The reactions which involve oxidation and reduction are called Redox reactions. Redox reactions are also called electrons transfer reactions.
Oxidation reduction reaction is the chemical reactions involves transmission of electrons from one kind of atom to another kind. Molecular Equations: It is the molecular form of reactants and products called molecular equation. Ionic Equations: The reaction consists of reactant and products in ionic form called ionic equations.
Molecular equation as follows
2FeCl3 + SnCl2 → 2FeCl2 + SnCl4
Ionic equation as follows
2Fe3+ + Sn2+ → 2Fe2+ + Sn4+
4HCl + MnO2 → MnCl2 + Cl2 + 2H2O
In this redox reaction process, HCl has been oxidized to Cl2 and MnO2 has been reduced to MnCl2.
Oxidation number is the charge on the atom of any element when it is in ionic or combined state. Oxidation number is also said oxidation state.
Rules for calculation of oxidation number for redox reaction process:
- The oxidation number is zero when any atom in its elementary state.
- Monoatomic ions oxidation number is equal to charge on it.
- H is in +1 oxidation number when combined with non-metals and in -1oxidation numberwhen combined with active metals like calcium,sodium, etc. Ex: In hydrides like NaH, CaH2.
- Oxygen is in -2 oxidation number excluding in peroxides like Na2O2,H2O2, etc. where it is -1 and OF2 where it is +2.
- Alkali and alkaline earth metals oxidation numbers have +1 and +2 respectively.
- Halogens have -1 oxidation number in metal halides.
- In metals and non-metal compounds, metals are in positive oxidation number while non-metals are in negative oxidation number.
- If the compounds having two various elements, the element which are more electronegative in nature has negative oxidation numbers although the other has positive oxidation number.
- In neutral molecules summation of oxidation numbers of all particles is zero.
- If the compounds containing complex ions, the summation of oxidation numbers of whole atoms is equivalent to the charge on the ion.
Example 1: Oxidation number of Cr in CrO5
By conventional method:
CrO5 i.e. Cr=x and O5= 5 x (-2)
So, x + 5 x (-2) = 0
or x = +10 (wrong)
Oxidation number 10 for Cr is wrong, because it cannot be more than +6, according to maximum number of valence electrons, 3d5, 4s1.
Oxidation number of Cr is calculated by chemical bonding method since CrO5 contains peroxide linkage, other than Cr=O
By Chemical bonding method
Structure of CrO5 is
Oxidation number of Cr in CrO5 is calculated as
For Cr = x
For Cr=O = 1 x (-2)
For O-O = 4 x (-1)
i.e. x + 1 x (-2) + 4 x (-1) = 0
x – 2 – 4 = 0 or x = -6
Some important terms for Redox reactions process:
Loss of electrons or increase in oxidation number of its atom is oxidation.
Oxidising agent –
It accepts electrons (electron acceptor) or oxidation number of atoms decreases.
Reduction (Electronation) –
Reduction is the gain of electrons or decrease in oxidation number of atoms.
Reducing agent –
It gives electron (electron donor) or oxidation number of atoms increases.
+6 +2 +3 +3
Cr2O7 + Fe → Cr + Fe
Here Oxidation number of Cr decreases by 3 as +6 to +3
And oxidation number of Fe increases by 1 as +2 to +3
Types of redox reaction process redox reaction steps:
- Intermolecular redox reaction
In this reaction one substance is oxidized and other is reduced.
2Al + Fe2O3 → Al2O3 + 2Fe
Al is oxidised to Al2O3; Fe2O3 is reduced to Fe
- Intra-molecular redox reaction
Redox reaction consist of one element of a compound is oxidized and another is reduced.
2KClO3 → 2KCl + 3O2
Cl(+5) in KClO3 is reduced to Cl(-1); O2(-2) in KClO3 is oxidised to O2(0)
- Disproportionation reaction (Auto-redox)
One molecule of the same substance is reduced at the expense of other which is oxidized.
Balancing Redox Equation by Oxidation number method:
This method is based on the principle that any increase in oxidation number must be compensated by a decrease. This method consists of the following steps .
