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

`Ce^(4+)-=Sn^(2+)(Sn^(2+)toSn^(4+)+2e)(n=2)`
`1mEq-=1mEq`
`N_1V_1-=N_2V_2`
`0.5xx`n-factor`xx40=1xx2`(n-factor)`xx10`
`20n=20`
`n=1`
n-factor of `Ce^(4+)` is 1, i.e., it is reduced to `Ce^(3+)`
`Ce^(4+)+e^(-)toCe^(3+)`