electrolysis of lead bromide experiment

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Identifying the products of electrolysis

In association with Nuffield Foundation

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Try this class experiment to carry out the electrolysis of various solutions and investigate the products formed

In this practical, students get the chance to practise doing electrolysis with a range of solutions, identifying the products that form at the electrodes. They should be able to link their practical experience with theory and learn how to construct simple ionic equations.

The class experiment is best done by students working in pairs or threes. There are plenty of tasks for each member of the team to complete.

It should be carried out in a well-ventilated laboratory as significant amounts of toxic chlorine, bromine and iodine can be produced in some cases, as well as highly flammable hydrogen.

• Eye protection
• Electrolysis apparatus (see note 1, and the diagram below)
• Graphite electrodes, about 5 mm diameter, x2 (see note 2)
• Large rubber bung to fit electrolysis cell, with holes to carry the graphite electrodes
• Small test tubes to fit over the electrodes, x2
• DC power supply, 6 V
• Small light bulb in holder, 6V 5W (optional)
• Leads and crocodile clips
• Wooden splints
• Small pieces of emery paper
• Strips of universal indicator paper
• Disposable plastic gloves
• Clamp and stand

Access to the following solutions, all approximately 0.5 M concentration (see note 3):

• Aqueous potassium bromide
• Aqueous sodium iodide
• Aqueous calcium nitrate
• Aqueous zinc chloride (IRRITANT at this concentration)
• Aqueous copper nitrate

Equipment notes

• The electrolysis apparatus shown below can be purchased ready-made. Alternatively, it can be made from thick glass tubing of 8–10 cm diameter, professionally cut into lengths of about 12 cm. A suitable glass bottle, with a wide-necked top and its base cut off, could be used instead (as shown in the diagram). The graphite rods should be well sealed into the holes, 2–3 cm apart, of the rubber bung, otherwise the electrolyte may leak onto the external wiring, causing it to corrode. Once made, this apparatus should last for several sessions, but the graphite rods tend to erode away quite quickly, particularly if students use larger than recommended voltages. The rods do eventually become thin and snap fairly easily, but they are cheap enough to replace.

Source: Royal Society of Chemistry

The equipment required for the electrolysis cell

• Once copper(II) nitrate has been electrolysed (preferably last), a deposit of copper will have formed on the cathode. This has to be removed before the cells can be used again. Immersing the plated part of the electrode in a small quantity of 50% concentrated nitric acid (CORROSIVE) in a small beaker can be used to do this. Gloves and eye protection should be worn and the cleaning done in a fume cupboard by a suitably qualified person.
• Depending on the volume of the electrolysis apparatus, each group of students needs enough solution to cover the electrodes plus about 2 cm to enable the full test tubes of liquid to be inverted over the electrodes.

Health, safety and technical notes

• Read our standard health and safety guidance.
• Wear eye protection throughout.
• The experiment should be carried out in a well-ventilated laboratory as it may produce significant quantities of toxic chlorine, bromine and iodine, as well as highly flammable hydrogen.
• Aqueous potassium bromide, KBr(aq) – see CLEAPSS Hazcard  HC047b .
• Aqueous sodium iodide, NaI(aq) – see CLEAPSS Hazcard HC047b.
• Aqueous calcium nitrate, Ca(NO 3 ) 2 (aq) – see CLEAPSS Hazcard  HC019B .
• Aqueous zinc chloride, ZnCl 2 (aq), (IRRITANT at concentration used) – see CLEAPSS Hazcard  HC108a .
• Aqueous copper nitrate, Cu(NO 3 ) 2 (aq) – see CLEAPSS Hazcard  HC027B .
• 50% concentrated nitric acid, HNO 3 (aq), (CORROSIVE) – see CLEAPSS Hazcard  HC067  and CLEAPSS Recipe Book RB061. This is used to clean copper off the electrodes, and is made by slowly adding concentrated nitric acid to an equal volume of water.
• Chlorine, Cl 2 (g), (TOXIC) – see CLEAPSS Hazcard  HC022a .
• Bromine, Br 2 (l), (TOXIC) – see CLEAPSS Hazcard  HC015a .
• Iodine, I 2 (s), (TOXIC) – see CLEAPSS Hazcard  HC054 .
• Set up a table for results like this:

