Question

Balance the reaction : ${\mathrm{H}}_2\mathrm{S}+{\mathrm{SO}}_2\to \mathrm{S}+{\mathrm{H}}_2\mathrm{O}$ and identify the Oxidizing and Reducing agents

Answer 1

1. The balanced reaction can be given as: $$\begin{gathered} 2{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)\to 3\mathrm{S}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Hydrogen}\mathrm{sulphide})(\mathrm{Sulphur}\mathrm{dioxide})\left(\mathrm{Sulphur}\right)\left(\mathrm{Water}\right) \end{gathered}$$.
2. Oxidizing agent: Here, Sulphur dioxide (${\mathrm{SO}}_2$) acts as Oxidizing agent because it loses Oxygen atom and get reduced to $\mathrm{S}$(Sulphur) therefore, it acts as Oxidizing agent.
3. Reducing agent: In this reaction, ${\mathrm{H}}_2\mathrm{S}$ (Hydrogen sulphide) by gaining Oxygen atom, it is oxidized to ${\mathrm{H}}_2\mathrm{O}$ (Water) therefore, it acts as Reducing agent.