`12.53cm^(3)0.051MseO_(2)`reacts completely with `25.5cm^(3)0.1MCrSO_(4)`to produce `Cr_(2)(SO_(4))_(3)`. What is the change in the oxidation number of Se in this redox reaction ?

Explanation

Let the oxidation number of Se in the newly produced compound be x
The redox reaction is as follows -
`[se^(4+)+xetoSe^(4-x)]xx1`
`underline([Cr^(2+)toCr^(3+)+e]xxx)`
`Se^(4+)+xCr^(2+)toSe^(4-x)+xCr^(3+)`
Now , `12.53Cm^(3)0.051SecO_(2)`=12.53xx0.051`25.5cm^(2)0.1McrSO_(4)`=25.xx0.1=2.55mmol `CrSO_(4)`
However according to the balanced equation , 1 mol `SeO_(2)`gets reduced by x mol `CrSO_(4)`
`therefore`2.55 mmol `CrSO_(4)^(-)`is reduced by `(2.55)/(x)`mmol `SeO_(2)`
But 0.64 mmol `SeO_(2)` actually gets reduced . So, `(2.55)/(x)`=0.64 or, x=4
`therfore` The change in oxidation number of Se-atom. =4-(4-x)=x=4.