Oxidation-reduction (redox) reactions Archives

Redox Reactions by Transfer of Electrons at a Distance

December 1, 2020December 1, 2020 by Veerendra

Redox Reactions by Transfer of Electrons at a Distance

In all redox reactions, electrons are transferred from the reducing agent to the oxidising agent.
When the reducing and oxidising agents are mixed together as in the previous reactions, the transfer of electrons occurs quickly and cannot be detected.
However, when the reducing and oxidising agents are separated by an electrolyte in a U-tube as shown in Figure 3.6, the transfer of electrons occurs through the connecting wires and can be detected by a galvanometer.
The reducing agent loses its electrons and hence undergoes oxidation. The electrode at which electrons are released by the reducing agent is called the negative terminal.
The electrons then flow through the connecting wires to the oxidising agent. The electrode at which electrons are accepted by the oxidising agent is called the positive terminal.
As the oxidising agent accepts the electrons, it undergoes reduction.
The electrolyte allows the movement of ions to take place, thus completing the electric circuit. This ensures a continuous flow of electrons in the external circuit.

When polyatomic ions such as manganate(VII) ion and dichromate(VI) ion are involved in redox reactions, the half-equations are more complex. The following tables illustrate how the half-equations are constructed.

Redox Reactions by Transfer of Electrons at a Distance 2

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Oxidation and reduction in the transfer of electrons at a distance experiment

Aim: To investigate oxidation and reduction in the transfer of electrons at a distance.
Materials: 2.0 mol dm^-3 sulphuric acid, 0.5 mol dm^-3 freshly prepared iron(II) sulphate solution, 0.2 mol dm^-3acidified potassium manganate(VII) solution, 0.5 mol dm^-3 potassium iodide solution, 0.2 mol dm^-3acidified potassium dichromate(VI) solution, 0.2 mol dm^-3 potassium thiocyanate solution, 1% starch solution.
Apparatus: U-tube, galvanometer, connecting wires with crocodile clips, carbon electrodes, retort stand and clamp, test tube, dropper, stoppers with one hole.
Procedure:

A U-tube is clamped to a retort stand.
Dilute sulphuric acid is poured into the U-tube until its levels are 6 cm away from the mouths of the U-tube.
Using a dropper, 0.5 mol dm^-3 iron(II) sulphate solution is carefully added to one of the arms of the U-tube until the layer of iron(II) sulphate solution reaches the height of 3 cm.
In a similar manner as in step 3, 0.2 mol dm^-3 acidified potassium manganate(VII) solution is added to the other arm of the U-tube.
A carbon electrode is placed in each arm of the U-tube.
The electrodes are connected to a galvanometer as shown in Figure. Based on the deflection of the galvanometer, the electrodes that act as the positive terminal and negative terminal are determined.
The set-up is left aside for 30 minutes. Any change is observed.
Using a clean dropper, 1 cm³ of iron(II) sulphate solution is drawn out and placed in a test tube. Then, a few drops of 0.2 mol dm^-3 potassium thiocyanate solution are added to the test tube. Any change is observed.
Steps 1 to 7 are repeated using 0.5 mol dm^-3 potassium iodide solution and 0.2 mol dm^-3 acidified potassium dichromate(VI) solution to replace the iron(II) sulphate solution and acidified potassium manganate(VII) solution. Step 8 is repeated to test the potassium iodide solution with 1 % starch solution.

Results:

1. Solutions used: Iron(II) sulphate solution and acidified potassium manganate(VII) solution

Observation	Inference
(a) The electrode in the iron(II) sulphate solution acts as the negative terminal while the electrode in the acidified potassium manganate(VII) solution acts as the positive terminal.	Electrons flow from iron(II) sulphate solution to acidified potassium manganate(VII) solution.
(b) Iron(II) sulphate solution changes from pale green to yellow. It gives blood-red colouration with potassium thiocyanate solution.	At the end of the reaction, iron(III) ions are present. Iron(II) ions have changed to iron(III) ions.
(c) The purple acidified potassium manganate(VII) solution decolourises.	Manganate(VII) ions that give the solution its purple colour are used up in the reaction.