• Clamp the electrolysis cell and pour in enough of the first electrolyte so that the tops of the electrodes are covered with about 1–2 cm of liquid. Fill the two test tubes with the same electrolyte. Wearing gloves, close the end of each test tube in turn with a finger and invert it over an electrode, so that no air is allowed to enter (see diagram). During electrolysis it may be necessary to lift the test tubes slightly to ensure that the electrodes are not completely enclosed, preventing the flow of current.
• Connect the circuit, and mark the polarity of each electrode on the bung. The circuit should be checked before being switched on.
• Observe whether or not the lamp lights up.
• Look for the substances produced at each electrode – ie gaseous, solid or in solution.
• Write down results after each observation, not when all the experiments are finished.
• Only carry out the electrolysis for long enough to make the necessary observations. Prolonging the electrolyses unnecessarily causes toxic gases such as chlorine and bromine to be produced in unacceptably hazardous quantities. Switch off the current immediately if this becomes apparent.
• After each electrolysis switch off the current and remove the test tubes from the cell to test any gases present by lifting them slowly in turn to let any remaining solution drain out before closing the end with a finger. Carry out the tests on the gases as instructed.
• (Optional) After removing the test tubes from the cell, quickly pour the liquid down the sink with plenty of water. Wipe a piece of universal indicator paper over each electrode and note any colour changes.
• Wash the cell with plenty of water and dry the outside with a paper towel before fixing it back into position and reconnecting the power supply. It is important to connect the leads according to the polarities marked on the bung.
• Repeat the experiment with each of the other four solutions, trying to keep to the order given in the table. Zinc chloride and copper nitrate should be the last electrolytes tested. This is because they deposit solids on the cathode. If zinc chloride is electrolysed first, the solid deposit on the cathode can be easily removed with a piece of emery paper or dipping the end of the electrode in some dilute hydrochloric acid in a beaker.

Teaching notes

The electrolysis of aqueous solutions, rather than molten salts, is easier and safer for students to do for themselves, Unfortunately the theory is more complicated, because the presence of water complicates what students may decide are the products formed at the electrodes.

More support

Use the animations included in our practical video Electrolysis of aqueous solutions to help learners think about which ions are present in an aqueous solution and what is happening at each electrode. 

Ensure that students do not attempt to smell directly any of the halogen fumes produced. It is important that you are aware of any students who are asthmatic or who might have an allergic reaction to these toxic gases. In this context do not allow the electrolyses of the halide solutions to proceed any longer than is absolutely necessary.

When testing for hydrogen or oxygen, the mouth of the test tube can be closed with a gloved finger, and the test tube transported to a central area, where a single naked flame has been set up, well away from the experiments. A supply of spills can also be kept in this area for the tests.

For the hydrogen test, students may well ask why there is little or no ‘pop’ or ‘squeak’. Explain that pure hydrogen – rather than a mixture of hydrogen and air – is being tested if the test tube was full of gas before it was removed.

For the oxygen test, care should be taken that the dampness at the mouth of the test tube does not extinguish the ‘glow’, causing the test to fail.

Once the electrolysis of zinc chloride or copper nitrate has been done, a deposit of metal will have formed on the cathode. This will have to be cleaned before the cell can be used again. These metal deposits can be removed using emery paper. Alternatively, small quantity of 50% concentrated nitric acid (CORROSIVE) in a small beaker can be used to remove the copper, providing gloves are worn and the operation is done in a fume cupboard by a suitably qualified person. Similarly dilute hydrochloric acid will remove the zinc.

