It requires `40.0mL` of `0.50M Ce^(4+)` to titrate `10.0 mL` of `1.0MSn^(2+) "to" Sn^(4+)`. What is the oxidation state of cerium in the reduced product?

Explanation

`Sn^(2+)rarrSn^(4+)+2e`
`"ne"+Ce^(4-)rarrCe^((4-n)+)`
`Meq .of Ce^(4+)=Meq. Of Sn^(2+)`
` 40xx0.5xxn=1xx2xx10.0 rArr n=1`
`:. Ce^(4+)+erarrCe^(3+)`
thus oxidation state of `Ce` in reduced state is `+3`.