Balance each redox reaction in acid solution.Mn 2+ + BiO3 - Æ MnO4 - + Bi 3+
Question
Balance each redox reaction in acid solution.Mn 2+ + BiO3 - Æ MnO4 - + Bi 3+
Solution
Sure, here is the step-by-step solution for balancing the given redox reaction in acidic solution:
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Identify the oxidation and reduction half-reactions: Mn2+ → MnO4- (Mn is oxidized from +2 to +7) BiO3- → Bi3+ (Bi is reduced from +5 to +3)
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Balance the atoms other than O and H in each half-reaction: For Mn: Mn2+ → MnO4- For Bi: BiO3- → Bi3+
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Balance the O atoms by adding H2O: Mn2+ → MnO4- + 4H2O BiO3- + 2H2O → Bi3+
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Balance the H atoms by adding H+: Mn2+ + 8H+ → MnO4- + 4H2O BiO3- + 2H2O → Bi3+ + 6H+
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Balance the charge by adding electrons: Mn2+ + 8H+ + 5e- → MnO4- + 4H2O BiO3- + 2H2O + 2e- → Bi3+ + 6H+
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Make the electron gain equal to the electron loss in the two half-reactions: Multiply the Mn half-reaction by 2 and the Bi half-reaction by 5: 2(Mn2+ + 8H+ + 5e- → MnO4- + 4H2O) 5(BiO3- + 2H2O + 2e- → Bi3+ + 6H+)
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Add the half-reactions and simplify: 2Mn2+ + 16H+ + 10e- + 5BiO3- + 10H2O + 10e- → 2MnO4- + 8H2O + 5Bi3+ + 30H+ Simplify to get the final balanced equation: 2Mn2+ + 5BiO3- + 14H+ → 2MnO4- + 5Bi3+ + 7H2O
So, the balanced redox reaction in acidic solution is: 2Mn2+ + 5BiO3- + 14H+ → 2MnO4- + 5Bi3+ + 7H2O
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