Explain why `Cr^(2+)` is strongly reducing while `Mn^(3+)` is strongly oxidizing.

Explanation

`Cr^(2+)` is strong reducing while Mn3+ is strongly oxidising,
`E^(0)_(Cr^(3+)//Cr^(2+))` is -0.41V
`Cr^(2+)+ 2e^(-) to Cr E^(0)= -091 V`
If the standard electrode potential `E^(0)` of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily.
`Mn to Mn^(3+)+ 3e^(-) Mn^(3+) [Ar]3d^(4)`
`Mn^(3+)+ e^(-) to Mn^(2+) E^(0) = +1.51V`
If the standard electrode potential Eº of a metal is large and positive, the metal is a powerful oxidising agent because it gains electrons easily,