Question

The metal cadmium tends to form Cd ${}^{2+}$ ions. The following observations are made: (i) When a strip of zinc metal is placed in ${\mathrm{CdCl}}_2(aq),$ cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in $\mathrm{Ni}{\left({\mathrm{NO}}_3\right)}_2(aq),$ 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?

Short answer

In summary, the net ionic equation for Observation (i) is: $C{d}^{2+}(aq)+Zn(s)\to Cd(s)+Z{n}^{2+}(aq)$, and for Observation (ii) is: $N{i}^{2+}(aq)+Cd(s)\to Ni(s)+C{d}^{2+}(aq)$. Based on these observations, we can determine that zinc is more reactive than cadmium, and cadmium is more reactive than nickel; i.e., $Zn>Cd>Ni$. To locate cadmium's precise position in the activity series, we need to perform experiments with other metal salts, such as MgSO4, CuSO4, Fe(NO3)3, and AlCl3, and the proper strips or by placing metal strips in a CdSO4 solution. This will help us better understand and accurately determine the position of cadmium in the activity series.

Step-by-step solution

Observation (i) Net Ionic Equation

Zinc (Zn) is placed in a solution with cadmium chloride (CdCl2). As a result, cadmium metal is deposited on the zinc strip. This implies that zinc displaces cadmium from its compound. It means that zinc is more reactive than cadmium, and thus it can reduce cadmium ions into cadmium metal. The reduction half equation for this reaction would be: $$C{d}^{2+}(aq)+2e^{-}\to Cd(s)$$ Meanwhile, Zn is being oxidized, losing two electrons to become zinc ions. So, the oxidation half-equation can be written as: $$Zn(s)\to Z{n}^{2+}(aq)+2e^{-}$$ Putting both half-equations together, eliminating the electrons, and balancing the equation, we get the net ionic equation: $$C{d}^{2+}(aq)+Zn(s)\to Cd(s)+Z{n}^{2+}(aq)$$

Observation (ii) Net Ionic Equation

A strip of cadmium metal (Cd) is placed in a solution of nickel nitrate (Ni(NO3)2). Here, nickel metal is deposited on the strip. This means that cadmium displaces nickel ions from its compound, suggesting that cadmium is more reactive than nickel. Cd is oxidized and changes into Cd2+ ions, releasing 2 electrons in the process. The half equation can be written as: $$Cd(s)\to C{d}^{2+}(aq)+2e^{-}$$ Simultaneously, Ni2+ ions accept 2 electrons and turn into Ni (s). The reduction half equation is: $$N{i}^{2+}(aq)+2e^{-}\to Ni(s)$$ Once again, putting both half-equations together, eliminating the electrons, and balancing the equation, we get the net ionic equation: $$N{i}^{2+}(aq)+Cd(s)\to Ni(s)+C{d}^{2+}(aq)$$

Position of Cadmium in Activity Series

Based on these two observations, we can conclude that zinc is more reactive than cadmium, and cadmium is more reactive than nickel. Thus, the activity series can be partially ordered as: $$Zn>Cd>Ni$$ According to the exercise, we should locate which elements more closely define the position of cadmium in the activity series. From the given information, cadmium's position is between zinc and nickel.

Experiments to Locate Cadmium's Position Precisely in the Activity Series

To locate the position of cadmium more precisely in the activity series, we should test it against other elements to see if it can displace their ions or vice versa. Below are some experiments that could be conducted: 1. Place a cadmium strip in the solutions of various metal salts, such as magnesium sulfate (MgSO4), copper sulfate (CuSO4), iron nitrate (Fe(NO3)3), and aluminum chloride (AlCl3). If cadmium displaces the metal ions, it means that it is more reactive than those particular metals. If not, it means that it is less reactive. 2. Conversely, place strips of these metals (Mg, Cu, Fe, Al) in a cadmium salt solution, like cadmium sulfate (CdSO4), and observe whether they can displace cadmium ions or not. This will provide further confirmation regarding the relative reactivity of cadmium compared to these metals. 3. Based on the results obtained, we can determine the position of cadmium more accurately in the activity series.

Key concepts

Redox Reactions

Redox reactions are chemical reactions that involve the transfer of electrons between two species. The name "redox" comes from two key concepts: reduction and oxidation. Reduction refers to the gain of electrons, while oxidation is the loss of electrons. These processes always occur simultaneously because when one species gains electrons, another must lose them.

In the context of the given exercise, we observe a redox reaction involving zinc and cadmium in the first observation, and cadmium and nickel in the second. Here's what happens:

• The zinc metal donates electrons, undergoing oxidation to become zinc ions: $$Zn(s)\to Z{n}^{2+}(aq)+2e^{-}$$
• Consequently, the cadmium ions gain these electrons, undergoing reduction to become cadmium metal: $$C{d}^{2+}(aq)+2e^{-}\to Cd(s)$$

By combining these half-reactions, we obtain the overall redox reaction, showing the interplay of electron transfer. The electrons are neither destroyed nor created, just transferred from one entity to another. Understanding this balance is crucial for mastering redox reactions, as it highlights the conservation of charge and matter in chemical reactions.

Net Ionic Equations

Net ionic equations are a simplified way of expressing chemical reactions, showing only those species that are directly involved in the chemical change. They are essential for understanding the essence of the reaction without the distractions of spectator ions which do not participate directly.

Consider the first observation with zinc and cadmium. The net ionic equation is expressed as:$$C{d}^{2+}(aq)+Zn(s)\to Cd(s)+Z{n}^{2+}(aq)$$

This succinct equation highlights that cadmium ions are reduced to solid cadmium, while zinc is oxidized to zinc ions. Similarly, for the nickel and cadmium interaction:$$N{i}^{2+}(aq)+Cd(s)\to Ni(s)+C{d}^{2+}(aq)$$

In each equation, we focus only on the ions and compounds undergoing a change. This approach offers a clearer view of which species are reacting and what their transformations are. It’s a straightforward method to visually capture the essence of redox processes, making complex reactions easier to analyze.

Reactivity of Metals

The reactivity of metals is determined by their ability to displace other metals from their compounds. This concept is crucial for establishing the activity series, which ranks metals based on their reactivity.

• A more reactive metal can displace a less reactive metal from its compound.
• In a displacement reaction, the metal that ends up in compound form is less reactive.

From the exercise, we learned that:

• Zinc can displace cadmium from cadmium chloride, proving zinc is more reactive than cadmium.
• Cadmium can displace nickel from nickel nitrate, indicating it's more reactive than nickel.

This positions cadmium between zinc and nickel in the activity series. However, discovering the exact placement of cadmium relative to other metals involves further experimentation. By conducting tests with a range of metal ions, we can determine if cadmium can displace metals like magnesium, copper, or iron. Such experiments refine our understanding of where cadmium fits in the grand reactivity scale, providing insight into its chemical behavior across different contexts.