2. Solutions used: Potassium iodide solution and acidified potassium dichromate (VI) solution

Observation	Inference
(a) The electrode in the potassium iodide solution acts as the negative terminal, whereas the electrode in the acidified potassium dichromate(VI) solution acts as the positive terminal.	Electrons flow from potassium iodide solution to acidified potassium dichromate(VI) solution.
(b) The colourless potassium iodide solution turns brown. It gives a dark blue colouration with starch solution.	At the end of the reaction, iodine is present. Iodide ions have changed to iodine.
(c) Potassium dichromate(VI) solution changes colour from orange to green.	Dichromate(VI) ions have changed to chromium(lll) ions.

Discussion:

1. Iron(II) sulphate solution and acidified potassium manganate(VII) solution
(a) Iron(II) ions act as the reducing agent, releasing electrons to become iron(III) ions. Thus, iron(II) sulphate solution changes colour from pale green to yellow.

(b) The electrons accumulate at the carbon electrode in the iron(II) sulphate solution and flow out to the connecting wires. This carbon electrode acts as the negative terminal.
(c) The electrons then flow to the positive terminal, which is the carbon electrode in the acidified potassium manganate(VII) solution.
(d) Manganate(VII) ions act as the oxidising agent, accepting the electrons and therefore, undergoing reduction to become colourless manganese(II) ions.

(e) The overall ionic equation is as follows:

2. Potassium iodide solution and acidified potassium dichromate(VI) solution
Redox Reactions by Transfer of Electrons at a Distance 7
(a) Iodide ions act as the reducing agent, releasing electrons to become iodine molecules. Thus, the colourless potassium iodide solution turns brown.

(b) The electrons accumulate at the carbon electrode in the potassium iodide solution and flow out to the connecting wires. This carbon electrode acts as the negative terminal.
(c) The electrons then flow to the positive terminal, which is the carbon electrode in the acidified potassium dichromate(VI) solution.
(d) Dichromate(VI) ions act as the oxidising agent, accepting the electrons and therefore, undergoing reduction to become chromium(III) ions.

(e) The overall ionic equation is as follows:

3. The continuous flow of electrons from the reducing agent at the negative terminal to the oxidising agent at the positive terminal produces an electric current that causes the indicator of the galvanometer to deflect.
4. Sulphuric acid has two functions:
(a) To separate the reducing agent from the oxidising agent
(b) To complete the circuit by .allowing the transfer of ions to occur
5. Other electrolytes such as potassium nitrate solution and sodium chloride solution can be used in place of sulphuric acid. The electrolyte should not react with either the reducing agent or the oxidising agent used.
6. There are few other pairs of reducing agent and oxidising agent that can be used in this activity. Here are some examples:
Redox Reactions by Transfer of Electrons at a Distance 11

Conclusion:
The transfer of electrons occurs from the reducing agent to the oxidising agent through the connecting wires.

Displacement of Halogen From Halide Solution

December 1, 2020December 1, 2020 by Veerendra

Displacement of Halogen from Halide Solution

Generally, halogens are good electron acceptors and therefore are good oxidising agents.
(a) When going down Group 17, the size of the halogen atoms increases. The nucleus is further away from the outermost occupied shell.
(b) Hence, the electronegativity of halogens or their ability to accept electrons to form negatively-charged halide ions decreases down the group.
(c) As a result, the strength of halogens as oxidising agents decreases down the group.
Chlorine, bromine and iodine are three
commonly used halogens in the laboratory. Each of them gives different colour in aqueous solution. However, the colour changes slightly with concentration.
Therefore, the presence of the halogens is confirmed using an organic solvent such as 1,1,1-trichloroethane, CH₃CCl₃.
This is done by mixing thoroughly 1,1,1-trichloroethane to an aqueous solution of a halogen. Two layers will be formed whereby the denser 1,1,1-trichloroethane layer will be at the bottom and the less dense aqueous layer will be at the top.

Based on the colour of the 1,1,1-trichloroethane layer, the halogen present is identified.

Halogen	Colour of halogen in aqueous solution	Colour of halogen in 1,1,1-trichloroethane
Chlorine	Pale yellow or colourless	Pale yellow or colourless
Bromine	Brown, yellowish-brown or yellow, depending on concentration	Brown, orange or yellow, depending on concentration
Iodine	Brown, yellowish-brown or yellow, depending on concentration	Purple

In a displacement of halogen, a more electronegative halogen displaces a less electronegative halogen from its halide solution.

