Question

In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced: a. \(\mathrm{Zn}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{ZnCl}_{2}(s)\) b. \(\mathrm{Cl}_{2}(g)+2 \mathrm{NaBr}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(l)\) c. \(2 \mathrm{PbO}(s) \longrightarrow 2 \mathrm{~Pb}(s)+\mathrm{O}_{2}(g)\) d. \(2 \mathrm{Fe}^{3+}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+\mathrm{Sn}^{4+}(a q)\)

Short answer

Zn is oxidized, Cl2 is reduced; NaBr is oxidized, Cl2 is reduced; PbO is reduced, O is oxidized; Sn2+ is oxidized, Fe3+ is reduced.

Step-by-step solution

Understand oxidation and reduction

Oxidation involves the loss of electrons, and the substance that loses electrons is known as the reducing agent. Reduction involves the gain of electrons, and the substance that gains electrons is known as the oxidizing agent.

Identify oxidation and reduction in reaction a

For the reaction \(\text{Zn(s) + Cl}_{2}\text{(g)} \rightarrow \text{ZnCl}_{2}\text{(s)}\), Zn starts in its elemental form (0 oxidation state) and moves to a +2 oxidation state in ZnCl2, indicating that Zn is oxidized. Cl2 moves from an elemental form (0 oxidation state) to -1 in ZnCl2, hence Cl2 is reduced.

Identify oxidation and reduction in reaction b

For the reaction \(\text{Cl}_{2}\text{(g) + 2 NaBr(aq) }\rightarrow \text{ 2 NaCl(aq) + Br}_{2}\text{(l)}\), Cl2 starts at 0 oxidation state and moves to -1 in NaCl, indicating reduction. Br in NaBr moves from -1 to 0 in Br2, indicating oxidation.

Identify oxidation and reduction in reaction c

For the reaction \(2 \text{PbO(s)} \rightarrow 2 \text{ Pb(s) + O}_{2}\text{(g)}\), Pb in PbO is in a +2 oxidation state and moves to 0 in elemental Pb, indicating reduction. O in PbO is in a -2 oxidation state and moves to 0 in \(\text{O}_{2}\text{(g)}\), hence, O is oxidized.

Identify oxidation and reduction in reaction d

For the reaction \(\text{2 Fe}^{3+}\text{(aq) + Sn}^{2+}\text{(aq) }\rightarrow \text{2 Fe}^{2+}\text{(aq) + Sn}^{4+}\text{(aq)}\), Fe moves from +3 to +2, indicating reduction. Sn moves from +2 to +4, indicating oxidation.

Key concepts

oxidation reactions

In chemistry, oxidation reactions involve the loss of electrons from a molecule, atom, or ion. When an element undergoes oxidation, its oxidation state increases. For example, in the reaction \(\text{Zn(s) + Cl}_{2}\text{(g)} \rightarrow \text{ZnCl}_{2}\text{(s)}\), zinc (Zn) loses electrons to chlorine (Cl), going from an oxidation state of 0 to +2. This makes Zn the reducing agent because it donates electrons. Elements or compounds that undergo oxidation are called 'oxidized substances'. It's important to remember that oxidation always involves a companion process: reduction.

reduction reactions

Reduction reactions are essentially the opposite of oxidation reactions. They involve the gain of electrons by a molecule, atom, or ion. This results in a decrease in the oxidation state. For instance, in the reaction \(\text{Cl}_{2}\text{(g) + 2 NaBr(aq) }\rightarrow \text{ 2 NaCl(aq) + Br}_{2}\text{(l)}\), chlorine (Cl) undergoes reduction by gaining electrons, which changes its oxidation state from 0 to -1. Substances that are reduced serve as oxidizing agents because they accept electrons. Hence, in every oxidation-reduction (redox) reaction, one species gets oxidized while another gets reduced.

electron transfer

Electron transfer is the key mechanism underlying both oxidation and reduction reactions. In redox reactions, electrons move from the species that gets oxidized to the species that gets reduced. For instance, look at the reaction \(\text{2 Fe}^{3+}\text{(aq) + Sn}^{2+}\text{(aq) }\rightarrow \text{2 Fe}^{2+}\text{(aq) + Sn}^{4+}\text{(aq)}\). Here, iron (Fe) reduces by gaining an electron and changes from a +3 to a +2 oxidation state. Simultaneously, tin (Sn) oxidizes by losing an electron, shifting from +2 to +4. Understanding how electrons transfer between reactants in these reactions explains the dynamic nature of redox processes.