Oxidation and Reduction , Oxidants and Reductants and Redox reaction

What is Oxidation and reduction ?

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New bonds are formed and old bonds are broken down during chemical reactions which causes the change in valence electrons of atoms or radicals. The chemical reactions in which the electronic states of the atoms or radicals change in reactants and products are termed as oxidation and reduction reactions.

Oxidation and reduction reactions are very common in our daily lives, eg. digestion of food, burning of fuel, rusting of iron articles, extraction of metals, electrolysis, electroplating, etc.

Green plants prepare carbohydrates by photosynthesis which is also oxidation and reduction. Here, reduction of carbon dioxide into glucose and oxidation of water into oxygen takes place.

These reactions can be defined with classical and electronic concepts.

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Classical concept of Oxidation and Reduction

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Oxidation :

Oxidation is a chemical process which involves addition of oxygen ( or any other electronegative species) or removal of hydrogen ( or other electropositive species). Examples :

Here, C is oxidized to CO₂.

Conversion of C into CO₂ is oxidation.

Here, H₂S is oxidized to S.

Conversion of H₂S into S is oxidation.

Reduction :

Reduction is a chemical process which involves the addition of hydrogen (or any other electropositive species) or removal of oxygen ( or any electronegative species). Examples:

Here, ZnO is reduced to Zn.

Conversion of ZnO into Zn is reduction.

Here, Cl₂ is reduced to HCl.

Conversion of Cl₂ into HCl is reduction.

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Modern or Electronic concept of Oxidation and Reduction

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Oxidation :

Oxidation is a process which involves loss of electrons. Examples:

Here, Na is oxidized to Na⁺ by losing one electron.

Here, Cl^– is oxidized to Cl by losing one electron.

Reduction :

Reduction is a process which involves gain of electrons. Examples:

Here, Cl is reduced to Cl^– by gaining one electron.

Here, Fe⁺⁺⁺ is reduced to Fe⁺⁺ by gaining one electron.

Oxidation = increase in positive charge or decrease in negative charge.

Reduction = increase in negative charge or decrease in positive charge.

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Oxidation and Reduction occurs simultaneously

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Oxidation involves loss of electron and reduction involves gain of electron. A species gains electron only when the electron is lost by other species and a species loses electron only when electron is gained by some other species. There is no oxidation without reduction and there is no reduction without oxidation. This means oxidation and reduction takes place simultaneously. For example:

This equation is written in ionic form as follows:

The oxidation half reaction is:

The reduction half reaction is:

The reaction involving both reduction and oxidation process is called redox reaction.

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Oxidation number or oxidation state

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Oxidation number (state) is the charge which an atom of the element has in its ion or appears to have when present in the combined state with other atoms.

Rules for determining oxidation number

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• Oxidation number of elements in free or uncombined state is always zero. For example- oxidation number of H₂, O₂, Mg, Al, etc. is zero.

• Algebraic sum of oxidation numbers of all elements in a chargeless molecule is equal to zero. Eg. In NaCl , O.N. of Na + O.N. of Cl = zero

• Algebraic sum of oxidation numbers of all elements in an ion is equal to the charge on the ion. Eg. In OH^– , O.N. of Oxygen + O.N. of H = -1

• Oxidation number of radicals is equal to the charge on them. Eg. oxidation number of Fe²⁺ = +2 , NH₄⁺ = +1 , OH^– = -1 , NO₃^– = -1 , CN^– = -1, SO₄^2- = -2 , etc.
• Oxidation number of alkali metal (Li, Na, K, Rb, Cs, Fr) is +1 in all compounds.
• Oxidation number of alkaline earth metal ( Be, Mg, Ca, Sr, Ba and Ra ) is +2 in all compounds.
• Oxidation number of metals in alloy is zero. Eg. O.N. of Na in sodium amalgam (Na.Hg) is zero.
• Oxidation number of hydrogen in compound is +1, except in metallic hydrides. In metallic hydrides, O.N. of hydrogen is -1 (eg. in NaH, CaH₂, etc. O.N. of H is -1)
• The oxidation number of halogens is always -1 in metal halides like MgBr₂, KF, AlCl₃, etc.
• Oxidation number of oxygen is -2 in compounds except in peroxides, suboxides and OF₂.
• In peroxides like H₂O₂, Na₂O₂, etc. the O.N. of oxygen is -1.

