Problem 1
Question
A student measures the potential of a cell made up with 1 M CuSO, in one solution reservoir and 1 M \(\mathrm{ZnSO}_{4}\) in the other. There is a metallic copper \(\left(\mathrm{Cu}^{0}\right)\) electrode in the \(\mathrm{CuSO}_{4}\) and a metallic zinc (Zn \(^{0}\) ) electrode in the ZnSO_, and the cell is set up as shown in Figure \(32.1 .\) She finds that the potential, or voltage, of the cell, \(E_{\text {call }}^{0}\) is \(1.076 \mathrm{V}\), and that the Zn electrode is negative. a. At which electrode is oxidation occurring? _____ b. Write the equation for the oxidation half-reaction in this cell. c. Write the equation for the reduction half-reaction in this cell. d. Write the net ionic equation for the spontaneous oxidation-reduction reaction that occurs in this cell.
Step-by-Step Solution
Verified Answer
Oxidation occurs at the Zn electrode. Oxidation half-reaction: \( \text{Zn} (s) \rightarrow \text{Zn}^{2+} (aq) + 2e^- \). Reduction half-reaction: \( \text{Cu}^{2+} (aq) + 2e^- \rightarrow \text{Cu} (s) \). Net ionic equation: \( \text{Zn} (s) + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu} (s) \).
1Step 1: Determine the Oxidation Process
Oxidation occurs at the anode, which is where the loss of electrons occurs. Given that the Zn electrode is negative, it indicates electrons are being released from the Zn electrode. Therefore, oxidation is happening at the Zn electrode.
2Step 2: Write the Oxidation Half-Reaction
In the zinc electrode (anode), the oxidation process can be expressed as: \[ \text{Zn} (s) \rightarrow \text{Zn}^{2+} (aq) + 2e^- \] This shows that solid zinc is losing electrons to form zinc ions.
3Step 3: Write the Reduction Half-Reaction
Reduction takes place at the cathode, where electrons are gained. In this cell, the cathode is the copper electrode. The reduction half-reaction is: \[ \text{Cu}^{2+} (aq) + 2e^- \rightarrow \text{Cu} (s) \] This indicates that copper ions gain electrons to become solid copper.
4Step 4: Write the Net Ionic Equation
Combine the oxidation and reduction half-reactions to find the net ionic equation. First, balance the electron transfer to ensure equal numbers of electrons are lost and gained: \[ \text{Zn} (s) + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu} (s) \] This is the net ionic equation for the reaction occurring in the cell.
Key Concepts
Oxidation-Reduction ReactionsOxidation Half-ReactionReduction Half-ReactionNet Ionic Equation
Oxidation-Reduction Reactions
Oxidation-reduction reactions, often referred to as redox reactions, are chemical processes where oxidation and reduction occur simultaneously. Understanding these reactions is essential for explaining how electrochemical cells work. In every redox reaction:
In the context of an electrochemical cell, these reactions generate electrical energy. Electrodes play a crucial role in facilitating these reactions:
Electrons move from the anode to the cathode, creating a flow of electric current. In our example with copper and zinc, zinc serves as the anode and copper as the cathode. This flow of electrons provides the cell voltage, which powers electronic devices.
- Oxidation involves the loss of electrons.
- Reduction means gaining electrons.
In the context of an electrochemical cell, these reactions generate electrical energy. Electrodes play a crucial role in facilitating these reactions:
- The anode is where oxidation happens.
- The cathode is where reduction takes place.
Electrons move from the anode to the cathode, creating a flow of electric current. In our example with copper and zinc, zinc serves as the anode and copper as the cathode. This flow of electrons provides the cell voltage, which powers electronic devices.
Oxidation Half-Reaction
An oxidation half-reaction shows how oxidation occurs at the molecular level. It focuses on the substance that loses electrons. In our cell example:
The half-reaction is written as:\[ \text{Zn} (s) \rightarrow \text{Zn}^{2+} (aq) + 2e^- \]
This equation demonstrates that solid zinc loses two electrons, transforming into zinc ions in solution. The lost electrons travel through an external circuit, which is essential for powering devices. Recognizing the oxidation process helps readers understand how metals can lose stability through losing electrons and turn into ionic forms.
- The zinc electrode is responsible for oxidation.
The half-reaction is written as:\[ \text{Zn} (s) \rightarrow \text{Zn}^{2+} (aq) + 2e^- \]
This equation demonstrates that solid zinc loses two electrons, transforming into zinc ions in solution. The lost electrons travel through an external circuit, which is essential for powering devices. Recognizing the oxidation process helps readers understand how metals can lose stability through losing electrons and turn into ionic forms.
Reduction Half-Reaction
The reduction half-reaction complements the oxidation process in a redox reaction. It details how a substance gains electrons. Within our cell example, the following occurs:
The reaction is expressed as:\[ \text{Cu}^{2+} (aq) + 2e^- \rightarrow \text{Cu} (s) \]
This means that copper ions in solution accept two electrons to form solid copper. These acquired electrons, from the zinc electrode, allow copper to regain its natural metallic state. Understanding reduction is vital for grasping how certain elements can stabilize by electron acceptance, completing the cycle of electric flow in the cell.
- Copper ions undergo reduction at the copper electrode.
The reaction is expressed as:\[ \text{Cu}^{2+} (aq) + 2e^- \rightarrow \text{Cu} (s) \]
This means that copper ions in solution accept two electrons to form solid copper. These acquired electrons, from the zinc electrode, allow copper to regain its natural metallic state. Understanding reduction is vital for grasping how certain elements can stabilize by electron acceptance, completing the cycle of electric flow in the cell.
Net Ionic Equation
The net ionic equation encapsulates the entirety of redox reactions, highlighting the reaction that actually occurs without spectator ions. It combines balanced oxidation and reduction half-reactions to showcase the full electron transfer.
In our cell example, the net ionic equation is:\[ \text{Zn} (s) + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu} (s) \]
This equation is important because it:
Understanding the net ionic equation enhances comprehension of overall cell reactions, linking conceptually to the conversion of chemical energy into electrical energy.
In our cell example, the net ionic equation is:\[ \text{Zn} (s) + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu} (s) \]
This equation is important because it:
- Illustrates the chemical change taking place.
- Keeps focus on key reacting substances without unnecessary components.
Understanding the net ionic equation enhances comprehension of overall cell reactions, linking conceptually to the conversion of chemical energy into electrical energy.