The question states that an oxidized atom will increase its oxidation state, which means that the atom will lose electrons and become more positively charged. Since the atom will lose electrons, the number of electrons will be less than the number of protons. Electrons are negatively charged and protons are positively charged; therefore, the presence of more protons in an oxidized atom will make the atom more positive. An oxidized atom will usually become a cation because its number of electrons will decrease.

Reduced atoms will gain electrons and become more negatively charged (decrease oxidation state). Notice that the reactions given in the passage are not redox reactions because the atoms in both reactions retain their oxidation states.

The atom that is oxidized will have a higher oxidation number as a reactant than as a product. Since reactant phosphorus (P₄) is in elemental form, it has an oxidation number of 0. As a product, phosphorus (PO₄^3-) has an oxidation number of +5.

Total charge = (P) + 4(O) = –3

–3 = (P) + 4(–2) = (P) + (–8)

p = +5

Because it is now more positive, we say that the phosphorus has been oxidized.

The oxidizing reagent is the reactant, not the atom, that is responsible for receiving electrons from another atom in order to oxidize it. The oxygen gas goes from having an oxidation number of 0 to –2. This means that oxygen gas is reduced, but because it oxidized another atom by taking its electrons, we call it the oxidizing reagent.

Reactant oxygen is elemental, and has oxidation number 0.

Product oxygen is in H₂O, with net charge of 0.

Total charge = 2(H) + (O) = 0

2(+1) + (O) = 2 + (O) = 0

O = –2

Decrease in oxidation number indicates reduction, and also identifies the oxidizing agent.

In the reaction, solid potassium (K) is initially in elemental form, so it has an oxidation state of 0. The potassium ions (K⁺) have a charge of +1 in the product. Since the potassium has lost an electron, it has been oxidized.

Since each hydrogen in reactants (NH₃) has an oxidation state of +1, and the hydrogen gas (H₂) has a charge of 0 as a product, the hydrogen has been reduced. Since it was reduced, we conclude that NH₃ is the oxidizing reagent for the reaction.

In this reaction, the oxidation number of oxygen goes from , in diatomic gaseous oxygen, to , in both carbon dioxide and water. This indicates that it has gained electrons; a gain of electrons indicates that oxygen has been reduced. Since it is reduced, it is the oxidizing agent.

The oxidation number of carbon goes from  in cyclohexane to  in carbon dioxide. This indicates that it has lost electrons; a loss of electrons indicates that carbon has been oxidized. Since it is oxidized, cyclohexane is the reducing agent.

Oxidation is defined by a loss of electrons, generally resulting in a more positive charge on an atom. Reduction is defined by a gain of electrons, generally resulting in a more negative charge on an atom. To identify the compound being oxidized, we need to find which compound is becoming more positive.

Zinc is initially neutral, but gains a positive two charge as a product. Hydrogen initially has one positive charge, but becomes neutral in the product. The distinction becomes even clearer if we break the overall reaction into two half-reactions.

We can see that zinc is losing electrons and hydrogen is gaining electrons. Zinc is thus being oxidized.

We often hear of combustion reactions producing carbon dioxide and water. This is true when the reactants contain all three elements (hydrogen, oxygen, and carbon) needed to make carbon dioxide and water. The trick here is to understand that although combustion is just an oxidation reaction (ie, creating more carbon-oxygen bonds), all reactions must still be balanced. One answer choice fails to balance hydrogen in the reaction.

Although we do see an oxidation reaction with the formation of carbon dioxide, the reaction is NOT balanced. This reaction cannot possibly take place as written.

An oxidizing agent is one that oxidizes another element and, becomes reduced in the process. Oxidizing and reducing agents are found in the reactants of the equation, eliminating Al³⁺ and Cr(s) as answer choices. Cr³⁺ gains three electrons to become Cr(s), which has an oxidation number of zero. Cr³⁺ becomes reduced, while Al(s) is oxidized, therefore, Cr³⁺ is the oxidizing agent. Remember "OIL RIG": Oxidation Is Loss of electrons and Reduction Is Gain of electrons.

There are a couple options to solve this problem. First, we could simply remember that anything which combines with oxygen in a redox reaction is oxidized, and the associated oxygen is reduced.

Another option is to think about the oxidation numbers of each element (an element whose oxidation number increases is oxidized, and one whose oxidation number decreases is reduced). If we choose this method, recall that free elements (Fe and O₂ in this case) have oxidation number 0. Also, oxygen in compounds has oxidation number -2. Oxygen's oxidation number decreases from 0 to -2, meaning that oxygen is reduced. Since oxidation and reduction occur at the same time in different elements of the redox reaction, iron's oxidation number must increase, so iron is oxidized.

This problem requires us to determine the oxidation number of carbon as a reactant and as a product. The oxidation number of carbon as a reactant is –4, because it is attached to four hydrogens each with a charge of +1. The overall charge on the molecule is zero, the carbon must cancel out the charges contributed by hydrogen.

Carbon as a product has an oxidation number of +4, because it is attached to two oxygens, each with an oxidation number of –2. Again, we know that the molecule is neutral, and carbon must balance the charges from oxygen

Since the oxidation number of carbon went from –4 to +4, we conclude that carbon has been oxidized in the reaction. Any atom that loses electrons is oxidized, while any atom to gain electrons is reduced.