Notes On Balancing Redox Equations - CBSE Class 11 Chemistry
A redox equation can be balanced using two methods - oxidation number method and half-reaction method. The oxidation number method is based on the change in the oxidation number of the reducing and oxidising agent. The half-reaction method is based on splitting the reaction into two halves - the oxidation half and the reduction half. The half-equation method is based on the principle that electrons lost during the oxidation half reaction are equal to the electrons gained during the reduction-half of the reaction. Steps to balance the redox equation using the oxidation number method: Ex: Permanganate ion reacts with bromide ion in basic medium to give Manganese dioxide and bromate ion. (i) Write down the correct formula of all the reactants and products in the reaction. MnO-4(aq) + Br-(aq) → MnO2(s) + BrO3-(aq) (ii) Assign the oxidation numbers to all the elements in the reaction and also identifying the atoms which undergo change in the oxidation number. (iii) Calculate the total increase or decrease in oxidation number per atom and for the entire molecule or ion in which it occurs. If the increase or decrease in oxidation number is not the same then equalise them by multiplying with suitable coefficients.                3(Decrease) x 2 = 6(Increase) (iv) Identify the involvement of ions and balance the ionic charges. If the reaction takes place in water, then add H+ or OH- ions on the appropriate side, so that the total ionic charge of reactants and products is equal. If the reaction takes place in an acidic solution, then add H+ ions. If the reaction takes place in a basic solution, then add OH- ions. (v) Balance the number of hydrogen atoms. This can be done by adding H2O molecules to the reactants or products. Steps to balance the redox equation using the half reaction method: Ex: Dichromate ions in aqueous acidic medium react with ferrous ions to give ferric and chromium three ions. (i) Write the skeletal equation for the reaction in the ionic form. (ii) Then identify the species which have been oxidised and reduced, and split the reaction into two half reactions. The oxidation number of chromium decreases from '+6' to '+3', thus dichromate ion is reduced. And the oxidation number of iron increases from '+2' to '+3', thus iron is oxidized. The two half reactions can be written to show the oxidation half reaction and the reduction half reaction. (iii) Balance all atoms except Oxygen and Hydrogen. (iv) Balance the Hydrogen and Oxygen atoms. If the reaction takes place in an acidic medium add H2O to balance Oxygen atoms and add H+ to balance Hydrogen atoms. If the reaction takes place in a basic medium, add H2O to balance hydrogen atoms and OH- ions to balance Oxygen atoms. (v) Balance the charge by adding electrons to one side of the each half reaction. If required, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate coefficients. Multiply Oxidation half-cell with 6 to balance the charge. (vi) Add the two half equations to obtain the overall reaction and cancel the electrons on each side. Now this is a balanced equation,                 6Fe2+(aq) + Cr2O72-(aq)+14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq)+7H2O(I)

#### Summary

A redox equation can be balanced using two methods - oxidation number method and half-reaction method. The oxidation number method is based on the change in the oxidation number of the reducing and oxidising agent. The half-reaction method is based on splitting the reaction into two halves - the oxidation half and the reduction half. The half-equation method is based on the principle that electrons lost during the oxidation half reaction are equal to the electrons gained during the reduction-half of the reaction. Steps to balance the redox equation using the oxidation number method: Ex: Permanganate ion reacts with bromide ion in basic medium to give Manganese dioxide and bromate ion. (i) Write down the correct formula of all the reactants and products in the reaction. MnO-4(aq) + Br-(aq) → MnO2(s) + BrO3-(aq) (ii) Assign the oxidation numbers to all the elements in the reaction and also identifying the atoms which undergo change in the oxidation number. (iii) Calculate the total increase or decrease in oxidation number per atom and for the entire molecule or ion in which it occurs. If the increase or decrease in oxidation number is not the same then equalise them by multiplying with suitable coefficients.                3(Decrease) x 2 = 6(Increase) (iv) Identify the involvement of ions and balance the ionic charges. If the reaction takes place in water, then add H+ or OH- ions on the appropriate side, so that the total ionic charge of reactants and products is equal. If the reaction takes place in an acidic solution, then add H+ ions. If the reaction takes place in a basic solution, then add OH- ions. (v) Balance the number of hydrogen atoms. This can be done by adding H2O molecules to the reactants or products. Steps to balance the redox equation using the half reaction method: Ex: Dichromate ions in aqueous acidic medium react with ferrous ions to give ferric and chromium three ions. (i) Write the skeletal equation for the reaction in the ionic form. (ii) Then identify the species which have been oxidised and reduced, and split the reaction into two half reactions. The oxidation number of chromium decreases from '+6' to '+3', thus dichromate ion is reduced. And the oxidation number of iron increases from '+2' to '+3', thus iron is oxidized. The two half reactions can be written to show the oxidation half reaction and the reduction half reaction. (iii) Balance all atoms except Oxygen and Hydrogen. (iv) Balance the Hydrogen and Oxygen atoms. If the reaction takes place in an acidic medium add H2O to balance Oxygen atoms and add H+ to balance Hydrogen atoms. If the reaction takes place in a basic medium, add H2O to balance hydrogen atoms and OH- ions to balance Oxygen atoms. (v) Balance the charge by adding electrons to one side of the each half reaction. If required, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate coefficients. Multiply Oxidation half-cell with 6 to balance the charge. (vi) Add the two half equations to obtain the overall reaction and cancel the electrons on each side. Now this is a balanced equation,                 6Fe2+(aq) + Cr2O72-(aq)+14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq)+7H2O(I)

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