balance the equation by oxidation no MnO3- + C2 H5OH equal to Mn2+. + CH3COOH
Answers
Explanation:
How will I balance the given reaction MnO4- + C2H5OH ---> (Mn2+) + CH3COOH?
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Step 1: Decide what gets oxidised and what gets reduced…
Oxidation = CH3CH2OH → CH3CO.OH (removal of H and addition of O to one C atom)
Reduction = MnO4^(-) → Mn^(2+) (Mn(VII) to Mn(II))
Step 2: Balance any oxygens with water…
Ox = CH3CH2OH + H2O → CH3CO.OH
Red = MnO4^(-) → Mn^(2+) + 4H2O
Step 3: Balance any outstanding Hydrogens with hydrogen ions
Ox = CH3CH2OH + H2O → CH3CO.OH + 4H^(+)
Red = MnO4^(-) + 8H^(+) → Mn^(2+) + 4H2O
Step 4: Balance outstanding charges with electrons…
Ox = CH3CH2OH + H2O → CH3CO.OH + 4H^(+) + 4e^(-)
Red = MnO4^(-) + 8H^(+) + 5e^(-) → Mn^(2+) + 4H2O
This gives us the half reactions for the reaction.
Step 5: Combine both equations…
CH3CH2OH + H2O + MnO4^(-) + 8H^(+) + 5e^(-) → CH3CO.OH + 4H^(+) + 4e^(-) + Mn^(2+) + 4H2O
Step 6: Cancel down any species that can be cancelled…
(After checking all balances): Final, balanced, ionic equation for the oxidation of ethanol in acidified manganate(VII) solution…
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