• In superoxides like KO₂, NaO₂, RbO₂, SnO₂, etc. the O.N. of oxygen is -1/2.

• In OF₂, oxidation number of oxygen is +2.

Q) Calculate the oxidation number (oxidation state) of N in HNO₃.

Here, oxidation number of H = +1

Oxidation number of O = -2

Suppose oxidation number of N = x

So, +1 + x + (3 x -2) = 0

x = +6 – 1 = +5

Therefore, oxidation number of N is +5.

Q) Calculate the oxidation number of P in PO₄^{– – –}

x + (4 x -2) = -3

Thus, x = -3+8 = +5

Q) Calculate the oxidation state of Fe in Fe₂(SO₄)₃.

Here, O.N. of SO₄ = -2

Suppose O.N. of Fe = x

Then, 2 x + (3 x -2) =0

Therefore, x = +3

Q) Calculate the oxidation number of underlined element.

(1) KMnO₄  (2) NH₄⁺ (3) Cr₂O₇^{– –}

(4) K₂Cr₂O₇  (5) HNO₃

(6) H₃PO₄ (7) H₂SO₄ (8) C₆H₁₂O₆

(9) Cu(OH)₂  (10) K₄[Fe(CN)₆]

(11) NH₄NO₃

Answer :

(1) +7  (2) -3  (3) +6  (4) +6

(5) +5  (6) +5  (7) +6  (8) 0

(9) +2  (10) +2  (11) -3

Note : NH₄NO₃ = NH₄⁺  NO₃^–

Therefore, x + (4 x +1) = +1

   or, x = -3.

Q) Calculate the oxidation number of Ag in [Ag(NH₃)₂] Cl.

NH₃ is a molecule. So, oxidation number of NH₃ = 0

Let, oxidation number of Ag = x

x + (0x2) + (-1) = 0

x + 0 – 1 = 0

Therefore , x = +1

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Differences between oxidation number and valency

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Oxidation and reduction in terms of oxidation number

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• Oxidation is a chemical process which involves increase in oxidation number.
• Reduction is a chemical process which involves decrease in oxidation number. Example :

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Oxidizing agents and Reducing agents

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Oxidizing agents or oxidants :

An oxidizing agent or oxidant is a substance which oxidizes the other substance and itself gets reduced. Eg.

Here, FeCl₃ is oxidizing agent because it has oxidized SnCl₂ into SnCl₄ and itself gets reduced to FeCl₂.

Reducing agents or reductants :

A reducing agent or reductant is a substance which reduces the other substance and itself gets oxidized. Eg.

Here, SnCl₂ is reducing agent because it has reduced FeCl₃ into FeCl₂ and itself gets oxidized to SnCl₄.

Note : Oxidizing agent is a species whose oxidation number decreases and Reducing agent is a species whose oxidation number increases.

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Minimum and maximum oxidation number

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Minimum possible oxidation number of an element is the number of electrons required to complete octet whereas the maximum oxidation number is the number of valence electrons. For example: in case of nitrogen,

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Oxidizing agent or Reducing agent or both

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• A species having an element in maximum oxidation state can act as oxidizing agent only.
• A species having an element in minimum oxidation state can act as reducing agent only.
• A species having an element neither in maximum nor in minimum oxidation state can act as oxidizing as well as reducing agent. Examples :

• A good reducing agent have a good capacity of losing electrons.
• A good oxidizing agent have a good capacity of gaining electrons.

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Redox reaction

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• A chemical reaction in which oxidation and reduction process take place simultaneously is called a redox reaction. Redox reaction involves both loss and gain of electrons.
• Redox reactions are the chemical reactions in which the reactants undergo a change in their oxidation states. Example :

In this reaction, C is oxidized to CO and ZnO is reduced to Zn. So this reaction is a redox reaction.

Role of redox reaction in daily life :

Redox reactions occur in each step of our daily lives. Majority of natural phenomena involve redox reactions. Eg. digestion of food; combustion of fuels like wood, gas, petrol, diesel, etc.; photosynthesis in green plants where CO₂ is reduced into carbohydrates and H₂O is oxidized into oxygen; extraction of metals where metal oxides are reduced into metals; generation of energy from different electrochemical cells; electrolysis; etc.