The halide ions of the less electronegative halogen act as the reducing agent. They lose their electrons and are oxidised to form halogen molecules.
The electrons are accepted by the more electronegative halogen which acts as the oxidising agent. By doing so, the halogen undergoes reduction to form its halide ions.
In short, there is an electron transfer from the halide ions of the less electronegative halogen to the more electronegative halogen.

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Displacement of Halogen from Halide Solution Experiment

Aim: To investigate the oxidation and reduction in displacement of halogen.
Materials: Chlorine water, bromine water, iodine solution, 0.5 mol dm^-3 potassium chloride solution, 0.5 mol dm^-3potassium bromide solution, 0.5 mol dm^-3 potassium iodide solution, 1,1,1-trichloroethane.
Apparatus: Test tubes, test tube rack.
Procedure:

2 cm³ of 0.5 mol dm^-3 potassium bromide solution is poured into a test tube.
2 cm³ of chlorine water is added to the test tube and the mixture is shaken thoroughly.
2 cm³ of 1,1,1 -trichloroethane is added to the mixture. The mixture is shaken thoroughly.
After a few seconds, the colour of the aqueous and the 1,1,1-trichloroethane layers are observed.
Steps 1 to 4 are repeated using the halogens and halide solutions as shown in Table 3.11.

Results:

Mixture	Colour of aqueous layer	Colour of 1,1,1- trichloroethane layer	Inferences
Chlorine + potassium bromide	Yellow	Orange	Bromine is present. Displacement of bromine has occurred.
Chlorine + potassium iodide	Yellow	Purple	Iodine is present. Displacement of iodine has occurred.
Bromine + potassium chloride	Yellow	Orange	Bromine is present. No displacement reaction has occurred.
Bromine + potassium iodide	Yellow	Purple	Iodine is present. Displacement of iodine has occurred.
Iodine + potassium chloride	Yellow	Purple	Iodine is present. No displacement reaction has occurred.
Iodine + potassium bromide	Yellow	Purple	Iodine is present. No displacement reaction has occurred.

Discussion:

The results can be summarised as follows.
Chlorine is more electronegative than bromine and iodine. Therefore,
(a) chlorine displaces bromine from potassium bromide solution. Chlorine acts as the oxidising agent, whereas bromide ions act as the reducing agent.

(b) chlorine displaces iodine from potassium iodide solution. Chlorine acts as the oxidising agent, whereas iodide ions act as the reducing agent.
(a) Bromine is less electronegative than chlorine. Therefore, bromine cannot displace chlorine from potassium chloride solution.
(b) Bromine is more electronegative than iodine. Therefore, bromine displaces iodine from potassium iodide solution. Bromine acts as the oxidising agent, whereas iodide ions act as the reducing agent.
Iodine is less electronegative than chlorine and bromine. Therefore,
(a) iodine cannot displace chlorine from potassium chloride solution.
(b) iodine cannot displace bromine from potassium bromide solution.

Conclusion:
A more electronegative halogen can displace a less electronegative halogen from its halide solution whereby the more electronegative halogen acts as the oxidising agent and the halide ions of the less electronegative halogen act as the reducing agent.

Redox reaction in the displacement of metals from its salt solution

December 1, 2020December 1, 2020 by Veerendra

Redox reaction in the displacement of metals from its salt solution

Generally, metals are good electron donors and therefore are good reducing agents. However, different metals have different strength as reducing agents.
The strength of metals as reducing agents can be compared by using the electrochemical series.
The electrochemical series lists metals according to their electropositivity, that is, according to their ability to lose electrons to form positive ions.
The higher the position of a metal in the electrochemical series, the more electropositive the metal is, the easier it is for the metal to lose its electrons. Thus, the better reducing agent the metal is.
On the other hand, the ability of a metal ion to accept electrons increases down the series. Thus, the strength of a metal ion as an oxidising agent increases down the electrochemical series.
In a displacement of metal, a more electropositive metal will displace a less electropositive metal from its salt solution.
(a) The more electropositive metal acts as the reducing agent. It loses electrons and undergoes oxidation to form positive ions.
(b) The ions of the less electropositive metal act as an oxidising agent by accepting the electrons. While doing so, the ions are reduced to metallic atoms.
(c) In short, there is an electron transfer from the more electropositive metal to the ions of the less electropositive metal.

