Perform a first-hand investigation to identify the conditions under which a galvanic cell is produced
This experiment is relatively easy, with simple results that are relatively easy to obtain. The only problem that may occur is incorrectly setting up the electrodes, so take care to ensure that the more reactive metal is being used as the anode. If not, the galvanic cell will not generate a current.
Set up the galvanic cell as shown in the diagram in dotpoint 1.4.2 on page 24.
There are a number of conditions which must be met in order for a galvanic cell to function.
- There must be an anode and cathode, each within their respective electrolytic solutions and physically
- A conducting circuit must connect the two half-cells to provide the only path by which electrons can
- A salt bridge must exist. The purpose of the salt bridge is to maintain electrical neutrality, as ions migrate from half-cell to half-cell in order to maintain a balance in the charges. If a charge imbalance built up (as would occur if a salt bridge was not present), the cell would stop functioning, as ions will accumulate in both electrolytes until the potential difference due to the ions is exactly opposite to the potential difference from the reaction. Thus the flow of electrons would be negated, and there would be no
Remember- The salt bridge exists to maintain electrical neutrality. As such, given that ions will be passing through the bridge and into the relevant electrolytes, ensure that a precipitate will not form. For this reason, is often used to soak the salt bridge, as both and will not form any precipitates regardless of which ions are present in the electrolyte.
Perform a first-hand investigation and gather first-hand information to measure the difference in potential of different combinations of metals in an electrolyte solution
This experiment is fairly useful for gaining an idea of how a galvanic cell works. Simply arrange the cell as shown in the diagram above, and connect a voltmeter to the external circuit between the two electrodes. Although this dotpoint does suggest a practical approach to determining the difference in potential of the different combinations of metal, for the purpose of clarity, a theoretical approach will be adopted here. You will be required to write both half equations and full equations, so take the time to learn how. Fortunately, the HSC data sheet lists pretty much all combinations you will be required to know, so it is simply a matter of choosing the right half equations.
With any given reaction that occurs in a galvanic cell, there will be two half-equations. Each half- equation has a standard potential E◦ (pronounced ‘e-naught’) value, a list of which will usually be provided at the back of the exam paper where necessary.
Although not essential, it may be useful to know that all E◦ values are calculated as the potential differences using a standard hydrogen electrode. This allows for a more convenient calculation of figures rather than recalculating the difference in potential between a pair of electrodes every time. You can see this quite easily by referring to the HSC data sheet, where the hydrogen electrode half equation has a potential of 0V.
Flipping the oxidation half-equation around, i.e. at the zinc electrode, and keeping the positions of the reduction reaction as is:
Why did we flip the oxidation half-equation around? Because the more reactive metal displace the less reaction metal and releases electrons. Hence the electrons are on the right side of the equation in the oxidation equation, while on the left side of the equation in the reduction equation.
Adding together the two half equations, the difference in potential of the two metals in this case is 1.10V. This should correspond to the value obtained when conducting the experiment. The full equation is then
Remember- The more reactive metal must thus be the anode, and is found above the cathode on the list of standard potentials in the data sheet, as the data sheet is sorted in descending order of reactivity.
In addition, remember to balance the half equations by their number of electrons. This is important as for some pairs such as sodium and zinc, sodium will have one electron in the products, yet zinc will have two electrons in the reactants. As such, double all species in the sodium half equation before writing the full equation.
Explain the displacement of metals from solution in terms of transfer of elec- trons
If you have difficulty understanding what is happening at this point, especially with the different relative activities, don’t worry too much as the next few dotpoints do make it clearer. For now, simply understand that a displacement reaction involves a transfer of electrons.
A displacement reaction, where metals are displaced from a solution, is a reaction in which there is a transfer of electrons between a metal and a metal ion. This occurs because each metal has a different relative activity.
Remember- Ionisation simply means that an atom is gaining or losing electrons. As such, if a metal ion becomes an atom, and a metal atom becomes an ion, then clearly all that has occurred is a transfer of electrons.
Identify the relationship between displacement of metal ions in solution by other metals to the relative activity of metals
Exactly which metal displaces what is often the main question that students will find themselves asking. This is conveniently summarised at the end of this dotpoint.
Of two different metals, the more reactive metal, X, will displace the other metal, Y, when it is an ionic solution. How reactive a metal is can be measured by how easily it oxidises. Potassium and sodium are highly reactive as they oxidise very easily.
Electrons will transfer from the metal X to the solution Y, resulting in X becoming a positive ion in solution, and Y turning back into a metal atom.
Remember- The more reactive metal displaces the less reactive metal.
Account for changes in the oxidation state of species in terms of their loss or gain of electrons
This dotpoint is really just introducing the concept behind redox reactions. Note that when one species loses electrons, another gains the same amount. The mnemonic at the end of this dotpoint is a simple way of remembering which process is what.
A change in the oxidation state of a species simply refers to its loss or gain in electrons. When a species is oxidised, it loses electrons (And thus its charge becomes more positive). When a species is reduced, it gains electrons (And thus its charge becomes more negative).
Reduction and oxidation reactions, or ‘redox’ reactions, occur simultaneously rather than indepen- dently of one another. As such, the electrons lost by one species are gained by another.
Remember- A simple mnemonic commonly used to remember this is ‘OILRIG’, or ‘Oxidation is Loss, Reduction Is Gain’.
Describe and explain galvanic cells in terms of oxidation/reduction reactions
Creating a galvanic cell is a great way of understanding what is happening here. .
