Given below are two statements: Statement I : CeO _ 2 can be used...

The $+ 3$ oxidation state of lanthanide is most stable and therefore lanthanide in $+ 4$ oxidation state has strong tendency to gain ${e}^{-}$ and converted into $+ 3$ and therefore act as strong oxidizing agent. $e . g . {Ce}^{+ 4}$ And therefore ${CeO}_{2}$ is used to oxidized alcohol aldehyde and ketones. Lanthanide in $+ 2$ oxidation state has strong tendency to loss ${e}^{-}$ and converted into $+ 3$ oxidation state therefore act as strong reducing agent. $∴ {EuSO}_{4}$ act as strong reducing agent.