Question

In each of the following reactions, identify which element is oxidized and which is reduced. a. \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) b. \(2 \operatorname{CuI}(s) \rightarrow \operatorname{CuI}_{2}(s)+\operatorname{Cu}(s)\) c. \(6 \mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \rightarrow\) \(6 \mathrm{Fe}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

In the given reactions: a. Zn is oxidized (0 to +2) and H is reduced (+1 to 0). b. Cu is both oxidized (+1 to +2) and reduced (+1 to 0) in different species. c. Fe is oxidized (+2 to +3) and Cr is reduced (+6 to +3).

Step-by-step solution

Reaction (a) - Identify changes in oxidation states

In the first reaction, we have: \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) First we must determine the oxidation states of the elements. In their elemental form, such as Zn(s) and H2(g), their oxidation state is 0. In ZnCl2, Zn has an oxidation state of +2 while Cl has an oxidation state of -1 (there are two Cl, so it balances out).

Reaction (a) - Determine oxidation and reduction

Zn goes from an oxidation state of 0 to +2 in ZnCl2, so it has lost 2 electrons. This means Zn is being oxidized. H in HCl goes from an oxidation state of +1 to 0 in H2, so it has gained an electron. This means H is being reduced.

Reaction (b) - Identify changes in oxidation states

In the second reaction, we have: \(2 \operatorname{CuI}(s) \rightarrow \operatorname{CuI}_{2}(s)+\operatorname{Cu}(s)\) In CuI, Cu has an oxidation state of +1 and I has an oxidation state of -1. In CuI2, Cu has an oxidation state of +2 while I have an oxidation state of -1 (there are two I, so it balances out). In Cu(s), the oxidation state is 0.

Reaction (b) - Determine oxidation and reduction

Cu goes from an oxidation state of +1 in CuI to 0 in Cu(s), so it has gained 1 electron. This means Cu is being reduced. Cu also goes from an oxidation state of +1 in CuI to +2 in CuI2, so it has lost 1 electron. This means Cu is also being oxidized.

Reaction (c) - Identify changes in oxidation states

In the third reaction, we have: \(6 \mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \rightarrow 6 \mathrm{Fe}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\) Fe2+ has an oxidation state of +2, which changes to +3 in Fe3+. Cr2O72- has Cr with an oxidation state of +6, which changes to +3 in Cr3+. H+ has an oxidation state of +1, which changes to 0 in H2O.

Reaction (c) - Determine oxidation and reduction

Fe goes from an oxidation state of +2 to +3, so it has lost 1 electron. This means Fe is being oxidized. Cr goes from an oxidation state of +6 to +3, so it has gained 3 electrons. This means Cr is being reduced.

Key concepts

Oxidation States

Understanding oxidation states is crucial when analyzing redox reactions. An element's oxidation state (also known as oxidation number) is a theoretical charge it would have if the compound was composed of ions. It helps keep track of the electrons during a reaction and identify which elements are oxidized or reduced.

For pure elements, the oxidation state is always 0. In ionic compounds, the oxidation state is equal to the charge of the ion, and in covalent compounds, it is determined based on a set of rules. For example, hydrogen typically has an oxidation state of +1 and oxygen -2, but there are exceptions. In the provided reactions, analyzing oxidation states indicates changes that occur: Zn goes from 0 to +2, showing a loss of electrons, and similarly, Fe from +2 to +3. These changes signify oxidation. On the other hand, elements like H in HCl and Cr in Cr2O7^2- show a decrease in their oxidation state, indicating reduction.

Oxidation and Reduction Identification

Identifying oxidation and reduction in chemical reactions is based on the changes in oxidation states. The element that increases in oxidation state is being oxidized, while the element that decreases in oxidation state is being reduced.

In redox reactions, there are some key identifiers for oxidation and reduction. If an element loses electrons (leading to a higher positive charge or less negative charge), it is undergoing oxidation. Conversely, when an element gains electrons (leading to a lower positive charge or more negative charge), it is undergoing reduction. For example, in reaction (a), Zn loses two electrons and is oxidized, while H gains an electron and is reduced. It's also possible to have an element that both gains and loses electrons in disproportionation reactions, such as Cu in reaction (b).

Chemical Reaction Analysis

Analysis of a chemical reaction involves closely examining both the reactants and products to understand the process of the reaction. It includes identifying the reactants, products, and the types of reactions occurring. With a focus on redox reactions, analyzing the changes in oxidation states helps to identify the overall electron transfer.

For instance, reviewing the species involved in the reactions and their oxidation numbers before and after the reaction enables the determination of which substance is oxidized or reduced. This process often involves writing half-reactions that separately show oxidation and reduction. In reaction (c), writing half-reactions would illustrate how Fe transitions from +2 to +3 (oxidation) and how Cr transitions from +6 to +3 in Cr^3+ (reduction), providing a clear picture of the electron transfer in the overall redox process.