Results and conclusions

Notes on the theory

Cathode reactions:

KBr NaI Ca(NO 3 ) 2 : 2H 2 O(l) + 2e – → H 2 (g) + 2OH –  (aq)

ZnCl 2 Cu(NO 3 ) 2 : M 2+ (aq) + 2e – → M(s)

Anode reactions:

KBr NaI ZnCl 2 : 2X – (aq) → X 2 (g/aq) + 2e –

Ca(NO 3 ) 2 Cu(NO 3 ) 2 : 2H 2 O(l) → O 2 (g) + 4H + (aq) + 4e –

For KBr NaI and Ca(NO 3 ) 2 it is likely that students will ask why hydrogen is the gas evolved rather than the metal. Students could then be asked to imagine what would happen if one of these metals is formed given that this occurs in the presence of water. The ensuing reaction produces hydrogen as one of the products and the metal hydroxide as the other. Students of higher ability could be introduced to the concept of electrode potentials and be given details of the probable reaction occurring at the cathode shown earlier.

With chlorine in particular and with bromine too students will find that the indicator paper is bleached as well as showing signs of acidity. Iodine usually stains the paper brown.

Some students may ask about the relative volumes of gases produced at the electrodes. While this practical is not designed to investigate this they can be told the following. The volume ratio of hydrogen and chlorine gases produced during the electrolysis of NaCl is actually 1:1. Nothing like this is observed in practice because chlorine is slightly soluble in the aqueous solutions and the gas does not begin to collect until the electrolyte solution has become saturated with it.

Less advanced students could be asked to concentrate on simple observations, eg Is a gas formed? What pH changes occur at the electrodes? The main principle to emphasise is that the conduction of electricity by aqueous solutions is due to the movement of ions (not electrons) and that these travel to the electrodes of opposite charge.

Less advanced students should simply note that:

• The solution around the cathode tends to become alkaline.
• The solution around the anode tends to become acidic.
• Metals low in the reactivity series appear to be deposited at the negative electrode.
• A gas is evolved at the negative electrode if the metal is high in the reactivity series. If appropriate they can be told that this gas is hydrogen.
• Non-metals are formed at the positive electrode: chloride ions produce gaseous chlorine bromide and iodide ions form bromine and iodine respectively which dissolve to form coloured solutions.
• The electrolysis of copper nitrate produces a colourless gas at the positive electrode. If appropriate students can be told that this is oxygen.

Additional information

This is a resource from the  Practical Chemistry project , developed by the Nuffield Foundation and the Royal Society of Chemistry.

Practical Chemistry activities accompany  Practical Physics  and  Practical Biology .

The experiment is also part of the Royal Society of Chemistry’s Continuing Professional Development course: Chemistry for non-specialists .

© Nuffield Foundation and the Royal Society of Chemistry

• 14-16 years
• 16-18 years
• Practical experiments
• Electrochemistry
• Equations, formulas and nomenclature

Specification

• (f) electrolysis of molten ionic compounds e.g. lead(II) bromide (including electrode equations)
• (n) electrolysis of aqueous solutions such as copper(II) chloride (including electrode equations)
• (o) electrolysis of aqueous solutions involving competing ions such as sodium chloride (including electrode equations)
• C3.3d describe neutralisation as acid reacting with alkali or a base to form a salt plus water
• C3.4c describe competing reactions in the electrolysis of aqueous solutions of ionic compounds in terms of the different species present
• 3 Investigate what happens when aqueous solutions are electrolysed using inert electrodes. This should be an investigation involving developing a hypothesis.
• During electrolysis, at the cathode (negative electrode), positively charged ions gain electrons and so the reactions are reductions. At the anode (positive electrode), negatively charged ions lose electrons and so the reactions are oxidations.
• Reactions at electrodes can be represented by half equations. Eg 2H⁺ + 2e⁻ → H₂ and 4OH⁻ → O₂ + 2H₂O + 4e⁻ or 4OH⁻ − 4e⁻ → O₂ + 2H₂O
• The ions discharged when an aqueous solution is electrolysed using inert electrodes depend on the relative reactivity of the elements involved.
• At the negative electrode (cathode), hydrogen is produced if the metal is more reactive than hydrogen. At the positive electrode (anode), oxygen is produced unless the solution contains halide ions when the halogen is produced.
• Students should be able to predict the products of the electrolysis of aqueous solutions containing a single ionic compound.
• RP21 Investigate what happens when aqueous solutions are electrolysed using inert electrodes. This should be an investigation involving developing a hypothesis.
• 4.7.5.2 Electrolysis
• 9 Investigate what happens when aqueous solutions are electrolysed using inert electrodes. This should be an investigation involving developing a hypothesis.
• 3.25 Explain the formation of the products in the electrolysis, using inert electrodes, of some electrolytes, including: copper chloride solution, sodium chloride solution, sodium sulfate solution, water acidified with sulfuric acid, molten lead bromide…
• 3.31 Investigate the electrolysis of copper sulfate solution with inert electrodes and copper electrodes
• 2 Electrolysis of aqueous sodium chloride or aqueous copper sulfate solution testing for the gases produced
• C3.3.8 describe the technique of electrolysis of an aqueous chloride solution of a salt
• C1 Electrolysis of aqueous sodium chloride or aqueous copper sulfate solution testing for the gases produced
• Electrolysis is the decomposition of an ionic compound into its elements using electricity.
• A d.c. supply must be used if the products of electrolysis are to be identified.
• Positive ions gain electrons at the negative electrode and negative ions lose electrons at the positive electrode.

