`12.53 mL` of `0.0509MSeO_(2)` reacted with `25.52 mL 0.1M CrSO_(4)` solution. In the raeaction `Cr^(2+)` was oxidised to `Cr^(3+)`. To what oxidation state selenium was converted in the reaction? Write the redox change for `SeO_(2)`.

Explanation

`Se^(4+) + (a-4)e rarr Se^(a+)`
`n = a-4` ("where `n` is valence factor")
or `Cr^(2+) rarr Cr^(3+) + e`
`"Meq. of" SeO_(2) = "Meq. of" CrSO_(4)`
`0.0509 xx 12.53 = 0.1 xx 1 xx 25.52`
`therefore n = 4`
`a-4=4`
or `a = 0`
Thus redox change is : `Se^(4+) + 4e rarr Se^(0)`