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For the reaction, M^(x+) + MnO(4)^(-)rar...

For the reaction, `M^(x+) + MnO_(4)^(-)rarrMO_(3)^(-) + Mn^(2+) + 1//2 O_2` if one mole of `MnO_4^(-)` oxidises 1.67 moles of `M^(x+) " to " MO_(3)^(-)` then the value of x in the reaction is?

Text Solution

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The correct Answer is:
2
`overset(+7) (MnO_4^(-))+5e^(-) rarrMn^(2+)`
Since 1 mole of `MnO_(4)^(-)` accepts 5 moles of electrons , therefore , 5 moles electrons are lost by 1.67 moles of `M^(x+)`
`:. 1` Mole of `M^(x+)` will lose electrons = 5/1.67 = 3 moles ( approx.)
Since `M^(x+)` changes to `MO_(3)^(-)` ( where O.N. of M = +5) by accepting 3 electrons
`:. x = +5-3 =+2`
`:. Mn^(7+)+5e rarr Mn^(2+)`
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