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Disproportionation products of one mole ...

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

A

`(1)/(3)` mol of `MnO_(4)^(-), (2)/(3)` mol of `MnO_(2)`

B

`(2)/(3)` mol of `MnO_(4)^(-)` mol of `MnO_(2)`

C

`(1)/(3)` mol of `Mn_(2)O_(7), (1)/(3)` mol of `MnO_(2)`

D

`(2)/(3)` mol of `Mn_(2)O_(7), (1)/(3)` mol of `MnO_(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`2e^(-) + 4H^(+) + MnO_(4)^(2-) rarr MnO_(2) + 2H_(2)O`
`(MnO_(4)^(2-) rarr MnO_(4)^(-) + e^(-) ) xx 2 `
overall reaction,
`3MnO_(4)^(2-) + 4H^(+) rarr MnO_(2) + 2MnO_(4)^(-) + 2 H_(2)O `
`therefore` 3 mol `MnO_(4)^(2-)` gives 1 mol `MnO_(2)`
`therefore` 1 mol `MnO_(4)^(2-) rarr (1)/(3)" mol of " MnO_(2)`
Moreover,
3 mol of `MnO_(4)^(2-)` gives 2 mol `MnO_(4)^(-)`
`therefore` 1 mol `MnO_(4)^(2-) rarr (2)/(3) " mol " MnO_(4)^(-)`
Hence, option (b) is correct.
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