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How many mole of electrons are involved ...

How many mole of electrons are involved in the reduction of one mole of `MnO_(4)^(-)` ion in alkaline medium to `MnO_(3)^(-)` ?

A

2

B

1

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
A
`{:("MnO_4^(-),MnO_3^(-)),(+2H_2O,+H_2O),(+2e^(-),+2OH^(-)):}`
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Knowledge Check

  • How many coulombs of electricity are required for the reduction of 1 mol of MnO_4^(-) " to " Mn^(2+) ?

    A
    96500 C
    B
    `1.93 xx 10^(5) C`
    C
    ` 4.83 xx 10^(5) C `
    D
    `9.65 xx 10^(6) C `

  • The product of oxidation of I^(-) with MnO_(4)^(-) in alkaline medium is

    A
    `IO_(3)^(-)`
    B
    `I_(2)`
    C
    `IO^(-)`
    D
    `IO_(4)^(-)`

  • The product of oxidation of I with MnO_(4) in alkaline medium is

    A
    `IO_(3)^(-)`
    B
    `I_2`
    C
    `IO^(-)`
    D
    `IO_(4)^(-)`

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    Disproportionation products of one mole of MnO_(4)^(-2) in aqueous acidic medium are

    The charge required for the oxidation of one mole of Mn_3O_4 " to " MnO_4^(2-) in alkaline medium is_____

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