Equilibrium electrochemistry
The principles of thermodynamics can be applied to solutions of electrolytes. For that we
need to take into account activity coefficients: they differ significantly from 1 on account of the
strong ionic interactions in electrolyte solutions. These coefficients are best treated as empirical
quantities, but it is possible to estimate them in very dilute solutions. This chapter describes
thermodynamic properties of reactions in electrochemical cells, in which, as the reaction
proceeds, it drives electrons through an external circuit. Thermodynamic arguments can be used
to derive an expression for the electric potential of such cells and the potential can be related to
their composition. Two major topics: (1) the definition and tabulation of standard potentials; (2)
the use of these standard potentials to predict the equilibrium constants and other thermodynamic
properties of chemical reactions.
Electrochemical cells
An electrochemical cell consists of two electronic conductors (metal or graphite) dipping into
an electrolyte (an ionic conductor), which may be a solution, a liquid, or a solid. The electronic
conductor and its surrounding electrolyte – electrode. The physical structure containing them –
an electrode compartment. The two electrodes may share the same compartment (left). If the
electrolytes are different, then the two compartments may be joined by a salt bridge – and
electrolyte solution that completes the electrical circuit by permitting ions to move between the
compartments (right). Alternatively, the two solutions may be in direct physical contact (through
a porous membrane) and form a liquid junction.
A galvanic cell – an electrochemical cell that produces electricity as a result of the
spontaneous reaction occurring inside it. An electrolytic cell – and electrochemical cell in
which a non-spontaneous reaction is driven by an external source of direct current. The
commercially available dry cells, nickel-cadmium cells, mercury cells, and lithium ion cells
used to power electrical equipment are all galvanic cells and produce electricity as a results of
the spontaneous chemical reaction between the substances built into them at manifacture. A
fuel cell – a galvanic cell in which the reagents, such as hydrogen and oxygen, are supplied
from outside. They are used in manned spacecraft. Electric eels and electric catfish –
biological versions of fuel cells. Electrolytic cells include the arrangement used to electrolyze
water into hydrogen and oxygen and to obtain aluminum from its oxide in the Hall process.
Electrolysis is the only commercially available means for the production of fluorine.
Half-reactions and electrodes
A redox reaction – the outcome of the loss of electrons (and perhaps atoms) from one
species and their gain by another species. Loss of electrons – oxidation – an element has
undergone an increase in oxidation number. Gain of electrons – reduction – an element has
undergone a decrease in oxidation number. The requirement to break and form covalent bonds
in some redox reactions is one of the reasons why they often achieve equilibrium quite slowly,
often much more slowly than acid-base proton transfer reactions. The reducing agent
(‘reductant’) – the electron donor; the oxidizing agent (‘oxidant’) – the electron acceptor.
Any redox reaction may be expressed as the difference of two reduction half-reactions:
Reduction of Cu2+:
Cu2+(aq) + 2 e- → Cu(s)
Reduction of Zn2+:
Zn2+(aq) + 2 e- → Zn(s)
Difference:
Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s)
Example 1. Expressing a reaction in terms of half-reactions
Express the dissolution of silver chloride as the difference of two reduction half-reactions:
Write the overall chemical equation. Then select one of the reactants and write a half-reaction
in which the reactant is reduced to one of the products. Next, subtract that half-reaction from
the overall reaction to identify the second half-reaction. In practice, it is easier to reverse the
half-reaction and add it to the overall reaction. Write the resulting half-reaction as a reduction
by reversing it.
AgCl(s) → Ag+(aq) + Cl-(aq)
overall
The reduction of the Ag(I) in AgCl to Ag(0):
AgCl(s) + e- → Ag(s) + Cl-(aq)
Reversing this half-reaction gives: Ag(s) + Cl-(aq) → AgCl(s) + eAdding to the overall reaction gives:
AgCl(s) + Ag(s) + Cl-(aq) → Ag+(aq) + Cl-(aq) + AgCl(s) + eAg(s) → Ag+(aq) + eReverse this half-reaction to obtain the second reduction half-reaction:
Ag+(aq) + e- → Ag(s)
The reduced and oxidized species in a half-reaction – redox couple: Cu2+/Cu, Zn2+/Zn. In
general, we write a couple as Ox/Red:
Ox + ν e- → Red
Reaction quotient Q for a half-reaction (electrons are ignored):
Cu2+(aq) + 2 e- → Cu(s)
Q = 1/a(Cu2+)
O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l)
aH2 2O
pΘ
Q= 4
= 4
The reaction quotient for the reduction of O2 to H2O in acid:
Θ
aH + pO2
aH + fO2 p
(
)
Reactions at electrodes
The reduction and oxidation processes responsible for the
overall reaction in a cell are separated in space: oxidation
takes place in one electrode compartment and reduction
takes place in the other compartment. As the reaction
proceeds, the electrons released in the oxidation
Red1 → Ox1 + ν eat one electrode travel through the external circuit and reenter the cell through the other electrode
Ox2 + ν e- → Red2
The electrode at which oxidation occurs – anode; the electrode at which reduction occurs – the
cathode. In a galvanic cell – the cathode has a higher potential than the anode: the species
undergoing reduction, Ox2, withdraws electrons from its electrode (the cathode), so leaving a
relative positive charge on it (a high potential). At the anode, oxidation results in the transfer of
electrons to the electrode, so giving it a relative negative charge (a low potential). In an
electrolytic cell, the anode is still the location of oxidation (by definition), but now the electrons
must be withdrawn from the species in that compartment because the oxidation does not occur
spontaneously, and at the cathode there must be a supply of electrons to drive the reduction.
