AP Chemistry
Second & Third Laws of Thermodynamics
Balancing Redox Equations
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Why?
The three laws of thermodynamics describe restrictions on the behavior of virtually the
entire physical world we can experience. Everything that is possible or impossible in a
physical, chemical, or biological system is in some way related to these laws. We have
previously talked about the First Law of Thermodynamics, which is concerned with the
conservation of matter and energy. The Second and Third Laws are concerned with
disorder and its relationship to spontaneous and non-spontaneous changes.
Oxidation-reduction reactions, also called redox reactions, involve the transfer of
electrons from one species to another. These kinds of reactions are at the heart of energy
producing devices such as batteries and fuel cells. They are also involved in many
electrochemical processes by which we obtain useful materials.
Learning Objectives
 Understand the Second and Third Laws of Thermodynamics
 Understand the significance of the thermodynamic functions of entropy and Gibbs
free energy
 Understand the relationships between both entropy and Gibbs free energy and the
spontaneity of a physical or chemical process
 Understand the relationship between ΔG and K
 Know the definitions of oxidation, reduction, oxidizing agent, and reducing agent
 Know the systematic procedure for balancing redox reactions by the ion-electron
method
Success Criteria
 Be able to carry out calculations using ΔG = ΔH – TΔS and interpret the meaning
of the results
 Be able to calculate ΔG0 and ΔS0 values from tabulated standard free energy data
and absolute entropy data
 Be able to calculate approximate values of ΔG at non-standard temperatures
 Be able to use ΔG = ΔG0 + RT lnQ to calculate free energy under non-standard
conditions
 Be able to calculate K from ΔG
 Be able to separate a redox reaction into an oxidation and a reduction half reaction
 Be able to balance any skeletal redox reaction by the ion-electron method
Information (Second Law of Thermodynamics)
The Second Law of Thermodynamics is concerned with a thermodynamic function called
entropy, S. Entropy is a measure of disorder. For example, water vapor has a higher value
of S (188.83 J/mol-K) than liquid water (69.91 J/mol-K), because the molecules move
more freely in the vapor than in the liquid, making the gaseous state more disordered than
the liquid state.
When a system undergoes a chemical reaction or physical change, the overall entropy
changes. This overall entropy change, ΔStotal, is the sum of the change in entropy of the
AP Chemistry
Second & Third Laws of Thermodynamics
Balancing Redox Equations
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system and of its surroundings:
ΔStotal = ΔSsystem + ΔSsurroundings
Some changes occur spontaneously, and others must be forced. A change to a more
disordered state is a more probably event, so when a change occurs spontaneously there
is an increase in the total entropy. This is the essence of the Second Law of
Thermodynamics:
Every spontaneous change results in an increase in total entropy.
The great physical chemist Rudolph Clausius (1822-1888) famously summarized both the
First and Second Laws as follows:
The energy of the world is a constant; the entropy strives for a maximum.
Unlike enthalpy, H, which is governed by the First Law of Thermodynamics, a change in
the entropy of the system, which is governed by the Second Law, does not require an
equal and opposite change in the surroundings. Some changes only involve an entropy
change of the system, but more often we encounter processes that involve entropy
changes for both the system and its surroundings. The Second Law only requires that the
total entropy must increase for a spontaneous change. Therefore, in a spontaneous
process, it is possible for the system to decrease in entropy so long as the surroundings
undergo a greater increase in entropy, and vice versa. For example, water freezes
spontaneously below its freezing point temperature, even though ice is a more ordered
state (ΔSsystem < 0). As freezing occurs, the water liberates heat (equal to its heat of fusion)
to the surroundings, causing greater thermal motion and more disorder in the molecules
of the surroundings (ΔSsurroundings > 0). This represents a greater increase in entropy for the
surroundings than the decrease in entropy from the ordering of water molecules in the ice
(the system), so the total entropy change is positive (ΔStotal > 0).
