Question #863c8

Jul 7, 2016

Hydrogen peroxide is the oxidizing agent and lead(II) sulfite is the reducing agent.

Explanation:

Lead(II) sulfite, $\text{PbS}$, will react with hydrogen peroxide, ${\text{H"_2"O}}_{2}$, to form lead(II) sulfate, ${\text{PbSO}}_{4}$, and water.

In this reaction, the lead(II) sulfite will be oxidized to lead(II) sulfate and the hydrogen peroxide will be reduced to water.

You can thus say that thee hydrogen peroxide acts as an oxidizing agent because it oxidizes lead(II) sulfite to lead(II) sulfate.

Similarly, the lead(II) sulfite acts as a reducing agent because it reduces the hydrogen peroxide to water.

The balanced chemical equation that describes this redox reaction looks like this

${\stackrel{\textcolor{b l u e}{+ 2}}{\text{Pb") stackrel(color(blue)(-2))("S")_ ((s)) + 4 stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(-1))("O")_ (2(aq)) -> stackrel(color(blue)(+2))("Pb") stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_ (4(aq)) + 4 stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(-2))("O}}}_{\left(l\right)}$

Notice that the oxidation state of oxygen goes from $\textcolor{b l u e}{- 1}$ on the reactants' side to $\textcolor{b l u e}{- 2}$ on the products' side. This tells you that oxygen is being reduced, since its oxidation state is decreasing.

On the other hand, the oxidation state of sulfur is going from $\textcolor{b l u e}{- 2}$ on the reactants' side to $\textcolor{b l u e}{+ 6}$ on the products' side. This tells you that sulfur is being oxidized, since its oxidation state is increasing.

Therefore, you have

$\textcolor{w h i t e}{a} \text{PbS} \to$reducing agent

${\text{H"_2"O}}_{2} \to$ oxidizing agent