Oxidation and reduction in the displacement of metals experiment

Aim: To investigate oxidation and reduction in the displacement of metals.
Materials: Zinc strip, copper strip, 0.5 mol dm^-3 copper(II) sulphate solution, 0.1 mol dm^-3 silver nitrate solution, sandpaper.
Apparatus: Test tubes, test tube rack.
Procedure:

2 cm³ of 0.5 mol dm^-3 copper(II) sulphate solution and 2 cm³ of 0.1 mol dm^-3 silver nitrate solution are poured into two separate test tubes.
A strip of zinc and a strip of copper are cleaned with sandpaper. The strips are then dropped into the test tubes as shown in Figure.
Any change in colour and whether any metal is deposited are observed.

Observations:

Test tube	Observations
X	The blue colour of the solution slowly fades until it becomes colourless. The zinc strip dissolves. A brown solid is deposited.
Y	The colourless solution slowly turns blue. The copper strip dissolves. A silvery grey solid is deposited.

Discussion:

In test tube X, zinc displaces copper from its salt solution.
(a) Zinc is more electropositive than copper. Thus, zinc acts as the reducing agent, losing electrons to form zinc ions. By doing so, zinc is oxidised. This explains why the zinc strip dissolves.

(b) The electrons are accepted by copper(II) ions in the solution. Thus, copper(II) ions act as the oxidising agent and are reduced to metallic copper. The brown solid deposited in test tube X is copper metal.

(c) The decreasing amount of copper(II) ions in the solution causes the solution to slowly change colour from blue to colourless.
(d) The redox reaction that occurs can be represented by the following equation.
In test tube Y, copper displaces silver from its salt solution.
(a) Copper is more electropositive than silver. So, copper acts as the reducing agent, losing electrons to form copper(II) ions. In other words, copper is oxidised. This explains why the copper strip dissolves.

(b) The increasing amount of copper(II) ions in the solution causes the solution to slowly change colour from colourless to blue.

(c) The electrons are accepted by silver ions in the solution. Thus, silver ions act as the oxidising agent and are reduced to silvery grey silver.
(d) The redox reaction that occurs can be represented by the following equation.

Conclusion:
A more electropositive metal can displace a less electropositive metal from its salt solution whereby the more electropositive metal acts as the reducing agent and the ions of the less electropositive metal act as the oxidising agent.

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Changing of iron(II) ions to iron(III) ions and vice versa

December 1, 2020December 1, 2020 by Veerendra

Changing of iron(II) ions to iron(III) ions and vice versa

Iron exhibits two oxidation numbers
(a) +2 as iron(II) ion, Fe²⁺(b) +3 as iron(III) ion, Fe³⁺
An aqueous solution containing iron(II) ions, Fe²⁺ is pale green in colour, whereas that containing iron(III) ions, Fe³⁺is yellow/yellowish-brown/ brown in colour.
Changing iron(II) ions to iron(III) ions is an oxidation and therefore requires an oxidising agent.
On the other hand, changing iron(III) ions to iron(II) ions is a reduction and therefore requires a reducing agent.

Table: Detecting the presence of iron(II) ions and iron(III) ions

Reagent	With iron ions	Observation
Sodium hydroxide solution or ammonia solution	Fe²⁺	Green precipitate, insoluble in excess alkali
Sodium hydroxide solution or ammonia solution	Fe³⁺	Brown precipitate, insoluble in excess alkali
Potassium hexacyanoferrate(ll) solution	Fe²⁺	Light blue precipitation
Potassium hexacyanoferrate(ll) solution	Fe³⁺	Dark blue precipitation
Potassium hexacyanoferrate(lll) solution	Fe²⁺	Dark blue precipitation
Potassium hexacyanoferrate(lll) solution	Fe³⁺	Greenish-brown solution
Potassium/ammonium thiocyanate solution	Fe²⁺	Pale red colouration
Potassium/ammonium thiocyanate solution	Fe³⁺	Blood-red colouration

The following are other oxidising agents that can replace bromine water in changing iron(II) ions to iron(III) ions.
Changing of iron(II) ions to iron(III) ions and vice versa 3

