Also watch on Youtube @InoopChem How to balance Redox Chemical Equation in 5 Easy Steps. by Ritik Bhola on Inoopchem YouTube Channel
BALANCING OF CHEMICAL EQUATIONS OF REDOX REACTION.
According to the law of Conservation of mass each chemical equation must be arithmetically balanced, that means the number of atoms of each elements on both sides of the chemical equation must be equal.
Two methods which have been used to balance all types of chemical equations are
1. Oxidation Number method
2. Half equation method Or Ion electron method
1. Oxidation number method.
The various steps are involved in the balancing of redox reactions by oxidation number method.
Step 1. write the skeletal equation of all the reactants and products of the reaction.
Step 2. Indicate the oxidation number of each element above its symbol and identify the elements which undergoes a change in the oxidation number.
Step 3. Calculate the increase or decrease in oxidation number per atom and identify the oxidizing or reducing agents.
Step 4. Multiply the formulae of oxidizing agent and the reducing agents by suitable Integers so as to equalize the total increase or decrease in the oxidation number as Calculated in step3.
Step 5. Balance all atoms other than oxygen and Hydrogen
Step 6. Finally balance Hydrogen and Oxygen atoms by adding H2O molecules using hit and trial method.
Step 7. In case of ionic reactions.
a) For Acidic medium ,
First balance Oxygen atoms by adding H2O molecules to whatever sides deficient in Oxygen atoms and then balance Hydrogen atoms by adding H+ ions to whatever side deficient Hydrogen atoms.
b)For Basic medium
First balance Oxygen atoms by adding H2O molecules to whatever side Deficient in Oxygen atoms. The Hydrogen atoms are then balanced by adding H2O molecules equal in number To the deficiency of Hydrogen atoms and an equal number of OH– ions are added to the opposite side of the equation, remove the duplication if any.
For Example,
when magnesium is react with nitric acid it gives magnesium nitrate along with Nitrous oxide and water molecule, the reaction is shown below
Mg + HNO3 → Mg(NO3)2 + N2O + H2O
Step 1. Indicate the oxidation number of each element above its symbol and identify the Elements which undergoes a change in the oxidation number.
Oxidation number of magnesium is Increases from 0 to 2 in magnesium nitrate.
Mg + HNO3 → Mg(NO3)2 + N2O + H2O
Oxidation number of Nitrogen decreases from +5 to +1 in nitrous oxide
Step2. Calculate the increase or decrease in oxidation number per atom and identify the oxidizing or reducing agents.
Oxidation number of magnesium is Increases from 0 to 2 in magnesium nitrate.
Oxidation number of Nitrogen decreases from +5 to +1 in nitrous oxide
Therefore Magnesium will act as Reducing agent and whereas
NITROGEN will act as Oxidizing agent.
Step3. Multiply the formulae of oxidizing agent and the reducing agents by suitable Integers so as to equalize the total increase or decrease in the oxidation number as Calculated in step2.
Mg + HNO3 → Mg(NO3)2 + N2O + H2O
Total Increase in OXIDATION NUMBER of Magnesium is +2
Total decrease in OXIDATION NUMBER of NITROGEN per atom Is 2 into 4 = 8
Balance increase or decrease in OXIDATION NUMBER.
On Right Hand Side there are 2 Nitrogen atoms and Only 1 in Left Hand Side, for equalization multiply HNO3 On Left Hand Side By 2.
Total increase in OXIDATION NUMBER of Magnesium is +2
Total decrease in OXIDATION NUMBER of Nitrogen per atom is 2 into 4 = 8, multiply Magnesium on Left Hand Side by 4 we get,
4Mg + 2HNO3 → Mg(NO3)2 + N2O + H2O
Step4. Balance all atoms other than Oxygen and Hydrogen
To balance Magnesium on either side, multiply Magnesium nitrate by 4, we get
4Mg + 2HNO3 → 4Mg(NO3)2 + N2O + H2O
Now, there are 10 nitrogen atoms on Right Hand Side in above equation and only 2 in Left Hand Side . For equalization of Nitrogen multiply 2HNO3 by 5, we get, the following reaction.
4Mg + 10HNO3 → 4Mg(NO3)2 + N2O + H2O
Step5. Finally balance Hydrogen and Oxygen atoms by adding H2O molecules using hit and trial method. We get,the following equation.
4Mg + 10HNO3 → 4Mg(NO3)2 + N2O + H2O
There, are 30 oxygen atoms on left Hand Side and on Right Hand Side there are only 26 oxygen atoms. Therefore to balance Oxygen atoms, Change the coefficient of H2O by 5 we get the Final Balanced equation,
4Mg + 10HNO3 → 4Mg(NO3)2 + N2O + 5H2O
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