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Mercuric oxide dry-cell batteries are often used where a high energy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{aligned} &\mathrm{HgO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}--\rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ &\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{aligned} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text {red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

Short Answer

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(a) The overall cell reaction for the Mercuric oxide dry-cell battery is: $$ \mathrm{HgO}(\mathrm{s}) + \mathrm{Zn}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) $$ (b) The standard reduction potential for the anode reaction is \(-1.252 V\). (c) The potential of the anode reaction is different than expected in an acidic medium because the half-cell reactions involve hydroxide ions (OH-) instead of hydrogen ions (H+), altering the reaction and its potential. Additionally, the pH of the medium influences the potential, leading to different results than in an acidic environment.

Step by step solution

01

(a) Determine the overall cell reaction

To find the overall cell reaction, we need to add the two half-cell reactions together. Remember to ensure that both reactions have the same number of electrons to cancel out. In this case, both reactions have two electrons already, so we are ready to add them. $$ \begin{aligned} \mathrm{HgO}(\mathrm{s}) + \mathrm{H}_{2}\mathrm{O}(l) + 2\mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l) + 2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s) + 2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \end{aligned} $$ Adding the reactions together and canceling out the electrons and hydroxide ions, we get the overall cell reaction. $$ \mathrm{HgO}(\mathrm{s}) + \mathrm{Zn}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) $$
02

(b) Determine the standard reduction potential for the anode reaction

We are given the standard reduction potential for the cathode reaction, \(E_{\text{red}}^{\circ} = +0.098 V\) and the overall cell potential, \(E_{\text{cell}}^{\circ} = +1.35 V\). The standard reduction potential for the anode reaction can be found using the formula: $$ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} $$ We need to find the value of \(E_{\text{anode}}^{\circ}\). Rearrange the equation to solve for \(E_{\text{anode}}^{\circ}\): $$ E_{\text{anode}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{cell}}^{\circ} $$ Plug in the given values and solve for \(E_{\text{anode}}^{\circ}\): $$ E_{\text{anode}}^{\circ} = +0.098 V - (+1.35 V) = -1.252 V $$ So, the standard reduction potential for the anode reaction is \(-1.252 V\).
03

(c) Explain why the potential of the anode reaction is different than expected in an acidic medium

In an acidic medium, the concentration of hydrogen ions (H+) is higher, whereas, in this battery, the half-cells reactions involve hydroxide ions (OH-) instead. The presence of hydroxide ions alters the anode reaction, which may lead to changes in the potential. Additionally, the potential of a half-cell reaction can also be influenced by the pH of the medium, which is different in an acidic environment. Therefore, the potential of the anode reaction is different than what would be expected if it occurred in an acidic medium.

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Key Concepts

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

Redox Reactions
Redox reactions are chemical processes involving the transfer of electrons between two substances. These reactions are fundamental in electrochemistry, serving as the basis for how batteries function.
In a redox reaction, there are two main parts: oxidation and reduction. Oxidation refers to the loss of electrons, while reduction is the gain of electrons. The substance that loses electrons is said to be oxidized, while the one that gains electrons is reduced.
In the context of a mercuric oxide battery, the redox reactions involve the transfer of electrons between metallic zinc and mercuric oxide. The zinc undergoes oxidation as it loses electrons, becoming zinc oxide, whereas the mercuric oxide gains electrons to form liquid mercury, thus undergoing reduction. This electron transfer is what generates electrical energy.
Standard Reduction Potential
The standard reduction potential, often represented as \(E^0\), is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. It is a crucial concept in understanding electrochemical cells.
These potentials are measured in volts and provide information about the relative strength of different oxidizing and reducing agents under standard conditions, which are typically 1 M concentrations, 1 atm pressure, and 25°C.
In electrochemical cells, the potential difference between the two electrodes is the driving force for electron flow. For instance, in the mercuric oxide dry cell battery problem, the given standard reduction potential of the cathode reaction is \(+0.098 \, V\). By knowing the overall cell potential, which is \(+1.35 \, V\), the standard reduction potential for the anode can be determined using the relationship \( E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0 \). The calculated anode potential \(-1.252 \, V\) indicates how the conditions in the battery affect its performance compared to other environments.
Battery Chemistry
Battery chemistry involves the study of materials and reactions used to store and deliver energy in batteries.
In a battery, chemical energy is converted into electrical energy through electrochemical reactions. This involves redox reactions that occur in separate compartments called half-cells.
The internal chemistry of a battery, like the mercuric oxide dry cell, determines various performance characteristics such as energy density, cycle life, and discharge voltage. High energy density is crucial for applications like watches and cameras, where power needs to be stored in a small space.
Understanding the chemistry within a battery helps in designing cells that are efficient and safe for particular applications. The choice of materials for the anode and cathode, as well as the electrolyte, affects how the battery operates and interacts with external circuits.
Half-Cell Reactions
Half-cell reactions are the individual oxidation and reduction reactions that occur in the two halves of an electrochemical cell.
Each half-cell consists of a conductive electrode and an electrolyte solution where either oxidation or reduction takes place. In the mercuric oxide battery example, one half-cell involves the reduction of \( \mathrm{HgO} \) to \( \mathrm{Hg} \), while the other involves the oxidation of \( \mathrm{Zn} \) to \( \mathrm{ZnO} \).
The half-cell reactions are balanced in terms of electrons, ensuring that the same number of electrons are lost and gained, which is essential for the flow of electric current.
Understanding the half-cell reactions provides insight into the overall efficiency and capability of a battery to convert chemical energy to electrical energy, thereby determining its viability for a given application. Half-cell reactions are thus integral to analyzing and optimizing battery performance.

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Most popular questions from this chapter

A \(1 M\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of Cu metal. A \(1 M\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which electrode gains mass and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

A voltaic cell is based on \(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(s)\) 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 \(\mathrm{C}\) 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 \(25^{\circ} \mathrm{C}\).

During a period of discharge of a lead-acid battery, \(402 \mathrm{~g}\) of \(\mathrm{Pb}\) from the anode is converted into \(\mathrm{PbSO}_{4}(s) .\) What mass of \(\mathrm{PbO}_{2}(s)\) is reduced at the cathode during this same period?

(a) The nonrechargeable lithium batteries used for photography use lithium metal as the anode. What advantages might be realized by using lithium rather than zinc, cadmium, lead, or nickel? (b) The rechargeable lithiumion battery does not use lithium metal as an electrode material. Nevertheless, it still has a substantial advantage over nickel-based batteries. Suggest an explanation.

This oxidation-reduction reaction in acidic solution is spontaneous: $$ \begin{array}{r} 5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)-\rightarrow \\ 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ A solution containing \(\mathrm{KMnO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is poured into one beaker, and a solution of \(\mathrm{FeSO}_{4}\) is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at \(298 \mathrm{~K}\) when the concentrations are the following: \(\mathrm{pH}=0.0, \quad\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M}, \quad\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

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