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 observations made above. (b) What can you conclude about 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?

Short answer

(a) The net ionic equations for the given observations are: (i) \(Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s)\) (ii) \(Cd(s) + Ni^{2+}(aq) \rightarrow Cd^{2+}(aq) + Ni(s)\) (b) Cadmium is less active than Zinc and more active than Nickel in the activity series. (c) To locate Cadmium's position more precisely, perform experiments with other metals and their salts to see if cadmium can displace them or be displaced by them, such as Cu, Fe, or Mg. Compare the results with the known activity series for metals.

Step-by-step solution

Observation (i):

A strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q)\), and cadmium metal is deposited on the strip. In this reaction, Zinc is displacing cadmium from its salt \(\mathrm{CdCl}_{2}\). Since only the metal ions are changing during the reaction, we can write the net ionic equation as follows: \[Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s)\]

Observation (ii):

A strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\), and nickel metal is deposited on the strip. In this reaction, Cadmium is displacing nickel from its salt \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\). Similar to the previous reaction, only the metal ions are changing, and we can write the net ionic equation as follows: \[Cd(s) + Ni^{2+}(aq) \rightarrow Cd^{2+}(aq) + Ni(s)\] (b) What can you conclude about the position of cadmium in the activity series?

Conclusion about Cadmium's position in the Activity Series:

Since Zinc is displacing Cadmium (from Observation (i)) and Cadmium is displacing Nickel (from Observation (ii)), we can conclude that the position of cadmium in the activity series is between zinc and nickel. In other words, Cadmium is less active than Zinc, but more active than Nickel. (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Propose Experiments:

To locate Cadmium's position in the activity series more precisely, we can perform experiments with other metals and their salts to see if cadmium can displace them or be displaced by them. Some of these could include: 1. Place a strip of cadmium metal in aqueous solutions of salts containing other metals such as copper, iron, magnesium, and see if cadmium displaces any of these metals from their salts. 2. Place strips of various metals (such as copper, iron, magnesium) in \(\mathrm{CdCl}_{2}(a q)\) and observe if any of these metals displace cadmium from its salt. By comparing the results of these experiments with the known activity series for metals, we can refine our understanding of Cadmium's position in the activity series.

Key concepts

Net Ionic Equations

Understanding net ionic equations is crucial when studying chemical reactions. These equations simplify reactions by highlighting only the species that actually participate in the reaction—the ions that change. By stripping away spectator ions, or ions that do not partake in the chemical change, we can focus on the heart of the chemical process.

For instance, in the provided example where zinc displaces cadmium from cadmium chloride, the net ionic equation is \[Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s)\]. Here, the chloride ions (\rCl^{-}) are spectator ions and do not appear in the net ionic equation. This simplification allows students to clearly see which elements are undergoing a change in oxidation state, hence capturing the essence of the displacement reaction.

Displacement Reactions

Displacement reactions involve an element displacing another element from a compound, usually through the process of oxidation and reduction. These are often visualized in chemistry when a more reactive metal can displace a less reactive metal from its compound. The activity series of metals helps predict whether a displacement reaction will occur.

In a displacement reaction, a metal in solid form will typically displace a metal ion in solution, resulting in a metal ion and solid metal product, respectively. For example, in our textbook problem, when cadmium is placed in a nickel nitrate solution, the following displacement reaction occurs: \[Cd(s) + Ni^{2+}(aq) \rightarrow Cd^{2+}(aq) + Ni(s)\]. The ability of cadmium to displace nickel indicates its relative reactivity, which is an important concept for students to grasp, as it underpins much of the predictive power in inorganic chemistry.

Experimental Determination of Reactivity

To better understand the experimental determination of reactivity and the activity series position, consider performing controlled reactions between metals and observing any evidence of displacement. This practical approach gives insights into the relative reactivity of metals, allowing us to arrange them in an activity series.

The reactivity of cadmium, for instance, can be discerned through experiments as suggested in the textbook exercise. By observing whether cadmium can displace or be displaced by other metals, we gather empirical evidence of its position in the series. You might see cadmium displacing copper from copper sulfate, but fail to displace magnesium from magnesium chloride, suggesting that cadmium is more reactive than copper but less so than magnesium.

These experimental observations provide the basis for constructing a more accurate activity series, which is an essential reference for predicting the course of displacement reactions.