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Electrolysis Experiments ( Edexcel IGCSE Chemistry )

Revision note.

Chemistry Lead

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Electrolysis of molten compounds

• E.g. lead(II) bromide
• When these compounds are heated beyond their melting point, they become molten and can conduct electricity as their ions can move freely and carry the charge
• These compounds undergo electrolysis and always produce their corresponding element
• To predict the products of any binary molten compound first identify the ions present
• The  positive  ion will migrate towards the  cathode  and the  negative  ion will migrate towards the  anode
• Therefore the  cathode  product will always be the metal and the product formed at the  anode  will always be the non-metal

The electrolysis of molten lead(II) bromide 

• Add lead(II) bromide into a crucible and heat so it will turn molten, allowing ions to be free to move and conduct an electric charge
• Add two graphite rods as the electrodes and connect this to a power pack or battery
• Turn on the power pack or battery and allow electrolysis to take place

Diagram showing the electrolysis of lead(II) bromide

Lead ions are attracted to the cathode, and bromide ions to the anode

What happens at the anode?

• Negative bromide ions move to the positive electrode (anode)
• At the anode, they lose two electrons to form bromine molecules
• There is bubbling at the anode as brown bromine gas is given off

What happens at the cathode?

• Positive lead ions move to the negative electrode (cathode)
• At the cathode they gain electrons to form grey lead metal 
• The lead deposits on the bottom of the electrode

Worked example

Identify the product formed during electrolysis at the anode and cathode for the following binary ionic compounds. 

• Molten copper chloride 
• Molten magnesium oxide
• Copper ions have a positive charge so are attracted to the cathode and form copper metal 
• Chloride ions have a negative charge so are attracted to the anode and form chlorine
• Magnesium ions have a positive charge so are attracted to the cathode and form magnesium metal
• Oxide ions have a negative charge so are attracted to the anode and form oxygen

Examiner Tip

Remember:   Opposites attract! 

Therefore, the positive ions will be attracted to the negative electrode and the negative ions to the positive electrode.

Electrolysis of aqueous solutions

• Aqueous solutions will always contain water molecules (H 2 O)
• In the electrolysis of aqueous solutions, the water molecules dissociate producing H +  and OH –   ions:

H 2 O   ⇌ H +  + OH –

• These ions are also involved in the electrolysis process and their chemistry must be considered
• We now have an electrolyte that contains ions from the compound plus ions from the water
• Which ions get discharged and at which electrode depends on the  relative reactivity   of the elements involved

What is produced at the anode?

• Negatively charged OH –   ions and non-metal ions are attracted to the positive electrode
• If halide ions (Cl - , Br - , I - ) and OH -   are present then the halide ion is discharged at the anode, loses electrons and forms a halogen (chlorine, bromine or iodine)
• If no halide ions are present, then OH -   is discharged at the anode, loses electrons and forms oxygen
• In both cases the other negative ion remains in solution

What is produced at the cathode?

