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Chapter 20: Problem 115

A voltaic cell is based on \(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(\mathrm{s})\) and \(\mathrm{Fe}^{3+}(a q) / \mathrm{Fe}^{2+}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode, and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(C\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The standard emf of the voltaic cell is \(0.03 \thinspace V\). The reaction at the cathode is Ag鈦�(aq) + e鈦� 鈫� Ag(s) and the reaction at the anode is Fe鲁鈦�(aq) + e鈦� 鈫� Fe虏鈦�(aq). To determine whether the standard cell potential increases or decreases with temperature, calculate the entropy change for the cell reaction and analyze it in the equation \(E_{cell}^{\circ} = -\frac{\Delta G^{\circ} - T\Delta S^{\circ}}{nF}\). If the entropy change is positive, the cell potential will decrease with temperature, and if it's negative, the cell potential will increase with temperature.

Step by step solution

01

Calculate the Standard Emf of the Cell

To calculate the standard emf of the cell, use the Nernst equation formula for standard conditions: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\) First, we need to find the standard reduction potentials of both half-cells. The standard reduction potentials for silver and iron are given as: Ag鈦�(aq) + e鈦� 鈫� Ag(s); \(E^{\circ} = 0.80 \thinspace V\) Fe鲁鈦�(aq) + e鈦� 鈫� Fe虏鈦�(aq); \(E^{\circ} = 0.77 \thinspace V\) Next, we need to identify which half-cell has the higher standard reduction potential. Since the Ag鈦�/Ag half-cell has a higher reduction potential (0.80 V compared to 0.77 V), it will be the cathode. Now, we can calculate the standard emf of the cell using the Nernst equation: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.80 \thinspace V - 0.77 \thinspace V = 0.03 \thinspace V\)
02

Identify the Reactions Occurring at the Cathode and Anode

We've already identified that the Ag鈦�/Ag half-cell is the cathode and the Fe鲁鈦�/Fe虏鈦� half-cell is the anode, based on their standard reduction potentials. Therefore, Cathode reaction: Ag鈦�(aq) + e鈦� 鈫� Ag(s) Anode reaction: Fe鲁鈦�(aq) + e鈦� 鈫� Fe虏鈦�(aq)
03

Analyze the Temperature Effect on the Standard Cell Potential

We need to determine whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\). We can use the following relationship between cell potential and Gibbs free energy change: \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\) Where \(\Delta G^{\circ}\) is the standard Gibbs free energy change, \(n\) is the number of electrons transferred in the reaction, and \(F\) is Faraday's constant. From the standard molar entropy values in Appendix C, we can calculate the entropy change for the cell reaction, \(\Delta S^{\circ}\). By combining the two equations, we can derive the following relationship between cell potential, free energy change, and entropy change: \(E_{cell}^{\circ} = -\frac{\Delta G^{\circ} - T\Delta S^{\circ}}{nF}\) If the entropy change is positive, the value of \(T\Delta S^{\circ}\) will be positive, and the numerator will be smaller, resulting in a decrease in cell potential (\(E_{cell}^{\circ}\)) when temperature increases. Conversely, if the entropy change is negative, the cell potential will increase when temperature increases. Calculate the entropy changes for both the cathode and the anode reactions using their respective standard molar entropies, and determine the net entropy change for the cell reaction. Then, you can deduce whether the standard cell potential will increase or decrease as the temperature increases above \(25^{\circ} \mathrm{C}\).

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Key Concepts

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

Standard EMF
In a voltaic or galvanic cell, the standard electromotive force (EMF) is a critical value that determines how much electrical energy can be obtained from a chemical reaction. Calculating the EMF involves using the standard reduction potentials of the cathode and anode half-cells involved in the cell reaction.

The formula for calculating the standard EMF, denoted by \(E_{cell}^{\circ}\), is:
  • \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\)
For the Ag鈦�/Ag and Fe鲁鈦�/Fe虏鈦� voltaic cell considered in this exercise, the standard reduction potentials are 0.80 V and 0.77 V, respectively.

Since the silver half-cell has a greater potential than the iron half-cell, it acts as the cathode. Plugging these values into our equation:
  • \(E_{cell}^{\circ} = 0.80 \thinspace V - 0.77 \thinspace V = 0.03 \thinspace V\)
Thus, the standard EMF of the cell is 0.03 V, indicating a relatively low energy yield under standard conditions.
Cathode and Anode Reactions
An essential part of understanding voltaic cells is identifying which reactions occur at the cathode and which occur at the anode. To do this, you look at the standard reduction potentials of the half-cells.

In the Ag鈦�/Ag and Fe鲁鈦�/Fe虏鈦� cell, the half-cell with the higher reduction potential, which is the Ag鈦�/Ag half-cell with 0.80 V, becomes the cathode. The reaction at the cathode is
  • \(\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\)
The cathode is where reduction occurs, which means a gain of electrons.
On the other hand, the Fe鲁鈦�/Fe虏鈦� half-cell with a lower potential of 0.77 V acts as the anode, where oxidation takes place:
  • \(\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq)\)
This reaction involves the loss of electrons. In short, the reaction at the cathode reduces ions, while the reaction at the anode oxidizes ions.
Temperature Effect on Cell Potential
The effect of temperature on cell potential can be understood through the relationship between cell potential, free energy, and entropy changes.
The formula linking these quantities is:
  • \(E_{cell}^{\circ} = -\frac{\Delta G^{\circ} - T\Delta S^{\circ}}{nF}\)
Here, \(\Delta G^{\circ}\) is the standard Gibbs free energy change, \(T\) is the temperature in kelvins, \(\Delta S^{\circ}\) is the standard entropy change, \(n\) is the number of moles of electrons exchanged, and \(F\) is Faraday's constant.

If the entropy change \(\Delta S^{\circ}\) is positive, an increase in temperature will make the term \(T\Delta S^{\circ}\) more significant. This will lead to a smaller value for the numerator \((\Delta G^{\circ} - T\Delta S^{\circ})\), causing the cell potential \(E_{cell}^{\circ}\) to decrease as temperature increases. Conversely, a negative entropy change implies an increase in the standard cell potential with rising temperature.
Thus, to predict how temperature affects this cell's potential, you need to consider the net entropy change of the whole reaction. This involves calculating individual entropy changes for both cathode and anode reactions and summing them up. Depending on whether this sum is positive or negative, the potential will either increase or decrease at higher temperatures.

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Most popular questions from this chapter

At \(900^{\circ} \mathrm{C}\) titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chloride. (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?

A voltaic cell is constructed that uses the following halfcell reactions: $$ \begin{gathered} \mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{gathered} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=3.5 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) was equal to \(0.15 \mathrm{M}\), at what concentration of \(\mathrm{I}^{-}\) would the cell have zero potential?

An iron object is plated with a coating of cobalt to protect against corrosion. Does the cobalt protect iron by cathodic protection? Explain.

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g)-\mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow\) \(2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) (c) \(3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)-\cdots\) \(3 \mathrm{~S}(s)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q)-\mathrm{-} \rightarrow\) \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) Oxidizing agents can convert \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).
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