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Chapter 20: Problem 105

The capacity of batteries such as the typical AA alkaline battery is expressed in units of milliamp-hours (mAh). An AA alkaline battery yields a nominal capacity of 2850 mAh. (a) What quantity of interest to the consumer is being expressed by the units of mAh? (b) The starting voltage of a fresh alkaline battery is 1.55 V. The voltage decreases during discharge and is 0.80 \(\mathrm{V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

Short Answer

Expert verified
a) The quantity of interest to the consumer expressed by the units of milliamp-hours (mAh) is the battery's capacity, which indicates how long a device powered by the battery will last. b) The maximum electrical work the AA alkaline battery could perform during discharge, considering the average voltage of 1.175 V, is approximately 3.35 watt-hours.

Step by step solution

01

Understanding the meaning of milliamp-hours (mAh)

The unit of milliamp-hours (mAh) is used to express the capacity of batteries. It describes the amount of current a battery can supply for a specific period. Specifically, it tells us how many milliamperes (mA) of current the battery can supply continuously for 1 hour. In this exercise, we are given that a typical AA alkaline battery has a capacity of 2850 mAh, which means the battery can supply a current of 2850 mA for 1 hour. This is of interest to the consumer because it indicates how long a device powered by this battery will last.
02

Calculate the average voltage during discharge

We are given that the starting voltage of the battery is 1.55 V, and the final voltage is 0.80 V when the battery has delivered its rated capacity. We can calculate the average voltage during discharge by finding the midpoint of the starting and final voltage: Average voltage = \(\frac{1.55\,\text{V} + 0.80\,\text{V}}{2}\) Average voltage = \(1.175\,\text{V}\)
03

Calculate the total charge delivered by the battery

To find the total charge (Q) delivered by the battery, we can use the formula: Q = capacity × time We know the capacity is 2850 mAh and since our battery is rated for 1 hour, the time is 1 hour. However, to make our units consistent, we need to convert the capacity from milliamperes to amperes: 2850 mA = 2.85 A Now we can calculate the total charge: Q = 2.85 A × 1 hour = 2.85 Ah (ampere-hours)
04

Calculate the maximum electrical work performed by the battery

Finally, to find the maximum electrical work (W) the battery can perform during discharge, we can use the formula for electrical work: W = Q × ΔV We know the total charge and the average voltage. So, we can calculate the electrical work: W = 2.85 Ah × 1.175 V W = 3.34875 Wh (watt-hours) The maximum electrical work the AA alkaline battery could perform during discharge is approximately 3.35 watt-hours.

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Key Concepts

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

milliamp-hours (mAh)
A core concept in understanding battery capacity is milliamp-hours, abbreviated as mAh. This unit tells us about the amount of electric charge a battery can store and deliver. Simply put, it indicates how many milliamperes (mA) of current a battery can supply continuously for 1 hour.
For example, a battery with a capacity of 2850 mAh can theoretically supply a current of 2850 mA for 1 hour, or it could supply a smaller current for a longer period of time. This is an appealing figure for consumers because it gives an estimate of how long the battery will power their devices.
  • A higher mAh value usually means a longer runtime for the device.
  • This specification is particularly important in devices that require a lot of power such as cameras or toys.
Understanding this unit helps users compare different batteries' capabilities and predict usage times.
average voltage
The average voltage of a battery is an essential factor in determining the quality of its discharge. Initially, a new AA alkaline battery might start at a voltage of 1.55 V, but this voltage decreases over time as the battery discharges. To evaluate the total discharge behavior of a battery, we need to calculate the average voltage throughout its discharge period.
In our exercise, the voltage decreases linearly from 1.55 V to 0.80 V. To find the average voltage, we take the midpoint of these two values: i.e., \[\text{Average voltage} = \frac{1.55\, \text{V} + 0.80\, \text{V}}{2} = 1.175\, \text{V}\] This value helps in calculating other important factors such as the electrical work the battery can perform.
By understanding the role of average voltage, users can better determine the efficiency and performance longevity of their batteries.
electrical work
Electrical work is the energy that the battery can deliver during its discharge. It's calculated using the formula for electrical work, which involves multiplying the total charge (Q) by the average voltage available during discharge. In formula terms, it can be expressed as:
\[W = Q \times \Delta V\] Where - \(W\) is the electrical work in watt-hours. - \(Q\) is the total charge, expressed in ampere-hours (Ah). - \(\Delta V\) is the average voltage.
In the given problem, the total charge is 2.85 Ah and the average voltage is 1.175 V, which results in:\[W = 2.85 \, \text{Ah} \times 1.175 \, \text{V} = 3.35 \, \text{Wh}\] Thus, the AA alkaline battery can perform approximately 3.35 watt-hours of electrical work. This measurement informs us of the total energy output and potential efficiency the battery delivers to your device.
discharge
Discharge refers to the process of a battery releasing its stored energy to power a device. Understanding the discharge curve is crucial because it shows how the voltage of the battery changes over time until it is fully spent.
In the context of our exercise, discharge implies the gradual reduction of voltage from its initial value as it delivers electrical work. The linear decline from 1.55 V down to 0.80 V represents this discharge phase. It is crucial to note:
  • The rate of discharge can affect how long the battery will last.
  • Knowing when a battery will reach its end voltage can help optimize performance and efficiency.
Properly managing discharge ensures that the battery life is used effectively, maximizing usage times before a replacement or recharge is necessary.

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Most popular questions from this chapter

If the equilibrium constant for a one-electron redox reaction at 298 \(\mathrm{K}\) is \(8.7 \times 10^{4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8}\) mol of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ., \(1973,\) Vol. \(50,61 )\) . (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage \((2\) \(\mathrm{mi}\) is 300 \(\mathrm{atm} ?(\mathbf{c})\) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

In some applications nickel-cadmium batteries have been replaced by nickel- zinc batteries. The overall cell reaction for this relatively new battery is: $$ \begin{aligned} 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{Zn}(s) & \\ & \longrightarrow 2 \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Zn}(\mathrm{OH})_{2}(s) \end{aligned} $$ (a)What is the cathode half-reaction? (b)What is the anode half-reaction? (c) A single nickel-cadmium cell has a voltage of 1.30 \(\mathrm{V}\) . Based on the difference in the standard reduction potentials of \(\mathrm{Cd}^{2+}\) and \(\mathrm{Zn}^{2+},\) what voltage would you estimate a nickel-zinc battery will produce? (d) Would you expect the specific energy density of a nickel-zinc battery to be higher or lower than that of a nickel-cadmium battery?

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?

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \operatorname{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 \mathrm{M},\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cl}^{-}\right]\) will increase, decrease, or stay the same as the cell operates.
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