• Positively charged H +   and metal ions are attracted to the negative electrode but only one will gain electrons
• Either hydrogen gas or the metal will be produced
• If the metal is  above hydrogen   in the reactivity series, then hydrogen will be produced and bubbling will be seen at the cathode
• This is because the more reactive ions will remain in solution, causing the least reactive ion to be discharged
• Therefore at the cathode,  hydrogen gas  will be produced unless the positive ions from the ionic compound are less reactive than hydrogen, in which case the  metal   is produced

The electrolysis of aqueous solutions

• The apparatus can be modified for the collection of gases by using inverted test tubes over the electrodes
• The electrodes are made from graphite which is inert and does not interfere with the electrolysis reactions

When answering questions on this topic, it helps if you first write down all of the ions present first. Only then you should start comparing their reactivity and deducing the products formed.

You must be able to identify the products formed at each electrode for the following aqueous solutions:

• Sodium chloride  
• Dilute sulfuric acid 
• Copper(II) sulfate

These can be found here . 

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Extraction of Metals

Electrolysis of Lead Bromide .

Lead bromide must be heated until it is molten before it will conduct electricity . Electrolysis separates the molten ionic compound into its elements .

The reactions at each electrode are called half equations . The half equations are written so that the same number of electrons occur in each equation .

Lead ions gain electrons ( reduction ) to form lead atoms . Bromide ions lose electrons ( oxidation ) to form bromine atoms . The bromine atoms combine to form molecules of bromine gas .

The overall reaction is

See some other examples of electrolysis .

