Step 1 : The skeletal ionic equation is :
`Cr_(2)O_(7)^(2-) (aq) + SO_(3)^(-2) (aq) to 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)(S)overset(-2)(O_(3)^(2-)) (aq) to overset(+3)(Cr)(aq) + overset(+6)(S)overset(-2)(O_(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 notice 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)(3S)overset(-2)(O_(3)^(2-)) (aq) to overset(+3)(2) Cr^(3+) (aq) + 3SO_(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) + 8 H^(+) to 2 Cr^(3+) (aq) + 3 SO_(4)^(2-)(aq)`
Step : 5 Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., 4`H_(2)O`) on the right to achieve balanced redox change.
`Cr_(2)O_(7)^(2-) (aq) + 3SO_(3)^(2-) (aq) + 8H^(+) (aq) to 2 Cr^(3+) (aq) + 3 SO_(4)^(2-) (aq) + 4H_(2)O (1)`
