_4 = 14` units
Total oxidation state of Mn in product `MnO_2 = + 8` units
Change in oxidation state of Mn = 14 - 8 =6 units
` therefore ` Equivalent weight of `KMnO_4`
` = (2M)/(6) =M/3 = 158/3 = 52.66`
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Knowledge Check

  • Potassium permanganate gives the following reactions in neutral medium MnO_4)^(-)+2H_2O+3e^(-)toMnO_2+4OH^(-) . The equivalent weight of KMnO_4 is (atomic mass of Mn=55u)

    A
    158
    B
    79
    C
    52.66
    D
    31.6

  • The equivalent weight of MnO_(4)^(-) ion in acidic medium is

    A
    `31.6`
    B
    `23.8`
    C
    55
    D
    119

  • In strongly alkaline medium, MnO_4^- MnO_4^(-) + e^(-) to MnO_4^(2-) In this medium, equivalent weight of KMnO_4 is reduced as:

    A
    31.6
    B
    52.67
    C
    79
    D
    158

    • Similar Questions

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    Consider the following reaction MnO_4^(-) + 8 H^(+) 5e^(-) to Mn^(2+) + 4H_2 O , E^@ = 1.51 V . The quantity of electricity required in Faraday to reduce five moles of MnO_4^- is _____________ .

    In acidic medium,potassium manganate undergoes disproportionation according to equation. 3MnO_(4)^(2-)+4H^(+)longrightarrow2MnO_(4)^(-)+MnO_(2)+2H_(2)O Equivalent weight of K_(2)MnO_(4) will be (Mol.wt.of K_(2)MnO_(4) =M)

    The equivalent weight of KMnO_4 in a redox reaction in a neutral medium is .

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