A galvanic cell converts chemical energy into electrical energy. Several things must be noted regarding galvanic cells:
- Two electrodes (An oxidant and a reductant) , The anode releases electrons through oxidation reactions, which flow into the cathode through reduction reactions. These two electrodes are kept physically separated, but are joined by an external circuit so as to allow a charge to
A common mix-up that tends to confuse some students is what exactly an oxidant or reductant is. Simply keep in mind that an oxidant is a substance that is reduced, and a reductant is a substance that is oxidised. Oxidants and reductants are also known as oxidising agents and reducing agents respectively.
- Appropriate electrolytes to allow the flow of electrons within the two half-cells.
- A salt bridge, used to allow the migration of ions to maintain a balance of negative and positive charges in each half-cell. Anions migrate to the anode, and cations migrate to the cathode. This occurs in order to offset the flow of electrons to and from the cathode and anode respectively, thereby maintaining electrical
Remember- A galvanic cell is powered by the flow of electrons, i.e. redox reactions. So long as the two metal electrodes are present (physically separated by joined by an external circuit), appropriate electrolytes are present, and a salt bridge spans between the two electrolytes, then a current will flow.
Outline the construction of galvanic cells and trace the direction of electron flow
Although this dotpoint seeks only for a brief explanation of the structure of a galvanic cell. Learn the significance of each part, as well as how to draw it, and the construction of the galvanic cell will be simple to describe.
If you have difficulty remembering what reaction occurs at which electrode, use the term ‘AnOx’ to remember that oxidation occurs at the anode. Once you remember this, it follows that at the cathode, the reduction reaction occurs. Otherwise, ‘Red Cat’ can be used to remember that reduction occurs at the cathode.
Define the terms anode, cathode, electrode and electrolyte to describe galvanic cells
Again, learn how to draw the galvanic cell, and you will find recounting each part of the galvanic cell a relatively simple task.
- The anode is the electrode at which the oxidation reaction occurs, releasing electrons into the external
- The cathode is the electrode at which the reduction reaction occurs, consuming electrons which have been released into the external
- An electrode is a conductor connected to the external circuit through which electrons may
- An electrolyte is a substance, either a solution or a solid in a molten state, which accommodates the flow of electrons and hence conducts
Solve problems and analyse information to calculate the potential requirement of named electrochemical processes using tables of standard potentials and half- equations
Calculating the E◦ value, pronounced ‘E-naught’, is simply a matter of finding the difference in potential.
Gather and present information on the structure and chemistry of a dry cell or lead-acid cell and evaluate it in comparison to one of the following: button cell, fuel cell, vanadium redox cell, lithium cell, liquid junction photovoltaic device (eg. the Gratzel cell) in terms of: chemistry, cost and practicality, impact on society, environmental impact
Unfortunately this dotpoint, even if you understand the concepts behind the galvanic cell, is ultimately an exercise in memorisation. Choosing any of the above cells is fine, but for the purposes of this guide, the dry cell and the silver oxide button cell will be examined.
The Dry Cell (Also known as the Leclanch´e Cell)
Zinc casing and a graphite rod serve as the anode and cathode respectively, and the electrolyte consists of a paste of ammonium chloride and zinc chloride is the electrolyte. Manganese dioxide is also present within the cell.
The zinc anode oxidises according to:
The electrons produced reduce the ammonium ion at the cathode.
Manganese dioxide converts the hydrogen gas to water as it is an oxidising agent.
The ammonia gas reacts with the ion.
These reactions have a net potential of 1.5 volts.
Cost and practicality
The dry cell is one of the most common and cheapest of the commercially available cells and is widely used in small or portable devices. Disadvantages of the cell are that it does not give a very large output given its size (both voltage and currents) and can leak when the zinc casing anode is gradually corroded. Relative to other cells, the dry cell also has a short shelf life, as the acidic ammonium ions reacts with the zinc casing. As this occurs, the voltage drops rapidly, after which the battery is said to go ‘flat’.
Impact on society
The dry cell was the first commercially available battery, and as such most of the earliest devices were modelled around it. Offering a small, portable power source, the dry cell had an enormous impact on society, and still offers a useful power source for low-current devices such as torches.
The dry cell is relatively safe for the environment. Manganese (III) oxide oxidises to the harmless manganese (IV) oxide, ammonium salts are harmless, and carbon is also relatively harmless. Although large quantities of zinc may pose environmental problems, particularly when it leaches into the soil, small quantities generally do pose a problem.
The Silver Oxide Button Cell
The silver oxide button cell has essentially the same structure as the dry cell, but uses different chemicals inside a smaller case.
In this cell, zinc and silver oxide serve as the anode and cathode respectively. An alkaline substance, usually either sodium hydroxide or potassium hydroxide, serves as the electrolyte. The following oxidation half-equation takes place:
Silver oxide then reduces to silver:
These reactions have a net potential of 1.5 volts.
Cost and practicality
The silver oxide button cell has found its ideal use within cameras and watches, due to its small size but relatively high (1.5V) output. Its ability to produce a very steady output is also idea for medical equipment such as pacemakers and hearing aids. In such instruments, its size and reliability take precedence over cost.
Impact on society
Considerably smaller that the dry cell while matching its output, the silver oxide button cell has had a large impact on society as smaller devices are able to make use of the batteries. The high reliability of the cell has also had a large impact upon the medical industry, as the life spans of various medical devices such as pacemakers have been extended. The size of the cell has also drastic implications, most noticeably through everyday items such as watches.
As with the dry cell, small amounts of zinc do not pose an environmental problem. Similarly, silver oxide cells do not produce any highly toxic wastes. This is in contrast with the mercury button cell, which is more effective through its control of zinc corrosion, yet has a much more harmful effect upon the environment due to the use of mercury.
All batteries still have an anode, cathode, and various electrolytes even if they look markedly different from other batteries. You will be required to write down all relevant half equations, so take care to learn them thoroughly.