(i) On the basis of the standard electrode potential values stated for acid solutions, predict whether `Ti^(4+)` species may be used to oxidise Fe(II) to Fe(III)
`{:(Ti^(4+) + e^(-) to Ti^(3+), E^(@) = +0.01V), (Fe^(3+) + e^(-) to Fe^(2+), E^(@)= +0.77V):}`
(ii) Based on the data arrange `Fe^(3+), Mn^(2+) " and " Cr^(2+)` in the increasing order of stability of +2 oxidation state. (Give a brief reason)
`E_(Cr^(3+)//Cr^(2+))^(@) = -0.4V`
`E_(Mn^(3+)//Mn^(2+))^(@) = +1.5V`
`E_(Fe^(3+)//Fe^(2+))^(@) = +0.8V`

Explanation

(i) State `Ti^(4+)//Ti^(3+)` has lower reduction potential than `Fe^(3+)//Fe^(2+)`, it cannot be reduced in comparison with `Fe^(3+)//Fe^(2+)` ions.
Hence, `Ti^(4+)` cannot oxidise `Fe^(2+) " to " Fe^(3+)`.
(ii) As the value of reduction potential increases the stability of +2 oxidation increases. Therefore, correct order of stability is :
`Cr^(3+)//Cr^(2+) lt Fe^(3+)//Fe^(2+) lt Mn^(3+)//Mn^(2+)`