Question

$C{e}^{4+}$ is a good oxidising agent while $S{m}^{2+}$ is a good reducing agent. Why?

Answer 1

Electronic configuration of $Ce$ is $4f^{1}5d^{1}6s^2$

Ce is most stable in +3 state. In +4 state it is most oxidizing because in +4 state it can give an electron to become more stable at +3 state.

Whereas in $S{m}^{+}2$, the lanthanides are more stable in +3 state and hence in +2 state it gets oxidized easily (good reducing agent) to get to +3 oxidation state.