For the following reaction, identify the substance being oxidized and reduced CH4 + 2 O2 = CO2 + 2 H2O
The substance oxidized is the one that loses electrons. On the left C is -4 and on the right +4. On the left O2 is zero and on the right -2.
To identify the substance being oxidized and reduced in a chemical reaction, we need to examine the changes in oxidation numbers of the elements involved.
Step 1: Determine the oxidation numbers of the elements in the reactants and products.
In CH4 (methane), carbon is assigned an oxidation number of -4, and hydrogen has an oxidation number of +1.
In O2 (oxygen), the oxidation number of oxygen is 0 since it is a diatomic molecule.
In CO2 (carbon dioxide), the oxidation number of carbon is +4, and oxygen remains at -2.
In H2O (water), the oxidation number of hydrogen is +1, and oxygen remains at -2.
Step 2: Compare the oxidation numbers of the elements before and after the reaction.
In CH4, the carbon atom's oxidation number changes from -4 to +4, indicating that carbon has been oxidized.
In O2, the oxidation number of oxygen remains unchanged, indicating that it does not undergo oxidation or reduction.
In CO2, the carbon atom's oxidation number changes from +4 to +4 (no change), indicating that carbon does not undergo oxidation or reduction.
In H2O, the hydrogen atom's oxidation number changes from +1 to +1 (no change), indicating that hydrogen does not undergo oxidation or reduction.
Step 3: Determine the substance being oxidized and reduced.
From the above analysis, we can conclude that in the given reaction:
The substance being oxidized is methane (CH4), as the carbon atom's oxidation number increases from -4 to +4.
The substance being reduced is oxygen (O2), as it does not undergo any change in oxidation number.
Therefore, in the reaction CH4 + 2 O2 = CO2 + 2 H2O, methane is oxidized, and oxygen is reduced.