Key Questions
1. Does the entropy of the system increase or decrease for the following changes?
Indicate whether ΔSsystem > 0 or ΔSsystem < 0.
a. Water is boiled
b. CaCO3(s)  CaO(s) + CO2(g)
c. N2(g) + 3 H2(g)  2 NH3(g)
Information (Gibbs Free Energy)
For a chemical reaction run at constant temperature and pressure, the reaction’s effect on
the entropy of the surroundings can be calculated by the equation
ΔSsurroundings = –ΔH/T
The ΔH is the enthalpy change of the system, which transfers heat to or from the
surroundings. The negative sign is inserted in the equation above to show the effect on
the surroundings. Note that if ΔH is negative, the entropy change for the surroundings
AP Chemistry
Second & Third Laws of Thermodynamics
Balancing Redox Equations
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will be positive, favoring a spontaneous process. This is consistent with the observation
that many exothermic reactions occur spontaneously. However, not all exothermic
reactions occur spontaneously, and not all endothermic reactions occur nonspontaneously. Recall that it is the total entropy change, ΔStotal, that determines whether
or not a reaction will be spontaneous. As we have seen, ΔStotal = ΔSsystem + ΔSsurroundings.
Substituting the expression –ΔH/T for ΔSsurroundings, we may write
ΔStotal = ΔSsystem – ΔH/T
Multiplying through by –T and defining ΔSsystem = ΔS, this may be rewritten
–TΔStotal = –TΔSsystem + ΔH = ΔH – TΔS
We can replace –TΔStotal with a function first proposed by the American mathematician J.
Willard Gibbs (1839-1903), called the Gibbs free energy, G:
G = H – TS
For a chemical reaction at constant pressure and temperature, we define the change in
Gibbs free energy as ΔG = –TΔStotal. Thus, our previous equation, –TΔStotal = ΔH – TΔS,
becomes
ΔG = ΔH – TΔS
where all terms refer to the system.
The sign convention for ΔG is consistent with what we have seen for other energy terms
(ΔE, ΔH), by which heat is liberated when the sign is negative. Because ΔG is defined on
the basis of –TΔStotal,, and because ΔStotal indicates whether or not a reaction is
spontaneous, we can make the following generalizations regarding the sign of ΔG:



If ΔG < 0, the reaction is spontaneous as written.
If ΔG > 0, the reaction is non-spontaneous as written, but is spontaneous in the
reverse direction.
If ΔG = 0, the reaction is at equilibrium.
From ΔG = ΔH – TΔS, we can see the factors that favor a spontaneous reaction:


Reactions with ΔH < 0 (exothermic) favor spontaneity.
Reactions that increase randomness (ΔS > 0) favor spontaneity.
These two factors may work in opposition in certain cases, and the spontaneity
determined from calculating ΔG = ΔH – TΔS depends upon the relative magnitudes of
ΔH and TΔS.
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Second & Third Laws of Thermodynamics
Balancing Redox Equations
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Key Questions
2. Fill in the missing values for the following reactions occurring at 25 oC, and
determine if the reaction is spontaneous or non-spontaneous.
ΔH (kJ/mol)
Reaction
ΔS (J/mol-K)
H2(g) + Br2(g)  2 HBr(g)
–72.46
+114.09
2 H2(g) + O2(g)  2 H2O(l)
–571.66
–326.34
2 N2(g) + O2(g)  2 N2O(g)
+163.2
–147.99
N2O4(g)  2 NO2(g)
+58.02
+176.61
ΔG (kJ/mol)
Spontaneous?
Information (Standard Gibbs Free Energy of Formation)
Like enthalpy, we can define a standard Gibbs Free Energy of formation, ΔG0f, which is
the value of ΔG0 when one mole of the substance in its standard state is formed from the
stoichiometric amounts of its component elements, each in their standard states. As with
ΔH0f , the standard state is 25C and 1 atm. These values of ΔG0f can be used in the same
way we used ΔH0f values to calculate the change for an overall reaction; i.e.,
ΔG0 = nΔG0f(products) – mΔG0f(reactants)
where n and m are the stoichiometric coefficients for each product and reactant,
respectively. As with ΔH0f , ΔG0f = 0 for all elements in their standard states.
