Notes on Redox Reactions (Chapter - 8)

Oxidation and Reduction

Oxidation‐reduction (redox) reactions

At different times, oxidation and reduction (redox) have had different, but complimentary, definitions. Compare the following definitions:

Oxidation is:                                                 Reduction is:

=>Gaining oxygen                                           =>Losing oxygen

=>Losing hydrogen                                         =>Gaining hydrogen

=>Losing electrons (+ charge increases)                =>Gaining electrons (+ charge decreases)

Oxidation and reduction are opposite reactions. They are also paired reactions: in order for one to occur, the other must also occur simultaneously.

The oxidation number describes the oxidation state of the element in a compound, and these numbers are assigned following a relatively simple set of rules.

Rules for assigning oxidation numbers

1.      The oxidation number of an element in its elemental form is 0 (zero).

2.      The oxidation number of a simple ion is equal to the charge on the ion. Both the size and the polarity of the charge are part of the oxidation number: an ion can have a +2 oxidation number or a -­‐2 oxidation number. The “+” and “-­‐“ signs are just as important as the number!

3.      The oxidation numbers of group 1A and 2A (group 1 and group 2) elements are +1 and +2 respectively.

4.      In compounds, the oxidation number of hydrogen is almost always +1. The most common exception occurs when hydrogen combines with metals; in this case the oxidation number of hydrogen is typically –1.

5.      In compounds, the oxidation number of oxygen is almost always –2. The most common exception is in peroxides, when the oxidation number is –1. Peroxides are compounds having two oxygen atoms bonded together. 

      For example, hydrogen peroxide is H‐O‐O‐H. In hydrogen peroxide, each oxygen atom has a ‐1 oxidation number. When oxygen is bonded to fluorine, as in hypofluorous acid (HOF), the oxidation number of oxygen is 0. Oxygen- fluorine compounds are relatively rare and not too terribly important for our studies.

6.      In compounds, the oxidation number of fluorine is always –1. The oxidation number of other halogens (Cl, Br, I) is also –1, except when they are combined with oxygen. The oxidation number of halides (except fluorine) combined with oxygen is typically positive. For example, in ClO^-, chlorine’s oxidation number is +1.

7.  For a complex ion, the sum of the positive and negative oxidation numbers of all elements in the ion equals the charge on the ion.

8.      For an electrically neutral compound, the sum of the positive and negative oxidation numbers of all elements in the compound equals zero.

Identifying redox reactions

Now that you can assign proper oxidation numbers to the elements in substances, we can use changes in the oxidation numbers to identify oxidation and reduction reactions.

Consider the following chemical equation:

Zn(s) + 2HCl(aq) → Zn⁺²(aq) + 2C^­‐(aq) + H₂ (g)

The oxidation number for elemental zinc is 0. The oxidation number for zinc ion is +2. The oxidation number for hydrogen in hydrogen chloride is +1. The oxidation number for elemental hydrogen, H2, is 0. 

The oxidation number for chlorine in hydrogen chloride is –1. The oxidation number for chloride ion is –1.

Zinc has lost two electrons, and therefore developed a +2 charge. Zinc has been oxidized – its oxidation number has become more positive (from 0 to +2).

Hydrogen has gained an electron, and its positive charge has decreased. Hydrogen has been reduced – its oxidation number has become less positive (from +1 to 0).

Chloride has not changed its oxidation state. It is a spectator ion in this reaction.

Figure 1 may be useful in deciding if an element has been oxidized or reduced. If an elements oxidation number increases (moves towards the right), then the element is oxidized. If an elements oxidation number decreases (moves towards the left), then the element is reduced.

 NOTE: an element doesn’t have to become positive or negative for oxidation or reduction to occur. Instead, the element has to become more positive or more negative. A change from -­‐3 to -­‐1 is still oxidation, while a change from +3 to +1 is still reduction.

Substances that cause changes in the oxidation state are called oxidizing agents or reducing agents. An oxidizing agent causes oxidation number to occur.

How does the oxidizing agent cause oxidation?

To increase the positive oxidation number of an element, the oxidizing agent must take one or more electrons from the element. As the element being oxidized loses electron(s), its oxidation number becomes more positive.

 However, the electrons don’t disappear! The oxidizing agent has taken these electrons, and therefore the oxidizing agent becomes more negative – it is reduced!

In our reaction above, hydrogen in hydrogen chloride takes an electron from the zinc metal. The zinc metal becomes more positive; it is oxidized. 

By taking the electron from the zinc metal, the hydrogen in hydrogen chloride becomes less positive; it is reduced. Hydrogen chloride is the oxidizing agent, because it contains the element that causes oxidation to occur.

Similarly, the zinc metal donates electrons to the hydrogen in hydrogen chloride, causing the oxidation state of hydrogen to decrease from +1 to 0. By providing the electrons necessary to reduce the oxidation number of hydrogen, the oxidation number of zinc increases from 0 to +2; zinc is oxidized. While being oxidized, zinc reduced the oxidation number of hydrogen. Therefore, zinc is a reducing agent.

Oxidizing agents are substances containing the element(s) that accept electrons, allowing another element(s) to be oxidized. By accepting electrons, the element(s) in the oxidizing agent are reduced.

Reducing agents are substances containing the element(s) that donate electrons, allowing another element(s) to be reduced. By donating electrons, the element(s) in the reducing agent are oxidized.

As you can see, there is a great deal of symmetry between oxidizing and reducing agents, and between oxidation and reduction. 

Whenever one process happens, the other process MUST also happen, because we are transferring electrons from one material to another material. Electrons must come from someplace, and must go someplace; electrons cannot simply appear and disappear.