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	This electrolysis cell is fully demonstrated and explained at the end using comprehensive illustrations. Brown bromine gas is observed at the anode and grey molten lead observed on the cathode. 9v dc is used, with a bulb in the circuit to show when the lead bromide begins to conduct electricity     =  Br   =  Pb Français Preparatory Lesson Video: Electrolysis of Molten Salts Chemistry • Third Year of Secondary School Join nagwa classes. Attend live Chemistry sessions on Nagwa Classes to learn more about this topic from an expert teacher! Remaining Seats: 3 In this video, we will learn how to describe the components of an electrolytic cell and predict the products of the electrolysis of molten salts. Video Transcript In this video, we will learn how to describe the components of electrolytic cells and predict the products of the electrolysis of molten salts. This video is about the electrolysis of molten salts. The electro- means we’re using electricity, and -lysis means separation. What are we separating? Well, molten means that something has been heated up to its liquid state. And a salt is another word for an ionic compound made up of positive and negative ions. During electrolysis, we will use electricity to separate these two types of ions. Let’s take a look at the electrolysis of lead(II) bromide as an example. In its solid state, the ions in lead bromide are bonded to one another and cannot float freely. If we heat it up so that the solid reaches the molten or liquid phase, the ions will no longer be bonded to one another and instead float freely in the liquid. In fact, a sample of molten liquid lead(II) bromide contains only lead ions and bromide ions floating around to make up the liquid. When we heat lead(II) bromine till liquid phase, it becomes an electrolyte. An electrolyte is a substance made of ions that can conduct electricity. Electricity is the flow of charged particles. So liquid lead(II) bromide, with its free-floating ions, can conduct electricity, while solid lead(II) bromide, with its bonded ions, cannot. We could also turn solid lead(II) bromide into an electrolyte by adding water to create an aqueous solution. We will learn about the electrolysis of aqueous salt solutions in another video. The electricity in electrolysis comes from a battery or another similar power source. The battery is connected by wires to two electrodes, usually made out of an inert substance like platinum or carbon. With an inert electrode, electrons can pass through the electrode without the atoms of the electrode taking part in the reaction. We will sometimes see a battery diagrammed like this, as two parallel lines connected to the wires. The longer parallel line represents the battery’s positive terminal, and the shorter parallel line represents the battery’s negative terminal. During electrolysis, when the circuit is turned on, the negative ions, also known as anions, move to the positive electrode. The positive ions, known as the cations, move to the negatively charged electrode. In fact, the names of the electrodes and the ions that are attracted to them correspond to one another. The anions are attracted to the anode, and the cations are attracted to the cathode. Soon after turning on the circuit, we will see a brown gas bubble up at the anode and waft up into the air. In addition, we will see a liquid silver-colored metal bead form beneath the cathode. What is going on to produce these two substances? Well, the bromide ions gather at the anode. At the anode, each bromide ion donates an electron. A pair of bromide ions can donate their electrons and then come together to form a bromine gas molecule. That brownish gas that we see is bromine gas. The half reaction for this process is two bromide ions produce a Br2 gas molecule and two electrons. The electrons donated by the bromide ions flow through the wire to the cathode. The lead ions gather around the cathode. There, each lead ion is able to accept two electrons, turning into a solid lead atom, which then sinks to the bottom of the vessel, forming this silver-colored liquid bead. The bead is made of molten metal lead. The half reaction for this process is a lead ion plus two electrons produces lead. Overall, this is the process of electrolysis. Electricity separates the positive ions and the negative ions, which then form new products at the electrodes. We sometimes refer to electrolysis as the decomposition of an electrolyte using electricity. To put this in simpler terms, it means the breaking down of a substance that conducts electricity. In our example, we broke down lead bromide to form lead and bromine gas. If we had instead electrolyzed aluminum oxide, the products would be aluminum and oxygen gas. We can extend this pattern and say that for the electrolysis of molten salts, the products will be the elemental form of the cation and the anion. The cation will tend to form a metal, and the anion will tend to form a gas. Earlier, we wrote down the chemical equations that show what is happening at each electrode. At the cathode, a lead two plus ion combines with two electrons to form lead. At the anode, two bromide ions form Br2 gas and two electrons. These chemical equations might look a little different than the ones we’re used to seeing because they are half reactions. Half reactions show the formation of one product and the electrons involved. There are two main categories of half reactions. Either the atoms or ions gain electrons or the atoms or ions lose electrons. For the reaction on the left, the lead ion gains electrons to form lead. In the reaction on the right, each bromide ion loses an electron during the formation of bromine gas. When the atom or ion gains electrons, we call this a reduction. In this case, the lead ion is being reduced. On the other hand, when an atom or ion loses electrons, we call this an oxidation. The bromide ions are being oxidized to form bromine gas. Because the full reaction here is the combination of a reduction half reaction and an oxidation half reaction, we call this full reaction an oxidation–reduction reaction. While this oxidation–reduction reaction