Key Question
3. Given the following ΔG0f values, calculate the standard free energy for the
combustion of one mole of C2H6(g), and determine if the reaction is spontaneous
or non-spontaneous:
C2H6(g)
H2O(l)
CO2(g)
–32.9 kJ
–237.1 kJ
–394.4 kJ
Information (Absolute Entropies and the Third Law)
If we lower the temperature of a substance, molecular motion will be diminished and
greater ordering will occur. At a temperature of absolute zero we might suppose that a
perfect crystal, representing the ultimate order, would have an absolute entropy of zero (S
= 0). This reasoning lead Walther Nernst in 1906 to formulate what is known as the
Third Law of Thermodynamics:
At the absolute zero of temperature, a perfect crystalline substance would have an
absolute entropy of zero.
But absolute zero is an unattainable temperature and no substance forms a perfect crystal,
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Second & Third Laws of Thermodynamics
Balancing Redox Equations
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so all substances have non-zero absolute entropies at all real temperatures.
Unlike enthalpy and free energy, absolute entropies, S, can be defined and calculated.
Values are obtained from the temperature variation of heat capacities. However, when
using these, it is important to realize that absolute entropies are not changes in entropy,
ΔS.
The standard absolute entropy of a substance, S0, is the entropy of the substance in its
standard state at 25C and 1 atm. By the Third Law of Thermodynamics, these values are
always positive numbers; i.e., S0 > 0. The change in entropy under standard condition for
a reaction , ΔS0, can be calculated from absolute standard entropy data as
ΔS0 = nS0(products) – mS0(reactants)
where n and m are the stoichiometric coefficients for each product and reactant,
respectively. Note, that the absolute entropy of an element is not zero, and the absolute
entropy of a compound cannot be calculated from the absolute entropies of its elements.
Standard absolute entropies of substances are routinely tabulated along with ΔH0f and
ΔG0f data.
Key Questions
4. Calculate ΔH0, ΔS0, and ΔG0 for the following reaction at 25C.
2 H2O2(l)  2 H2O(l) + O2(g)
Given the following data:
Substance
H2O(l)
O2(g)
H2O2(l)
H0f (kJ/mol) S0f (J/mol-K)
–285.83
+69.91
0
+205.0
–187.8
+109.6
Information (Gibbs Free Energy and Temperature)
From the relationship ΔG = ΔH – TΔS, we can see that the value of the Gibbs free energy
of a reaction depends upon the absolute temperature, T. But the values for ΔH and S
generally show only small changes with temperature. This allows us to use data for ΔH0
and S0 to estimate ΔG values at non-standard temperatures (i.e., TH 298 K). When we use
ΔH0 and S0 values at nonstandard temperatures, however, we should realize that the ΔG
values obtained are only estimates. Nonetheless, these values usually lead to correct
deductions about a reaction’s spontaneity at or near the chosen temperature.
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Second & Third Laws of Thermodynamics
Balancing Redox Equations
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Key Questions
5. Consider the reaction
CaCO3(s)  CaO(s) + CO2(g)
for which ΔH0 = +178.1 kJ/mol, ΔG0 = +130.2 kJ/mol, and ΔS0 = +160.5 J/K-mol.
a. Is this reaction spontaneous at 25C?
b. Assuming that ΔH and ΔS do not change significantly with changing
temperature, is this reaction spontaneous at 1200 K?
6. For the vaporization of cyclohexane,
C6H12(l) = C6H12(g)
ΔH0 = +33.1 kJ/mol and ΔS0 = 93.84 J/K-mol. Assuming that these values do not
change significantly with increasing temperature, estimate the boiling point
temperature of cyclohexane. [Hint: Recall that boiling means that the liquid and
vapor are in equilibrium, and therefore ΔG = 0.]
