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The standard reduction potentials of the following half-reactions are given in Appendix E: $$ \begin{array}{l} \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) \\ \mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Cd}(s) \\\ \mathrm{Sn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Sn}(s) \\\ \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s) \end{array} $$ (a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell potential and calculate the value. (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell potential and calculate the value.

Short Answer

Expert verified
The combination of half-cell reactions that leads to the largest positive cell potential is Fe(s) with Ag(s), and the value can be calculated using \( E°_{Ag} - E°_{Fe} \). The combination with the smallest positive cell potential is Fe(s) with Cd(s), and the value can be calculated using \( E°_{Cd} - E°_{Fe} \).

Step by step solution

01

List all possible half-cell reaction combinations

To find the combination of half-cell reactions that produce the largest and smallest positive cell potentials, start by listing all the possible combinations. Ignore same element combinations as they result in no reaction. There are four different elements, and hence, six possible combinations: 1. Fe(s) with Cd(s) 2. Fe(s) with Sn(s) 3. Fe(s) with Ag(s) 4. Cd(s) with Sn(s) 5. Cd(s) with Ag(s) 6. Sn(s) with Ag(s)
02

Find the standard cell potentials for each combination

Using the standard reduction potentials from Appendix E, calculate the standard cell potential E° of each combination by subtracting the lower potential from the higher potential, ensuring that the resultant potential is positive. (Note: You can find the standard reduction potentials of these half-cell reactions in the AIME table). 1. Fe(s) with Cd(s): \( E°_{Cd} - E°_{Fe} \) 2. Fe(s) with Sn(s): \( E°_{Sn} - E°_{Fe} \) 3. Fe(s) with Ag(s): \( E°_{Ag} - E°_{Fe} \) 4. Cd(s) with Sn(s): \( E°_{Sn} - E°_{Cd} \) 5. Cd(s) with Ag(s): \( E°_{Ag} - E°_{Cd} \) 6. Sn(s) with Ag(s): \( E°_{Ag} - E°_{Sn} \)
03

Determine the combination with the largest positive cell potential

After calculating the standard cell potentials for each combination, you'll find that Fe(s) with Ag(s) has the largest positive cell potential. 3. Fe(s) with Ag(s)
04

Determine the combination with the smallest positive cell potential

Similar to step 3, you'll find that Fe(s) with Cd(s) has the smallest positive cell potential. 1. Fe(s) with Cd(s)
05

Calculate the values of the largest and smallest positive cell potentials

By comparing the calculated E° values, you'll be able to find the largest and smallest positive cell potentials: (a) The largest positive cell potential value occurs when Fe(s) and Ag(s) react together, and its E° value can be calculated as follows: \( E°_{Ag} - E°_{Fe} \) (b) The smallest positive cell potential value occurs when Fe(s) and Cd(s) react together, and its E° value can be calculated as follows: \( E°_{Cd} - E°_{Fe} \)

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Key Concepts

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

Standard Reduction Potentials
Understanding standard reduction potentials is key in electrochemistry. These are the voltages associated with reduction reactions, measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C). They tell us how easily a species gains electrons to become reduced. The higher the reduction potential, the more likely the substance will gain electrons and get reduced. For instance, in our exercise, you'd look up values from the table and see that silver ions (\(\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}(s)\)) might have a higher potential than, say, cadmium ions.
This means in a reaction, silver ions are more likely to get reduced than cadmium ions. Using these values, comparisons between half-cells can be made to determine which reactions will occur spontaneously.
Cell Potential
Cell potential, also known as electromotive force (EMF), expresses the energy difference between two half-cells in an electrochemical cell. It's calculated using the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
This equation involves subtracting the anode's potential from the cathode's potential. A positive value means the reaction is spontaneous, indicating which half-reaction will proceed.
Calculating these in our exercise allows us to determine the largest and smallest positive cell potentials by using the given standard reduction potentials.
  • The combination with the highest positive potential usually involves the strongest oxidizing and reducing agents.
  • The combination resulting in the lowest positive potential usually involves weaker agents.
Half-Reactions
In electrochemistry, a full reaction is broken down into two half-reactions: one for oxidation and one for reduction. These reactions display how electrons transfer between substances.
The importance of half-reactions lies in their ability to provide a clear view of the electron movement. For example, in the exercise, iron might act in one half-reaction:\( \text{Fe}^{2+} + 2\text{e}^- \rightarrow \text{Fe}(s) \), indicating reduction. Whereas another substance may undergo oxidation in its half-reaction.
Finding these pairs is essential to determine the overall reaction direction. The half-reactions are summed to yield the full electrochemical reaction that the exercise aims to explore.
Electrochemical Cells
Electrochemical cells are devices that convert chemical energy into electrical energy. They consist of two half-cells, each containing an electrode and electrolyte. The cells can generate electricity through spontaneous chemical reactions.
In our exercise, identifying the right combination of half-reactions constructs a cell. For example, combining silver and iron reactions in a cell sets up a flow of electrons. These electrons move through an external circuit, driven by the difference in standard reduction potentials.
  • The anode is where oxidation occurs, and it loses electrons.
  • The cathode gains electrons, where reduction happens.
The overall potential of the cell reflects its ability to perform work, making it critical for applications like batteries and galvanic cells.

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Most popular questions from this chapter

Heart pacemakers are often powered by lithium-silver chromate "button" batteries. The overall cell reaction is $$ 2 \mathrm{Li}(s)+\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s) \longrightarrow \mathrm{Li}_{2} \mathrm{CrO}_{4}(s)+2 \mathrm{Ag}(s) $$ (a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode? (b) Choose the two half-reactions from Appendix \(\mathrm{E}\) that most closely approximate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half-reactions? (c) The battery generates an emf of \(+3.5 \mathrm{~V}\). How close is this value to the one calculated in part (b)? (d) Calculate the emf that would be generated at body temperature, \(37^{\circ} \mathrm{C}\). How does this compare to the emf you calculated in part (b)?

The purification process of silicon involves the reaction of silicon tetrachloride vapor \(\left(\mathrm{SiCl}_{4}(g)\right)\) with hydrogen to \(1250^{\circ} \mathrm{C}\) to form solid silicon and hydrogen chloride. \((\mathbf{a})\) Write a balanced equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant?

Indicate whether each statement is true or false: (a) The cathode is the electrode at which oxidation takes place. (b) A galvanic cell is another name for a voltaic cell. (c) Electrons flow spontaneously from anode to cathode in a voltaic cell.

(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? (d) Why are active metals such as Al obtained by electrolysis using molten salts rather than aqueous solutions?

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{HNO}_{3}(a q)\) (b) \(\mathrm{FeS}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\) (c) \(\mathrm{Fe}(s)+2 \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+ 2 \mathrm{NO}_{2}(g)+\mathrm{FeO}(s)\)

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