Question on Balancing Chemical Equations

Mn²⁺ + BiO₃^- + H⁺ = Bi³⁺ + MnO₄^- + H₂O

What will the coefficient of H⁺ be when the above equation is balanced with whole number coefficients?

Answer:

In the above reaction Mn gets oxidized from +2 state to +7 state by losing 5 moles of electrons per mole of it. Similarly Bi is getting reduced from +5 state to +3 state by gaining 2 moles of electrons per mole of it. In a redox reaction the number of electrons lost by the reducing agent should be equal to the number of electrons gained by the oxidising agent. So, 2 moles of Mn²⁺ will have to react with 5 moles BiO₃^- in order to have same number of electrons (in this case 10 electrons) lost and gained in this redox reaction. Note that the number of moles of Mn²⁺ & BiO₃^- used are the inverse ratio of the electrons exchanged by them in this reaction. To balance the atoms on the product side we need 5 moles of Bi³⁺ and 2 moles of MnO₄^-. This brings 15 moles and 8 moles of O atoms on the reactant side and product side respectively. To balance the remaining O atoms on the product side we need 7 moles of H₂O. This brings 14 moles of H atoms on the product side and to balance it we need 14 moles of  H⁺ on the reactant side. Finally the balanced equation is

2Mn²⁺ + 5BiO₃^- + 14H⁺ = 5Bi³⁺ + 2MnO₄^- + 7H₂O