Question 27 of 31intermediate🔧 ApplyShort Answer2 marks

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Correct Answer

The balanced redox equation for the reaction between sulphur dioxide and excess chlorine in water is:

SO2(aq)+Cl2(g)+2H2O(l)→H2SO4(aq)+2HCl(aq)SO_2(aq) + Cl_2(g) + 2H_2O(l) \rightarrow H_2SO_4(aq) + 2HCl(aq)

Alternatively, in ionic form:

SO2(aq)+Cl2(g)+2H2O(l)→SO42−(aq)+2Cl−(aq)+4H+(aq)SO_2(aq) + Cl_2(g) + 2H_2O(l) \rightarrow SO_4^{2-}(aq) + 2Cl^-(aq) + 4H^+(aq)
Exercise: EXERCISES | Q: 7.23 | (Page Page 20)
For More Understanding

Explanation

In this redox reaction, sulphur dioxide (SO2SO_2) acts as the reducing agent and is oxidised to sulphate (SO42−SO_4^{2-}), while chlorine (Cl2Cl_2) acts as the oxidising agent and is reduced to chloride (Cl−Cl^-). By assigning oxidation numbers, we see that the oxidation state of S increases from +4 to +6, and the oxidation state of Cl decreases from 0 to -1. Balancing the atoms and charges in the aqueous medium gives the final balanced equation.

Solution Steps

  1. Step 1: Identify the oxidation and reduction half-reactions. Sulphur in SO2SO_2 is oxidised from +4 to +6 (forming SO42−SO_4^{2-} or H2SO4H_2SO_4), and chlorine in Cl2Cl_2 is reduced from 0 to -1 (forming Cl−Cl^- or HClHCl).

  2. Step 2: Balance the atoms and charges for the half-reactions in water. Oxidation: SO2+2H2O→SO42−+4H++2e−SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^- Reduction: Cl2+2e−→2Cl−Cl_2 + 2e^- \rightarrow 2Cl^-

  3. Step 3: Add the two half-reactions together to get the balanced ionic equation:

    SO2(aq)+Cl2(g)+2H2O(l)→SO42−(aq)+2Cl−(aq)+4H+(aq)SO_2(aq) + Cl_2(g) + 2H_2O(l) \rightarrow SO_4^{2-}(aq) + 2Cl^-(aq) + 4H^+(aq)

  4. Step 4: Represent in molecular form if required:

    SO2(aq)+Cl2(g)+2H2O(l)→H2SO4(aq)+2HCl(aq)SO_2(aq) + Cl_2(g) + 2H_2O(l) \rightarrow H_2SO_4(aq) + 2HCl(aq)
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