Other reducing agents that can replace zinc powder in changing iron(III) ions to iron(II) ions are as follows.
Changing of iron(II) ions to iron(III) ions and vice versa 4

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Changing of iron(II) ions to iron(III) ions and vice versa experiment

Aim: To investigate oxidation and reduction in the change of iron(II) ions to iron(III) ions and vice versa.
Materials: 0.5 mol dm^-3 freshly prepared iron(II) sulphate solution, 0.5 mol dm^-3 iron(III) sulphate solution, bromine water, zinc powder, 2.0 mol dm^-3 sodium hydroxide solution, filter paper.
Apparatus: Dropper, spatula, test tubes, test tube holder, Bunsen burner, filter funnel, test tube rack.
Procedure:
A. Changing of iron(II) ions to iron(III) ions

2 cm³ of 0.5 mol dm^-3 iron(II) sulphate solution is poured into a test tube.
Using a dropper, bromine water is added to the solution drop by drop.
The test tube is warmed gently.
2.0 mol dm^-3 sodium hydroxide solution is added slowly to the mixture until in excess.

B. Changing of iron(III) ions to iron(II) ions

2 cm³ of 0.5 mol dm^-3 iron(III) sulphate solution is poured into a test tube.
Half a spatula of zinc powder is added to the solution.
The mixture is filtered.
2.0 mol dm^-3 sodium hydroxide solution is added slowly to the filtrate until in excess.

Observations:

Activity	Reagent	Observations
A	Bromine water	Bromine water decolourises. The solution changes colour from pale green to yellow.
A	Sodium hydroxide solution	Brown precipitate is formed. It is insoluble in excess alkali.
B	Zinc powder	Some of the zinc powder dissolves. The solution changes colour from brown to pale green.
B	Sodium hydroxide solution	Green precipitate is formed. It is insoluble in excess alkali.

Discussion:
A. Changing of iron(II) ions to iron(III) ions

Bromine water oxidises iron(II) ions, Fe²⁺ to iron(III) ions, Fe³⁺. The presence of Fe³⁺ ions is confirmed by the formation of brown precipitate with sodium hydroxide solution.
Fe²⁺ ions lose their electrons and are oxidised to Fe³⁺ ions.
Bromine molecules, which give the bromine water its brown colour, gain the electrons and are reduced to colourless bromide ions, Br^–. This explains why the bromine water is decolourised.
In this reaction, bromine water acts as the oxidising agent, where as Fe²⁺ ions act as the reducing agent.

B. Changing of iron(III) ions to iron(II) ions

Zinc powder reduces iron(III) ions, Fe³⁺ to iron(II) ions, Fe²⁺. The presence of Fe²⁺ ions is confirmed by the formation of green precipitate with sodium hydroxide solution.
Zinc atoms lose their electrons and are oxidised to zinc ions, Zn²⁺. This explains why zinc powder dissolves in iron(III) sulphate solution.
Fe³⁺ ions accept these electrons and are reduced to Fe²⁺ ions.
In this reaction, Fe³⁺ ions act as the oxidising agent, whereas zinc acts as the reducing agent.

Conclusion:

Bromine water acts as an oxidising agent, changing iron(II) ions to iron(III) ions.
Zinc acts as a reducing agent, changing iron(III) ions to iron(II) ions.

What is a redox reaction?

December 1, 2020December 1, 2020 by Veerendra

What is a redox reaction?

1. Many processes that occur around us are redox reactions. These include combustion, rusting, photosynthesis, respiration and decomposition.

Redox reactions are chemical reactions involving oxidation and reduction occurring simultaneously.
Therefore, redox reaction is also known as oxidation-reduction reaction.
It is interesting to note that oxidation is always accompanied by reduction. Both oxidation and reduction have to occur simultaneously.
Redox reactions can be explained based on:
(a) Loss or gain of oxygen
(b) Loss or gain of hydrogen
(c) Transfer of electrons
(d) Changes in oxidation number

Types of redox reactions

Not all chemical reactions are redox reactions. For example, acid-base reactions and double decomposition reactions (as in the precipitation method) are non-redox reactions.
Four examples of redox reactions are as follows:
(a) Changing of iron(II) ions to iron(III) ions and vice versa
(b) Displacement of metal from its salt solution
(c) Displacement of halogen from its halide solution
(d) Transfer of electrons at a distance

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What is oxidation and reduction reaction?

