Describe how you might construct batteries with each of the following voltages: (a) \(0.10 \mathrm{V} ;\) (b) \(2.5 \mathrm{V} ;\) (c) \(10.0 \mathrm{V}\). Be as specific as you can about the electrodes and solution concentrations you would use, and indicate whether the battery would consist of a single cell or two or more cells connected in series.

Short Answer

Expert verified
To create batteries with specific voltages, consider the type of battery cells, the cell reaction, and the number of cells connected in series. For 0.1V, manipulate the concentration levels in a Zinc-Copper cell. For 2.5V, use two Copper-Zinc cells in series and adjust standard states or add additional cells as needed. For 10V, connect seven Alkaline cells in series, manipulate concentrations or use a resistor to meet the exact voltage.

Step by step solution

01

Determine the cell reaction and voltage

Different types of battery materials result in different voltages. The standard cell potential (E0), usually measured in volts, of a cell is the difference between the reduction potentials of the cathode and the anode. If we consider a simple battery-like a Zinc-Copper (Zn-Cu) cell, the cell reaction is \( Zn + Cu^{2+} -> Zn^{2+} + Cu \) and this gives around 1.1V. To achieve the desired voltage, batteries can be combined together in series.
02

Creating a 0.1V Battery

A Zinc-Copper cell can be manipulated to achieve 0.1V by using different concentration levels for the solutions in the half-cells. This will move the Nernst equation (\( E=E0-(RT/nF)*lnQ \)) away from standard conditions, reducing the total cell potential.
03

Creating a 2.5V Battery

To achieve 2.5V, one could use two Copper-Zinc cells in series, which would generate around 2.2V. To reach 2.5V, slightly different standard states or additional cells may be needed.
04

Creating a 10.0V Battery

To reach 10.0V, it would be easier to use a battery type with a naturally higher voltage. For example, an Alkaline (AA, AAA) battery gives out around 1.5V. Therefore, connecting seven of these cells in series would provide \( 1.5V*7=10.5V \). The voltage could be reduced to exactly 10V by manipulating the concentrations in some of the cells or using a resistor.

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Key Concepts

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

Cell Potential and Reaction
Understanding the cell potential is crucial when building a battery. The cell potential, also referred to as electromotive force (EMF), is fundamentally determined by the chemical reaction that occurs within the battery. In simple terms, it's the force that moves electrons from the anode (negative electrode) to the cathode (positive electrode).

For a basic battery, such as the Zinc-Copper (Zn-Cu) cell mentioned in the exercise, the reaction is: \( Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu \). This standard reaction under standard conditions provides an EMF of approximately 1.1 volts. To construct batteries with voltages other than the standard output, we must alter certain conditions such as the concentrations of reactants, which affects the cell potential as described by the Nernst equation.
Batteries in Series
Connecting batteries in series is a strategy to increase the overall voltage output. When batteries are connected in series, the voltage of each battery adds up while the current remains the same. This is essential when a specific voltage higher than what a single cell can provide is required.

For example, to create a 2.5V battery, one option could be to connect two standard Zinc-Copper cells, which typically yield 1.1V each, in series. This would provide a combined voltage of 2.2V. To fine-tune to the exact desired voltage (2.5V), we might need to adjust conditions such as the electrode materials or solution concentrations or add an extra cell into the series.
Nernst Equation
The Nernst equation plays a key role in battery construction as it relates to cell potential. The Nernst equation is given by: \( E = E^0 - \frac{RT}{nF} \ln Q \), where \( E \) is the cell potential under non-standard conditions, \( E^0 \) is the standard cell potential, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient reflecting the conditions of the current state of the reaction.

The Nernst equation tells us that by changing the solution concentrations, we can directly influence the cell potential. This can be utilized to adjust the voltage to meet specific requirements, like achieving a battery voltage of 0.10V, by shifting the reaction away from its standard state.
Electrode Materials
Selecting the appropriate electrode materials is fundamental in dictating the overall voltage and efficiency of a battery. Different materials come with varying standard electrode potentials, which contribute to the cell's total voltage. Some common materials used for electrodes include zinc, copper, lead, and lithium, each with its own unique properties and standard potentials.

