3.7: Redox Reactions

Learning Objectives 

• Use given guidelines to determine oxidation numbers of atoms in a compound. 
• Differentiate between oxidation numbers and charges 
• Use oxidation number to determine which species in a reaction is oxidized and which is reduced, which is the oxidizing agent and reducing agent. 
• Recognize and identify redox reactions 
• Write balanced equations for redox reaction 
• Determine if single displacement redox reactions are spontaneous or not based on activity series. 

Oxidation/Reduction Reactions

In addition to precipitation reactions and acid/base neutralization reactions, we will often encounter oxidation-reduction reactions, in which an overall exchange of electrons takes place.

Oxidation and reduction must occur simultaneously, that is, electrons are conserved in a chemical reaction and so for a balanced chemical reaction the number of electrons lost equals the number of electrons gained. So, that is why these types of reactions are called oxidation-reduction, or Redox Reactions, for short. So, let us define what it actually means for a species to be oxidized or reduced.  

Oxidation: is the loss of electrons
Reduction: is the gain of electrons

To help you remember, think about OIL RIG - Oxidation Is Loss (of electrons), and Reduction Is Gain (of electrons). 

Since the electrons lost by a species in a reactions must be gained by another species in the same reaction, you can quickly realize that charge must be conserved (the total charge on both sides must be the same). We will balance oxidation/reduction reactions in the second semester. For now, lets look at the following redox reaction:

 

2Fe(s) + 3O₂(g) --> 2Fe⁺³ + 3O^-2 = Fe₂O₃(s)

 

In the above equation, you may have spotted that iron on the left side has zero charge, whereas on the right side, it has a +3 charge. The fact that iron became more positive (moving to the right on a number line) indicates that electrons were lost, and so iron is oxidized. Anytime you notice that the charge becomes more positive, you can conclude the species is oxidized.  Iron then becomes the reductant(reducing agent) because it reduces the oxygen, while becoming oxidized (2Fe -> 2Fe⁺³ + 6e^-). So, iron is losing electrons (being oxidized itself) and it supplies those electrons to reduce another species, the oxygen (therefore, the reducing agent).  

By the same token, the charge on oxygen on the left is zero and on the right it is -2.  You may realize that means that it become more negative (moving left on the number line), meaning that it gained electrons, and so is reduced. So, you can use the strategy that if the charge goes down, that species is reduced. Oxygen then becomes the oxidant (oxidizing agent) because it oxidizes the iron, while becoming reduced (3O₂ +6e^- -> 3O^-2). So, oxygen is gaining electrons (being reduced itself) and is, therefore,  being the agent that oxidizes the other species, the iron (therefore, the oxidizing agent). 

 

The equation above makes it clear to see "charge" changes so you track what is oxidized and what is reduced. It is clear that a metal typically loses electrons and nonmetal gains them. However, more often than not, reactions can be elusive because many redox reactions are not as straightforward as a metal combining with nonmetal. In those cases, we need to determine the oxidation number , the "charge" that the atoms appears to have. Here are some guidelines for determining the oxidation number of elements in a compound. 

 

 

Determining Oxidation Numbers

Assign each element an Oxidation Number (State). If any element changes it’s oxidation state during a reaction, it is a redox reaction

[0] Oxidation States indicate element is neutral

[+] Oxidation States indicate element is “electron poor”

[-] Oxidation States indicate element is “electron rich”

 

Rules for Identifying Oxidation Numbers

1. Oxd # = 0 for pure elements

2. Oxd # = charge of monatomic ion

3. Oxd # of F = -1 in compounds with other elements

4. Oxd # of Cl, Br & I = -1 in compounds except with Oxygen & Fluorine

5. Oxd # of H= +1, except for Metal Hydrides (-1)

6. Oxd # of O = -2, except with fluorides, Peroxides (-1) and Superoxides (-1/2)

7. The Sum of the oxd # ‘s of all elements in a compound = 0, and = the charge of a polyatomic ion

Example \(\PageIndex{1}\)

Assign an oxidation number to the sulfur atom in Na₂S₂O_3. 

Given: we know the oxidation number of  [Na] = +1 and [O]=-2 according to the rules above. 

Strategy: Set up equation, note here the sum equals zero, but if it were a polyatomic ion, it would equal the charge of the ion: (2)[Na] + (2)[S] + (3)[O] = 0

Solution

Solve Equation for unknown

(2)[1] + (2)[x] + (3)[-2] = 0

2 + 2x - 6 = 0

2x = 4

x = 2

 

Let's try another example with video tutor. 