Redox reactions based on loss or gain of oxygen

Originally, chemists explain redox reactions in terms of loss or gain of oxygen.
Oxidation is a gain of oxygen.
Therefore, when a substance gains oxygen, it is said to be oxidised.
The substance that causes oxidation is called the oxidising agent or oxidant.
Reduction is the opposite of oxidation. Reduction is a loss of oxygen.
Therefore, when a substance loses its oxygen, it is said to be reduced.
The substance that causes reduction is called the reducing agent or reductant.
The following is another redox reaction involving oxygen.

(a) Zinc undergoes oxidation because it gains oxygen to form zinc oxide.
(b) Lead(II) oxide causes zinc to be oxidised. So, lead(II) oxide is the oxidising agent in this redox reaction.

(c) At the same time, lead(II) oxide undergoes reduction because it loses its oxygen to zinc. It is reduced to metallic lead.
(d) Zinc causes lead(II) oxide to be reduced. So, zinc acts as the reducing agent.

Redox reactions based on loss or gain of hydrogen

Not all redox reactions involve oxygen. For redox reactions involving hydrogen gas or substances containing hydrogen, it is easier to explain the oxidation and reduction in terms of loss or gain of hydrogen.
Oxidation is a loss of hydrogen, whereas reduction is a gain of hydrogen.
Therefore, when a substance loses its hydrogen, it is said to be oxidised.
Take the following reaction as an example.

(a) Ammonia is oxidised as it loses its hydrogen to form nitrogen gas.
(b) Bromine is the oxidising agent as it causes ammonia to be oxidised.

(c) At the same time, bromine is reduced as it gains the hydrogen lost by ammonia to form hydrogen bromide.
(d) Ammonia is the reducing agent as it causes bromine to be reduced.
Consider the reaction between hydrogen sulphide and chlorine to produce sulphur and hydrogen chloride.

(a) Hydrogen sulphide is oxidised to sulphur as it loses its hydrogen to chlorine.
(b) Chlorine gains the hydrogen and therefore is reduced to hydrogen chloride.
(c) Chlorine acts as the oxidising agent because it causes hydrogen sulphide to be oxidised.
(d) Hydrogen sulphide acts as the reducing agent because it causes chlorine to be reduced.

Redox reactions based on transfer of electrons

Many redox reactions do not involve oxygen or hydrogen. These reactions can be explained based on the transfer of electrons that occurred.
Oxidation is a loss of electrons and reduction is a gain of electrons.
Thus, the electron acceptor acts as the oxidising agent and the electron donor acts as the reducing agent.
Consider the following reaction.

(a) In this reaction, there is a transfer of electrons from sodium to chlorine.
(b) This reaction can be taken as two separate changes occurring at the same time. Each change is called a half-reaction and its equation is called a half-equation.
Oxidation half-equation:

Each sodium atom is oxidised as it loses one electron to form a sodium ion.
Reduction half-equation:

Each molecule of chlorine is reduced as it accepts two electrons from sodium atoms to form two chloride ions.
(c) Sodium is the electron donor and therefore is the reducing agent. On the other hand, chlorine is the electron acceptor. Thus, chlorine acts as the oxidising agent.
(d) We can get the overall equation by adding up the two half-equations.
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
The two half-reactions can be represented as follows.

(a) Zinc undergoes oxidation as it loses electrons to form zinc ions.
(b) The electrons are accepted by copper(II) ions to form metallic copper.
(c) Zinc acts as the reducing agent, whereas copper(II) ions act as the oxidising agent.
(d) The overall equation is given below.

Note: This overall equation is also an ionic equation (an equation without the spectator ions).

Redox reactions based on changes in oxidation number

Complete transfer of electrons occurs during redox reactions involving ions. There are, however, redox reactions involving molecules that do not involve complete transfer of electrons.
Redox reactions such as these can be explained in terms of changes in oxidation number.
Oxidation number or oxidation state of an element is the charge that the atom of the element would have if complete transfer of electrons takes place.
Each element in a substance can be assigned with an oxidation number. Chemists assign the numbers according to a set of rules.