As seen in the original exercise, a Zinc-Copper cell can produce around 1.1V. However, if a different electrode material was chosen with a higher or lower standard potential, this would consequently alter the potential of the battery. This is why understanding the electrochemistry of materials is vital when constructing batteries for specific applications.
Solution Concentrations
Concentrations of the solutions in the half-cells are paramount when fine-tuning the voltage output of batteries. According to the Nernst equation, variations in concentration levels of electrolytes directly impact the cell potential. This is especially useful when the desired voltage is significantly lower than what the standard cell potential offers.

In practice, to construct a 0.1V battery, one can manipulate the concentrations of copper and zinc ions in a Zinc-Copper cell to lower the EMF. This involves creating a reaction quotient, \( Q \), that shifts the equilibrium, according to the Nernst equation, thus reducing the cell potential to the targeted 0.1V. This illustrates how minute changes in solution concentrations can have a profound effect on the performance characteristics of a battery.

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Most popular questions from this chapter

\(\mathrm{Ni}^{2+}\) has a more positive reduction potential than \(\mathrm{Cd}^{2+}\) (a) Which ion is more easily reduced to the metal? (b) Which metal, Ni or Cd, is more easily oxidized?

A common reference electrode consists of a silver wire coated with \(\mathrm{AgCl}(\mathrm{s})\) and immersed in \(1 \mathrm{M} \mathrm{KCl}\) $$\mathrm{AgCl}(\mathrm{s})+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}^{-}(1 \mathrm{M}) E^{\circ}=0.2223 \mathrm{V}$$ (a) What is \(E_{\text {cell }}^{\circ}\) when this electrode is a cathode in combination with a standard zinc electrode as an anode? (b) Cite several reasons why this electrode should be easier to use than a standard hydrogen electrode. (c) By comparing the potential of this silver-silver chloride electrode with that of the silver-silver ion electrode, determine \(K_{\mathrm{sp}}\) for \(\mathrm{AgCl}\).

Use the Nernst equation and data from Appendix D to calculate \(E_{\text {rell for each of the following cells. }}\) (a) \(\operatorname{Mn}(\mathrm{s}) | \mathrm{Mn}^{2+}(0.40 \mathrm{M}) \| \mathrm{Cr}^{3+}(0.35 \mathrm{M})\) \(\mathrm{Cr}^{2+}(0.25 \mathrm{M}) | \mathrm{Pt}(\mathrm{s})\) (b) \(\operatorname{Mg}\left(\text { s) } | \operatorname{Mg}^{2+}(0.016 \mathrm{M}) \|\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}(0.25 \mathrm{M})\right.\) \(\mathrm{OH}^{-}(0.042 \mathrm{M}) | \mathrm{Al}(\mathrm{s})\)

The theoretical \(E_{\text {cell }}^{\circ}\) for the methane-oxygen fuel cell is \(1.06 \mathrm{V} .\) What is \(E^{\circ}\) for the reduction half-reaction \(\mathrm{CO}_{2}(\mathrm{g})+8 \mathrm{H}^{+}(\mathrm{aq})+8 \mathrm{e}^{-} \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) ?\)

For the reaction \(\operatorname{Co}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Co}^{2+}(\mathrm{aq})+\) \(\mathrm{Ni}(\mathrm{s}), E_{\mathrm{cell}}^{\circ}=0.03 \mathrm{V} .\) If cobalt metal is added to an aqueous solution in which \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M},\) (a) the reaction will not proceed in the forward direction at all; (b) the displacement of \(\mathrm{Ni}(\mathrm{s})\) from the \(\mathrm{Ni}^{2+}(\mathrm{aq})\) will go to completion; (c) the displacement of \(\mathrm{Ni}(\mathrm{s})\) from the solution will proceed to a considerable extent, but the reaction will not go to completion; (d) there is no way to predict how far the reaction will proceed.

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