1.

Us the data of Q.8 and find out ht emost stable oxidised species.

Answer»

`Cr^(3+)`
`MnO_(4)^(-)`
`Cr_(2)O_(7)^(2-)`
`Mn^(2+)`

Solution :The velue of `E^(Theta)Cr^(3+)//Cr` is -0.74 V, which is minimum value AMONG given FOUR.
reduction :`Cr^(3+)+3e^(-)toCr`
Oxidation :`Crto Cr^(3+)+3e^(-)`
`Cr^(3+)//Cr` has most NEGATIVE value of standard reduction potential .Hence `Cr^(3+)` is the most oxidized spedcies.
So,Cr is oxidation performing and `Cr^(3+)` is a product of oxidation .
Cr has LEAST standard reduction potential and hence ,`Cr^(3+)` is highest STABLE among four.


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