Worked solutions to textbook questions 13
Chapter 27 Cells and batteries
Q1.
One major disadvantage of the dry cell was that the zinc casing would begin to disintegrate if it was left operating in an appliance for several weeks, potentially causing damage to the appliance. Explain why this occurs.
A1.
In dry cells the zinc forms the casing. Since zinc is oxidised during the cell reaction, holes may develop in the case and the contents leak out.
Q2.
Although very small alkaline cells can be made, they are not used in watches. Why are button cells preferred for this application?
A2.
The electronic circuitry in watches requires an almost constant voltage for its operation. Alkaline cells do not provide a constant voltage, unlike silver–zinc and mercury–zinc button cells.
Q3.
Write overall equations for the cell reactions in the dry cell, alkaline cell, and lithium and silver–zinc button cells using the anode and cathode reactions given in the text.
A3.
dry cell: Zn(s) + 2MnO2(s) + 2NH4+(aq) ® Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l)
alkaline cell: Zn(s) + 2MnO2(s) + H2O(l) ® Zn(OH)2(s) + Mn2O3(s)
lithium cell: Li + MnO2 → LiMnO2
silver–zinc cell: Zn(s) + Ag2O(s) + H2O(l) ® Zn(OH)2(s) + 2Ag(s)
Q4.
Describe the energy transformations that occur when a galvanic cell discharges.
A4.
chemical energy ® electrical energy ® other forms of energy, depending upon the application for the cell
Q5.
A galvanic cell is constructed by connecting a Cl2/Cl– half cell to a Zn2+/Zn half cell using a salt bridge at standard conditions. You will need to refer to Table 26.2 (p.418).
a Write an overall equation for the cell.
b What is the polarity (positive or negative) of the zinc electrode?
c Calculate the potential difference of the cell.
d What happens to the concentrations of Cl– and Zn2+ ions when electric current is drawn from the cell?
e Explain the purpose of the salt bridge in the cell.
A5.
a Zn(s) + Cl2(g) ® Zn2+(aq) + 2Cl–(aq)
b Zinc is oxidised, generating negatively charged electrons. Hence the zinc electrode is negative.
c cell potential difference = higher half cell Eo – lower half cell Eo
= Cl2/Cl– half cell Eo – Zn2+/Zn half cell Eo
= 1.36 – (–0.76) V
= 2.12 V
d Both increase as the reaction proceeds in the forward direction until equilibrium is reached.
e Movement of ions through the salt bridge balances charges that are formed in the half cells during the cell reaction. This movement of charge completes the flow of current in the circuit.
E1.
Write equations for the reactions that occur at the anode and the cathode of the sodium–sulfur cell as it discharges.
AE1.
Anode reaction Na(l) ® Na+(l) + e–
Cathode reaction S(l) + 2e– ® S2–(l)
E2.
Write an overall equation for the cell reaction during:
a discharge
b recharge
AE2.
The overall reaction that occurs when the cell discharges is obtained by adding the anode and cathode reactions. In this case the anode reaction must first be doubled so that there equal number of electrons on both sides of the equation
2Na(l) + S(l) + 2e– ® S2–(l) + 2Na+(l) + 2e–
This equation simplifies to
2Na(l) + S(l) ® S2–(l) + 2Na+(l)
The reverse reaction occurs when the cell is recharged.
S2–(l) + 2Na+(l) ® 2Na(l) + S(l)
E3.
Why do you think it is necessary for the cells to be heated in order for them to produce a voltage?
AE3.
The operation of the cells requires that sodium anode, sulfur cathode and electrolyte are all in the molten state.
E4.
Provided the cells are discharged and charged every day, they remain at working temperatures. Suggest the reason for this.
AE4.
During recharging, some of the electrical energy entering the cell is transformed into chemical energy and some is transformed into heat. Heat energy is also released in the cell when it discharges. This heat is sufficient to maintain the cell at working temperatures if a discharge–recharge cycle is carried out every day.
E5.
According to the article, what are the disadvantages of electric vehicles compared with petrol-driven vehicles at present?
AE5.
Electric vehicles do not generate carbon dioxide, which is a greenhouse gas.
E6.
Suggest the meaning of the term ‘energy density’. Why is energy density an important factor in the construction of electric vehicles?
