For the List of Chemicals Below Write Complete Balanced Equations

Redox Reactions

Overview: This section reviews reduction-oxidation (redox) reactions, a common type of chemical reaction.

Skills:

• Determining oxidation states
• Balancing redox reaction equations

New terms:

• Reduction
• Oxidation
• Half-reaction
• Oxidation State
• Oxidation Number

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Redox reactions are reactions in which one species is reduced and another is oxidized. Therefore the oxidation state of the species involved must change. These reactions are important for a number of applications, including energy storage devices (batteries), photographic processing, and energy production and utilization in living systems including humans.

Reduction: A process in which an atom gains an electron and therefore decreases (or reduces its oxidation number). Basically the positive character of the species is reduced.
Oxidation: A process in which an atom loses an electron and therefore increases its oxidation number. In other words, the positive character of the species is increased.

Historically, the term "oxidation" was used because the redox reactions that were first systematically investigated took place in oxygen, with oxygen being reduced and the other species being oxidized, hence the term oxidation reaction. However, it was later realized that this case (oxidation reactions involving oxygen) was just one possible scenario. For example consider the redox reaction shown below.

In this process the Fe²⁺ ion is oxidized, but there is no oxygen involved in this reaction. The Ce⁴⁺ ion, which is reduced acts as the oxidizing agent. So oxidation reactions need not involve oxygen. This redox reaction is actually the sum of two separate half-reactions (a reduction half-reaction and an oxidation half-reaction).

Example 1.

In the following redox reaction, which species is being oxidized? Which one is being reduced?

Al(s) is being oxidized.
Ag⁺ (aq) is being reduced.

A mnemonic you might find helpful to remember the definitions of oxidation and reduction is: Leo the lion goes ger. Leo: lose electron(s) = oxidation. Ger: gain electron(s) = reduction.

Oxidation State: The condition of a species with a specified oxidation number. An element with a given oxidation number exists in the corresponding oxidation state.

Assigning Oxidation Numbers

The following rules for assignment of oxidation numbers are listed in hierarchical order.

1. Pure elements (in their natural, standard state): ox. # = 0.
2. Monatomic ions: ox. # = ionic charge.
3. F is always F (-I) in compounds.
4. Alkali metals (those in the 1st column of the periodic table): ox. # = I.
5. Alkaline-earth metals (those in the 2nd column of the periodic table): ox # = II.
6. Hydrogen is almost always H (I). The exception is in metal hydrides (MH_x).
7. Oxygen is almost always O (-II) in compounds. Exceptions are O-O and O-F.

The sum of all oxidation numbers in the species will equal the total charge of that species.

In Chemistry 111, you will learn how to determine oxidation numbers in compounds without using the rules. You will learn how these rules were derived.

Example 2.

Assign oxidation numbers for the following elements in these compounds.

Element	Oxidation Number	Reason
O in O2	0	Standard state of oxygen. See Rule 1.
H in HF	I	See Rule 6.
F in HF	-I	See Rule 3
F in F2	0	Standard state of fluorine. See Rule 1.
Mg2+	II	This is a monatomic ion. See Rule 2.
Na in NaCl	I	This is an alkali metal.
O in H2O	-II	See Rule 7.
Mg in MgO	II	This is an alkaline-earth metal. See Rule 5.

When assigning oxidation numbers for molecules use the following equation:

For instance take HBrO₂. We know that O has an oxidation number of -2 from Rule 7 and hydrogen is H (I) from Rule 6. The total charge on HBrO₂ is zero. If we use the equation above to solve for the oxidation number of Br we get the following result.

Guidelines for Balancing Redox Equations:

1. Determine the oxidation states of each species.
2. Write each half reaction and for each:
   1. Balance atoms that change oxidation state.
   2. Determine number of electrons gained or lost
   3. Balance charges by using H⁺ (in acidic solution) or OH^- (in basic solution).
   4. Balance the rest of the atoms (H's and O's) using H₂O.
3. Balance the number of electrons transferred for each half reaction using the appropriate factor so that the electrons cancel.
4. Add the two half-reactions together and simplify if necessary.

Example 3.

Balance the following redox reaction.

Step 1. Determine the oxidation states of the species involved.

The charges don't match yet so this is not a balanced equation. We can use each half-reaction to balance the charges. Notice that the Cl^- ions drop out, as they are spectator ions and do not participate in the actual redox reaction.

Step 2. Write the half reactions.

Step 2a. Balance the atoms that change their oxidation states.

Step 2b. Determine the number of electrons gained or lost.

Aluminum changes from 0 to III, so three electrons are lost. For hydrogen, the case is a little different. Hydrogen is going from I to 0. This means that for each H⁺ ion that reacts, one electron is needed. Since there are two H⁺ ions that react, two electrons are needed.

Steps 2c and 2d are not needed in this case as the equations are balanced.

Step 3. Balance the number of electrons transferred.

The common factor for the electrons transferred is 6, so the above multiplication is performed.

Step 4. Now the charges and atoms are balanced. To verify this add all of the charges and atoms on each side. Both the charges and number of atoms must balance. Note that this reaction is not neutral. Remember that the spectator ions, Cl^-, neutralize the solution.

Example 4.

Balance the following reaction, which occurs in acidic solution.

Step 1. Determine the oxidation states of the species involved.
To determine the oxidation state of Mn in MnO₄ ^-, apply Equation 1 (see Equation 1 above): x + 4(-2) = -1.

The 4 is from the number of oxygen atoms, -2 is the oxidation state of oxygen and -1 is the overall charge of the molecule.
Which species is oxidized and which species is reduced? Answer

Step 2. Write the half reactions.

Step 2a. Mn and Cl are balanced.

Step 2b. Mn changes from VII to II, so five electrons are needed. Cl^- loses two electrons as it goes from I to -I.

Step 2c. The charges are not balanced on this example. Since this is in acidic solution, use H⁺ to balance these charges.

Remember that the electrons carry a negative charge and must be considered whenever balancing the charges. Verify that the charges are balanced on each side of the equation.

Step 2d. Now the oxygen and hydrogen atoms need to be balanced.

Step 3. Balance the number of electrons transferred.

Step 4. Write the net reaction.

Now all charges and number of atoms balance.
Finally, two terms you may run across in the future are oxidizing agent (or oxidant) and a reducing agent (reductant). An oxidizing agent causes oxidation and is reduced in the reaction. A reducing agent causes the reduction in the redox reaction. The reducing agent is oxidized in the reaction. In Example 4 above, MnO₄ ^- is the oxidizing agent and Cl^- is the reuducing agent.

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Advanced Applications: Redox Chemistry in Molecular Electronics and Photosynthesis.

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Summary

Now you should have a good working understanding of the fundamentals of redox reactions. You should be able to determine oxidation numbers as well as balance redox reaction equations.

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Practice Quiz: Redox Reactions

Note: You will need a pencil, scratch paper, calculator, periodic table and equation sheet to work the practice quiz. Quizzes are timed (approximately 4 minutes per question).

We suggest that you print out the periodic table and the constants/equations page before starting the quizzes.

Open Periodic Table in a separate browser window.

Open Equations/Constants Page in a separate browser window.

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Return to the Chemistry Subject Index © Washington University in St. Louis, 2005.

For the List of Chemicals Below Write Complete Balanced Equations

Source: http://www.chemistry.wustl.edu/~coursedev/Online%20tutorials/Redox.htm

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