describes the entire electrolysis, each half reaction gives us some extra information about what’s happening to the electrons at each electrode. During electrolysis, electrons flow through the wire from the anode to the cathode, where they are donated to ions. So the reduction half reaction, where an ion gains electrons, will occur at the cathode. Conversely, since ions give up their electrons at the anode, the anode is where the oxidation reaction takes place. If we have trouble remembering the definitions of oxidation and reduction, there’s a handy acronym we can use, OIL RIG. Oxidation involves a loss of electrons. Reduction involves a gain of electrons. Another acronym we can use is “Leo the lion says GER.” Lose, electrons, oxidation. Gain, electrons, reduction. As a summary, during electrolysis, one ion will gain electrons and be reduced at the cathode. The other ion will lose electrons and be oxidized at the anode. Now that we’ve learned about the electrolysis of molten salts, let’s do some practice problems to review. In the electrolysis of molten lead bromide, shown in the picture, which electrode is on the right? (A) The cathode, as it is oxidizing the bromine. (B) The cathode, as it is reducing the bromide ions. (C) The anode, as it is reducing the bromide ions. (D) The anode, as it is oxidizing the bromide ions. Or (E) the anode, as it is oxidizing the bromine. This picture depicts the electrolysis of molten lead bromide with a brown gas wafting up from one of the electrodes. This brown gas is bromine gas. This question is asking us to describe how bromine gas forms and at which electrode. Each choice has three parts, meaning there’s three pieces of information we need to gather in order to answer the question. Is bromine gas produced at the anode or the cathode? When bromine gas is produced, does it involve an oxidation or a reduction? And does that oxidation or reduction happen to bromide ions or bromine atoms? Let’s start with the third piece of information. Does this process involve bromide ions or bromine atoms? Molten lead bromide is made up of free-floating lead ions and bromide ions. Electrolysis requires an electrolyte. An electrolyte is a substance made up of ions or able to release ions that can conduct electricity. When the circuit is turned on, the bromide ions are drawn to one of the electrodes to begin the formation of bromine gas. Since the reactants in this process are bromide ions and not bromine atoms, we can eliminate choice (A) and choice (E) from consideration. Next, let’s consider the name of the electrode. The negatively charged bromide ions are known as anions. This is in contrast to the positively charged lead cations. Which electrode, the anode or the cathode, are anions drawn to? Well, we can remember that their names correspond with one another. Anions are drawn to the anode, and cations are drawn to the cathode. In this example, bromide anions are drawn to the anode, so we can eliminate choice (B) from consideration as well. The last question to consider is whether these bromide ions are being oxidized or reduced at the anode. Let’s take a look at what happens to the bromide ions at the anode in order to answer this question. First, each bromide ion donates an electron to the anode. Then, the ions that have donated their extra electrons can pair up to form a bromine gas molecule. The half reaction for this process is 2Br- ions produce Br2 and two electrons. With this process in mind, are the bromide ions being oxidized or reduced? In this situation, since the bromide ions are giving up or losing their electrons, it’s an oxidation. On the other hand, a reduction is when an atom or ion gains electrons. For example, at the cathode in this experiment, the lead ions will gain electrons or be reduced to form elemental lead. Since the bromide ions are losing electrons or being oxidized, we can identify (D) as the correct answer. The electrolysis of molten lead bromide produces the brown bromine gas as a product. The bromide ions are attracted to the anode, where they donate electrons to form bromine gas. Since the bromide ions lose electrons, we also call this process an oxidation. So in the electrolysis of molten lead bromide, shown in the picture, which electrode is on the right? That’s choice (D), the anode, as it is oxidizing the bromide ions. Barium metal can be obtained through electrolysis of its molten salt. Which of the following equations shows the reaction occurring at the negative electrode? (A) Barium plus two electrons produces barium two plus. (B) Barium produces a barium two plus ion plus two electrons. (C) A barium two plus ion plus two electrons produces barium. (D) Barium two plus ion plus two electrons produces two barium atoms. Or (E) a barium two plus ion produces barium and two electrons. This question is asking about the electrolysis of a molten salt. This process occurs when we dip two electrodes connected to a battery into the liquid form of a salt. The liquid salt is made of free-floating positive and negative ions. In this question, the positive ions or cations will be barium ions. The identity of the negative ions isn’t important to answering the question, so let’s just use chloride as an example. When the circuit is turned on, the ions will be attracted to the electrodes of the opposite charge. When the ions reach each electrode, a reaction will occur. This question is asking, what happens to the barium ions on the surface of the electrode shown here on the left? Each choice is essentially the same three pieces of information rearranged. In order to answer this question, we need to ask, when are barium ions present? When are barium atoms present? And are electrons absorbed or released during this process? Let’s look at the wording of the question for some clues about what’s going on. As we’ve just explained, a molten salt involves ions. And the electrolysis of a molten salt starts with ions. Similarly, if we are trying to obtain barium metal, that must mean that the atomic form of barium is a product of the reaction. That means we can eliminate choice (A) and (B) from consideration. We want the ion to be on the left side of the equation as a