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Magazine Chapter 12: Problem 14
Write the equivalent voltaic cells for the following reactions (assume all concentrations are \(1 M\) ): (a) \(6 \mathrm{Fe}^{2+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+} \rightleftharpoons 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{IO}_{3}^{-}+5 \mathrm{I}^{-}+6 \mathrm{H}^{+} \rightleftharpoons 3 \mathrm{I}_{2}+3 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{Zn}+\mathrm{Cu}^{2+} \rightleftharpoons \mathrm{Zn}^{2+}+\mathrm{Cu}\) (d) \(\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{SeO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{Cl}^{-}+\mathrm{SeO}_{4}^{2-}+4 \mathrm{H}^{+}\)
Short Answer
Expert verified
Each equation broken into oxidation and reduction half-reactions forms different voltaic cells.
Step by step solution
01
Determine Half-Reactions for (a)
Identify and separate the oxidation and reduction half-reactions for the given equation (a):- **Oxidation Half-Reaction:** \[ 6 \mathrm{Fe}^{2+} \rightarrow 6 \mathrm{Fe}^{3+} + 6e^- \]- **Reduction Half-Reaction:** \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]Combine these to identify the process occurring in a voltaic cell.
02
Write the Voltaic Cell for (a)
Using the half-reactions, construct the corresponding voltaic cell:- **Anode (Oxidation):** \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^- \)- **Cathode (Reduction):** \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \)- Overall reaction reflects the flow of electrons from anode to cathode.
03
Determine Half-Reactions for (b)
Identify and separate the half-reactions for reaction (b):- **Oxidation Half-Reaction:** \[ 5\mathrm{I}^{-} \rightarrow \frac{5}{3}\mathrm{I}_2 + \frac{10}{3}e^- \]- **Reduction Half-Reaction:** \[ \mathrm{IO}_{3}^{-} + 6\mathrm{H}^{+} + 6e^- \rightarrow 3\mathrm{I}_2 + 3\mathrm{H}_2\mathrm{O} \]Write these balanced reactions to formulate the corresponding voltaic cell.
04
Write the Voltaic Cell for (b)
Construct the voltaic cell using the half-reactions identified:- **Anode (Oxidation):** \( \mathrm{I}^{-} \rightarrow \mathrm{I}_2 + e^- \)- **Cathode (Reduction):** \( \mathrm{IO}_{3}^{-} + 6\mathrm{H}^{+} + 6e^- \rightarrow 3\mathrm{I}_2 + 3\mathrm{H}_2\mathrm{O} \)This reflects the process in a voltaic cell as the electrons flow from anode to cathode.
05
Determine Half-Reactions for (c)
Identify and separate the oxidation and reduction half-reactions for the given equation (c):- **Oxidation Half-Reaction:** \[ \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \]- **Reduction Half-Reaction:** \[ \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu} \]Use these half-reactions to describe the activity in a voltaic cell.
06
Write the Voltaic Cell for (c)
Write the equivalent voltaic cell setup:- **Anode (Oxidation):** \( \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \)- **Cathode (Reduction):** \( \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu} \)The reactions show electron movement from the zinc anode to the copper cathode.
07
Determine Half-Reactions for (d)
Identify and separate the half-reactions for reaction (d):- **Oxidation Half-Reaction:** \[ \mathrm{H}_{2}\mathrm{SeO}_3 \rightarrow \mathrm{SeO}_{4}^{2-} + 2e^- + 2\mathrm{H}^{+} + \mathrm{H}_2\mathrm{O} \]- **Reduction Half-Reaction:** \[ \mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^{-} \]These half-reactions complete the setup for a voltaic cell.
08
Write the Voltaic Cell for (d)
Construct the voltaic cell using the identified half-reactions:- **Anode (Oxidation):** \( \mathrm{H}_{2}\mathrm{SeO}_3 \rightarrow \mathrm{SeO}_{4}^{2-} + 2e^- \)- **Cathode (Reduction):** \( \mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^{-} \)Illustrates electron flow from anode to cathode in the voltaic cell.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Voltaic Cells
Voltaic cells, also known as galvanic cells, are devices that convert chemical energy into electrical energy through redox reactions. These cells are fundamental in the study of electrochemistry. In a voltaic cell, two different metals or metal ions are involved in two distinct half-reactions. Each half-reaction occurs in its own half-cell and is connected by a wire allowing electrons to flow between them.
The main components of a voltaic cell are:
The main components of a voltaic cell are:
- An anode, where oxidation occurs, releasing electrons.
- A cathode, where reduction occurs, gaining electrons.
- An electrolyte that allows ions to move, completing the circuit and balancing the charge flow.
- A salt bridge or porous disk which separates the half-cells to prevent the direct mixing of different solutions while maintaining electrical neutrality by allowing ions to pass.
Half-Reactions
In the construction of voltaic cells, understanding half-reactions is crucial. A half-reaction is a representation of either the oxidation part or the reduction part of a redox reaction.
To identify half-reactions, one must:
To identify half-reactions, one must:
- Divide the full chemical reaction into two parts, one showing loss of electrons (oxidation) and the other showing gain of electrons (reduction).
- Each half-reaction must be balanced separately by adjusting the coefficients to ensure mass and charge are conserved.
- Oxidation: \( 6 \mathrm{Fe}^{2+} \rightarrow 6 \mathrm{Fe}^{3+} + 6e^- \)
- Reduction: \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \)
Oxidation and Reduction
The processes of oxidation and reduction are key to understanding how electrons are transferred in redox reactions within voltaic cells. Oxidation involves the loss of electrons and occurs at the anode. Conversely, reduction involves the gain of electrons and occurs at the cathode.
Important properties:
Important properties:
- In the electronic equation, the substance that is oxidized loses electrons and its oxidation state increases.
- The substance that is reduced gains electrons, resulting in a decrease in its oxidation state.
- Oxidation: \( 5\mathrm{I}^{-} \rightarrow \frac{5}{3}\mathrm{I}_2 + \frac{10}{3}e^- \)
- Reduction: \( \mathrm{IO}_{3}^{-} + 6\mathrm{H}^{+} + 6e^- \rightarrow 3\mathrm{I}_{2} + 3 \mathrm{H}_{2} \mathrm{O} \)
Electrode Reactions
Electrode reactions are the heart of the electrochemical processes occurring in voltaic cells. These are chemical reactions at the anode and cathode that drive the flow of electrons and ultimately produce electrical current.
Here鈥檚 how it works:
Here鈥檚 how it works:
- At the anode, oxidation reactions occur. For example, in reaction (c) \( \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \), electrons are released.
- At the cathode, reduction reactions occur, such as in \( \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu} \) where electrons are gained.
