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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 $$ \begin{aligned} \mathrm{HgO}_{g}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} & \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) & \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{aligned} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text {ied }}\) 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
(a) The overall cell reaction is: HgO(s) + H₂O(l) + Zn(s) → Hg(l) + 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 what we'd expect in an acidic medium because the reaction occurs in the presence of hydroxide ions (OH⁻) in the battery, which affects the concentrations of H⁺ and OH⁻ ions and, in turn, the potential of the anode reaction.

Step by step solution

01

(a) Overall Cell Reaction

To find the overall cell reaction, we simply add the two given half-cell reactions together. Remember to cancel out any species that appear on both sides of the reaction. HgO(s) + H₂O(l) + 2 e⁻ → Hg(l) + 2 OH⁻(aq) Zn(s) + 2 OH⁻(aq) → ZnO(s) + H₂O(l) + 2 e⁻ Now, add the reactions together and cancel out the common terms (2 e⁻, 2 OH⁻, and H₂O): HgO(s) + H₂O(l) + Zn(s) → Hg(l) + 2 OH⁻(aq) + ZnO(s) + H₂O(l) The overall cell reaction is: HgO(s) + H₂O(l) + Zn(s) → Hg(l) + ZnO(s)
02

(b) Standard Reduction Potential of the Anode Reaction

We are given the following information: - Standard reduction potential for the cathode (E_cathode) = +0.098 V - Overall cell potential (E_cell) = +1.35 V We can determine the standard reduction potential of the anode reaction (E_anode) using the formula for the overall cell potential: E_cell = E_cathode - E_anode Rearrange the formula to find E_anode: E_anode = E_cathode - E_cell Now, substitute the given values: E_anode = (+0.098 V) - (+1.35 V) E_anode = -1.252 V So, the standard reduction potential for the anode reaction is -1.252 V.
03

(c) Anode Reaction Potential in an Acidic Medium

The potential of the anode reaction is different than what we'd expect in an acidic medium because the environment of the reaction is different. In an acidic medium, the concentration of protons (H⁺) is higher, which can affect the potential of the reaction. In this problem, the anode reaction occurs in the presence of hydroxide ions (OH⁻) in the battery, representative of a basic medium. The concentration of H⁺ ions is lower, and the concentration of OH⁻ ions is higher, compared to the acidic medium. This difference in the concentrations of H⁺ and OH⁻ ions affects the equilibrium and the potential of the anode reaction. Thus, the potential of the anode reaction is different than what we'd expect 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, short for reduction-oxidation reactions, are chemical processes that involve the transfer of electrons between two substances. These reactions are crucial in electrochemistry, as they underpin the operation of batteries and electrochemical cells. In a redox reaction, one substance loses electrons (oxidation) and another gains electrons (reduction).
Every redox reaction is made up of two half-reactions: an oxidation half-reaction and a reduction half-reaction. The oxidation half-reaction involves the loss of electrons from the substance being oxidized, while the reduction half-reaction involves the gain of electrons by the substance being reduced. For example, in the original exercise, the reaction of mercuric oxide with water captures electrons and gets reduced to mercury, while zinc undergoes oxidation by losing electrons and forming zinc oxide.
Key points to remember about redox reactions include:
  • Redox reactions always happen in pairs.
  • In a battery, or an electrochemical cell, the reduction reaction occurs at the cathode, and the oxidation takes place at the anode.
  • Electrons flow from the anode to the cathode outside the cell, enabling the electrical work to occur.
Understanding redox reactions is important for comprehending how batteries provide power through electron transfer.
Standard Reduction Potential
Standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is crucial in determining the direction and feasibility of redox reactions.
The standard reduction potential (denoted as \(E^0\)) is usually given in volts (V) and is determined under standard conditions, which typically involve solute concentrations of 1 M, a pressure of 1 atm, and a temperature of 25 °C (298 K). Each half-reaction involved in a redox process has its own standard reduction potential. The overall cell potential can be determined by using these potentials.
For example, in the original exercise, the potential of the cathode reaction is given as \(+0.098 V\). Using the formula \(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\), we can compute the anode potential. This potential indicates the ease of oxidation reactions occurring at the anode compared to standard hydrogen conditions.
Important notes about standard reduction potential:
  • Standard potentials are measured relative to the standard hydrogen electrode, which has a potential of 0 V by definition.
  • A more positive standard reduction potential means a greater tendency to be reduced, while a negative potential indicates a greater tendency to be oxidized when compared with the hydrogen electrode.
  • The calculated cell potential informs us about the spontaneity of the redox reaction; a positive cell potential indicates a spontaneous reaction.
Battery Chemistry
Battery chemistry involves the conversion of chemical energy into electrical energy through electrochemical reactions. Batteries consist of one or more electrochemical cells, each containing two electrodes and an electrolyte. The electrochemical reactions at the electrodes are integral to battery operation.
In a typical battery setup, the anode, where oxidation occurs, releases electrons that travel through an external circuit to the cathode, where reduction takes place, completing the circuit through the electrolyte.
Mercuric oxide dry-cell batteries, as in the original exercise, are praised for their stable discharge voltage and long lifespan. Such batteries are often found in compact devices like watches or cameras. They consist of a zinc anode and a mercuric oxide cathode, where zinc is oxidized and mercury oxide is reduced during battery discharge.
Key insights into battery chemistry include:
  • The choice of materials for the anode and cathode determine the voltage and capacity of the battery.
  • Environmental factors, such as temperature and medium (acidic or basic), impact battery life and performance.
  • Batteries should be disposed of properly as they contain chemicals that can be harmful to the environment.
Mastering battery chemistry principles helps us understand how different batteries operate and how they can be optimized for specific applications.

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Most popular questions from this chapter

By using the data in Appendix E, determine whether each of the following substances is likely to serve as an exidant or a reductant: (a) \(\mathrm{Cl}_{2}\) (g), (b) \(\mathrm{MnO}_{4}^{-}\)(aq, acidic solution), (c) \(\mathrm{Ba}\) (s), (d) \(\mathrm{Zn}(\) s).

Given the following reduction half-reactions: $$ \begin{gathered} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) \quad E_{\text {red }}^{\infty}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \quad E_{\mathrm{ret}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I) \quad \mathrm{E}_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) \quad E_{\mathrm{red}}^{0}=+1.00 \mathrm{~V} \end{gathered} $$ (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q)\), by \(\mathrm{N}_{2} \mathrm{O}(a q)\), and by \(\mathrm{VO}_{2}^{+}\)(aq). (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

The Haber process is the principal industrial route for converting nitrogen into ammonia: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ (a) What is being oxidized, and what is being reduced? (b) Using the thermodynamic data in Appendix \(\mathrm{C}\), calculate the equilibrium constant for the process at room temperature. (c) Calculate the standard emf of the Haber process at room temperature.

(a) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metal hydride batteries over nickel-cadmium batteries?

Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\), (a) What mass of calcium can be preduced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the minimum voltage needed to cause the electrolysis?

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