reactant and the atom to be on the right side of the equation as a product. We know that barium metal will form as a product of the reaction. It will do so by plating on the electrode. Our next question is, what is going on with the electrons to make this happen? In electrolysis, electrons flow from the anode to the cathode. They are taken from the ions at the anode and given to the ions at the cathode. In our example here, the electrons at the cathode are donated to the barium two plus ions. The two plus charge of the barium ion and the combined two minus charge of the two electrons balance out. As a result, barium metal atoms form. We can simplify this process by saying that the barium ion has gained electrons to form the atom. Choice (E), where electrons are a product of the reaction, is the opposite situation, where electrons are released from the ion. We want the electrons to combine with the ion to form the atom like they do in choice (C) and (D). The last thing we need to consider is whether the combination of a barium ion and two electrons would produce one barium atom or two barium atoms. Simply put, the correct answer is choice (C). One ion combines with electrons to form one atom. This reaction describes what occurs at the negative electrode when we electrolyze a molten salt containing barium. The barium ion will gain two electrons to form a barium metal atom. When an atom or ion gains electrons, we call that a reduction. Electrolysis is one way to isolate pure metals. In fact, barium was first isolated by British chemist Sir Humphry Davy in 1808 when he electrolyzed molten barium oxide. So when we obtain barium metal through electrolysis of its molten salt, the equation that shows the reaction occurring at the negative electrode is choice (C), a barium ion plus two electrons produces a barium atom. Now that we’ve done some practice problems, let’s review the key points of the lesson. Electrolysis is essentially the electrical separation of ions. And a molten salt is the liquid form of an ionic compound. A molten salt is comprised entirely of free-floating, positive and negative ions. The products of the electrolysis of molten salt will be the elemental forms of the ions present. Typically, the cation will form a metal, and the anion will form a gas. When the circuit is turned on, the negative anions will flow to the electrode known as the anode, where they will give up their electrons. When an atom or ion loses electrons, we call this oxidation. Conversely, the positively charged cations will flow to the cathode, where they gain electrons. The process of gaining electrons is known as a reduction. Lastly, half reactions show the formation of one product and the electrons involved. For example, the half reaction that shows the formation of lead is a lead two plus ion plus two electrons forms lead. Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Interactive Sessions Chat & Messaging Realistic Exam Questions Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy 152 COMMENTS Electrolysis of molten lead(II) bromide | Experiment | RSC ... Introduce your students to the study of electrolysis by demonstrating how conduction is only possible where lead (II) bromide is molten, and that metallic lead and bromine are the products of the molten electrolyte. This demonstration does not need too much preparation and the apparatus involved is very straightforward. Practical electrolysis | CPD | RSC Education Figure 3: Apparatus set up for the microscale electrolysis of molten lead bromide. The positive electrode is a carbon fibre rod and the negative electrode is an iron nail. A small spirit burner provides sufficient heat to melt the lead bromide. Identifying the products of electrolysis | Experiment | RSC ... Try this class experiment to carry out the electrolysis of various solutions and investigate the products formed. In this practical, students get the chance to practise doing electrolysis with a range of solutions, identifying the products that form at the electrodes. Electrolysis Experiments | Edexcel IGCSE Chemistry Revision ... The electrolysis of molten lead(II) bromide Method. Add lead(II) bromide into a crucible and heat so it will turn molten, allowing ions to be free to move and conduct an electric charge; Add two graphite rods as the electrodes and connect this to a power pack or battery; Turn on the power pack or battery and allow electrolysis to take place Electrolysis of Molten Lead Bromide - YouTube Electrolysis of Molten Lead Bromide. We discuss the effect of passing an electric current through liquid lead bromide. The equations at the electrodes, the overall equations and what... Electrolysis of Molten Lead Bromide - GCSE Chemistry ... Visit www.KayScience.com for access to 800+ GCSE science videos, quizzes, exam resources AND daily science and maths LIVE TUITION!!! In this video you will l... Electrolysis of molten Lead (II) Bromide using inert graphite ... This short flash animation outlines the process in which molten lead (II) bromide is broken down in... GCSE CHEMISTRY - Electrolysis of Lead Bromide - Ionic ... Electrolysis of Lead Bromide. Lead bromide must be heated until it is molten before it will conduct electricity. Electrolysis separates the molten ionic compound into its elements. The reactions at each electrode are called half equations. The half equations are written so that the same number of electrons occur in each equation. GCSE Chemistry Electrolysis Lead Bromide Lithium Chloride ... Calculations. Formulae, Equations, RMM, Moles, Reacting Masses. Well done. Keep on learning. An overview of this video-cast is given below. Molten Lead Bromide (PbBr 2) using inert graphite electrodes. This electrolysis cell is fully demonstrated and explained at the end using comprehensive illustrations. Lesson Video: Electrolysis of Molten Salts | Nagwa We could also turn solid lead (II) bromide into an electrolyte by adding water to create an aqueous solution. We will learn about the electrolysis of aqueous salt solutions in another video. The electricity in electrolysis comes from a battery or another similar power source. essay homework paper phd project resume review speech thesis