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Chapter 21: Problem 146

The zinc-air battery is a less expensive alternative to silver batteries for use in hearing aids. The cell reaction is $$2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)$$ A new battery weighs \(0.275 \mathrm{~g}\). The zinc accounts for exactly \(\frac{1}{10}\) of the mass, and the oxygen does not contribute to the mass because it is supplied by the air. (a) How much electricity (in C) can the battery deliver? (b) How much free energy (in J) is released if \(E_{\text {cell }}\) is \(1.3 \mathrm{~V}\) ?

Short Answer

Expert verified
The battery can deliver 81.19 C of electricity. The free energy released is -105.55 J.

Step by step solution

01

Determine the mass of Zinc in the Battery

The mass of the battery is given as 0.275 g. Since Zinc accounts for \( \frac{1}{10} \) of this mass: \[ \text{Mass of Zinc} = \frac{1}{10} \text{ of } 0.275 \text{ g} = 0.0275 \text{ g} \]
02

Convert the Mass of Zinc to Moles

The molar mass of Zinc (Zn) is approximately 65.38 g/mol. The number of moles of Zinc can be calculated using the formula: \[ \text{Moles of Zn} = \frac{\text{Mass of Zn}}{\text{Molar mass of Zn}} = \frac{0.0275 \text{ g}}{65.38 \text{ g/mol}} \approx 0.000421 \text{ mol} \]
03

Relate Moles of Zinc to Moles of Electrons Transferred

From the cell reaction: \[ 2 \text{Zn} (s) + \text{O}_2 (g) \rightarrow 2 \text{ZnO} (s) \] Each Zinc atom transfers 2 electrons when it forms ZnO.Thus, the total number of moles of electrons transferred would be: \[ \text{Moles of Electrons} = 2 \times \text{Moles of Zn} = 2 \times 0.000421 = 0.000842 \text{ mol} \]
04

Calculate the Total Electric Charge Delivered

The total charge (Q) can be calculated using the formula: \[ Q = n \times F \] where \( n \) is the moles of electrons and \( F \) (Faraday's constant) is approximately 96485 C/mol. Thus, \[ Q = 0.000842 \times 96485 \approx 81.19 \text{ C} \]
05

Calculate the Free Energy Released

The free energy (ΔG) released can be calculated using the formula: \[ \Delta G = - n F E_{\text{cell}} \] where \( E_{\text{cell}} \) is the cell potential (1.3 V). \( n \) is the number of moles of electrons (0.000842) and \( F \) is 96485 C/mol. Thus, \[ \Delta G = - 0.000842 \times 96485 \times 1.3 = -105.55 \text{ J} \]

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Key Concepts

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

Zinc
Zinc, a chemical element with the symbol Zn, is a critical component in many batteries. It's a bluish-white metal known for its ability to effectively manage and release electrical charge when participating in redox reactions. In the context of the zinc-air battery, zinc reacts with oxygen from the air to form zinc oxide. This simple yet efficient reaction is at the core of how the zinc-air battery produces electricity. Understanding zinc's role is crucial for grasping the essence of the reaction and the battery's functionality. The mass and reactivity of zinc in the battery directly influence the overall performance and capacity of the device.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical reactions. In a zinc-air battery, electrochemistry principles explain how zinc and oxygen produce electrical energy. The cell reaction 2 Zn(s) + O_2(g) → 2 ZnO(s) involves the transfer of electrons from zinc to oxygen. Every zinc atom releases two electrons as it reacts. These electrons then flow through an external circuit, creating electric current. Understanding these principles helps in analyzing how batteries work and predicting their behavior under various conditions.
Moles Calculation
To fully utilize the mass of zinc in a battery, we need to calculate the number of moles involved. The mole is a fundamental unit in chemistry representing a specific number of particles (6.022 x 10^23 particles). By converting the mass of zinc (0.0275 g) to moles, using its molar mass (65.38 g/mol), we find: Moles of Zn = 0.0275 g / 65.38 g/mol ≈ 0.000421 mol This basic calculation is vital as it allows us to determine how many zinc atoms are available to react and how many electrons will be transferred during the reaction. This step lays the groundwork for further calculations like determining the total charge and free energy.
Faraday's Constant
Faraday's constant (F) is a crucial figure in electrochemistry, representing the charge of one mole of electrons, approximately 96485 Coulombs per mole. This constant plays a pivotal role in converting moles of electrons to electric charge. In the zinc-air battery, we calculated the moles of electrons as 0.000842 mol. Using Faraday's constant, we can find the total electric charge delivered: Q = n × F = 0.000842 mol × 96485 C/mol ≈ 81.19 C This calculation demonstrates how Faraday's constant links the microscopic scale of chemical reactions to measurable electric quantities.
Gibbs Free Energy
Gibbs Free Energy (ΔG) is a thermodynamic quantity that indicates the amount of energy available to do work during a chemical reaction. For the zinc-air battery, the free energy change can be calculated using the formula: ΔG = -nFE_cell where n is the number of moles of electrons (0.000842), F is Faraday's constant (96485 C/mol), and E_cell is the cell potential (1.3 V). Plugging in these values, we get: ΔG = -0.000842 × 96485 × 1.3 ≈ -105.55 J This negative value indicates that the reaction releases energy, driving the battery and providing the power needed for devices such as hearing aids.

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Most popular questions from this chapter

Write the cell notation for the voltaic cell that incorporates each of the following redox reactions: (a) \(\mathrm{Al}(s)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Cr}(s)\) (b) \(\mathrm{Cu}^{2+}(a q)+\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$\mathrm{Cu}(s)+\mathrm{SO}_{4}^{2-}(a q)+4 \mathrm{H}^{+}(a q)$$

Can one half-reaction in a redox process take place independently of the other? Explain.

The \(\mathrm{MnO}_{2}\) used in alkaline batteries can be produced by an electrochemical process of which one half-reaction is $$\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}$$ If a current of \(25.0 \mathrm{~A}\) is used, how many hours are needed to produce \(1.00 \mathrm{~kg}\) of \(\mathrm{MnO}_{2} ?\) At which electrode is the \(\mathrm{MnO}_{2}\) formed?

A voltaic cell consists of two \(\mathrm{H}_{2} / \mathrm{H}^{+}\) half-cells. Half-cell \(\mathrm{A}\) has \(\mathrm{H}_{2}\) at 0.95 atm bubbling into \(0.10 \mathrm{M} \mathrm{HCl}\). Half-cell \(\mathrm{B}\) has \(\mathrm{H}_{2}\) at 0.60 atm bubbling into \(2.0 \mathrm{M} \mathrm{HCl}\). Which half-cell houses the anode? What is the voltage of the cell?

To examine the effect of ion removal on cell voltage, a chemist constructs two voltaic cells, each with a standard hydrogen electrode in one compartment. One cell also contains a \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell; the other contains a \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell. (a) What is \(E^{\circ}\) of each cell at \(298 \mathrm{~K} ?\) (b) Which electrode in each cell is negative? (c) When \(\mathrm{Na}_{2} \mathrm{~S}\) solution is added to the \(\mathrm{Pb}^{2+}\) electrolyte, solid \(\mathrm{PbS}\) forms. What happens to the cell voltage? (d) When sufficient \(\mathrm{Na}_{2} \mathrm{~S}\) is added to the \(\mathrm{Cu}^{2+}\) electrolyte, CuS forms and \(\left[\mathrm{Cu}^{2+}\right]\) drops to \(1 \times 10^{-16} \mathrm{M} .\) Find the cell voltage.
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