Ethanol, C2H5OH, will combust in air according to the equation below. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) ΔHo = –1270 kJ/molrxn (a) Is O2(g) oxidized in the reaction, or is it reduced? Justify your answer in terms of oxidation numbers.

The answer to whether the Oxygen (O₂) is oxidized or reduced is that;

The oxygen (O₂) is being reduced.

We are given the equation;

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g) ΔH° = –1270 kJ/mol_rxn

Now, Oxidation in this context will be an increase in oxidation number. This means that we consider a chemical element in a chemical reaction to be undergoing oxidation if there is a positive increase in its oxidation number from left to right in the reaction. Otherwise, it is said to undergo reduction.

Now, when we assign oxidation numbers to the elements in the chemical equation, the oxygens in O₂ (g) would be zero. Meanwhile the oxygens in both CO₂ (g) and H₂O (g) are -2.

In conclusion, since the oxygens in both CO₂ (g) and H₂O (g) are -2, it means oxidation number has reduced and as such there is a reduction in oxygen.

Read more about Oxidation and Reduction at; link

O2(g) was Reduced

Explanation:

It is necessary at this point to restate the working definitions of oxidation and reduction.

Oxidation refers to increase in oxidation number. A chemical specie is oxidized in a chemical reaction if there is a positive increase in its oxidation number from left to right in the reaction.

A chemical specie is said to be reduced when there is a decrease in its oxidation number from left to right in a reaction. Hence reduction refers to a decrease in oxidation number.

Now let us focus on O2(g). Its oxidation number on the left hand side is zero. On the right hand side, its oxidation number decreases to -2. This shows a decrease in oxidation number. From our premises above, we can safely conclude that O2(g) was reduced in the reaction.

give me abc and i will get it for u.

catya that's not even a b or c