AE6.
The energy density of a cell is the energy available per unit mass of the cell. The greater the energy density of a cell, the greater the range of the electric vehicle that is using the cell.
E7.
‘Electric vehicles cause no pollution.’ Comment on the accuracy of this statement.
AE7.
While electric vehicles cause almost no pollution at the site at which they are used, the electrical energy that is used to charge their batteries is produced by a series of energy transformations that involve the production of pollutants. Typically, fossil fuels will be burnt in order to produce the electrical energy.
E8.
Suppose your family operated an electric car that used lead–acid batteries. In what ways would this affect your family’s lifestyle?
AE8.
Owning an electric car, rather than a petrol-driven car, might restrict your range and frequency of travel. You would save money by not having to buy petrol, and maintenance costs (e.g. oil changes, tune-ups) could also decrease, but cost would be involved in replacing your battery every few years.
Q6.
The pH of the electrolyte in a lead–acid battery varies according to the state of charge. Describe how the pH would change as a battery discharged and was recharged.
A6.
As a lead–acid battery is discharged, H+ ions are consumed. As the concentration of H+ drops, the pH of the electrolyte will increase. The opposite occurs when a lead–acid battery is recharged. As the concentration of H+ increases, the pH will decrease.
Q7.
A lead–acid battery uses 50 g of lead to provide the electrical energy needed to start a typical car. Calculate the mass of lead oxide consumed at the positive electrode and the total mass of lead sulfate produced in the battery while starting the car.
A7.
Step 1 Write a balanced equation.
Pb(s) + PbO2(s) + 2SO42–(aq) + 4H+(aq) ® 2PbSO4(s) + 2H2O(l)
Step 2 From the equation, 1 mol of PbO2 reacts with 1 mol Pb. Use stoichiometry to calculate the amount of PbO2.
=
n(PbO2) = n(Pb)
=
= 0.24 mol
Step 3 Calculate the mass of PbO2.
m(PbO2) = 0.24 mol ´ 239.19 g mol–1
= 57.51 g
= 58 g
Step 4 From the equation, 2 mol PbSO4 is produced by 1 mol Pb. Use stoichiometry to calculate the amount of PbSO4.
=
n(PbSO4) = 2 ´ n(Pb)
= 2 ´
= 0.48 mol
Step 5 Calculate the mass of PbSO4.
m(PbSO4) = 0.48 mol ´ 303.19 g mol–1
= 145.5 g
= 1.5 ´ 102 g
Q8.
A mechanic wishes to recharge a car battery using a battery charger.
a Which battery terminals should the positive and negative terminals of the battery charger be connected to?
b What would be the effect on the battery if the charger were connected incorrectly?
A8.
a The positive terminal of the charger is connected to the battery’s positive terminal and the negative terminal of the charger is connected to the battery’s negative terminal.
b Connecting the battery charger to the battery so that terminals of opposite polarity are linked will force the spontaneous cell reaction to occur at a faster rate, further flattening the battery and possibly damaging it irreparably.
Q9.
Why is large-scale production of electricity using fuel cells suggested as part of a solution to the greenhouse problem?
A9.
Fuel cells are about twice as efficient as coal-fired power stations. Consequently, fuel cells produce the same quantity of energy using about half as much fuel. Less fuel means less carbon dioxide gas is produced. As carbon dioxide is a major greenhouse gas, the use of fuel cells has the potential to reduce the greenhouse effect.
Q10.
The use of hydrogen-powered fuel cells in public transport is being considered in many cities around the world. What are the advantages and disadvantages of using fuel cells for this purpose?
A10.
Hydrogen powered fuel cells generate water as the only product and do not generate noise when in operation. Wide spread use of fuel cells requires extensive new infrastructure such a generation plant, storage facilities, specialised transport and distribution outlets.
Chapter review
Q11.
When constructing a galvanic cell in the laboratory, two separate containers are generally used for the half cells. However, in commercial cells such as the dry cell, all the reactants are placed in the one container.
a How are commercial cells constructed so that the use of two containers is avoided?
b Design a cell that could be constructed in one container in the laboratory and which uses the same cell reaction as in the Daniell cell (p. 415).
A11.
a Most commercial cells contain an oxidant and reductant in the form of a solid or moist powder; each is unable to migrate and react directly with the other. The electrolyte is usually a paste, allowing ions to move.
b
Q12.
Explain the difference between the terms:
a positive electrode and negative electrode
b cell and battery
c primary cell and secondary cell
d discharge and recharge
A12.
a The negative electrode anode in a galvanic cell or is the electrode at which oxidation occurs. The positive electrode or cathode is the electrode at which reduction occurs.
b A battery consists of a number of cells connected together in series.
c A primary cell cannot be recharged and is discarded once it no longer supplies electrical energy. A secondary cell can be connected to a battery charger to reverse the cell reaction. Once the cell has been recharged, it can again be used to supply electrical energy.
d When a cell discharges chemical energy is converted to electrical energy. When it is recharged, electrical energy is converted to chemical energy.
Q13.
What feature enables secondary cells such as lithium ion cells to be recharged?
A13.
In secondary or rechargeable cells such as lithium ion cells or a lead acid battery, the products of cell reaction remain in contact with the electrodes and are in forms that can be converted back to reactants.
Q14.
Explain what happens to the oxidant, reductant, anode and cathode when a secondary cell is switched from discharge to recharge.
A14.
When a cell discharges, oxidation occurs at the anode which has a negative polarity. Reduction occurs at the cathode which has positive polarity. The oxidation and reduction reactions in the cell generate electricity. When the cell is recharged the direction of the electron flow is reversed. An external supply of electricity is required to reverse these reactions. Oxidation now occurs at the positive electrode while reduction occurs at the negative electrode.
Q15.
Both dry and alkaline cells use powdered graphite mixed with a manganese dioxide oxidant. From your knowledge of the properties of graphite, suggest what its role might be in these cells.
A15.
Graphite is a good conductor of electricity because it contains delocalised electrons between the layers of carbon atoms. Graphite is incorporated in the cells to conduct electrons to the manganese dioxide oxidant.
Q16.
Describe an application where each of the following would be used:
a a nickel–metal hydride cell
b a lead–acid accumulator
c a lithium button cell
d a dry cell
e an alkaline cell
A16.
a an appliance that draws a moderate current, e.g. camera flash, portable cassette player, remote controlled car
b starting internal combustion engines in motor bikes, cars, trucks, and buses
c an appliance with a small current drain that requires a constant voltage, e.g. watch, camera, hearing aid, calculator
d an infrequently used device that requires a cheap power source, e.g. glove box torch, transistor radio
e an appliance with a moderate to high current drain that is not used extensively, e.g. electric toothbrush, portable cassette player
Q17.
A manual car with a flat battery can often be started by pushing, but this is not done with an automatic car. Find out why a manual car can be started in this way but an automatic cannot.
A17.
The role of the battery is to turn the starter motor which, in turn, cranks the engine. Pushing a manual car that has its gears engaged will cause the motor to turn and, subsequently, cause the alternator to produce an electric current to fire the spark plugs and start the engine.
An automatic car has no direct coupling between the engine and transmission; a fluid coupling system is used to drive the transmission when the engine is running.
Q18.
Mains electricity costs about 5 cents per MJ of energy, or less, depending on the tariff. The cost of the same amount of electrical energy produced by a cell is far more—about $1300 for a dry cell and even more for a button cell. Why are people prepared to use cells and pay such relatively high prices for electricity?
A18.
While the cost of electrical energy purchased in the form of a dry cell or button cell is far higher than the cost of mains electricity, people are prepared to pay the higher price for the convenience and flexibility of the portable equipment powered by these cells. Furthermore, the price of individual cells is regarded as relatively low.
Q19.
A car has been powered using an experimental ten-cell aluminium–air battery. Each of the ten cells contains aluminium plates and sodium hydroxide solution, and air is fed into the battery. New aluminium plates and water must be added regularly to the battery.
a Write half equations for the anode and cathode reactions in the battery.
b Draw a diagram to show how each cell might be constructed.
c Give two reasons why this battery weighs less than a similar lead–acid battery.
d Why can this battery be described as a fuel cell?
A19.
a anode: Al(s) ® Al3+(aq) + 3e–; cathode: O2(g) + 2H2O(l) + 4e– ® 4OH–(aq)
b