Information (Calculating ΔG Under Non-Standard Conditions)
Tabulated data for Gibbs free energies are values under standard conditions. Very often
we are interested in a chemical system under non-standard conditions. The value of ΔG
under nonstandard conditions can be calculated from ΔGo by the equation
ΔG = ΔG0 + RT lnQ
in which R = 8.314 J/K-mol, T is in units of kelvin, and Q is the reaction quotient, which
we defined in our discussions of the equilibrium constant. Recall that Q has the same
form as K, except the concentrations or pressures are not assumed to be equilibrium
values.
Under standard conditions, all substances have unit activities; i.e., their effective
concentrations are 1. This means that under standard conditions, all concentrations and
pressures in the Q expression are 1, so ln Q = ln (1) = 0. Thus, under standard conditions
the equation ΔG = ΔG0 + RT lnQ reduces to ΔG = ΔG0, as it should. But most often we
deal with chemical systems in which the concentrations and pressures of reactant and
product species are not 1, even at 25C.To calculate ΔG under such non-standard
conditions of concentration and pressure, we need to evaluate Q and use ΔG = ΔG0 + RT
lnQ.
Key Questions
7. Under standard conditions, ΔG0 = –32.74 kJ/mol for the reaction
N2(g) + 3 H2(g) = 2 NH3(g)
What is the value of ΔG for the reaction at 298 K when the partial pressures of a
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Second & Third Laws of Thermodynamics
Balancing Redox Equations
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mixture are 0.0100 atm for N2(g), 0.0100 atm for H2(g), and 5.00 atm for NH3(g)?
Is the reaction spontaneous or non-spontaneous under these conditions?
Information (Free Energy and the Equilibrium Constant)
At equilibrium, ΔG = 0 and Q = K. We can use these restrictions to derive the
relationship between ΔG and K from
ΔG = ΔG0 + RT lnK = 0.
Rearranging, we have
ΔG0 = –RT lnK
from which we obtain
𝑲 = 𝒆−∆𝑮
𝟎 /𝑹𝑻
We see the following generalizations from these equations:
ΔG0 > 0 K < 1
ΔG0 = 0 K = 1
ΔG0 < 0 K > 1
K in this equation is the thermodynamic equilibrium constant, defined in terms of the
activities of participants in their standard states. This rigorous form of K inherently has
no units. For gas phase reactions, K is approximately Kp, because the standard state of
gas species at 25C is defined in terms of one atmosphere. Otherwise, the type of K
calculated from ΔG0 depends upon the definition of standard states used in the
determination of the Gibbs free energy. If all reactants and products are ions in solution,
K approximately corresponds to Kc, because the standard state of ions in solution is
defined at 25C in terms of mol/L.
Key Question
8. Under standard conditions, ΔG0 = -32.74 kJ for the reaction
N2(g) + 3 H2(g) = 2 NH3(g)
What is the value of K, the thermodynamic equilibrium constant, at 25C?
Information (Electron Transfer Reactions)
A reaction in which one species transfers electrons to another is called an oxidationreduction reaction, also called a redox reaction. For example, we can think of the reaction
of metallic iron with chlorine gas to form ionic iron(III) chloride as the net transfer of six
electrons from two iron atoms to three chlorine molecules:
AP Chemistry
Second & Third Laws of Thermodynamics
Balancing Redox Equations
2 (Fe0  Fe3+ + 3e-)
3 (Cl20 + 2e-  2Cl-)
2Fe(s) + 3Cl2(g)  2FeCl3(s)
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electrons "pushed"  oxidation
electrons "pulled"  reduction
redox
In essence, the Fe "pushes" electrons and the Cl2 "pulls" electrons, thereby effecting
electron transfer. On this basis, we have the following definitions:
Oxidation - loss of electrons by a substance
Reduction - gain of electrons by a substance
As this example shows, we can separate the overall redox reaction into two half reactions,
one for the oxidation and one for the reduction. Notice that in the oxidation half reaction,
the electrons appear on the right, and in the reduction half reaction they appear on the
left. Each half reaction is multiplied by a factor so that the number of electrons produced
by the oxidation is equal to the number consumed by the reduction.
