Table of Contents
- Introduction to Redox Reactions
- Types of Redox Reactions
- Balancing Redox Reactions
- Application of Redox Reactions
The term redox divides into two words ‘Reduction’ and ‘Oxidation’ thus implying a chemical reaction that has both reduction and oxidation processes.
To understand the concept let us first focus on following terms:
Oxidation occurs if one of the following conditions are satisfied
- Oxygen is added to a compound.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
Methane is oxidised due to addition of oxygen.
- An electronegative element is added to compound
Mg (s) + F2 (g) → MgF2 (s)
Magnesium is oxidised due to addition of an electronegative element, fluorine.
- Hydrogen is removed from the substance.
2H2S (g) + O2 (g) → 2S (s) + 2H2O (l)
Sulphur is oxidised due to removal of hydrogen
- An electropositive element is removed from compound
2K4 [Fe (CN)6] (aq) + H2O2 (aq) → 2K3 [Fe (CN)6](aq) + 2KOH (aq)
Potassium ferricyanide is oxidised due to removal of the electropositive element, Potassium
- Increase in oxidation number of a substance
- Loss of electrons from a substance
2Na (s) → 2Na+ (g) + 2e-
Sodium loses an electron and is oxidised.
Quite opposite to oxidation Reduction occurs when following conditions are satisfied
- Oxygen is removed from a compound.
2HgO (s) à 2Hg (l) + O2 (g)
Mercury is reduced due to removal of oxygen from mercuric oxide.
- An electronegative element is removed from compound
2FeCl3 (aq) + H2 (g) →2FeCl2 (aq) + 2HCl (aq)
Removal of electronegative element, chlorine from ferric chloride reduces Fe.
- Hydrogen is added to a compound.
CH2 = CH2 (g) + H2 (g) → H3C-CH3 (g)
Addition of hydrogen
- An electropositive element is added to the compound
2HgCl2 (aq) + SnCl2 (aq) → Hg2Cl2 (s) + SnCl4 (aq)
Addition of mercury to mercuric chloride
- Decrease in oxidation number of substance
- Gain of electrons to a substance
Cl2 (g) + 2e- → 2Cl- (g)
Chlorine gains electrons and is reduced
Fig. Process explaining Redox Reaction
Oxidising Agent: The species that undergo reduction thus enabling other species to undergo oxidation is known as an Oxidising Agent.
Reducing Agent: The species that undergo oxidation thus enabling other species to undergo reduction is known as Reducing Agent.
There is also a concept of half reaction; oxidation and reduction reactions when separately mentioned are known as half reactions and together they form a redox reaction.
A redox reaction, can hence, be easily identified by breaking the reaction into two half reactions or accounting for the change of electron or change in oxidation number.
A + B → C shows a combination reaction. This reaction can be redox if either A and B or both A and B are in the elemental form.
C(s) + O2 (g) → CO2 (g)
3Mg(s) + N2 (g) → Mg3N2 (s)
Decomposition Reactions are the opposite of combination reactions. During a decomposition, a compound breaks into two or more components. If any one of the components is in its elemental form, the said decomposition reaction is a redox reaction
2H2O (l) → 2H2 (g) + O2 (g)
2KClO3 (s) → 2KCl (s) + 3O2 (g)
Reactions of the form, XY + Z → XZ + Y where an atom or ion of a singular species is replaced with that of another species is a displacement reaction. The further classification includes:
a. Metal Displacement Redox Reactions
Another metal in the uncombined state displaces a metal in a compound.
V2O5 (s) + 5Ca (s) → 2V (s) + 5CaO (s)
Cr2O3 (s) + 2Al (s) → Al2O3 (s) + 2Cr(s)
b. Non Metal Displacement redox reactions
Hydrogen or Oxygen (non-metals) are displaced
2Na(s) + 2H2O (l) → 2NaOH (aq) + H2 (g)
2H2O (l) + 2F2 (g) → 4HF (aq) + O2 (g)
Reaction in which an element shows both reduction and oxidation then that reaction is called Disproportionation Reaction. A requirement for this type of reaction is that reacting substance can exist in at least three oxidation states. Thus, the reacting substance has intermediate oxidation number and the products formed have both higher and lower oxidation states.
In above example, focusing on oxygen, we can see that oxygen has its intermediate oxidation number of -1 in reacting substance and a higher and lower oxidation state of 0 and -1 respectively in products formed.
These reactions are precisely opposite of Disproportionation reaction. Here two recanting substances, each containing the same element but with a different oxidation number, will form a product in which the mentioned element is in another oxidation state.
This method uses half reactions involving oxidation and reduction and then balancing them individually only to add them together again to form the complete reaction.
As an example, we will balance a reaction between Fe2+ and dichromate ions (Cr2O7)2- in acidic medium. The following steps are involved in this task:
Step 1: Write a skeleton equation for the reaction in ionic form to form a basic idea:
Fe2+ (aq) + Cr2O72- (aq) → Fe3+ (aq) + Cr3+ (aq)
Step 2: Separate into half- reactions:
Step 3: The key is to balance the atoms other than O and H in each of half reaction individually, here Fe and Cr. In our example, oxidation half takes care of itself while for balancing the reduction half we multiply it by 2 thus balancing the Cr atoms.
