By looking at the reaction, we see that the cations for the reactants have their anions switched in the products. This means that the reaction follows the general outline:

This is an example of a double-replacement reaction.

Addition reactions convert two reactants to a single product, while decomposition reactions convert a single reactant to multiple products. Single-replacement reactions only switch one cation-anion pair.

In a double displacement reaction, or double replacement reaction, opposite ions combine. What this means is that the cation of one compound will recombine and bond to the anion of the other compound and vice versa. 

In this example, we have to remember the ideal convention of cations being written before the anion in the compound. So in AB, we can presume A being the cation and B being the anion. The same goes for CD - C is the cation and D is the anion.

With this in mind, we can now easily see how the replacement would be possible. 

If A is a cation, and was originally bound to B (anion), the only other anion it is left with to bind is D. If C is a cation, and was originally bound to D (anion), it is only left with the option of binding to the now-free B anion. 

That's why it is:

The process illustrated is the reduction of the chromium atom. Reduction occurs when electrons are gained, causing the oxidation number (charge) of the atom to decrease. We can examine this process by looking at the oxidations numbers of each atom involved.

Oxygen always has an oxidation number of , and the ion has a total charge of . This allows us to find the oxidation number for chromium. Note that there are two chromium atoms and seven oxygen atoms in the ion.

The initial oxidation state of chromium is .

Following the reaction, chromium has a charge of , meaning that it also has an oxidation number of .

During the process, the chromium atom went from an oxidation state of  to an oxidation state of . This means that it must have gained electrons (reducing the oxidation state). This is a reduction process.

The oxidizing agent is the element that causes the other element to be oxidized, and is itself being reduced in the equation. Reduction is defined as the gain of electrons, meaning the element would get "more negative."

In this reaction, aluminum transitions from an oxidation state of 3+ to an oxidation state of 0. This indicates a reduction, as the atom has gained more electrons and reduced charge.  is thus the oxidizing agent in this reaction.

Use the rules for assigning oxidation numbers. We know that the sum of oxidation numbers of all atoms present in the compound must equal zero because the compound is neutral.

The oxidation number of hydrogen is  except in compounds with less electronegative elements (which is not the case here). Because there are two hydrogens, the overall oxidation is .

The oxidation number of oxygen is always . Multiply that by the total number of oxygens to get .

From here, simple math tells us that the oxidation number of sulfur must be  to make the total oxidation state equal to zero.

A combustion reaction is characterized as an oxidation-reduction reaction because the oxidation number of the reactants changes, and oxygen is used to burn a fuel molecule. Carbon dioxide is often produced as a product. The following is an example of a combustion reaction:

In an oxidation reaction, one species is oxidized (loses electrons). During a reduction reaction, one species is reduced (gains electrons). These transfers of electrons results in the change of oxidation states. One way to help you remember which reaction is which, is by using the mnemonic: OIL RIG - Oxidation Is Loss of electrons; Reduction Is Gain of electrons. 

Balancing redox reactions involves the following steps:

1. Divide the reaction into oxidation and reduction half reactions and balance all atoms that are not oxygen and hydrogen:

2. Balance the oxygens by adding water molecules on the opposite side of the reactions:

3. Balance the hydrogens by adding protons to the opposite side of the equation:

4. Add electrons in order to equal the charges on both sides of the equation:

5. In order to make the electron exchange equal in each half step, we must multiply the top half reaction by 3:

6. Add up the reactants and products while cancelling out substances on opposite sides of the reactions. For example: we will cancel the 6 water molecules as reactants and be left with only one water molecule as a product. In addition, only 2 protons will be left on the reactant's side after canceling the 12 from the product's side.

In the balanced redox reaction, the coefficient on sulfur dioxide is 3.

To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of . Phosphorus begins with an oxidation number of  and ends with an oxidation number of . Note that the oxidation numbers for fluorine and iodine reamain constant at  for each.

Now we can begin to look at the half-reactions.

Balance the atoms.

Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.

We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.

An oxidation-reduction (redox) reaction is a reaction where electrons are transferred between two substances. When an atom is oxidized, it is called the reducing agent, and it loses a number of electrons. Similarly, a reduced atom is called the oxidizing agent, and gains the same number of electrons. A popular mnemonic to help remember is OIL RIG, or Oxidation is Loss of electrons, Reduction is Gain of electrons.