Oxidation and reduction always involve transfer of electrons. Therefore, there is never
oxidation without reduction and vice versa in a redox reaction. Oxidizing something must
cause something else to be reduced and vice versa. Therefore, the substance oxidized is
seen to be the agent of the other substance's reduction, and the substance reduced is seen
to be the agent of the other substance's oxidation. This leads to the following definitions:
Oxidizing agent (oxidant):
Reducing agent (reductant):
a substance that causes another
substance to be oxidized and is itself
reduced.
a substance that causes another
substance to be reduced and is itself
oxidized.
In these terms, all redox reactions take on the general form
Ox1 + Red2 = Red1 + Ox2
Redox also causes a change in the oxidation numbers of the reductant and oxidant. In a
reduction, one element in a species experiences a lowering of its oxidation number, while
in an oxidation the opposite occurs.
2 (Fe0  Fe3+ + 3e-)
Fe oxidation number increases, 0  +3  oxidation
0
3 (Cl2 + 2e  2Cl )
Cl oxidation number decreases, 0  –1  reduction
2Fe(s) + 3Cl2(g)  2FeCl3(s)
redox
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Second & Third Laws of Thermodynamics
Balancing Redox Equations
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Key Questions
9. For each of the following, separate the skeletal (unbalanced) equation into two
half reactions. For each half reaction, balance the elements, and then add electrons
to the right or left side to make a net charge balance. Identify which half reaction
is the oxidation and which is the reduction. Then, multiply each half reaction by
an appropriate factor so that the two multiplied half reactions add together to
make a balanced redox equation.
a. Hg22+ + S2O32-  Hg + S4O62b. Al + Cr3+  Al3+ + Cr2+
c. Au3+ + I-  Au + I2
Information (Balancing Redox Equations by the Ion-Electron Method)
There are two principal methods for balancing redox equations: (1) oxidation state
method, and (2) ion-electron method. The latter is easier to use with redox reactions in
aqueous solution and if necessary can be adapted to many situations that are not in
aqueous solution. Our primary interest will be in aqueous-solution redox; therefore, we
will use the ion-electron method. One of the major advantages of this method is that it
makes it completely unnecessary to assign individual oxidation numbers.
To balance a redox equation by the ion-electron method, carry out the following steps in
sequence:
1. Separate the skeletal equation into two half reactions. One half reaction will
be a reduction and the other will be an oxidation. It is not necessary at this stage
to identify which is which.
2. Balance each half reaction separately. Balance atoms on each side of a half
reaction by inspection. If the reaction occurs in acidic medium, you may add H2O
and/or H+ to balance oxygen and/or hydrogen. If the reaction occurs in basic
medium, you may add H2O and/or OH- to balance oxygen and/or hydrogen. Do
not add any other new species (e.g., O2, H2) unless already a part of the skeletal
half reaction.
3. Balance the net charge across each half reaction by adding electrons to the
side with the more positive net ionic charge. If by this process electrons are
added on the left side of a half reaction, the half reaction is a reduction. If
electrons are added to the right side, the half reaction is an oxidation. (If you add
electrons to the same side in both half reactions, something is wrong!)
4. Multiply both half-reactions by appropriate whole number factors, so that
the number of electrons is the same in both half reactions and will cancel
when the two are added together.
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Second & Third Laws of Thermodynamics
Balancing Redox Equations
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5. Add the two multiplied half reactions together to obtain the overall redox
equation.
