Identify the correct balanced redox reaction by using ion-electron method: BiO−3+Mn2+→Bi3++MnO−4+H2O
A
5BiO−3+22H++Mn2+→5Bi3++7H2O+MnO−4
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B
5BiO−3+14H++2Mn2+→5Bi3++7H2O+2MnO−4
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C
2BiO−3+4H++Mn2+→2Bi3++2H2O+MnO−4
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D
6BiO−3+12H++3Mn2+→6Bi3++6H2O+3MnO−4
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Solution
The correct option is B5BiO−3+14H++2Mn2+→5Bi3++7H2O+2MnO−4 Reduction half cell : +5BiO−3→Bi3+ Oxidation half cell : Mn2+→+7MnO−4 Taking reduction half cell Bi is balanced. So add 3H2O in product side. Now to balance hydrogen 6 H+ have to add in reactant side. BiO−3+6H+→Bi3++3H2O Now balance the charge by adding 2e− in reactant side: BiO−3+6H++2e−→Bi3++3H2O
For oxidation half cell : Mn is balanced. So add 4H2O in reactant side. Now to balance hydrogen 8H+ have to add at product side. Mn2++4H2O→MnO−4+8H+ Now balance the charge by adding 5e− at product side: Mn2++4H2O→MnO−4+8H++5e−
BiO−3+6H++2e−→Bi3++3H2O−−(i)
Mn2++4H2O→MnO−4+8H++5e−−−(ii) __________________________________ (i)×5+(ii)×2, we get 14H++5BiO−3+2Mn2+→5Bi3++2MnO−4+7H2O Hence, (b) is the correct balanced reaction.