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:
SO32- (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.
SO32- (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.