Rules for assigning oxidation number

Rule 1 : The oxidation number of an atom in its elemental state is zero. For example, C, Na, Mg, Al, H₂, O₂, Cl₂ and Br₂ have the oxidation number of 0.
Rule 2 : The oxidation number of a monoatomic ion is equal to its charge.
What is a redox reaction 12
Rule 3 : In compounds, the more electronegative elements are given a negative oxidation number. The sequence of electronegativity of some elements is shown below.
What is a redox reaction 13
Therefore, the following rules are adopted.
(a) The oxidation number of fluorine in all its compounds is -1 as it is very electronegative.
(b) The oxidation numbers of other halogens (chlorine, bromine and iodine) in their compounds are -1 except when they combine with more electronegative elements such as oxygen and nitrogen.
For example,
What is a redox reaction 14
(c) The oxidation number of hydrogen in a compound is always +1 except when hydrogen combines with reactive metals in metal hydrides, where it is -1.
For example,
What is a redox reaction 15
(d) The oxidation number of oxygen in a compound is always -2 except in peroxides and when oxygen combines with a more electronegative element such as fluorine.
For example,
What is a redox reaction 16
Rule 4 : The sum of the oxidation numbers of all the elements in the formula of a compound must be zero.
For example,
What is a redox reaction 17
Rule 5 : The sum of the oxidation numbers of all the elements in the formula of a polyatomic ion must be equal to the charge of the ion.
What is a redox reaction 18

Oxidation Number Example 1. Determine the oxidation number of nitrogen in N₂O₅.
Solution:

Oxidation Number Example 2. Determine the oxidation number of tin in SnF₄.
Solution:

Oxidation Number Example 3. Determine whether each of these reactions is a redox reaction or a non-redox reaction.
I: 2KClO₃(s) → 2KCl(s) + 3O₂(g)
II: CaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂0(l)
Solution:
What is a redox reaction 21

Oxidation number and nomenclature of compounds

Many elements, especially transition metals, exhibit more than one oxidation number in their compounds. To avoid confusion, oxidation numbers are included in the nomenclature or naming of their compounds.

For example, oxidation numbers are included in names of simple ionic compounds in the IUPAC nomenclature. The oxidation number of a metal ion is represented by a Roman numeral in brackets, immediately following the name of the metal. Table illustrates some examples.

Formula	Oxidation number of metal ion	IUPAC name
FeO	Iron(II) oxide	+2
Fe₂O₃	Iron(III) oxide	+3
CuCl	Copper(I) chloride	+1
CuCl₂	Copper(II) chloride	+2
PbO	Lead(II) oxide	+2
PbO₂	Lead(IV) oxide	+4
MnO	Manganese(II) oxide	+2
Mn₂O₃	Manganese(III) oxide	+3
MnO₂	Manganese(IV) oxide	+4

For elements that have only one oxidation number such as those in Groups 1, 2 and 13, their oxidation numbers are not included in their names.
Oxidation numbers are also included in the systematic naming of anions containing metals that, can take more than one oxidation number. The oxidation number is represented by a Roman numeral in brackets, immediately following the name of the anion.
Formula Systematic name Traditional name
KMnO₄ Potassium manganate(VII) Potassium permanganate
K₂CrO₄ Potassium chromate(VI) Potassium chromate
K₂Cr₂O₇ Potassium dichromate(VI) Potassium dichromate

For non-metal elements that exhibit more than one oxidation number, the oxidation numbers are written as Roman numerals in brackets, immediately following the name of ions containing them.

Formula	Systematic name	Traditional name
KNO₂	Potassium nitrate(III)	Potassium nitrite
KNO₃	Potassium nitrate(V)	Potassium nitrate
HNO₂	Nitric(III) acid	Nitrous acid
HNO₃	Nitric(V) acid	Nitric acid
H₂SO₃	Sulphuric(IV) acid	Sulphurous acid
H₂SO₄	Sulphuric(VI) acid	Sulphuric acid
NaOCl	Sodium chlorate(I)	Sodium hypochlorite
NaClO₃	Sodium chlorate(V)	Sodium chlorate

Some of the systematic names shown in Tables were originally introduced by the Stock system. The traditional names, however, are adopted in the IUPAC nomenclature as they already existed for a long time and they are much easier to use.

