The reaction M^(n+)MnO(4)^(-)toMO(3)^(-)...

The reaction `M^(n+)MnO_(4)^(-)toMO_(3)^(-)+Mn^(2+)+`occurs in and acid medium . In the reaction , If 1 mol of `MnO_(4)^(-1)` oxidised 1.67 mol `M^(n+)`,then calculate the value of n.

Text Solution

Verified by Experts

Oxidation reactions:
`[M^(n+)+3H_(2)OtoMO_(3)^(-)+6H^(+)+(5-n)e]xx5`
Reduction reaction : `underline([MnO_(4)^(-)+8H^(+)+5etoMn^(2+)+4H_(2)O]xx(5-n))`
Net reaction :
`5M^(n+)MnO_(4)^(-)+(4n-5)H_(2)Oto(5-n)Mn^(2+)+5MO_(3)^(-)+(8n-10)`
Therefore ,(5-n) mol of `MnO_(4)^(-)-=5mol of Mn^(n+)`
or,1 mol of `MnO_(4)^(-)=((5)/(5-n))mol of Mn^(n+)`
Given : 1 mol of `MnO_(4)^(-)` oxidises 1.67 mol of `M^(n+)`
`therfore(5)/(5-n)=1.67ornN=2`

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