Video Tutor: Determining Oxidation Numbers for Atoms in a Compound

The following video explains oxidation states of chlorine, and sodium thiosulfate

Exercise \(\PageIndex{1}\)

For the following, calculate the oxidation state of the element that is not hydrogen or oxygen:

a. HClO

b. HClO₂

c. HClO₃

d. HClO₄

e. ClO^-

f. ClO₂^-

g. ClO₃^-

h. ClO₄^-

i. H₂SO₄

j. H₂SO₃

k. SO₃

l. SO₃^-2

m. SO₄^-2

n. S₂O₃^-2

Answer

a. 1

b. 3

c. 5

d. 7

e. 1

f. 3

g. 5

h. 7

i. 6

j. 4

k. 6

l. 4

m. 6

n. 2

 

Tell-Tale Signs of a Redox Reaction

 

There are a few strategies you can use to quickly identify if a reaction is a redox reaction or not. The following guideline indicate the presence of a redox reaction. 

1. Any substance combusting with oxygen 

2. When metal reacts with a non-metal

3. Single Displacement reactions 

4. When there is an overall electron electron. 

Exercise \(\PageIndex{2}\)

Which reactions are redox reactions? For those that are redox reactions, identify the oxidizing and reducing agents.

1. 2 SO₂ + O₂ → 2 SO₃ 
2. CuSO₄ + 2 AgNO₃ → Ag₂SO₄ + Cu(NO₃)₂ 
3. 2K + 2H₂O → 2KOH + H

Answer

a. Redox reaction. Combustion with oxygen. O₂ is reduced and is the oxidizing agent. sulfur is oxidized (on left [S] = +4 and on the right [S]= +6, and SO₂ is the reducing agent.

b. Not a redox reaction. This is double displacement reaction in which ions are simply swapping places. None of them change in oxidation numbers.

c. Redox. Single displacement reactions. potassium is in its elemental form on the left ([K]=0), and bonded to hydroxide on the right (K=+1), so it is oxidized and is the reducing agent. The hydrogen in water has an oxidation number of +1 and is in its elemental form on the right [H₂] = 0, so hydrogen is reduced and water is the oxidizing agent. 

 

So how can we determine if a single displacement reaction will proceed?

We use the concept of activity, and state that more active substances would rather be ionic compounds, and less active substances prefer not to form ionic compounds, and stay as their pure substances. Before going over the activity series though, lets take a quick look at what is going on in a single displacement reaction.

Single displacement reactions actually involve the transfer of electrons, and these belong to a class of reactions called oxidation-reduction reactions. If we separate the ionic compound into its ions for the reaction of iron with gold(II)sulfate and ignore the phases, we get
Fe + [Au⁺² + SO₄⁺²] --> [Fe⁺² + SO₄^-2] + Au

we see the iron loses two electrons and the gold(II) gains two

Fe →Fe⁺²   + 2e-  and     Au⁺² +2e^- --> Au

Activity Series

The activity series (table 3.2a) lists the metals in terms of their activity, where the ones at the top are the most active metals (prefer to form ionic compounds) and the ones at the bottom are the least active metals (prefer not to react and do not tend to form ionic compounds).

Table 3.6a: Activity Series. The compounds at the top are the most active and prefer to form ions, while the ones at the bottom tend to not react, and are often stable a pure metals.

NOTE: If one species must go left to right it loses electrons, so another species must go right to left (and gain the electrons).

If we take a closer look at the activity series and the periodic table we can make some observations. The alkali metals are the most active, in fact they are so active, that they will react with water, forming hydroxide and hydrogen gas while giving off enough heat to combust the gas. In fact, they definition of alkali is something that is basic and the alkali metals are so reactive that they are not stable in the atmosphere if there is water, as they react and make it basic. Alkaline earths also react with water although less vigorously.

Video3.6a: 6'16" Youtube by the Royal Society of Chemistry showing the reactions of Alkali (active) metals with water.

How do we use the Activities Series to determine if a single displacement reaction is spontaneous?

One of the species is on the left hand column (getting oxidized), and the other is on the right hand column (getting reduced). If the species on the left is above the species on the right, the reaction will proceed. If the one on the right is above the one on the left, nothing will happen. (Note, iron is above gold, so iron metal will react with Au(II), but Au metal will not react with Iron (II).

Predicting Single Displacement Reactions with Activity Series

The following video describes how to predict if a Single Displacement Reaction will proceed.

3:14 video 3.6b

 

Exercise \(\PageIndex{3}\)

Is the following single displacement reaction spontaneous? 

Fe(NO₃)₂ + 2 Ag → 2 AgNO₃ + Fe

Answer

Non-spontaneous. Fe⁺²  + 2 e^- --> Fe           Ag --> Ag⁺ + e-
The oxidation of iron is higher on the activity series than silver meaning that iron is more likely to be oxidized than silver would be. Here, iron is being reduced and silver is being oxidized, which is unlikely to be spontaneous. 

 

 

 

Contributors

• Bob Belford (UALR) and November Palmer (UALR)
• Ronia Kattoum (UA of Little Rock)