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Magazine Chapter 25: Problem 7
The cell diagram for an electrochemical cell is given as $$ \operatorname{In}(s)\left|\operatorname{In}\left(\mathrm{ClO}_{4}\right)_{3}(a q) \| \mathrm{CdCl}_{2}(a q)\right| \operatorname{Cd}(s) $$ Write the equations for the half reactions that occur at the two electrodes and the net cell reaction.
Short Answer
Expert verified
The net cell reaction is: \( 2\operatorname{In}(s) + 3\operatorname{Cd}^{2+}(aq) \rightarrow 2\operatorname{In}^{3+}(aq) + 3\operatorname{Cd}(s) \).
Step by step solution
01
Identify the Anode and Cathode
In the cell diagram, the anode is the phase on the left and the cathode is the phase on the right. Thus, In(s) is at the anode and Cd(s) is at the cathode. The anode is where oxidation occurs, and the cathode is where reduction occurs.
02
Write the Oxidation Half-Reaction
Oxidation occurs at the anode. The reaction is: \[ \operatorname{In}(s) \rightarrow \operatorname{In}^{3+}(aq) + 3e^- \] In this reaction, solid indium (In) is oxidized to indium(III) ions \(\operatorname{In}^{3+}\).
03
Write the Reduction Half-Reaction
Reduction occurs at the cathode. The reaction is: \[ \operatorname{Cd}^{2+}(aq) + 2e^- \rightarrow \operatorname{Cd}(s) \] In this reaction, cadmium ions (\(\operatorname{Cd}^{2+}\)) in the solution are reduced to form solid cadmium (Cd).
04
Balance the Electron Transfer
To balance the electron transfer between the half reactions, multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3:- Oxidation: \[ 2(\operatorname{In}(s) \rightarrow \operatorname{In}^{3+}(aq) + 3e^-) \]- Reduction: \[ 3(\operatorname{Cd}^{2+}(aq) + 2e^- \rightarrow \operatorname{Cd}(s)) \]
05
Combine to Write the Net Cell Reaction
Combine the balanced half reactions:\[2\operatorname{In}(s) + 3\operatorname{Cd}^{2+}(aq) \rightarrow 2\operatorname{In}^{3+}(aq) + 3\operatorname{Cd}(s) \]This is the overall balanced reaction for the cell.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Oxidation-Reduction
In the world of electrochemistry, oxidation-reduction reactions, also known as redox reactions, are fundamental processes that describe how electrons are transferred between substances. In an electrochemical cell, these two processes occur at different electrodes. Oxidation is defined as the process where a substance loses electrons. Thus, it increases its oxidation state. On the other hand, reduction is when a substance gains electrons, leading to a decrease in its oxidation state.
It's important to remember:
It's important to remember:
- Oxidation occurs at the anode (left side of the cell diagram).
- Reduction takes place at the cathode (right side of the cell diagram).
- The mnemonic "OIL RIG" can help: Oxidation Is Loss, Reduction Is Gain (of electrons).
Half-Reactions
The concept of half-reactions is crucial in electrochemistry. Each half-reaction represents the oxidation or reduction process separately. By writing these half-reactions, we can see clearly how electrons are transferred, which helps in understanding the entire redox reaction.
For instance, in the given cell diagram:
For instance, in the given cell diagram:
- The oxidation half-reaction involves solid indium turning into indium(III) ions and releasing electrons: e.g., \( \operatorname{In}(s) \rightarrow \operatorname{In}^{3+}(aq) + 3e^- \)
- In contrast, the reduction half-reaction involves cadmium ions gaining electrons to become solid cadmium: e.g., \( \operatorname{Cd}^{2+}(aq) + 2e^- \rightarrow \operatorname{Cd}(s) \)
Cell Diagrams
Cell diagrams are shorthand notations that represent electrochemical cells. They provide a quick snapshot of what's happening within the cell. Each diagram follows a specific format, with components separated by lines, emphasizing the reaction sequence. In a cell diagram:
- The anode half-cell (oxidation) is listed on the left.
- Double vertical lines (||) separate the two different solutions, representing the salt bridge or membrane.
- The cathode half-cell (reduction) is on the right side.
Electron Transfer
In any electrochemical cell, electron transfer is the heart of the process. Electrons flow from one electrode to another, driving the cell's operation. To maintain neutrality, electrons lost in the oxidation half-reaction must equal those gained in the reduction half-reaction. This is known as balancing the electron transfer.
When balancing:
When balancing:
- Ensure the number of electrons lost equals the number of electrons gained.
- Multiply the half-reactions by appropriate factors to achieve balance.
- Multiply the oxidation reaction by 2 to have 6 electrons in total: e.g., \( 2(\operatorname{In} \rightarrow \operatorname{In}^{3+} + 3e^-) \)
- Multiply the reduction reaction by 3 to balance with 6 electrons: e.g., \( 3(\operatorname{Cd}^{2+} + 2e^- \rightarrow \operatorname{Cd}) \)
Electrochemistry Concepts
Electrochemistry is the study of chemical processes that cause electrons to move, leading to the generation of electricity. It forms the basis for understanding batteries, fuel cells, and more. Key concepts include the direction of electron flow, the purpose of the salt bridge, and the significance of electrode reactions.
Major points to consider:
Major points to consider:
- Electrons always flow from the anode to the cathode.
- Salt bridges maintain electrical neutrality by allowing ions to migrate between two half-cells.
- The energy produced in electrochemical cells comes from the potential difference between electrodes.
