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A chemistry student needs to determine the oxidation state changes of the atoms and identify the oxidizing and reducing agents in the following reaction: C2H5OH + 2 MnO4- → 2 CH3COOH + 2 MnO2 + 4 OH-

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To determine the oxidation state changes of the atoms and identify the oxidizing and reducing agents in the given reaction, we need to assign oxidation numbers to each atom in the reactants and products.C2H5OH  Ethanol :C: -2  each carbon atom, since it is bonded to more electronegative atoms, hydrogen and oxygen H: +1  each hydrogen atom O: -2MnO4-  Permanganate ion :Mn: +7O: -2CH3COOH  Acetic acid :C: -3  the carbon atom bonded to the oxygen atom C: -1  the carbon atom bonded to the hydrogen atom H: +1  each hydrogen atom O: -2MnO2  Manganese dioxide :Mn: +4O: -2OH-  Hydroxide ion :O: -2H: +1Now, let's analyze the changes in oxidation states:1. Ethanol  C2H5OH  to Acetic acid  CH3COOH :The carbon atom in ethanol changes its oxidation state from -2 to -1  loses one electron . This is an oxidation process.2. Permanganate ion  MnO4-  to Manganese dioxide  MnO2 :The manganese atom in MnO4- changes its oxidation state from +7 to +4  gains three electrons . This is a reduction process.Thus, the oxidizing agent is the MnO4- ion, and the reducing agent is the ethanol  C2H5OH .

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