Problem 57 - Chemistry: The Central Science

Question

The metal cadmium tends to form \(\mathrm{Cd}^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q)\), cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\), nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) Which elements more closely define the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Step-by-Step Solution

Answer

The net ionic equations for the two observed redox reactions are: 1. Zn (s) + Cd\(^{2+}\) (aq) → Zn\(^{2+}\) (aq) + Cd (s) 2. Cd (s) + Ni\(^{2+}\) (aq) → Cd\(^{2+}\) (aq) + Ni (s) Cadmium is positioned between zinc and nickel in the activity series. To determine its exact position, perform experiments involving other metals like copper, iron, or magnesium. Compare cadmium's reactivity against these metals and their ions in displacement experiments to establish its precise position in the activity series.

Step 1: Write net ionic equations for each observation

: The half-reactions for each observation are as follows. 1. Redox reaction between Zinc and Cd\(^{2+}\): Oxidation half-reaction: Zn (s) → Zn\(^{2+}\) (aq) + 2e\(^{-}\) Reduction half-reaction: Cd\(^{2+}\) (aq) + 2e\(^{-}\) → Cd (s) Combine both half-reactions to form the net ionic equation: Zn (s) + Cd\(^{2+}\) (aq) → Zn\(^{2+}\) (aq) + Cd (s) 2. Redox reaction between Cadmium and Ni\(^{2+}\): Oxidation half-reaction: Cd (s) → Cd\(^{2+}\) (aq) + 2e\(^{-}\) Reduction half-reaction: Ni\(^{2+}\) (aq) + 2e\(^{-}\) → Ni (s) Combine both half-reactions to form the net ionic equation: Cd (s) + Ni\(^{2+}\) (aq) → Cd\(^{2+}\) (aq) + Ni (s)

Step 2: Determine cadmium's position in the activity series

: Based on the given observations, we can say that Zn is more active than Cd as it displaces Cd from its solution (Observation 1). Similarly, Cd is more active than Ni because it displaces Ni from its solution (Observation 2). In the activity series, cadmium is placed between zinc and nickel.

Step 3: Suggest experiments to locate the position of cadmium in the activity series more precisely

: To determine cadmium's exact position in the activity series, we need to perform experiments involving other metals and their ions to see if cadmium can displace them from their solutions. For example, compare cadmium's reactivity with metals such as copper, iron, or magnesium. Once we determine which metals Cd can displace and which metals can displace Cd, we can locate its exact position in the activity series. Perform the following experiments: 1. Place a strip of cadmium in solutions of Cu\(^{2+}\), Fe\(^{2+}\), and Mg\(^{2+}\). If any of these metals are displaced, record the observation and move on to other metals. 2. Place a strip of Cu, Fe, and Mg in a solution of Cd\(^{2+}\). Observe if any of these metals can displace cadmium. By performing these experiments, we can determine if Cd is more or less active than the referenced metals and thus locate its position more precisely in the activity series.

Key Concepts

Redox Reactions

Redox reactions are a type of chemical reaction that involves the transfer of electrons between two species. The term 'redox' is a shorthand for 'reduction-oxidation.' In these reactions, one substance loses electrons and is oxidized, while another substance gains electrons and is reduced.

For example, in the interaction between zinc and cadmium chloride solution, zinc is oxidized to zinc ions (Zn²⁺), losing two electrons in the process. Meanwhile, cadmium ions in the solution (Cd²⁺) gain these two electrons to form cadmium metal. This can be represented by two half-reactions:

• Oxidation half-reaction: Zn (s) → Zn²⁺ (aq) + 2e⁻
• Reduction half-reaction: Cd²⁺ (aq) + 2e⁻ → Cd (s)

When combined, they form the net ionic equation for the redox reaction:
Zn (s) + Cd²⁺ (aq) → Zn²⁺ (aq) + Cd (s).

Understanding these processes is crucial for a variety of applications, including battery technology, metallurgy, and electrochemical devices.

Net Ionic Equations

Net ionic equations offer a simplified representation of redox reactions by focusing on the substances that undergo a change in oxidation state. They exclude the spectator ions, which do not participate in the reaction. This simplification allows us to see the essential chemical changes without additional complexity.

Consider the reaction between cadmium and a nickel nitrate solution. The net ionic equation would show only the atoms that participate directly in the redox reaction:

• Oxidation half-reaction: Cd (s) → Cd²⁺ (aq) + 2e⁻
• Reduction half-reaction: Ni²⁺ (aq) + 2e⁻ → Ni (s)

Combining these gives us the net ionic equation:
Cd (s) + Ni²⁺ (aq) → Cd²⁺ (aq) + Ni (s).

This equation highlights the electron transfer and shows that cadmium is oxidized while nickel is reduced, with cadmium metal and nickel ions being the reactants and the products of the reaction, respectively.

Electrochemical Activity

Electrochemical activity is essentially a measure of how readily a metal can lose electrons, which relates to its position in the activity series. This series is a chart of metals listed in order of decreasing reactivity or their tendency to be oxidized.

In the context of our exercise, experiments involving cadmium metal and other metals like zinc and nickel help us understand electrochemical activity. For example, because zinc displaces cadmium in a solution, it is more electrochemically active. Conversely, cadmium displaces nickel, indicating that cadmium is more electrochemically active than nickel.

These observations help place cadmium within the activity series. To locate cadmium more precisely, experimentation with additional metal ions like those of copper (Cu²⁺), iron (Fe²⁺), and magnesium (Mg²⁺) is suggested. Through these experiments, we observe whether cadmium can displace or be displaced by these metals, which gives us a benchmark of its electrochemical activity relative to others in the series.