Therefore, in an electrolytic cell the anode must be made relatively positive to the cathode.
In a gas electrode, a gas is in equilibrium with a solution of its
ions in the presence of an inert metal. The inert metal (Pt) acts as a
source or sink of electrons but takes no other part in the reaction
except acting as a catalyst.
One important example – the hydrogen electrode, in which
hydrogen is bubbled through an aqueous solution of hydrogen ions
and the redox couple is H+/H2. This electrode is denoted as
Pt(s)|H2(g)|H+(aq). In this electrode, the junctions are between the
platinum and the gas and between the gas and the
liquid containing its ions.
Pt(s)|H2(g)|HCl(aq)|AgCl(s)|Ag(s)
The cell reaction
The current produced by a galvanic cell arises from the spontaneous reaction taking place
inside it. The cell reaction – the reaction in the cell written on the assumption that the right-hand
electrode is the cathode and the reduction is taking place in the right-hand compartment. We’ll see
later how to predict if the right-hand electrode is in fact the cathode; if it is, then the cell reaction is
spontaneous as written. If the left-hand electrode turns out to be the cathode, then the reverse of the
cell reaction is spontaneous.
To write the cell reaction corresponding to the cell diagram, we first write the half-reactions
at both electrodes as reductions, and then subtract the left-hand equation from the right-hand
equation. The cell:
Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s)
Right-hand electrode:
Cu2+(aq) + 2 e- → Cu(s)
Left-hand electrode:
Zn2+(aq) + 2 e- → Zn(s)
The overall cell reaction:
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
The cell potential
A galvanic cell does electrical work as the reaction drives electrons through an external
circuit. The work done by a given transfer of electrons depends on the cell potential – the
potential difference between the two electrodes (measured in volts, 1 V = 1 J C-1). Large cell
potential – large amount of electrical work can be done by given number of electrons traveling
between the electrodes.
The maximum electrical work that a system (the cell) can do is given by ΔG. For a
spontaneous process at constant temperature and pressure
we,max = ΔG
We must ensure that the cell is operating reversibly – only then we can use this equation.
Moreover, the reaction Gibbs energy is a property relating to a specified composition of the
reaction mixture. Therefore, to measure ΔrG we must ensure that the cell is operating reversibly
at a specific constant composition – measuring the cell potential when it is balanced by an
exactly opposing sourse of potential so that the cell reaction occurs reversibly, the composition is
constant, and no current flows. In effect, the cell reaction is poised for change, but not actually
changing. The resulting potential difference – the electromotive force (emf), E, of the cell.
The relation between E and ΔrG
The relation between the reaction Gibbs energy and the emf:
-νFE = ΔrG
F – the Faraday constant, the magnitude of electric charge per mole of electrons:
F = eNA = 96.485 kC mol-1.
This equation – the key connection between electrical measurements and
thermodynamic properties.
Justification. We consider the change in G when the cell reaction advances by an infinitesimal
amount dξ at some composition:
dG = ∑ µ J dnJ = ∑ ν J µ J dξ
J
J
 ∂G 
dG = Δ r Gdξ
Δ rG =   = ∑ν J µ J
∂ξ
  p,T J
The maximum
non-expansion (electrical) work that the reaction can do as it advances by dξ at
€
dwe = Δ r Gdξ
constant T and p:
€
This work is infinitesimal and the composition
of the system is virtually constant when it
€
occurs. When the reaction advances by dξ, then νdξ electrons must travel from the anode to
the cathode. The total charge transferred is -νeNAdξ (because νdξ is the amount of the
€
electrons and the charge per mole of electrons is -eNA). Hence, the total charge transported is νFdξ because eNA = F.
The work done when an infinitesimal charge -νFdξ travels from the
anode to the cathode – the product of the charge and the potential
dwe = −νFEdξ = Δ r Gdξ
difference E:
-νFE = ΔrG
By knowing the reaction Gibbs energy at a specified composition, we can
state the cell emf at that composition. A negative Gibbs energy corresponding to a spontaneous cell reaction, corresponds to a positive cell
emf. The€driving power of a cell (emf) is proportional to the slope of Gibbs
energy with respect to the extent of reaction. A reaction far from equilibrium (a steep slope) has a large emf. Close to equilibrium – small emf.
