Problem 2 - Chemistry: The Central Science

Question

Consider the reaction in Figure 20.3. Describe what would happen if (a) the solution contained cadmium(II) sulfate and the metal was zinc, (b) the solution contained silver nitrate and the metal was copper. [Section 20.3]

Step-by-Step Solution

Answer

In reaction (a), zinc will dissolve into the solution as Zn²⁺ ions while cadmium ions will be reduced and plate onto the zinc electrode as metallic cadmium. In reaction (b), copper will dissolve into the solution as Cu²⁺ ions while silver ions will be reduced and plate onto the copper electrode as metallic silver.

Step 1: Reaction (a): Cadmium(II) sulfate solution with a zinc metal electrode

Step 1: Write the half-cell reactions for both the metal ions in the solution Cadmium(II) sulfate: \[Cd^{2+} + 2e^- \rightarrow Cd\] Zinc: \[Zn^{2+} + 2e^- \rightarrow Zn\] Step 2: Check the standard electrode potentials (E°) of both half-cell reactions to determine spontaneity Standard electrode potentials (E°) values: \(E°_{Zn}\) = -0.76 V \(E°_{Cd}\) = -0.40 V Step 3: Determine the spontaneous reaction Since the value of \(E°_{Zn}\) is lower than that of \(E°_{Cd}\), the zinc half-cell reaction proceeds spontaneously in the reverse direction, acting as the anode. Zinc will oxidize to form Zn²⁺ ions and release electrons while cadmium ions in the solution will be reduced in the cathode: Anode (oxidation): \[Zn \rightarrow Zn^{2+} + 2e^-\] Cathode (reduction): \[Cd^{2+} + 2e^- \rightarrow Cd\] Step 4: Describe the outcome Zinc metal will dissolve into the solution as Zn²⁺ ions, while the cadmium ions in the solution will be reduced and will plate onto the zinc electrode as metallic cadmium. The concentration of cadmium ions will decrease, and the concentration of zinc ions will increase in the solution.

Step 2: Reaction (b): Silver nitrate solution with a copper metal electrode

Step 1: Write the half-cell reactions for both the metal ions in the solution Silver nitrate: \[Ag^+ + e^- \rightarrow Ag\] Copper: \[Cu^{2+} + 2e^- \rightarrow Cu\] Step 2: Check the standard electrode potentials (E°) of both half-cell reactions to determine spontaneity Standard electrode potentials (E°) values: \(E°_{Cu}\) = +0.34 V \(E°_{Ag}\) = +0.80 V Step 3: Determine the spontaneous reaction Since the value of \(E°_{Cu}\) is lower than that of \(E°_{Ag}\), the copper half-cell reaction proceeds spontaneously in the reverse direction, acting as the anode. Copper will oxidize to form Cu²⁺ ions and release electrons while silver ions in the solution will be reduced in the cathode: Anode (oxidation): \[Cu \rightarrow Cu^{2+} + 2e^-\] Cathode (reduction): \[Ag^+ + e^- \rightarrow Ag\] Step 4: Describe the outcome Copper metal will dissolve into the solution as Cu²⁺ ions, while silver ions in the solution will be reduced and will plate onto the copper electrode as metallic silver. The concentration of silver ions will decrease, and the concentration of copper ions will increase in the solution.

Key Concepts

Redox Reactions

Redox reactions are chemical processes where oxidation and reduction occur simultaneously. In these reactions, electrons are transferred between two substances. Oxidation involves the loss of electrons by a molecule, atom, or ion, while reduction is the gain of electrons. These reactions are fundamental to understanding electrochemistry since they form the basis of how batteries and electrochemical cells operate.
For example, when cadmium(II) sulfate reacts with a zinc metal electrode, zinc undergoes oxidation by losing electrons and turning into zinc ions (\(Zn^{2+}\) ions). Meanwhile, cadmium ions (\(Cd^{2+}\)) gain electrons, becoming cadmium metal. This transfer of electrons allows the electrochemical cell to generate electrical energy from a chemical reaction.

Standard Electrode Potentials

Standard electrode potentials (\(E^°\)) are a measure of the intrinsic tendency of a chemical species to lose or gain electrons. This potential is measured against a reference electrode under standard conditions (1 molar concentration, 25°C, and 1 atm pressure).
The electrode potential values help predict which direction a redox reaction will proceed. In the reaction involving zinc and cadmium, the standard electrode potential for zinc (\(E°_{Zn} = -0.76 \, \text{V}\)) is more negative compared to cadmium (\(E°_{Cd} = -0.40 \, \text{V}\)). Therefore, zinc is more likely to lose electrons (oxidize) compared to cadmium, making zinc the anode.
Conversely, in a copper and silver nitrate reaction, the value for copper (\(E°_{Cu} = +0.34 \, \text{V}\)) is lower than silver (\(E°_{Ag} = +0.80 \, \text{V}\)). Thus, copper oxidizes (acts as the anode), and silver ions get reduced at the cathode.

• Negative potential: higher tendency to oxidize.
• Positive potential: higher tendency to reduce.

Understanding these potentials is crucial for predicting the feasibility and direction of redox reactions.

Metal Ion Solutions

Metal ion solutions contain dissolved metal ions that engage in redox reactions during electrochemical processes. When you submerge a metal in an electrolyte solution containing its ions, a dynamic equilibrium is established. This involves the metal dissolving into the solution as ions and the ions depositing back onto the metal surface.
For instance, with zinc dipped into cadmium(II) sulfate, zinc metal becomes ions (\(Zn^{2+}\)) in the solution, and the cadmium ions present are reduced to form cadmium metal. Similarly, when copper is in silver nitrate, copper becomes \(Cu^{2+}\) ions in the solution, while \(Ag^+\) ions are reduced to silver metal.
Metal ion solutions are integral to the functioning of electrochemical cells, dictating which metal will undergo oxidation or reduction based on ion concentration and standard electrode potentials. Recognizing these dynamics helps in understanding processes like electroplating, corrosion, and energy storage in batteries.