Cr2O72- (aq) → 2Cr3+ (aq)
Step 4: As water is neutral, to balance our oxygen atoms we use H2O. In addition, the reaction takes place in an acidic medium; hence, we can add H+ to balance out extra hydrogen introduced in the system by addition of H2O. Thus, we get:
Cr2O72- (aq) + 14H+ (aq) → 2Cr3+ (aq) + 7H2O (l)
Step 5: For a reaction to be stable, it must be electrically neutral. Thus, we have to balance charges in both of our half reactions and the overall reaction. Adding electrons (e-) or protons (H+) to sides of half reactions accordingly balances it. The oxidation half reaction is thus rewritten to balance the charge:
Fe2+ (aq) → Fe3+ (aq) + e-
The total positive charge on the left hand side of the equation is twelve while right hand side only have six, thus creating an unbalance in the charge. To nullify extra positive charge we add six electrons to left hand side with excess positive charge.
Cr2O72- (aq) + 14H+ (aq) + 6e- → 2Cr3+ (aq) + 7H2O (l)
Adding six electrons to left hand side causes imbalance of negative charge when both the half reactions are compared. Thus, we balance charges across both the equations by multiplying the oxidation half by six:
6Fe2+ (aq) → 6Fe3+ (aq) + 6e-
Step 6: Adding to half reactions, after all the above steps, we can get fully balanced ionic equation:
6Fe2+ (aq) + Cr2O72- (aq) + 14H+ (aq) → 6Fe3+ (aq) + 2Cr3+ (aq) + 7H2O (l)
Step 7: Now the last step involves manual verification; check whether both sides have the same number of a single type of atoms and the charges cancel out each other thus concluding the fully balanced equation.
The above example is discussed for reaction in acidic medium but for the reaction in a basic medium; first balance the atoms as is done in the acidic medium. You can balance each H+ ion by adding an equal amount of OH- ions on both sides of the equation. Where H+ and OH‑ appear on the same side of the equation, combine these to give H2O.
The following example illustrates the oxidation number method:
Step 1: Consider an ionic equation for example:
Cr2O72- (aq) + SO32- (aq) → Cr3+ (aq) + SO42- (aq)
Step 2: Note down the oxidation numbers for Cr and S on the top of reaction thus giving an idea for the oxidising and reducing agent
Here dichromate ion is the oxidising agent and sulphate ion is the reducing one.
Step 3: Oxidation number for Cr changes from +6 to +3 thus accounting for a change of -3 while that of S changes from +4 to +6, a difference of +2. We need to balance out these changes in oxidation state thus multiplying Cr3+ by 2 and SO32- by 3 we get:
Step 4: Considering the acidic medium and uneven distribution of ionic charges on both sides, we add 8H+ on the left to make ionic charges equal
Cr2O72- (aq) + 3SO32- (aq) + 8H+ → 2Cr3+ (aq) + 3SO42- (aq)
Step 5: Finally, balance the oxygen atoms by adding H2O. The addition of H2O will increase the number of H atoms in the equation, which can be easily rearranged by increasing the number of H+ ions as suited. Finally, we get a balanced equation as given by:
Cr2O72- (aq) + 3SO32- (aq) + 8H+ (aq) → 2Cr3+ (aq) + 3SO42- (aq) + 4H2O (l)
The redox reactions are very common in our daily lives. We have practically seen many of them; rusting of iron, for instance, is a redox reaction where iron oxidises and oxygen reduces. Redox reactions also find many industrial and biological applications. Below are some of them:
Oxygen being approximately 20% in the atmosphere forms compounds with most of the metals. That is why most metal ores exist as oxides. As it is known that removal or addition of oxygen can be a reduction or oxidation reaction most of the metallurgical reactions are redox reaction. Suitable reducing agents are used to reduce metal oxides and obtain metal from the ore.
Examples include metal ores of metals like aluminium, iron sodium, potassium, strontium, and other alkaline earth metals or alkali metals.
One of the famous examples is of reduction of Fe2O3 in a blast furnace by coke.
Photosynthesis converts CO2 and H2O into carbohydrates in presence of sunlight and chlorophyll. This reaction is typically a redox reaction where CO2 is reduced to carbohydrates and water is oxidised to oxygen. Thus, a redox reaction takes place in food chain itself as the carbohydrate produced acts as a food source for both plants and animals.
6CO2 (g) + 6H2O (l) → C6H12O6 (aq) +6O2 (g) in presence of chlorophyll and carbohydrate.
Burning of fuels to produce energy is a common day-to-day redox reaction that we experience. During burning of a fuel oxygen is added to the fuel compound, which is nothing but oxidation of fuel and reduction of oxygen; thus, a redox reaction.
Fuel + O2 → CO2 + H2O + Energy
The energy produced in this reaction can be used for various purposes, be it industrial, domestic or transportation.
Redox Reactions can be used in titrations to determine the amount of unknown species. With the help of indicator, we can figure out the equivalence point and use it for calculations. The reduction and oxidation reactions involved in the titration greatly helps in determination of unknown species in the solution.
Redox Reaction also finds its applications in preparation of many of the chemical compounds that are essential in industries and domestic uses. Some examples include sodium hydroxide, chlorine, fluorine etc.
The batteries are an indispensable part of industrial sectors now. In the era, where cars are now being run on batteries, they play a significant role in various fields. Our vehicles cars, trucks, buses, etc. are moving into the storage cells based on redox reactions.
At electrodes of a cell or battery metal ions loss or gain electrons thus giving rise to an overall redox reaction. Hydrogen cells, a redox reaction between hydrogen and oxygen, are now used to power up rockets as a fuel source.
Watch this Video for more reference