Redox reactions based on changes in oxidation number

Oxidation occurs when there is an increase in oxidation number.
Conversely, reduction occurs when there is a decrease in oxidation number.
These definitions can be used to explain many more redox reactions than the previous definitions.
Table shows how a few redox reactions are explained in terms of changes in oxidation number. First, each element in all substances is assigned with an oxidation number. Then, the changes in the oxidation numbers are analysed.

Reaction	Explanation
	The oxidation number of zinc increases from 0 to +2. So, zinc is oxidised to zinc ion. Since oxygen oxidises zinc, oxygen acts as the oxidising agent. The oxidation number of oxygen decreases from 0 to -2. So, oxygen is reduced to oxide ion. Since zinc reduces oxygen, zinc acts as the reducing agent.
	Magnesium is oxidised to magnesium ion whereby its oxidation number increases from 0 to +2. Since carbon dioxide oxidises magnesium, carbon dioxide acts as the oxidising agent. Carbon dioxide is reduced to carbon whereby the oxidation number of carbon in carbon dioxide decreases from +4 to 0. Magnesium acts as the reducing agent.
	Hydroiodic acid is oxidised to iodine as the oxidation number of iodine in hydroiodic acid increases from -1 to 0. Bromine acts as the oxidising agent. Bromine is reduced to hydrobromic acid as its oxidation number decreases from 0 to -1. Hydroiodic acid acts as the reducing agent.

In non-redox reactions, the oxidation numbers of all elements remain unchanged.
Determining whether a reaction is a redox reaction or a non-redox reaction:
(a) If one of the elements shows a change in oxidation number, it is sufficient to conclude that the reaction is a redox reaction. You need not work out the oxidation numbers of the other elements. This is because an increase in oxidation number is always accompanied by a decrease in oxidation number.
(b) The oxidation numbers of elements in polyatomic ions such as NH₄⁺, SO₄^2- and NO₃^– need not be determined if the ions appear both in the reactants and in the products.

Redox Reaction Experiment Discussion

Aim: To investigate redox reaction involving oxygen.
A. Combustion of metal in oxygen
Materials: Magnesium ribbon, sandpaper, gas jar containing oxygen gas.
Apparatus: Tongs, Bunsen burner.
Procedure:

A piece of 5 cm magnesium ribbon is cleaned with sandpaper.
Using a pair of tongs, the ribbon is lit and quickly placed into a gas jar filled with oxygen gas. Observation is made.

Observations:
The magnesium ribbon burns with a bright white flame producing a white ash.
Discussion:

Magnesium burns in oxygen to produce the white ash of magnesium oxide.
The equation representing the combustion of magnesium:
Magnesium gains oxygen to form magnesium oxide. Therefore, magnesium is said to be oxidised to magnesium oxide.
The oxidation process is caused by oxygen gas. Thus, oxygen gas is said to act as the oxidising agent (oxidant).

B. Heating of metal oxide with carbon

Materials: Copper(II) oxide, carbon powder.
Apparatus: Crucible, pipe-clay triangle, tripod stand, spatula, Bunsen burner.
Procedure:

A spatulaful of copper(II) oxide and a spatulaful of carbon powder are mixed thoroughly in a crucible.
The apparatus is set up as shown in Figure.
The mixture is heated strongly. Observation is made.

Observations:

The mixture burns brightly.
Reddish-brown globules are formed.

Discussion:

When copper(II) oxide is heated with carbon, it produces copper which is reddish-brown in colour.
The equation representing the reaction that occurs:
Copper(II) oxide loses its oxygen to form copper. Thus, copper(II) oxide is said to be reduced to copper.
The reduction is caused by carbon. Therefore, carbon is said to act as the reducing agent (reductant).
Simultaneously, carbon gains oxygen to form carbon dioxide. Carbon is oxidised to carbon dioxide.
The oxidation of carbon is brought about by copper(II) oxide. So, copper(II) oxide acts as the oxidising agent (oxidant).

Conclusion:

In the combustion of metal in oxygen, the metal is oxidised by oxygen to metal oxide.
In the heating of metal oxide with carbon, the metal oxide is reduced by carbon to metal. Simultaneously, carbon is oxidised by the metal oxide to carbon dioxide.