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-) (aq) + SO_(3)^(2-) (aq) rarr Cr^(3+) (aq) + SO_(4)^(2-) (aq)`
Step 2 : Assign oxidation numbers for Cr and S
`overset(+6)(Cr_(2))overset(-2)(O_(7)^(2-))(aq) + overset(+4-2)(SO_(3)^(2-))(aq) rarr overset(+3)(Cr)(aq) + overset(+6 -2)(SO_(4)^(2-))(aq)`
This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.
Step 3 : Calculate the increase and decrease of oxidation number, and make them equal : from step-2 we can notic that there is change in oxidation state of chromium and sulphur. Oxidation state of chromium changes form +6 + to +3. There is decrease of +3 in oxidation state of chromium on right hand side of the equation. Oxidation state of sulphur changes from +4 to +6. There is an increase of +2 in the oxidation state of sulphur on right hand side. To make the increase and decrease of oxidation state equal, place numeral 2 before cromium ion on right hand side and numeral 3 before sulphate ion on right hand side and balance the chromium and sulphur atoms on both the sides of the equation. Thus we get
`overset(+6)(Cr_(2))overset(-2)(O_(7)^(2-)) (aq) + overset(+4 -2)(3 SO_(3)^(2-)) (aq) rarr overset(+3)(2 Cr^(3+)) (aq) + overset(+6 -2)(3 SO_(4)^(2-)) (aq)`
Step 4 : As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add `8H^(+)` on the left to make ionic charges equal
`Cr_(2) O_(7)^(2-) (aq) + 3SO_(3)^(2-) (aq) + 8H^(+) rarr 2Cr^(3+) (aq) + 3SO_(4)^(2-) (aq)`
Step 5 : Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., `4H_(2)O`) on the right to achieve balanced redox change.
`Cr_(2)O_(7)^(2-) (aq) + 3SO_(3)^(2-) (aq) + 8H^(+) (aq) rarr 2Cr^(3+) (aq) + 3SO_(4)^(2-) (aq) + 4H_(2)O (I)`