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Chapter 19: Problem 45

Zinc reacts spontaneously with silver ion. $$\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)$$ Describe a voltaic cell using this reaction. What are the half-reactions?

Short Answer

Expert verified
Oxidation: \( \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2e^- \); Reduction: \( 2\mathrm{Ag}^+(aq) + 2e^- \rightarrow 2\mathrm{Ag}(s) \).

Step by step solution

01

Identify the Overall Reaction

The given reaction is \( \mathrm{Zn}(s) + 2 \mathrm{Ag}^+(aq) \longrightarrow \mathrm{Zn}^{2+}(aq) + 2 \mathrm{Ag}(s) \). This is the complete redox reaction that represents the chemical changes taking place during the reaction.
02

Determine the Oxidation Half-Reaction

In the overall reaction, \( \mathrm{Zn}(s) \) is oxidized to \( \mathrm{Zn}^{2+}(aq) \). So, the oxidation half-reaction is \( \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2e^- \). This shows that zinc loses two electrons.
03

Determine the Reduction Half-Reaction

In the overall reaction, \( \mathrm{Ag}^+(aq) \) is reduced to \( \mathrm{Ag}(s) \). Therefore, the reduction half-reaction is \( \mathrm{Ag}^+(aq) + e^- \rightarrow \mathrm{Ag}(s) \). Since there are two \( \mathrm{Ag}^+ \) ions, this half-reaction should appear twice as \( 2 \mathrm{Ag}^+(aq) + 2e^- \rightarrow 2 \mathrm{Ag}(s) \).
04

Describe the Voltaic Cell Setup

In a voltaic cell for this reaction, zinc metal is used as the anode, where oxidation takes place, and silver ions are present in the solution at the cathode, where reduction occurs. The two half-cells are connected by a salt bridge, which allows ions to flow and maintain electrical neutrality, while electrons flow through an external circuit from the anode (zinc) to the cathode (silver).

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Key Concepts

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

Oxidation Half-Reaction
The concept of oxidation involves the loss of electrons from a substance. In a redox reaction, the substance that gets oxidized loses electrons, resulting in an increase in oxidation state. For the given reaction, zinc (Zn) undergoes oxidation.

The oxidation half-reaction can be written as:
  • \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \)
In this equation, solid zinc \( (\text{Zn}(s)) \) loses two electrons and transforms into a zinc ion \( (\text{Zn}^{2+}(aq)) \). The release of electrons signifies that zinc is oxidized. This is an essential aspect of how a voltaic cell functions, as it forms one-half of the overall cell reaction.
Reduction Half-Reaction
Reduction is the gain of electrons by a substance. It results in a decrease in the oxidation state of the substance gaining the electrons. In the given chemical reaction, silver ions \( (\text{Ag}^+(aq)) \) undergo reduction to form silver \( (\text{Ag}(s)) \).

The reduction half-reaction is defined as:
  • \( \text{2Ag}^+(aq) + 2e^- \rightarrow \text{2Ag}(s) \)
Here, the silver ions accept electrons, forming solid silver. As each silver ion receives one electron, making it essential to account for both silver ions in the equation. Together with the oxidation half-reaction, it completes the redox process that occurs in the voltaic cell.
Redox Reaction
Redox reactions are chemical processes involving both reduction and oxidation, hence the name 'redox,' which is a combination of the two terms. In any redox reaction, one reactant is oxidized while another is reduced.

For the given reaction:
  • \( \text{Zn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s) \)
The redox reaction combines the two half-reactions: one for oxidation (loss of electrons) and one for reduction (gain of electrons). The electrons lost by zinc are the same ones gained by the silver ions. The transfer of electrons from zinc to silver is what ultimately powers the voltaic cell, generating electrical energy.
Anode and Cathode
In a voltaic cell, the anode and cathode are the two electrodes where the half-reactions take place. Each has a distinct role in the cell's function.

  • **Anode**: The anode is the electrode where oxidation occurs. In the voltaic cell using zinc and silver, the anode is made of zinc. Here, zinc undergoes oxidation, releasing electrons that travel through the external circuit to do electric work.
  • **Cathode**: The cathode is the site of reduction. In this reaction, silver serves as the cathode. The silver ions gain electrons at the cathode, reducing to form solid silver.
The flow of electrons from the anode to the cathode through an external circuit generates electricity. The two electrodes are connected by a salt bridge that allows ions to flow between the two solutions, maintaining charge balance and completing the electrical circuit.

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Most popular questions from this chapter

A voltaic cell has an iron rod in \(0.30 M\) iron(III) chloride solution for the cathode and a zinc rod in \(0.20 \mathrm{M}\) zinc sulfate solution for the anode. The half-cells are connected by a salt bridge. Write the notation for this cell.

Iron may be protected by coating with tin (tin cans) or with zinc (galvanized iron). Galvanized iron does not corrode as long as zinc is present. By contrast, when a tin can is scratched, the exposed iron underneath corrodes rapidly. Explain the difference between zinc and tin as protective coatings against iron corrosion.

You place a battery in a flashlight in which all of the electrochemical reactions have reached equilibrium. What do you expect to observe when you turn on the flashlight? Explain your answer.

The voltaic cell is represented as $$\operatorname{Zn}(s)\left|\mathrm{Zn}^{2+}(1.0 M) \| \mathrm{Cu}^{2+}(1.0 M)\right| \mathrm{Cu}(s)$$ Which of the following statements is not true of this cell? a. The mass of the zinc electrode, \(\mathrm{Zn}(s),\) decreases as the cell runs. b. The copper electrode is the anode. c. Electrons flow through the external circuit from the zinc electrode to the copper electrode. d. Reduction occurs at the copper electrode as the cell runs. e. The concentration of \(\mathrm{Cu}^{2+}\) decreases as the cell runs.

Half-cells were made from a nickel rod dipping in a nickel sulfate solution and a silver rod dipping in a silver nitrate solution. The half-reactions in a voltaic cell using these half-cells were $$ \begin{aligned} \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow & \mathrm{Ag}(s) \\ \mathrm{Ni}(s) & \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{e}^{-} \end{aligned} $$ Sketch the cell and label the anode and cathode, showing the corresponding electrode reactions. Give the direction of electron flow and the movement of cations.
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