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Chapter 20: Problem 37

The standard reduction potentials of the following halfreactions are given in Appendix E: $$ \begin{gathered} \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \\ \mathrm{Ni}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(s) \\ \mathrm{Cr}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(s) \end{gathered} $$ (a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell emf, and calculate the value. (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell emf, and calculate the value.

Short Answer

Expert verified
The combination of half-cell reactions that results in the largest positive cell emf is between Ag and Cr, with a cell emf of \(1.54 \ \text{V}\). The smallest positive cell emf is attained from the combination of Ni and Cu half-cell reactions, with a cell emf of \(0.59 \ \text{V}\).

Step by step solution

01

(a) Largest Positive Cell EMF:

Combine Ag and Cr reactions, noting that the Cr reaction has been reversed. The cell reaction will now be: \(\text{Cr}(s) \longrightarrow \text{Cr}^{3+}(aq) + 3e^{-} \) (\(E_{\text{Cr}/\text{Cr}^{3+}} = 0.74 \ \text{V}\)) as oxidation \(\text{Ag}^{+}(aq) + e^{-} \longrightarrow \text{Ag}(s) \) (\(E_{\text{Ag}^{+}/\text{Ag}} = 0.80 \ \text{V}\)) as reduction The overall reaction becomes: \(\text{Cr}(s) + 3\text{Ag}^{+}(aq) \longrightarrow \text{Cr}^{3+}(aq) + 3\text{Ag}(s)\) The largest positive cell emf is the sum of the two potentials: \(E_{\text{cell}}^{\circ} = E^{\circ} _{\text{Cr}/\text{Cr}^{3+}} + E^{\circ} _{\text{Ag}^{+}/\text{Ag}} = 0.74 \ \text{V} + 0.80 \ \text{V} = 1.54 \ \text{V}\)
02

(b) Smallest Positive Cell EMF:

Combine Ni and Cu reactions, noting that the Ni reaction has been reversed. The cell reaction will now be: \(\text{Ni}(s) \longrightarrow \text{Ni}^{2+}(aq) + 2e^{-} \) (\(E_{\text{Ni}/\text{Ni}^{2+}} = 0.25 \ \text{V}\)) as oxidation \(\text{Cu}^{2+}(aq) + 2e^{-} \longrightarrow \text{Cu}(s) \) (\(E_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \ \text{V}\)) as reduction The overall reaction becomes: \(\text{Ni}(s) + 2\text{Cu}^{2+}(aq) \longrightarrow \text{Ni}^{2+}(aq) + 2\text{Cu}(s)\) The smallest positive cell emf is the sum of the two potentials: \(E_{\text{cell}}^{\circ} = E^{\circ} _{\text{Ni}/\text{Ni}^{2+}} + E^{\circ} _{\text{Cu}^{2+}/\text{Cu}} = 0.25 \ \text{V} + 0.34 \ \text{V} = 0.59 \ \text{V}\) In summary, the largest positive cell emf is 1.54 V for the combination of Ag and Cr reactions, and the smallest positive cell emf is 0.59 V for the combination of Ni and Cu reactions.

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Key Concepts

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

Standard Reduction Potential
Standard reduction potential is a measure of the tendency of a chemical species to be reduced, meaning it gains electrons in a redox reaction. It's represented by the symbol \(E^\circ\) and is measured in volts (V). All standard reduction potentials are compared relative to the standard hydrogen electrode (SHE), which is assigned a potential of 0 volts.

In electrochemistry, these potentials are crucial for predicting the direction of redox reactions. A positive standard reduction potential indicates a strong tendency to gain electrons, making the species a good oxidizing agent. Conversely, a negative value suggests a lesser tendency to gain electrons.
  • For example, silver ions (\(\text{Ag}^+\)) have a standard reduction potential of 0.80 V, indicating a strong propensity to be reduced to metallic silver (\(\text{Ag}\)).
  • Chromium ions (\(\text{Cr}^{3+}\)) have a lower potential when reversed for oxidation, showcasing a strong reducing action in the opposite direction, i.e., converting \(\text{Cr}\) metal to \(\text{Cr}^{3+}\) ions.
Cell EMF
Cell EMF, or electromotive force, is the voltage produced by an electrochemical cell when no current flows. It reflects the energy difference between the electrodes due to their differing reduction potentials. The cell emf is calculated using the equation:\[E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ\]where \(E^\circ_{\text{cathode}}\) is the standard reduction potential of the substance being reduced (gain of electrons) and \(E^\circ_{\text{anode}}\) is the standard reduction potential of the substance being oxidized (loss of electrons).

For a spontaneous reaction in a galvanic cell, the emf is positive. It determines how much energy is available to do work:
  • For example, if silver and chromium reactions are combined, the total standard emf results from adding their potentials after adjusting for direction. The positive 1.54 V indicates a strong spontaneous reaction.
  • Conversely, nickel and copper give a lower cell emf of 0.59 V, still spontaneous but less energetically favorable.
Redox Reactions
Redox reactions, also known as oxidation-reduction reactions, are processes where electrons are transferred between two substances. Key concepts in these reactions include oxidation, where a substance loses electrons, and reduction, where a substance gains electrons:
  • Reduction occurs at the cathode (where electrons are gained).
  • Oxidation occurs at the anode (where electrons are lost).
  • In a galvanic cell, redox reactions are harnessed to generate electrical energy.
For example, in the pairing of reactions where silver and chromium are involved, silver ions are reduced to silver metal by gaining electrons. Simultaneously, chromium metal is oxidized to chromium ions by losing electrons. The total energy transfer in these redox processes calculates the cell emf, demonstrating a key principle of electrochemical cells: converting chemical energy to electrical energy.

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Most popular questions from this chapter

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules? (c) What is special about a standard cell potential?

A voltaic cell utilizes the following reaction and operates at \(298 \mathrm{~K}\) : $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s)-\rightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 \mathrm{M}\), \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M}\) ? (c) What is the emf of the cell when \(\left[\mathrm{Ce}^{4+}\right]=0.10 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=1.75 \mathrm{M}\) and \(\left[\mathrm{Cr}^{3+}\right]=2.5 \mathrm{M} ?\)

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3}\), and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\). The overall cell reaction is $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode, or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

(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?

At \(298 \mathrm{~K}\) a cell reaction has a standard emf of \(+0.17 \mathrm{~V}\). The equilibrium constant for the cell reaction is \(5.5 \times 10^{5} .\) What is the value of \(n\) for the cell reaction?
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