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

(a) The net ionic equations 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) In the activity series, cadmium is located between zinc and nickel. (c) To locate the position of cadmium more precisely, perform experiments with other metals in the activity series and observe displacements. This would involve testing the reactivity of cadmium towards a variety of metallic ions, such as Cu^2+, Ag^+, Pb^2+, and Mg^2+.

Step 1: (Step 1: Identify the reactions involved in each observation)

In order to write the net ionic equations, we have to identify the reactions occurring in these observations based on given information: Observation (i): Zinc strip is placed in CdCl2 (aq) Observation (ii): Cadmium strip is placed in Ni (NO3)2 (aq)

Step 2: (Step 2: Write the ionic equations for each reaction)

Now, we'll write the ionic equations for each reaction: For Observation (i): Zn (s) + Cd^2+ (aq) → Zn^2+ (aq) + Cd (s) For Observation (ii): Cd (s) + Ni^2+ (aq) → Cd^2+ (aq) + Ni (s)

Step 3: (Step 3: Answer part a - Write the net ionic equations)

Based on step 2, we can write the net ionic equations for each reaction since there are no spectator ions involved. Net ionic equations are the same as the ionic equations in this case: (i) Zn (s) + Cd^2+ (aq) → Zn^2+ (aq) + Cd (s) (ii) Cd (s) + Ni^2+ (aq) → Cd^2+ (aq) + Ni (s)

Step 4: (Step 4: Answer part b - Locate the position of cadmium in the activity series)

Activity series is a list of elements organized by their ability to undergo a redox reaction. More active metals readily lose electrons and form cations, whereas less active metals form cations with difficulty. Based on the reactions above: (i) Zn is more active than Cd, as Zn can displace Cd^2+ ions. (ii) Cd is more active than Ni, as Cd can displace Ni^2+ ions. So, in the activity series, cadmium is located between zinc and nickel.

Step 5: (Step 5: Answer part c - Experiments to locate the position of Cd more precisely)

To locate the position of cadmium more precisely, we can perform experiments with other metals in the activity series. We could add cadmium to solutions containing ions of other metals in the series (both more active and less active than zinc and nickel) and observe any possible displacements. This would require testing the reactivity of cadmium towards a variety of metallic ions, such as Cu^2+, Ag^+, Pb^2+, and Mg^2+. By observing whether cadmium can successfully displace these ions, we can further identify its position in the activity series with better accuracy.

Key Concepts

Redox Reactions

A redox reaction is a chemical reaction where the oxidation state of atoms are changed through the transfer of electrons. Redox is short for reduction-oxidation, where:

• Reduction involves the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.
• Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion.

In any redox reaction, one element is oxidized (loses electrons), and another is reduced (gains those electrons). This complementary process ensures that the number of electrons lost is equal to the number of electrons gained, maintaining charge balance.
An example can be found in the exercise with zinc and cadmium. Zinc undergoes oxidation:

• Oxidation: \( ext{Zn}(s) \rightarrow ext{Zn}^{2+}(aq) + 2e^- \)

and cadmium undergoes reduction:

• Reduction: \( ext{Cd}^{2+}(aq) + 2e^- \rightarrow ext{Cd}(s) \)

In these reactions, electrons are transferred from zinc to cadmium, allowing the formation of elemental cadmium as zinc forms its cationically charged state. This exchange highlights the fundamental nature of redox reactions.

Net Ionic Equations

Net ionic equations are used to express the essentials of a chemical reaction without showing all dissociated ions as we do in complete ionic equations. They focus on the chemical species that actually change or participate in the reaction. The key steps to write a net ionic equation include:

• Start by writing a balanced molecular equation for the reaction.
• Write the complete ionic equation, showing all strong electrolytes as ions.
• Identify and cancel out spectator ions, which are ions that appear on both sides of the equation but do not participate in the reaction.

In the given exercise, even though no spectator ions are present, the process can be applied as such. Let's illustrate this with the example of the zinc and cadmium reaction:

• The molecular and ionic equation are identical: \( ext{Zn}(s) + ext{Cd}^{2+}(aq) \rightarrow ext{Zn}^{2+}(aq) + ext{Cd}(s) \)

Here we simply keep track of the substances that undergo transformation, omitting surplus ions. Understandably, net ionic equations help clarify the true nature of a chemical reaction by focusing only on substances that change their state.

Metal Displacement Reactions

Metal displacement reactions are a type of redox reaction where a metal in a compound is replaced by a more active metal. This activity is often determined by the metal's position in the activity series, with more active metals capable of displacing less active metals.These reactions rely heavily on the principles of the activity series, which ranks elements based on their reactivity. In our specific exercise:

• Zinc displaces cadmium from cadmium chloride solution: \( ext{Zn}(s) + ext{Cd}^{2+}(aq) \rightarrow ext{Zn}^{2+}(aq) + ext{Cd}(s) \)
• Cadmium displaces nickel from nickel nitrate solution: \( ext{Cd}(s) + ext{Ni}^{2+}(aq) \rightarrow ext{Cd}^{2+}(aq) + ext{Ni}(s) \)

From these observations, it is clear that zinc is more reactive than cadmium, and cadmium is more reactive than nickel. Therefore, cadmium will appear in between zinc and nickel in the activity series. The ability to undergo such displacement reactions helps in understanding the comparative reactivity of metals, allowing chemists to align them properly in the activity series. Metal displacement reactions showcase the principle of competitive electron stealing, where one metal will forego its electrons to a more favorable (active) agent.