Balance the net equtation fro th 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 and sulphate ion.
Strategy : Follow the seven -step proceduce , one step at a time.

Explanation

Write the skeletal ionic equation :
`Cr_(2)O_7^(2-) (aq.) + SO_3^(2-) rarr Cr^(3+) (aq.) + SO_4^(2-) (aq.)`
Assing oxidation number to all the elements on both sides
`overset(+6)(Cr_2) overset(-2)(O_7^(2-)) (aq.) + overset(+4-2)(SO_3^(2-)) (aq.) rarr overset(+3)(Cr^(3+)) (aq) + overset(+6-2)(SO_(4)^(2-)) (aq.)`
This indicates that `Cr` and S are changing their oxidation numbers.
Step 3: Identify the oxidizing and reducing agents :

`Cr_(2)O_(7)^(2)` ion is the oxidizing agent as it is reduced while ` SO_(3)^(2-)` ion is the reducing agent as it is oxdized, Since there are two Cr atoms in ` Cr_(2)O_7^(2-)` ion, we must calculate the change in oxidation number of two `Cr` atoms. Thus, the total decrease in the oxidation number of `Cr` is of ` 6` units ,. Since there is only one S atom on either side, the total increase in the oxidation number of S is of two units .
Step 4: Make the total increase and decrease in oxiation numbers equal by multiplying ` SO_3^(2-)` ion (reducing agent ) by `3` . Also multiply ` SO_(4)^(2-)` ion by `3` to balance S atoms (Step 5).
`Cr_2O_(7)^(2-) (aq.) + 3 underset(3(2-))(SO_3^(2-) (aq.)) rarr underset(2(3+))(2 Cr^(3+) (aq.)) + underset(3(2-))(3SO_4^(2-) (aq.))`

Step 5: Balance the ionic chargesd on both the sides by adding `H^(+)` ions as the reactin is carried out in acidic condition. On the left side, there are 8 units od negative charge dur to one `Cr_(2)O_7^(2-)` ion and due to three `SO_3^(2-)` ions but on the right side, there in no charge due to two `Cr^(3+)` ions and three ` SO_4^(2-)` ions. Thus, we add ` 8 H^(+)` ions on the left to make ionic charges equal.
` Cr_2O_7^(2-) (aq.) +3SO_3^(2-) (aq.) + 8H^+ (aq) rarr 2 Cr^(3+) (aq.) + 3 SO_4^(2-) (aq)`
Step 6: Finally, balance H and O atoms by adding `H_2O` molecules. There are `8H` atoms and ` 16 O` atoms on the left side while there are zero H atoms and `12 O` atoms on the right side. Thus, add `4 H_(2)O` molecules on the right side to achieve thr balancced redox reaction :
` Cr_2O_7^(2-) (aq.) + 3 SO_3^(2-) (aq.) + 8H^+ (aq.) rarr 3Cr^(3+) (aq.) + 3SO_4^(2-) (aq.) + 4H_(2)O(1)`.