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tesla

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Balancing Redox Reactions - Oxidation Number

« on: June 19, 2011, 05:30:47 PM » I'm having a bit of trouble with balancing redox reactions, and I just want to confirm that I'm doing it the right way. MnO4- + H2C2O4 Mn2+ + CO2 + H2OI assigned oxidation numbers, and found that there is a transfer of 5e-/Mn as well as 5e-/MnO4, and 1e-/C and 4e-/H2C2O4 So when I balanced and added water and hydrogen ions I ended up with22H+ + 4MnO4- + 5H2C2O4 4Mn2+ + 10CO2 + 16H2OIs that correct?Thanks for looking Logged

Borek

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Re: Balancing Redox Reactions - Oxidation Number

« Reply #1 on: June 19, 2011, 06:07:50 PM » Too many electrons for oxalic acid, hence final answer is wrong. Logged ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

lillybeans

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Re: Balancing Redox Reactions - Oxidation Number

« Reply #2 on: June 19, 2011, 10:26:43 PM » Hi there, assuming this is in acidic conditions, thenMnO4- + H2C2O4 --> Mn2+ + CO2Step 1: Identify atoms that change oxidation number; balance them.MnO4 + H2C2O4 --> Mn2+ + 2CO2Step 2: Recalculate the total number of electrons transferred and make them equal (multiply Mn by 2, C by 5 which I see you did)2 MnO4- + 5 H2C2O4 --> 2 Mn2+ + 10 CO2Step 3: Add H+ to balance the charge (Left: -2; Right: +4)2 MnO4- + 5 H2C2O4 + 6 H+ --> 2Mn2+ + 10 CO2Step 4: Add waters to balance the H2 MnO4- + 5 H2C2O4 + 6 H+ ---> 2 Mn2+ + 10 CO2 + 8 H2OStep 5: Check to make sure the charges on both sides are the same. (Left: +4, right: +4). Check for atoms to make sure they equal, which they do. voila! Logged

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