The Nernst equation
ΔrG = ΔrG∅ + RT ln Q
E = -ΔrG/νF
E = -ΔrG∅/νF – (RT/νF) ln Q
E = E∅– (RT/νF) ln Q
E∅ = -ΔrG∅/νF
Nernst equation
Standard emf of the cell
According to the Nernst equation, there is a linear dependence of cell potential on
log Q. The standard emf can be interpreted as the emf when all the reactants and
products are in their standard states:
Q=1
ln Q = 0
∅
∅
E is simply ΔrG in a disguised form.
Because at 25°C, RT/F = 25.7 mV, a practical form of the Nernst equation is
E = E∅– (25.7 mV/ν) ln Q
For a reaction with ν = 1, if Q is decreased by a factor of 10, the emf becomes more positive by
59.2 mV. The reaction has a greater tendency to form products. If Q increases by a factor of 10,
then the cell potential falls by 59.2 mV and the reaction has a lower tendency to form products.
Concentration cells
We can use the Nernst equation to derive an expression for the emf of an electrolyte
concentration cell: M|M+(aq, L)|M+(aq, R)|M
M+(aq, R) + e- → M
Right
The standard emf of the cell is zero.
+
M (aq, L) + e → M
Left
RT aL
RT bL
+
+
E
=
−
ln
≈
−
ln
M (aq, R) → M (aq, L)
Q = aL/aR, ν = 1
F aR
F bR
If R is the more concentrated solution, E > 0. The positive potential arises because positive ions tend
to be reduced, so withdrawing electrons from the electrode. This process is dominant on the right.
€
One important example of a concentration cell – the biological cell membrane more
permeable to K+ ions than either Na+ or Cl- ions. In the enzyme (Na+-K+)-ATP, the hydrolysis
of ATP drives the pumping of ions across the membrane. The concentration of K+ inside an
inactive nerve cell is about 20 times that on the outside, whereas the concentration of Na+
outside the cell is about 10 times that on the inside. The difference in concentrations of ions
results in a transmembrane potential difference of about –62 mV. This negative potential
difference – the resting potential of the cell membrane.
The transmembrane potential difference – important for the transmission of nerve
impulses. Upon receiving an impulse (called an action potential), a site in the nerve cell
membrane becomes transiently permeable to Na+ and the transmembrane potential difference
changes. To propagate along a nerve cell, the action potential must change the transmembrane
potential by at least 20 mV, to values less negative than –40 mV. Propagation occurs when an
action potential in one site of the membrane triggers an action potential in an adjacent site,
while sites behind the moving action potential return to the resting potential.
Cells at equilibrium
A special case of the Nernst equation – reaction at equilibrium. Q = K, K – the equilibrium
constant of the cell reaction. Because chemical reaction at equilibrium cannot do work, it generates
zero potential difference between the electrodes. Q = K and E = 0 gives
ln K = νFE∅/RT
Using this equation we can predict equilibrium constants from standard cell potentials.
The standard potential of the Daniell cell is +1.10 V. Then we can calculate the
equilibrium constant for the cell reaction as
Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s)
ln K = 2×(9.6485×104 C mol-1)×(1.10 V)/{(8.3145 J K-1 mol-1) ×(298.15 K)} = 42.81
K = 1.5×1037
The displacement of copper by zinc virtually goes to completion.
If E∅ > 0, K > 1 – at equilibrium the cell reaction lies in favor of products. If E∅ < 0,
K < 1 – the reactants are favored at equilibrium.
Standard potentials
Each electrode in a galvanic cell makes a characteristic contribution to the overall cell
potential. It is not possible to measure the contribution of a single electrode – one electrode
can be assigned a value zero and the others assigned relative values on that basis. The
specially selected electrode – the standard hydrogen electrode (SHE):
Pt(s)|H2(g)|H+(aq)
E∅ = 0 at all T.
The standard potential, E∅(Ox/Red), is then measured by constructing a cell in which
the couple of interest form the right-hand electrode and the standard hydrogen electrode is on
the left. For example, the standard potential of the Ag+/Ag couple is the standard potential of
the cell
Pt(s)|H2(g)|H+(aq)||Ag+(aq)|Ag(s) – +0.80 V
AgCl/Ag,Cl- couple: the standard emf of the following cell:
Pt(s)|H2(g)|H+(aq)||Cl-(aq)|AgCl(s)|Ag(s)
E∅( AgCl/Ag,Cl-) = E∅ = +0.22 V
Although a standard potential is written like it refers to a half reaction
AgCl(s) + e- → Ag(s) + Cl-(aq) E∅(AgCl/Ag,Cl-) = +0.22 V
it should be understood as the potential for the overall reaction:
AgCl(s) + 1/2 H2(g) → Ag(s) + H+(aq) + Cl-(aq)
E∅ = +0.22 V
The standard potential is thus determined by properties of both the hydrogen electrode and the
couple to which the potential refers.
