Reduction and Oxidation Help (page 2)

By — McGraw-Hill Professional
Updated on Aug 28, 2011


In electrochemical reactions, you will sometimes see redox reactions written as half-reactions. The charge must be balanced overall and the electrons produced in one part of the reaction must be lost in another part. If this didn’t happen, electrical charges would build up and cause a big problem.

Look at the electrical ion transfer ( electrolysis ) of molten sodium chloride to produce chlorine gas and sodium metal. The sodium (Na + ) and chloride (Cl ) ions are in the electrolyte solution. When an electrical current is passed through the solution, the chloride ions react at the anode and the sodium ions react at the cathode as shown in the following reactions:

2Cl ⇒ Cl 2 + 2e (oxidation)

Na + + e ⇒ Na (reduction)

The same number of electrons begins at the cathode and leaves at the anode, so the overall reaction is found by multiplying the second half-reaction by two and adding the two reactions to get:

2Na + + 2Cl ⇒ 2Na + Cl 2

Oxidation Number

The number used to keep track of the electrons in a reaction is called the oxidation number . Elements, like the halogens, may have different oxidation numbers depending on the specific reaction and environment. Oxidation number also tells us something about the strength or ability of a compound to be oxidized or reduced or to serve as an oxidizing agent or reducing agent. Oxygen has an oxidation number of −2. Using this as a starting place, oxidation numbers are assigned to all other elements.

Since water is a neutral molecule, it is fairly simple to figure out the oxidation numbers for the elements of the H 2 O molecule. Oxygen has an oxidation number of −2, so then each hydrogen must have an oxidation number of +1, to allow the total charge to equal zero. Table 8.1 lists a few hints to help figure out oxidation numbers.

Table 8.1 When figuring out oxidation numbers, remembering a few hints can help.

General rules of oxidation and reduction

1. An uncombined element has an oxidation number of zero (K, Fe, H 2 , O 2 )

2. All oxidation numbers added together in a compound must equal zero.

3. In an ion of one atom, the oxidation number is equal to the ion’s charge.

4. In an ion of more than one atom, all the oxidation numbers add up to the ion’s charge.

5. When oxygen is part of a compound, the oxidation number is −2 (except peroxides H 2 O 2 (−1)).

6. Hydrogen’s oxidation number is equal to its +1 charge (except when combined with metals, then it is −1).

Oxidation numbers can be calculated by multiplying the number of elemental atoms present by the oxidation numbers and setting the entire equation equal to zero. The following example shows you how.


What is the oxidation number of chromium in Li 2 Cr 2 O 7 , lithium dichromate?

Li = +1, Cr = x , O = −2

Li 2 (2 × +1) = 2

O 7 (7 × −2) = −14

+2 + 2 x + (−14) = 0

2 x = 12

x = 6

So the oxidation number of chromium is 6.

Figuring out the oxidation and reduction of elements in a sample is fairly simple if you use the Periodic Table and the rules of reaction. Working with redox reactions is basically an accounting task. You need to be able to keep track of all of the electrons as they transfer from one ion form to another. The trick to balancing redox reactions is to balance the charge as well as the elements on each side of the reaction.

Practice problems for these concepts can be found at – The Hydrogen Atom Practice Test

View Full Article
Add your own comment

Ask a Question

Have questions about this article or topic? Ask
150 Characters allowed