Balancing of Redox Reaction by half-Reaction Method:

Step-1: First we write the skeletal ionic equation

\(\overset{+6}{Cr_2}\overset{2-}{O_7} (aq)+\overset{+4}{S}\,\overset{2-}{O_3}(aq)\rightarrow \overset{+3}{Cr^{3+}}(aq)+\overset{+6}{S}\,\overset{2-}{O_4}\)

Step-2: The two half reactions are: Oxidation half reaction

\(\overset{+4}{S}\,\overset{2-}{O_3}(aq)\rightarrow \overset{+6}{S}\,\overset{2-}{O_4}(aq)\)

Reduction half reaction

\(\overset{+6}{Cr_2}\overset{2-}{O_7}_{(aq)}\rightarrow \overset{+3}{Cr^{3+}}_{(aq)}\)

Step-3: Balance the Cr atoms in reduction half reaction:

Cr₂O₇^-2_(aq) → 2Cr³⁺ _(aq)

Step-4: To balance O-atoms in oxidation half reaction, we add one water molecule to the left side:

SO₃^2- (aq) + H₂O(I) → SO₄²⁻ (aq)

And to balance O-atoms in reduction-half reaction, we add seven water molecules to the right side:

Cr₂O₇^-2_(aq)→ 2Cr³⁺ _(aq) + 7H₂O(l)

To balance the H-atoms in oxidation half reaction, we add two H⁺ ions to the right side.

SO₃^2- (aq) + H₂O (l) → SO₄²⁻(aq) + 2H⁺(aq)

To, balance the H-atoms in reduction half reaction, we add 14H⁺ ions to be left side.

Cr₂O₇^-2_(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l)

Step-5: Balance the charges of two half reactions,

\(\overset{+4}{S}\overset{2-}{O_3}(aq) +H_2O(l)\rightarrow\overset{+6}{S}\overset{2-}{O_4}(aq) +2H^+(aq) +2e^-\)

\(\overset{+6}{Cr_2}\overset{2-}{O_7}(aq) + 14H^+(aq)+6e^- \) → \(2\overset{+3}{Cr^{3+}}\) (aq) + 7H₂O(I)

Now, to equalize the number of electrons, we multiply the oxidation half reaction by 3, then we get

3SO₃²⁻ (aq) + 3H₂O(I) → 3SO₄²⁻ (aq) + 6H⁺(aq) + 6e⁻

Cr₂O₇²⁻ (aq) + 14⁺ + 6e⁻ + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(I)

Step-6: Add two half reactions to obtain the net reaction after cancelling the electrons and other species on both sides. 

The net equation is:

3SO₃²⁻ (aq) + Cr₂O₇²⁻ (aq) + 8H⁺(aq) → 3SO₄²⁻ (aq) + 2Cr³⁺ (aq) + 4H₂O(I)

A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides.