One among the lanthanoides, `Ce(III)`, can be easily oxidized to Ce(IV) (At.No. of ce=58) explain why?

Why among the lanthanides Ce ,(III), can be easily oxidised to Ce(IV) (At. No. of Ce=58). Explain why? Or of the lanthanides,cerium (At.Nos.58) forms a tetrapositive ion Ce^(4+) in aqueous solution. Why?

Why among the lanthanides Ce ,(III), can be easily oxidised to Ce(IV) (At. No. of Ce=58). Explain why? Or of the lanthanides , cerium (At.Nos.58) forms a tetrapositive ion Ce^(4+) in aqueous solution. Why?

Among the lanthanoids, Ce(III) can be easily oxidised to Ce(IV) (Atomic number of Ce = 58). Give reason.

One among the lanthanoids, Ce(III) (Z=58) can easily be oxidised to Ce (IV). Explain why?

Give reasons for the following. Among the lanthanoid, Ce (III) is easily oxidised to Ce (IV).

Explain why: (i) E^0 for Mn^(3+)//Mn^(2+) couple is more positive than that for Fe^(3+)//Fe^(2+) . (At. Nos. Mn = 25, Fe = 26). (ii) Ce^(3+) can be easily oxidised to Ce^(4+) . (At. No. Ce = 58).