Write the net ionic equation for the reaction of potassium dichromate (VI), `K_(2) Cr_(2) O_(7)` with sodium sulphite, `Na_(2)SO_(3)`, in an acid solution to give chromium (III) ion and the sulphate ion.

Explanation

Step-1 :
The skeletal ionic equation is :
`Cr_(2)O_(7)^(2-)(ao)+SO_(3)^(-2)(ao)to Cr^(3+)(ao)-:SO_(4)^(2-)(ao)`
Step-2
Assign oxidation numbers for Cr and S +6 -2 +4 -2 +3 +6 -2
`Cr_(2)O_(7)^(2-)(ao)+SO_(3)^(_2)(ao) to Cr(ao)+SO_(4)^(2-)(ao)`
This indicates that the dichromate ion is the oxidant (it oxidises sulphite ion to sulphate ion) and the sulphite ion is the reductant. (it reduces dichromate ion to chromium (III).
Step -3 :
Calculate the increase and decrease of oxidation numbers of respective species and make them equal.
+6 -2 +4 -2 +3 +6-2
`Cr_(2)O_(7)^(2-)(ao)+3SO_(3)^(-2)(ao) to Cr^(3+)(ao)+SO_(4)^(2-)(ao)`
As reduction is total 6 units due to two `Cr^(3+)` formed, oxidation also must be 6 units. This is obtained by multiplying `SO_(3)^(-2)` with 3.
Step -4
Adjust the coefficients of the products accordingly
+6 -2 +4-2 +3 +6-2
`Cr_(2)O_(7)^(2-)(ao)+3SO_(3)^(-2)(ao)to 2Cr^(3+)(ao)+3SO_(4)^(2-)(ao)`
Step -5
(a) Add `H^(+)` ions in acid medium or `H_(2)O` molecules in basic medium in the required number to hydrogen atoms deficient side.
(b) Add `H_(2)O` molecules in acid medium or `OH^(-)` ions in basic medium in the required number to oxygen atoms difficient side. (a) and (b) may be repeated any number of times by hit and trial method until hydrogen and oxygen atoms are same in number on both the sides of the redox reaction.
The given reaction is in acid medium
`Cr_(2)O_(7)^(2-)(ao)+3SO_(3)^(2-)(ao)+8H^(+)`
(a) `to 2 Cr_(2-)^(3+)(ao)+3SO_(4)^(2-)(ao)`
(b) `Cr_(2)O_(7)^(2-)(ao)+3SO_(3)^(2-)(ao)+8H^(+)(ao)`
`to 2Cr^(3+)(ao)+3SO_(4)^(2-)(ao)+4H_(2)O(l)`