# When the gases sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction "SO"_2 + 2"H"_2"S" -> 2"H"_2"O" + 3"S" occurs. Here hydrogen sulphide is acting as?

## a) an oxidizing agent b) a reducing agent c) a dehydrating agent d) a catalyst

Jun 28, 2018

Hydrogen sulphide is a reducing agent.

#### Explanation:

A good starting point for any reaction is to look at oxidation numbers. If they change, a redox reaction has occurred.

Here are some thoughts.

A dehydrating agent will extract water from another reactant. I would say that you need a hydroxyl (OH) group for this to occur. A dehydrating agent could be an acid and protonate the OH group, which can then leave as water. A catalyst is not used up in the reaction, so is hydrogen sulphide used up?

Hydrogen is usually in a +1 oxidation state and oxygen a -2 state.

Reactants

${H}_{2}^{\textcolor{red}{+ 1}} {S}^{\textcolor{b l u e}{- 2}}$

${S}^{\textcolor{red}{+ 4}} {O}_{2}^{\textcolor{b l u e}{- 2}}$

Products

${S}^{\textcolor{red}{0}}$

${H}_{2}^{\textcolor{red}{+ 1}} {O}^{\textcolor{b l u e}{- 2}}$

Look at ${H}_{2} S$, the oxidation number of sulphur has gone from -2 to 0. This means that it has been oxidised and is therefore a reducing agent.

For $S {O}_{2}$, the oxidation number of sulphur has gone from +4 to 0. This is a reduction and makes $S {O}_{2}$ an oxidising agent. We can essentially ignore H and O in this case as they don't change.