Chemistry 30

1.3 Recognizing Redox Reactions

Oxidation numbers are a convenient way of identifying redox reactions and also indicating which element is oxidized and which is reduced. Here's an example - the reaction between sodium metal and chlorine gas:

2 Na

+

Cl2

→

2 NaCl

It is often useful to write the oxidation number for every element, in every compound, above the element in the equation. Thus for our reaction we have:

0		0		+1   -1
2 Na	+	Cl2	→	2 NaCl

Be sure to note that the balancing coefficients in the equation (the "2" in front of Na and in front of NaCl) do not affect the value of the oxidation numbers. We'll return to these coefficients soon.

A chart is a useful way for us to summarize the changes in oxidation number for each element:

element	initial ox no		final ox no	change in electrons (e-)	oxidized or reduced
Na	0	→	+1	lost 1 e-	oxidized
Cl	0	→	-1	gain 1 e-	reduced

We see several important things in our table -

Since oxidation numbers did change, this was a redox reaction Na's oxidation number increased - from 0 on the reactant side to +1 on the product side. An element becomes more positive by losing electrons. Loss of electrons is Oxidation (LEO) Cl's oxidation number decreased, from 0 to -1, as chlorine gained electrons. Gain of electrons is Reduction (GER)

An increase in oxidation number indicates oxidation A decrease in oxidation number indicates reduction

Let's try another example. Identifying what is oxidized and what is reduced - and how many electrons are involved - is very important to master now for we will expand upon this skill shortly.

Consider the reaction

2 Mg

+

O2

→

2 MgO

Determine oxidation numbers for all elements in every compound:

0		0		+2   -2
2 Mg	+	O2	→	2 MgO

Summarize the changes, determine the number of electrons transferred per atom, and identify what is oxidized and what is reduced:

element	initial ox no		final ox no	change in electrons (e-)	oxidized or reduced
Mg	0	→	+2	lost 2e-	oxidized
O	0	→	-2	gain 2e-	reduced

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Two new terms before we continue. You will recall that we mentioned in the first section of this unit that oxidation cannot occur without reduction, and vice versa. The substance losing electrons (undergoing oxidation) gives its electrons to the substance gaining electrons (undergoing reduction). If the reduced substance will not accept electrons, the oxidized substance could not give away electrons. Thus, one allows for the other to occur.

Reducing agent the substance that is oxidized. It allows another element to be reduced. Oxidizing agent the substance that is reduced. It allows another element to be oxidized.

In our last example above, magnesium was oxidized, therefore it was the reducing agent.
Oxygen was reduced, therefore it was the oxidizing agent.

By convention we often refer to the oxidizing agent and reducing agents as the entire compound the element is in, not just individual element. Consider the following reaction. Oxidation numbers are shown only for substances whose oxidation numbers undergo a change:

0		+5 (N)		+2  (Zn)		+4 (N)
Zn	+	HNO3	→	Zn(NO3)2	+	NO2	+	H2O

Summarize:

element	initial ox no		final ox no	e-	oxidized or reduced	Agent
Zn	0	→	+2	lost 2e-	oxidized	reducing agent: Zn
N	+5	→	+4	gain 1e-	reduced	oxidizing agent: HNO3

We see that HNO₃ is referred to as the oxidizing agent, not just N.

One final thing to note in this example - notice that the nitrogen in Zn(NO₃)₂ did not undergo a change in oxidation number.

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Here is a final example.

Consider the reaction

N2

+

2H2

→

2 NH3

Determine oxidation numbers for all elements in every compound:

0		0		-3   +1
N2	+	2H2	→	2 NH3

Summarize the changes, determine the number of electrons transferred per atom, and identify what is oxidized and what is reduced, and identify the oxidizing agent and reducing agent:

element	initial ox no		final ox no	e-	oxidized or reduced	Agent
N	0	→	-3	gain 3e-	reduced	oxidizing agent: N2
H	0	→	+1	lose 1e-	oxidized	reducing agent: H2

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Check your understanding with the practice questions, set 2, then complete Assignment 1.