6. Check the balance. No electrons should appear in the overall redox equation. Not
only should there be an element-by-element balance across the equation, but also
the net charge (the sum of both ionic charges and electron charges) on both sides
of the equation should be equal.
Note that this procedure does not involve assigning oxidation numbers. Nonetheless, if
oxidation numbers are assigned to the balanced equation, it will always occur that the
reduction involves lowering an oxidation state of some element, and the oxidation
involves raising an oxidation state of some element. The following examples illustrate the
ion-electron procedure, starting from the skeletal equation in either acidic or basic
solution.
Example: In acid, NO3- + Fe2+  HNO2 + Fe3+
2(Fe2+  Fe3+ + e-)
2e- + 3 H+ + NO3-  HNO2 + H2O
2 Fe2+ + 3 H+ + NO3-  2 Fe3+ + HNO2 + H2O
oxidation
reduction
redox
To make the oxygen balance in the NO3-/HNO2 half reaction, we added H2O to the right
and then added 3 H+ to make the hydrogen balance. These are the only allowable species
to use in acid medium.
Example: In base, NiO2 + Cd  Ni(OH)2 + Cd(OH)2 (NiCad battery reaction)
2 OH- + Cd  Cd(OH)2 + 2e2e + 2 H2O + NiO2  Ni(OH)2 + 2 OHNiO2 + Cd + H2O  Ni(OH)2 + Cd(OH)2
-
oxidation
reduction
redox
Because this is in base, we can only add H2O and/or OH– to make the oxygen and
hydrogen balances. The need to add OH- in the Cd/Cd(OH)2 half reaction is
straightforward. In the NiO2/Ni half reaction, think of H2O as an acid neutralizing basic
NiO2. Thus, we add two H2O to the left to neutralize the two O2- ions of NiO2, and then
we add two OH– to the right side to complete the balance.
Balancing oxygen and hydrogen in basic redox reactions sometimes can be difficult,
because both OH– and H2O contain both elements. A trick to get around this is to balance
any troublesome half-reaction or the entire redox reaction first as if it were in acid, using
H+ and H2O. Then, the acid-balanced equation is converted to its form in basic medium
by adding the same number of OH– to both sides of the equation that would be needed to
"neutralize" any H+ in the acid-balanced equation. Combine H+ and OH– pairs to become
H2O; i.e., H+ + OH– = H2O. The following example shows this technique for a redox
reaction to be balanced in base.
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Second & Third Laws of Thermodynamics
Balancing Redox Equations
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Example: In base, I- + MnO4- → IO3- + MnO2
We will balance this in acid first, then "neutralize" any H+ to convert the redox reaction to
basic conditions.
3 H2O + I- → IO3- + 6 H+ + 6e–
3(2e– + 4 H+ + MnO4- → MnO2 + 2 H2O)
6 H+ + I- + 3 MnO4- → IO3- + 3 MnO2 + 3 H2O
6 OH+ 6 OH3 H2O + I- + 3 MnO4- → IO3- + 3MnO2 + 6 OH-
in acid
“neutralizing”
in base
On the left, the six added OH- ions are combined with the six H+ ions of the acid-balanced
equation to make 6 H2O. Three of these cancel with the 3 H2O on the right in the acidbalanced equation. Thus, we have a net of 3 H2O on the left in the base-balanced
equation. All six OH- ions added on the right appear in the net redox reaction in base.
Key Questions
10. Use the ion-electron method to complete and balance the following skeletal redox
equations, occurring in either acidic or basic aqueous solution, as indicated.
Identify the oxidation and reduction half reactions in each case.
a. In acid, Cu + NO3- → Cu2+ + N2O4
b. In acid, XeO3 + BrO3- → Xe + BrO4-
c. In acid, MnO4- + CH3OH → Mn2+ + HCO2H
d. In acid, Cr2O72- + I2 → Cr3+ + IO3-
e. In base, Pb(OH)42- + ClO- → PbO2 + Cl-
f. In base, SO2 + MnO4- → SO42- + MnO2