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The half equation for the reduction of carbon dioxide to methanoic acid and its standard electrode potential value is given as:

CO2 + 2H+ + 2e- gives HCOOH EO = -0.11 V

The half equation for the reduction of ACIDIFIED potassium manganate(VII) and its standard electrode potential value is given as:

MnO4- + 8H+ + 5e- gives Mn2+ + 4H2O EO = +1.51 V

Thus, the standard electrode potential E for the overall reaction is:

E = 1.51 - (-0.11) = +1.62 V

Since it is positive (and well positive), the oxidation of methanoic acid by potassium manganate(VII) can take place under standard conditions.

The overall equation (obtained by balancing the number of electrons) is:

2MnO4- + 6H+ + 5HCOOH gives 2Mn2+ + 8H2O + 5CO2

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13y ago
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9y ago

2KMnO4=K2MnO4+MnO2+O2

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Q: What is the chemical equation for the reaction between methanoic acid and acidified potassium permanganate?
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