Question
Question: What happens when sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt?...
What happens when sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt?
Solution
As we know that Sulphur dioxide SO2 is a colourless, heavy gas. It is produced in large amounts in intermediate steps of sulphuric acid production. Sulphur dioxide acts as a reducing agent in chemical reaction.
Complete answer:
& S{{O}_{2}}+2{{H}_{2}}O\to SO_{4}^{2-}+4{{H}^{+}}+2{{e}^{-}} \\\ & \left[ F{{e}^{3+}}+{{e}^{-}}\to F{{e}^{2+}} \right]\times 2 \\\ & S{{O}_{2}}+2{{H}_{2}}O+2F{{e}^{3+}}\to S{{O}_{4}}^{2-}+4{{H}^{+}}+2F{{e}^{2+}} \\\ \end{aligned}$$ **> Hence, we can conclude that when sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt, then $S{{O}_{2}}$ acts as a reducing agent and it reduces Fe(III) to Fe(II) that is ferric ions to ferrous ions.** **Additional information:** \- Sulphur dioxide is found to be mild toxic and can be harmful in high concentration. It is also found that $S{{O}_{2}}$ is a major air pollutant and also has a dangerous effect on human health. In the atmosphere it also can influence the plant and animal habitat. **Note:** > We should not get confused in terms of oxidising and reducing agents. As an oxidising agent it gains electrons and is itself reduced in a reaction. It is also called an electron acceptor. > Whereas, reducing agent loses electrons and itself gets oxidized. It is also called an electron donor.When sulphur dioxide reacts with water the following reaction takes place:
SO2+2H2O→SO42−+4H++2e−
Consider the above reaction as 1. Here, we can see that sulphur is donating electrons and hence acts as electron donor (that is reducing agent).
- Now, further iron which is present in Fe3+ form will accept the electron and will form Fe2+. Here, we can see that change in oxidation state takes place from +3 to +2. We can see the following reaction that takes place, and consider it as reaction 2:
Fe3++e−→Fe2+
- Now, from reaction 1 and 2, we can write the complete reaction as: