Electrochemistry and batteries

You will remember from Chapter 4 that a galvanic cell (also known as a voltaic cell) is a type of electrochemical cell where a chemical reaction produces electrical energy. The electromotive force (emf) of a galvanic cell is the difference in voltage between the two half cells that make it up. Galvanic cells have a number of applications, but one of the most important is their use in batteries. You will know from your own experience that we use batteries in a number of ways, including cars, torches, sound systems and cellphones to name just a few.

How batteries work

A battery is a device in which chemical energy is directly converted to electrical energy. It consists of one or more voltaic cells, each of which is made up of two half cells that are connected in series by a conductive electrolyte. The voltaic cells are connected in series in a battery. Each cell has a positive electrode (cathode), and a negative electrode (anode). These do not touch each other but are immersed in a solid or liquid electrolyte.

Each half cell has a net electromotive force (emf) or voltage. The voltage of the battery is the difference between the voltages of the half-cells. This potential difference between the two half cells is what causes an electric current to flow.

Batteries are usually divided into two broad classes:

Primary batteries irreversibly transform chemical energy to electrical energy. Once the supply of reactants has been used up, the battery can't be used any more.
Secondary batteries can be recharged, in other words, their chemical reactions can be reversed if electrical energy is supplied to the cell. Through this process, the cell returns to its original state. Secondary batteries can't be recharged forever because there is a gradual loss of the active materials and electrolyte. Internal corrosion can also take place.

Battery capacity and energy

The capacity of a battery, in other words its ability to produce an electric charge, depends on a number of factors. These include:

Chemical reactions

The chemical reactions that take place in each of a battery's half cells will affect the voltage across the cell, and therefore also its capacity. For example, nickel-cadmium ( $NiCd$ ) cells measure about 1,2 V, and alkaline and carbon-zinc cells both measure about 1,5 V. However, in other cells such as Lithium cells, the changes in electrochemical potential are much higher because of the reactions of lithium compounds, and so lithium cells can produce as much as 3 volts or more. The concentration of the chemicals that are involved will also affect a battery's capacity. The higher the concentration of the chemicals, the greater the capacity of the battery.
Quantity of electrolyte and electrode material in cell

The greater the amount of electrolyte in the cell, the greater its capacity. In other words, even if the chemistry in two cells is the same, a larger cell will have a greater capacity than a small one. Also, the greater the surface area of the electrodes, the greater will be the capacity of the cell.
Discharge conditions

A unit called an Ampere hour (A·h) is used to describe how long a battery will last. An ampere hour (more commonly known as an amp hour) is the amount of electric charge that is transferred by a current of one ampere for one hour. Battery manufacturers use a standard method to rate their batteries. So, for example, a 100 A·h battery will provide a current of 5 A for a period of 20 hours at room temperature. The capacity of the battery will depend on the rate at which it is discharged or used. If a 100 A·h battery is discharged at 50 A (instead of 5 A), the capacity will be lower than expected and the battery will run out before the expected 2 hours.

The relationship between the current, discharge time and capacity of a battery is expressed by Peukert's law:

$\begin{matrix} C_{p} = I^{k} t \end{matrix}$ (1)

In the equation, ' $C_{p}$ ' represents the battery's capacity (A·h), I is the discharge current (A), k is the Peukert constant and t is the time of discharge (hours).

Lead-acid batteries

In a lead-acid battery, each cell consists of electrodes of lead ( $Pb$ ) and lead (IV) oxide ( ${PbO}_{2}$ ) in an electrolyte of sulfuric acid ( $H_{2} {SO}_{4}$ ). When the battery discharges, both electrodes turn into lead (II) sulphate ( ${PbSO}_{4}$ ) and the electrolyte loses sulfuric acid to become mostly water.

The chemical half reactions that take place at the anode and cathode when the battery is discharging are as follows:

Anode (oxidation): $Pb (s) + {SO}_{4}^{2 -} (aq) ⇋ {PbSO}_{4} (s) + 2 e^{-} (E^{0} = - 0.356 V)$

Cathode (reduction): ${PbO}_{2} (s) + {SO}_{4}^{2 -} (aq) + 4 H^{+} + 2 e^{-} ⇋ {PbSO}_{4} (s) + 2 H_{2} O (l) (E^{0} = 1.685 V)$

The overall reaction is as follows:

${PbO}_{2} (s) + 4 H^{+} (aq) + 2 {SO}_{4}^{2 -} (aq) + Pb (s) \to 2 {PbSO}_{4} (s) + 2 H_{2} O (l)$

The emf of the cell is calculated as follows:

$EMF = E^{0} (cathode) - E^{0} (anode)$

$EMF = + 1,685 V - (- 0,356 V)$

$EMF = + 2,041 V$

Since most batteries consist of six cells, the total voltage of the battery is approximately 12 V.

One of the important things about a lead-acid battery is that it can be recharged. The recharge reactions are the reverse of those when the battery is discharging.

The lead-acid battery is made up of a number of plates that maximise the surface area on which chemical reactions can take place. Each plate is a rectangular grid, with a series of holes in it. The holes are filled with a mixture of lead and sulfuric acid. This paste is pressed into the holes and the plates are then stacked together, with suitable separators between them. They are then placed in the battery container, after which acid is added (Figure 1).

**Figure 1:** A lead-acid battery

Lead-acid batteries have a number of applications. They can supply high surge currents, are relatively cheap, have a long shelf life and can be recharged. They are ideal for use in cars, where they provide the high current that is needed by the starter motor. They are also used in forklifts and as standby power sources in telecommunication facilities, generating stations and computer data centres. One of the disadvantages of this type of battery is that the battery's lead must be recycled so that the environment doesn't become contaminated. Also, sometimes when the battery is charging, hydrogen gas is generated at the cathode and this can cause a small explosion if the gas comes into contact with a spark.

The zinc-carbon dry cell

A simplified diagram of a zinc-carbon cell is shown in Figure 2.

**Figure 2:** A zinc-carbon dry cell

A zinc-carbon cell is made up of an outer zinc container, which acts as the anode. The cathode is the central carbon rod, surrounded by a mixture of carbon and manganese (IV) oxide ( ${MnO}_{2}$ ). The electrolyte is a paste of ammonium chloride ( ${NH}_{4} Cl$ ). A fibrous fabric separates the two electrodes, and a brass pin in the centre of the cell conducts electricity to the outside circuit.

The paste of ammonium chloride reacts according to the following half-reaction:

$2 {NH}_{4}^{+} (aq) + 2 e^{-} \to 2 {NH}_{3} (g) + H_{2} (g)$

The manganese(IV) oxide in the cell removes the hydrogen produced above, according to the following reaction:

$2 {MnO}_{2} (s) + H_{2} (g) \to {Mn}_{2} O_{3} (s) + H_{2} O (l)$

The combined result of these two reactions can be represented by the following half reaction, which takes place at the cathode:

Cathode: $2 {NH}_{4}^{+} (aq) + 2 {MnO}_{2} (s) + 2 e^{-} \to {Mn}_{2} O_{3} (s) + 2 {NH}_{3} (g) + H_{2} O (l)$

The anode half reaction is as follows:

Anode: $Zn (s) \to {Zn}^{2 +} + 2 e^{-}$

The overall equation for the cell is:

$Zn (s) + 2 {MnO}_{2} (s) + 2 {NH}_{4}^{+} \to {Mn}_{2} O_{3} (s) + H_{2} O + Zn {({NH}_{3})}_{2}^{2 +} (aq) (E^{0} = 1.5 V$ )

Alkaline batteries are almost the same as zinc-carbon batteries, except that the electrolyte is potassium hydroxide ( $KOH$ ), rather than ammonium chloride. The two half reactions in an alkaline battery are as follows:

Anode: $Zn (s) + 2 {OH}^{-} (aq) \to Zn {(OH)}_{2} (s) + 2 e^{-}$

Cathode: $2 {MnO}_{2} (s) + H_{2} O (l) + 2 e^{-} \to {Mn}_{2} O_{3} (s) + 2 {OH}^{-} (aq)$

Zinc-carbon and alkaline batteries are cheap primary batteries and are therefore very useful in appliances such as remote controls, torches and radios where the power drain is not too high. The disadvantages are that these batteries can't be recycled and can leak. They also have a short shelf life. Alkaline batteries last longer than zinc-carbon batteries.

Environmental considerations

While batteries are very convenient to use, they can cause a lot of damage to the environment. They use lots of valuable resources as well as some potentially hazardous chemicals such as lead, mercury and cadmium. Attempts are now being made to recycle the different parts of batteries so that they are not disposed of in the environment, where they could get into water supplies, rivers and other ecosystems.

Exercise 1: Electrochemistry and batteries

Problem 1:

A dry cell does not contain a liquid electrolyte. The electrolyte in a typical zinc-carbon cell is a moist paste of ammonium chloride and zinc chloride.

The paste of ammonium chloride reacts according to the following half-reaction:

$2 {NH}_{4}^{+} (aq) + 2 e^{-} \to 2 {NH}_{3} (g) + H_{2} (g)$ (a)

Manganese(IV) oxide is included in the cell to remove the hydrogen produced during half-reaction (a), according to the following reaction:

$2 {MnO}_{2} (s) + H_{2} (g) \to {Mn}_{2} O_{3} (s) + H_{2} O (l)$ (b)

The combined result of these two half-reactions can be represented by the following half reaction:

$2 {NH}_{4}^{+} (aq) + 2 {MnO}_{2} (s) + 2 e^{-} \to {Mn}_{2} O_{3} (s) + 2 {NH}_{3} (g) + H_{2} O (l)$ (c)

Explain why it is important that the hydrogen produced in half-reaction (a) is removed by the manganese(IV) oxide.

In a zinc-carbon cell, such as the one above, half-reaction (c) and the half-reaction that takes place in the $Zn$ | ${Zn}^{2 +}$ half-cell, produce an emf of 1,5 V under standard conditions.

Answer 1:

The hydrogen that is produced must be removed to drive the reaction in the correct direction.

Problem 2:

Write down the half-reaction occurring at the anode.

Answer 2:

$Zn(s) \to {Zn}^{2 +} + 2 e^{-}$

Problem 3:

Write down the net ionic equation occurring in the zinc-carbon cell.

Answer 3:

$Zn(s) + 2 {MnO}_{2} (s) + 2 {NH}_{4}^{+} \to {Mn}_{2} O_{3} (s) + H_{2} O + {Zn(NH}_{3})_{2}^{2 +}$

Problem 4:

Calculate the reduction potential for the cathode half-reaction.

Answer 4:

$E_{cell}^{0} = E_{cathode} - E_{anode}$

$E_{cathode} = 1, 5 + (- 0, 76) = 0, 74 V$

Problem 5:

When in use the zinc casing of the dry cell becomes thinner, because it is oxidised. When not in use, it still corrodes. Give a reason for the latter observation.

Answer 5:

The oxygen in the air causes the zinc to be oxidised and so corrodes.

Problem 6:

Dry cells are generally discarded when 'flat'. Why is the carbon rod the most useful part of the cell, even when the cell is flat?

(DoE Exemplar Paper 2, 2007)

Answer 6:

The carbon rod has not undergone any chemical reaction and so can easily be reused either in new batteries or in other applications.

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