Question

balance the following oxidation reduction in an acidic solution Cl2(g) = Cl– (aq) + OCl–(aq)

(hint: this reaction is a disproportionation, meaning the same element undergoes both oxidation and reduction. Use Cl2 as the reactant for each of your half reactions)

Please explain. Thank you!

Answer 1

Solution :-

Step 1 - Half equations

Cl2(g) ----- > Cl^-(aq)

Cl2(g) ------ > OCl^-(aq)

Step 2-balancing elements other than H and O

Cl2(g) -------- > 2Cl^- (aq)

Cl2(g) ------ > 2OCl- (aq)

Step 3 - Balancing the oxygens by adding H2O

Cl2(g) -------- > 2Cl^-(aq)

2H2O(l) + Cl2(g) ------ > 2OCl-(aq)

Step 4- Balancing the hydrogens by adding H+

Cl2(g) -------- > 2Cl^-(aq)

2H2O(l) + Cl2(g) ------ > 2OCl-(aq) + 4H^+(aq)

Step 5- Balancing the electronic charge by adding electrons

2e^- + Cl2(g) -------- > 2Cl^-(aq)

2H2O(l) + Cl2(g) ------ > 2OCl-(aq) + 4H^+(aq) +2e^-

Step 6- Now add equations by cancelling the similar species on the opposite sides

2e^- + Cl2(g) -------- > 2Cl^-(aq)

2H2O(l) + Cl2(g) ------ > 2OCl-(aq) + 4H^+(aq) +2e^-

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2Cl2(g) + 2H2O(l) ------- > 2Cl^-(aq) + 2OCl-(aq) + 4H^+(aq)

Therefore the balanced redox equation in the acidic medium is

2Cl2(g) + 2H2O(l) ------- > 2Cl^-(aq) + 2OCl-(aq) + 4H^+(aq)