- Note the elements which undergo change in oxidation numbers.
- Select the suitable coefficients for the oxidising and reducing agents so that the total decrease in oxidation number of the oxidising agent becomes equal to the total increase in the oxidation number of the reducing agent.
Example 2: CuO + NH3 → Cu + N2 + H2O
In the above equation oxidation number of Cu decreases from +2 (in CuO) to 0 (in Cu) while that of N increases from -3 (in NH3) to 0 (in N2) and hence:
In order to equalise the total increase in O.N (=3) to the total decrease in O.N. (=2), we should have three atoms of Cu for every two atoms of N and hence the equation should be written as:
3CuO + 2NH3 → 3Cu + N2 + H2O
Now in order to balance O-atoms we should add 3H2O molecule to the right hand side. Thus:
3CuO + 2NH3 → 3Cu + N2 + 3H2O
Balancing Redox Equations by Ion-electron Method – By the use of Half-reactions
- Divide the whole equation in two half reactions, in one-half reaction changesgo through reducing agent and the other-half changesgo through oxidising agent.
- Balance the both half reaction equal to the number of atoms of each element in the reaction. For this purpose:
- For each half-reaction balance the atoms other than H and O by using simple multiples.
- H2O and H+are added in neutral and acid solutionsfor balancing oxygen and hydrogen atoms. Oxygen atomgets first balanced andeveryextra oxygen atom on one side of the equation, add one H2O molecule to the other side equation. Now use H+ to balance hydrogen atoms.
- In alkaline solutions, OH– may be used. For each additionalatom on one side, balance the equation by adding one H2O to the same side and add 2H– to the other side equation. If hydrogen is still unbalanced, balance the equation by adding one OH– for each additional hydrogen atom on the same side and one H2O to the other side of the equation.
- Balance the charges on both sides of equation by additionof electrons to the side having inadequate negative charges.
- With suitable number multiply one or both half-reactions,so that on adding the both equations, the electrongets balanced.
- Both balanced half reactions are added and cancel any common termsif there to both sides. Also see that all electrons cancel.
Example 3: Fe2+ + MnO4– + H+ → Mn2+ + Fe3+ + H2O
Above redox reaction takes place in acidic medium and can be broken into the following two half reactions showing redox reaction process redox reaction steps:
MnO4– → Mn2+ ……Reduction half reaction
And Fe2+ → Fe3+ …….Oxidation half reaction
For reduction half reaction,
- For balancing O-atom add 4H2O to right hand side to get.
MnO4– → Mn2+ + 4H2O
- For balancing H-atoms add 8H+ to the left hand side to get.
MnO4– + 8H+→ Mn2+ + 4H2O
- For balancing the charges add 5e– to the left hand side to get.
MnO4– + 8H+ +5e– → Mn2+ + 4H2O (i)
For oxidation half reaction,
Balance the charges on both sides by adding 1e- to the left hand side to get,
Fe2+ → Fe3+ + e–
5Fe2+ → 5Fe3+ + 5e–(ii) or
On adding equation (i) and (ii) we get:
MnO4– + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
This is the balanced equation.
The cell or device produce electric current from a chemical (redox) reaction is electrochemical cell, i.e. alteration of chemical energy to electrical energy.
This redox reaction involves two half reaction, one is oxidation half reaction and other is reduction half reaction.
Example 4: Zn + CuSO4→ZnSO4 + Cu
Zn + CuSO4→ZnSO4 + Cu
Or Zn + Cu2+ → Zn2+ + Cu
Two half reactions of this redox reactions are-
Zn → Zn2+ +2e–(oxidation half reaction)
Cu2+ + 2e–→ Cu (reduction half reaction)
Electrochemical cell based on this reaction is called Daniel cell.
Electrochemical cell has two half reactions using two half cells which are joined to each other by a salt bridge.
A U-shaped tube enclosing a concentrated solution of an inactive electrolyte like K2SO4, KCl, KNO3, etc. is a Salt bridge.