Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l}{\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)} \\ {\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{znO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}}\end{array} $$ (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

Expert verified
The overall cell reaction for the mercuric oxide dry-cell battery is: \[ \mathrm{HgO}(s) + \mathrm{Zn}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \] The standard reduction potential for the anode reaction, based on the given cathode and cell potential, is -1.252 V. The potential of the anode reaction is different in an acidic medium because the species involved in the reactions are different, and the medium influences the electrochemical equilibrium. Activity and concentration of ions can vary based on the acidity, which can affect the electron transfer process, causing the standard reduction potential to vary depending on the medium.

Step by step solution

01

(a) Overall Cell Reaction:

For this step, we need to add up the two half-cell reactions provided. The overall cell reaction can be obtained by performing the following process: \[ \begin{array}{l}{\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)} \\ + {\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}}\end{array} \] Next, we can cancel out the same species appearing on both sides of the equation: \[ \mathrm{HgO}(s) + \mathrm{Zn}(s) + \cancel{\mathrm{H}_{2} \mathrm{O}(l)} + \cancel{2 \mathrm{e}^{-}} + \cancel{2 \mathrm{OH}^{-}(a q)} \longrightarrow \mathrm{Hg}(l) + \cancel{2 \mathrm{OH}^{-}(a q)} + \mathrm{ZnO}(s) + \cancel{\mathrm{H}_{2} \mathrm{O}(l)} + \cancel{2 \mathrm{e}^{-}}\] Thus, the overall cell reaction is: \[ \mathrm{HgO}(s) + \mathrm{Zn}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \]
02

(b) Standard Reduction Potential for the Anode Reaction:

We are given the standard reduction potential for the cathode reaction, \(E_{\mathrm{red}}^{\circ} = +0.098\text{ V}\), and the overall cell potential, \(E_{\text{cell}}^{\circ} = +1.35\text{ V}\). To find the standard reduction potential for the anode reaction, we can use the equation for overall cell potential: \[ E_{\text{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ} \] Rearranging the equation to solve for the anode reduction potential: \[ E_{\mathrm{anode}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\text{cell}}^{\circ} \] Substitute the given values into the equation: \[ E_{\mathrm{anode}}^{\circ} = (+0.098\text{ V}) - (+1.35\text{ V}) \] \[ E_{\mathrm{anode}}^{\circ} = -1.252\text{ V} \] Therefore, the standard reduction potential for the anode reaction is -1.252 V.
03

(c) Explanation for difference in Anode Reaction Potential:

The potential of the anode reaction is different in an acidic medium because the species involved in the reactions are different, and the medium influences the electrochemical equilibrium. The activity and concentration of ions can vary based on the acidity, which can affect the electron transfer process. Consequently, the standard reduction potential will vary depending on the medium in which the reaction is taking place.

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Key Concepts

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

Half-Cell Reactions
In electrochemistry, half-cell reactions play a pivotal role as they are the basic units of a chemical reaction in a battery. A half-cell is a structure that includes a conductive electrode and a surrounding conductive electrolyte. When two half-cells are connected, they form a full electrochemical cell, which is the basis for batteries.

Each half-cell has an associated reaction where either oxidation or reduction occurs. Oxidation involves the loss of electrons, whereas reduction involves the gain of electrons. In the case of a mercuric oxide dry-cell battery, one half-cell involves the reduction of mercuric oxide and the other involves the oxidation of zinc. When combined, they deliver the energy needed to power electronic devices such as watches and cameras.

Understanding half-cell reactions is crucial, as it allows one to deduce the overall cell reaction, balance chemical equations, and ultimately understand how a battery functions at the molecular level. In the self-contained reaction environment of the half-cell, one is able to focus on either the oxidation or the reduction processes, get insights into electron flow, and measure individual potentials, which then serve as a foundation for calculating the cell’s voltage.
Standard Reduction Potential
The standard reduction potential, denoted as red, is the measure of the tendency of a chemical species to acquire electrons and be reduced. It is measured under standard conditions: a 1M concentration for each ion participating in the reaction, a pressure of 1 atmosphere for any gases involved in the reaction, and a temperature of 25°C (298 K).

Each half-cell in an electrochemical cell has its own standard reduction potential, and the difference between the two potentials is what drives the overall reaction in the cell. The more positive the standard reduction potential, the greater the species’ affinity for electrons. For example, in a mercuric oxide dry-cell battery, the half-cell reaction involving mercuric oxide has a standard reduction potential of +0.098 V, indicating a moderate tendency to gain electrons.

Standard reduction potentials are essential for predicting the direction of electron flow and the feasibility of a reaction. They are also fundamental in determining cell potential through calculations, which can help in designing batteries with desired properties such as voltage and longevity.
Cell Potential Calculations
Calculating the potential of an electrochemical cell, often called the cell potential, involves understanding the standard reduction potentials of the half-cells that make up the battery. The cell potential, represented as cell, is calculated by subtracting the anode’s (negative electrode) standard reduction potential from the cathode’s (positive electrode) standard reduction potential:

cell = cathode - anode.

For a battery to function, it must have a positive cell potential, meaning that the cathode’s standard reduction potential must be higher than the anode’s. In the example with the mercuric oxide dry-cell battery, the standard reduction potential of the cathode is +0.098 V, and the overall cell potential is +1.35 V. By using the cell potential equation, we can deduce the anode’s standard reduction potential, which in this case equates to -1.252 V, showing that zinc has a high propensity to lose electrons (oxidize).

Cell potential calculations are not only theoretical but also have practical applications such as estimating the maximum voltage that a cell can deliver and consequently its suitability for various applications.

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Most popular questions from this chapter

Li-ion batteries used in automobiles typically use a LiMn_ \(_{2} \mathrm{O}_{4}\) cathode in place of the LiCoO \(_{2}\) cathode found in most Li- ion batteries. (a) Calculate the mass percent lithium in each electrode material. (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with \(\mathrm{LiMn}_{2} \mathrm{O}_{4}\) cathodes deliver less power on discharging? (c) In a battery that uses a LiCoO\(_{2}\) cathode, approximately 50\(\%\) of the lithium migrates from the cathode to the anode on charging. In a battery that uses a LiMn_ \(_{2} \mathrm{O}_{4}\) cathode, what fraction of the lithium in LiMn_ \(_{2} \mathrm{O}_{4}\) would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

Disulfides are compounds that have \(S-\) S bonds, like peroxides have \(O-O\) bonds. Thiols are organic compounds that have the general formula \(R-S H,\) where \(R\) is a generic hydrocarbon. The SH \(^{-}\) is the sulfur counterpart of hydroxide, OH \(^{-} .\) Two thiols can react to make a disulfide, \(\mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R}\) (a) What is the oxidation state of sulfur in a thiol? (b) What is the oxidation state of sulfur in a disulfide? (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols? (d) If you wanted to convert a disulfide to two thiols, should you add a reducing agent or oxidizing agent to the solution? (e) Suggest what happens to the H's in the thiols when they form disulfides.

Metallic gold is collected from below the anode when a mixture of copper and gold metals is refined by electrolysis. Explain this behavior.

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{K} :\) $$ \begin{array}{l}{\text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s)} \\ {\text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g)} \\ {\text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \rightarrow} \\ {\quad 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l)}\end{array} $$

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{K},\) and calculate the equilibrium constant \(K\) at 298 \(\mathrm{K}\) (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q)\) . (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(l) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)

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