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Disproportionation reaction

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A chemical reaction in which a same species is oxidized as well as reduced is called disproportionation. Example:

When chlorine gas is passed through water, Cl₂ undergoes disproportionation reaction.

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Objective Questions and Answers

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1. Which of the following can act as both oxidizing as well as reducing agent ?

a. NH₃       b. HNO₃

c. HNO₂    d. NaNO₃

2. Oxidation state of oxygen in ozone is

a. +3   b. +1

c. -2    d. 0

3. Which of the following is the most powerful oxidizing agent ?

a. F₂     b. Cl₂

c. Br₂   d. I₂

4. Hydrogen sulphide is

a. Oxidizing agent only

b. Reducing agent only

c. Both

d. None

5. Oxidation number of manganese in potassium permanganate is

a. +7   b. -7

c. +5   d. +3

6. Oxidation number of phosphorus in H₃PO₄ is

a. +7   b. -7

c. +5   d. 0

7. In which of the following reactions, SO₂ acts as oxidizing agent ?

a. With H₂S

b. With acidified KMnO₄

c. With acidified K₂Cr₂O₇

d. With acidified C₂H₅OH

8. In the reaction, P₄ + 3KOH + 3H₂O → 3KH₂PO₂ + PH₃. Phosphorus is

a. Reduced

b. Oxidized

c. Oxidized as well as reduced

d. Neither oxidized nor reduced

9. When K₂Cr₂O₇ is converted to K₂Cr₂O₄, the change in oxidation number of Cr is

a. 0   b. 4

c. 6   d. 3

10. SO₂ acts as

a. Oxidizing agent only

b. Reducing agent only

c. Both oxidizing and reducing agent

d. Neither oxidizing nor reducing agent

11. Oxygen has an oxidation state of +2 in

a. H₂O   b. H₂O₂

c. SO₂    d . OF₂

12. The strongest oxidizing agent of the following is

a. O₃    b. O₂

c. Cl₂   d. Br₂

13. The strongest reducing agent is

a. HNO₂    b. H₂S

c. H₂SO₄   d. SnCl₂

14. Oxidation number of Fe in K₃[Fe(CN)₆] is

a. +1   b. +2

c. +3   d. +4

15. The highest oxidation state of Mn is shown by

a. K₂MnO₄   b. KMnO₄

c. MnO₂   d. Mn₂O₃

16. Oxidation state of oxygen atom in potassium superoxide is

a. 0    b. -1

c. -2   d. -1/2

17. Better reducing agents have a good capacity of :

a. Losing electrons

b. Gaining electrons

c. Being reduced

d. All of the above

18. Oxidation number of Zn in Cu-Zn is :

a. +2   b. -2

c. 0   d. +4

19. Oxidation number of C in HCN is :

a. +2   b. -2

c. 0     d. +4

20. Oxidation number of Na in sodium amalgam is

a. +5   b. +1

c. -1   d. 0

Answers :

1-c   2-d   3-d  4-b

5-a  6-c  7-a  8-c

9-a  10-c 11-d 12-a

13-b  14-c  15-b  16-d

17-A  18-C  19-A  20-d

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References

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• Sthapit, M.K., Pradhananga, R.R., Foundations of Chemistry, Vol 1 and 2, Fourth edition, Taleju Prakashan, 2005.
• Gurtu, J.N., Snehi, H., Advanced Physical Chemistry, Seventh Edition, Pragati Prakashan India, 2000.
• Madan, R.L., Tuli, G.D., Physical Chemistry, S. Chand and company, New Delhi, 2012.
• Negi, A.S., Anand, S.C., A Text Book of Physical Chemistry, Seventh Edition, New Age International Pvt. Ltd. Publishers, 1999.
• http://www.chemistry-assignment.com/oxidation-and-reduction-in-terms-of-oxidation-number/
• https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation-Reduction_Reactions
• https://www.khanacademy.org/science/ap-chemistry/chemical-reactions-ap/types-of-chemical-reactions-ap/a/oxidation-number