There are two general considerations to keep in mind when discussing oxidizing/reducing agents.

First, in most cases there will only be one element in each agent that is oxidized or reduced. In our reaction, only hydrogen in hydrogen chloride was reduced – the chloride wasn’t changed. 

Second, the oxidizing or reducing agent is the compound containing the element that is oxidized or reduced.

The reducing agent is hydrogen chloride, not just hydrogen ion.

 Similarly, the oxidizing agent is the compound containing the element that is reduced. Oxidizing and reducing agents are ALWAYS reactants, NEVER products!

Balancing redox reactions.

We can use our knowledge of redox reactions to help us balance chemical reactions. The simplest process to use is the half-­‐reaction method. Consider the

following chemical reaction (unbalanced):

MnO4^‐ + C2O4^‐2 → Mn⁺² + CO2

First, we assign oxidation numbers to all elements shown in the reaction:

+7  ‐2	+3  ‐2	+2	+4 -2
MnO4-   +	C2O4‐	→  Mn+2   +	CO2

Comparing oxidation numbers, we see that manganese goes from +7 to +2, and has been reduced, while carbon goes from +3 to +4 and has been oxidized. 

Oxygen doesn’t change its oxidation state. The oxidizing agent is permanganate, and the reducing agent is oxalate.

The half‐reaction method involves balancing the oxidation reaction as if it were an isolated reaction. Then the reduction half-reaction is balanced as if it were an isolated reaction. Finally, the two half‐reactions are combined.

The oxidation half‐reaction is:

C₂O₄^‐2 → CO₂

The first step is to balance elements, other than oxygen and hydrogen, using the normal methods of balancing chemical equations. We have two carbons in oxalate, so we need a 2 in front of carbon dioxide to balance the carbons:

C₂O₄^‐2 → 2CO₂

By comparing reactants and products, we see that we have the same number of carbon and oxygen atoms on each side of the reaction. There aren’t any other elements present, so we don’t need to change any more coefficients.

Charge balance requires that the net charge of reactants and products must be equal. We equalize the charges by adding 2 electrons to the products side:

C₂O₄^‐2 → 2CO₂ + 2e^‐

All elements are identical on both sides, the number of atoms of each element is the same on both sides, and the total charge is the same on both sides: ‐2 for oxalate and ‐2 for the two electrons. This is a balanced oxidation half‐reaction.

The reduction half‐reaction is:

MnO₄^‐ → Mn⁺²

The manganese atoms are balanced, but what are we to do about oxygen? To balance oxygen, we add water molecules. Four oxygen atoms in permanganate require four water molecules as products:

MnO₄^‐ → Mn⁺² + 4H₂O

Adding water molecules balanced oxygen, but now we have hydrogen atoms. We balance 8 hydrogen atoms from 4 water molecules by adding 8 hydrogen ions to the reaction:

8H⁺ + MnO₄^‐ → Mn⁺² + 4H₂O

Now we have the same number and kinds of atoms on both sides. However, charge balance requires that the net charge of reactants and products must be equal. We have a net electrical charge of +7 for the reactants, and a net electrical charge of +2 for the products. We add 5 electrons to the reactant side of the equation:

8H⁺ + MnO₄^‐ + 5e^- → Mn⁺² + 4H₂O

This is a balanced reduction half-reaction.

(NOTE: in an oxidation half‐ reaction, electrons are produced, while in a reduction half‐reaction, electrons are consumed.)

We are ready to combine the two half‐reactions. To combine these reactions, the same number of electrons must be produced as are consumed.

 When different numbers of electrons are produced and consumed, you need to find the least common multiple. 

Since 5 electrons are consumed in the reduction reaction, and 2 electrons are produced in the oxidation reaction, the least common multiple is 10. We need to multiply the reduction reaction by 2, and the oxidation reaction by 5:

16H⁺ + 2MnO₄^‐ + 10e^‐ → 2Mn⁺² + 8H₂O

5C₂O₄^‐2 → 10CO₂ + 10e^-

Now that we have the same number of electrons produced and consumed, and we can add the reactions:

16H⁺ + 2MnO₄^- + 10e^- + 5C₂O₄^‐2 → 2Mn⁺² + 8H₂O + 10CO₂ + 10e^-

We eliminate substances that are identical on both sides of the equation:

16H⁺ + 2MnO₄^- + 5C₂O₄^‐2 → 2Mn⁺² + 8H₂O + 10CO₂

Finally, we check the equation to insure that mass and charge balance have been achieved.

Reactants        Products

16 H        16 H

2 Mn        2 Mn

28 O        28 O

10 C       10 C

+4       +4

This is an example of balancing a redox reaction in acid solution, by the half-­‐ reaction method. We can also balance the reaction in basic solution. 

The initial work is the same as shown above. Once the reaction is balanced in acid solution, we neutralize hydrogen ion by adding an equal number of hydroxide ions to both sides of the reaction:

16OH^- + 16H⁺ + 2MnO₄^- + 5C₂O₄^‐2 → 2Mn⁺² + 8H₂O + 10CO₂ + 16OH^‐

Hydroxide ion combines with hydrogen ion, forming water molecules, and we eliminate equal numbers of water molecules from both sides of the equation:

16H₂O + 2MnO₄^- + 5C₂O₄^-2 → 2Mn⁺² + 8H₂O + 10CO₂ + 16 OH^-

8H₂O + 2MnO₄^- + 5C₂O₄^-2 → 2Mn⁺² + 10CO₂ + 16OH^-

Finally, we check to insure mass and charge balance.

Reactants       Products

16 H       16 H

2 Mn       2 Mn

36 O       36 O

10 C      10 C

‐12    ‐12