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Chapter 20: Problem 2

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]

Short Answer

Expert verified
In reaction (a), zinc will dissolve into the solution as Zn虏鈦� 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 Cu虏鈦� ions while silver ions will be reduced and plate onto the copper electrode as metallic silver.

Step by step solution

01

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掳) of both half-cell reactions to determine spontaneity Standard electrode potentials (E掳) values: \(E掳_{Zn}\) = -0.76 V \(E掳_{Cd}\) = -0.40 V Step 3: Determine the spontaneous reaction Since the value of \(E掳_{Zn}\) is lower than that of \(E掳_{Cd}\), the zinc half-cell reaction proceeds spontaneously in the reverse direction, acting as the anode. Zinc will oxidize to form Zn虏鈦� 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 Zn虏鈦� 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.
02

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掳) of both half-cell reactions to determine spontaneity Standard electrode potentials (E掳) values: \(E掳_{Cu}\) = +0.34 V \(E掳_{Ag}\) = +0.80 V Step 3: Determine the spontaneous reaction Since the value of \(E掳_{Cu}\) is lower than that of \(E掳_{Ag}\), the copper half-cell reaction proceeds spontaneously in the reverse direction, acting as the anode. Copper will oxidize to form Cu虏鈦� 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 Cu虏鈦� 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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (\(E^掳\)) 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掳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 (\(E掳_{Zn} = -0.76 \, \text{V}\)) is more negative compared to cadmium (\(E掳_{Cd} = -0.40 \, \text{V}\)). 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 (\(E掳_{Cu} = +0.34 \, \text{V}\)) is lower than silver (\(E掳_{Ag} = +0.80 \, \text{V}\)). 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.

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Most popular questions from this chapter

(a) What is a standard reduction potential? (b) What is the standard reduction potential of a standard hydrogen electrode?

A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s)--\rightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode compartment is \(1.00 M\) and the cell generates an emf of \(+0.22 \mathrm{~V}\), what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode compartment? (b) If the anode compartment contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 M\) in equilibrium with \(\mathrm{PbSO}_{4}(s)\), what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

A voltaic cell is based on the reaction $$ \mathrm{Sn}(s)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{Sn}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$ Under standard conditions, what is the maximum electrical work, in joules, that the cell can accomplish if \(75.0 \mathrm{~g}\) of \(\mathrm{Sn}\) is consumed?

(a) What is an electrolytic cell? (b) The negative terminal of a voltage source is connected to an electrode of an electrolytic cell. Is the electrode the anode or the cathode of the cell? Explain. (c) The electrolysis of water is often done with a small amount of sulfuric acid added to the water. What is the role of the sulfuric acid?

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \mathrm{AgCl}(s)+\mathrm{e}^{-\longrightarrow} \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two cell compartments have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=2.55 M\), respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cl \(^{-}\) ] will increase, decrease, or stay the same as the cell operates.
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