An important feature of standard emf of cells and standard potentials of electrodes – they
are unchanged if the chemical equation for the cell reaction or a half-reaction is multiplied by
a numerical factor. A numerical factor increases ΔrG∅ but it also increases the number of
electrons transferred by the same factor, so the value of E∅ remains unchanged.
The cell emf in terms of individual standard potentials
To calculate the standard emf of a cell formed from any pair of electrodes:
E∅ = ER∅ – EL∅
ER∅ – the standard potential of the right-hand electrode.
EL∅ – the standard potential of the left-hand electrode.
Ag(s)|Ag+(aq)||Cl-(aq)|AgCl(s)|Ag(s)
E∅ = E∅(AgCl/Ag+,Cl-) – E∅(Ag+/Ag) = +0.22 – 0.80 = -0.58 V
Because ΔrG∅ = -νFE∅, if E∅ > 0, then the corresponding cell reaction is spontaneous in the
direction written (K > 1). Once we have the value of E∅, we can use the Nernst equation for
equilibrium to calculate the equilibrium constant of the cell reaction.
To calculate the equilibrium constant for the disproportionation:
2 Cu+(aq) → Cu(s) + Cu2+(aq) at 298 K
Right-hand electrode: Cu(s)|Cu+(aq)
Cu+(aq) + e- → Cu(s)
E∅ = +0.52 V
Left-hand electrode: Pt(s)|Cu2+(aq),Cu+(aq) Cu2+(aq) + e- → Cu+(aq)
The standard emf of the cell: E∅ = +0.52 V – 0.16 V = +0.36 V
ln K = 0.36 V / 0.025693 V
K = 1.2×106
E∅ = +0.16 V
Example 2. Calculating an equilibrium constant
Evaluate the solubility constant of silver chloride (the equilibrium constant for the
dissolution of AgCl(s)) and its solubility from cell potential data at 298.15 K.
We need to find an electrode that reproduces the solubility equilibrium. Then we identify
the solubility constant with the equilibrium constant of the cell reaction. We calculate the
standard emf of the cell from the standard potentials and use it to obtain K. For the solubility,
S, which is the concentration of solute at equilibrium, we express the solubility K in terms of S
and solve the resulting equation for S.
The solubility equilibrium: AgCl(s) → Ag+(aq) + Cl-(aq)
Right
AgCl(s) + e- → Ag(s) + Cl-(aq) E∅ = +0.22 V
Left
Ag+(aq) + e- → Ag(s)
E∅ = +0.80 V
The standard emf is –0.58 V, ν = 1
ln K = νFE∅/RT = νE∅/(RT/F) = -0.58 V / 0.025693 V
K = 1.6×10-10
K = aAg+aCl- ≈ S2
S = 1.3×10-5 mol kg-1
The measurement of standard potentials
Consider a specific case – the silver chloride electrode. The measurement is made on the
Harned cell: Pt(s)|H2(g)|HCl(aq)|AgCl(s)|Ag(s)
1/2 H2(g) + AgCl(s) → HCl(aq) + Ag(s)
aH + aCl−
RT
Θ
−
E
=
E
AgCl
/
Ag,Cl
−
ln
The Nernst equation:
1/2
F
f pΘ
(
)
(
H2
)
For simplicity, we set f = p∅ and express the activities in terms of the molality b and the mean
RT
RT
2RT
2RT
activity coefficient γ+: E = E Θ −
ln b2 −
ln γ ±2
E+
ln b = E Θ −
ln γ ±
F
F
F
F
€
2RT
From the Debye-Hückel limiting law:
γ+ ∝ -b1/2
E+
ln b = E Θ + Cb1/2
F
1/2
1/2
The expression on
the
left
is
plotted
against
b
and
extrapolated
to
b
=
0.
The
intercept
at
b
€
€
= 0 is the value of E∅ for the silver/silver chloride electrode.
€
Example 3. Determining the standard emf of a cell
The emf of the cell Pt(s)|H2(g,p∅)|HCl(aq,b)|AgCl(s)|Ag(s) at 25°C is:
b/(10-3 b∅)
3.215
5.619
9.138
25.63
E/V
0.52053
0.49257
0.46860
0.41824
2RT
E+
ln b = E Θ + Cb1/2
F
-3 ∅ 1/2
(b/(10 b ))
1.793
2.370
3.023
5.063
E/V + (2RT/F)lnb
0.2256 0.2263
0.2273
0.2299
∅
The intercept of the plot gives E = 0.2232 V.
€
The measurement of activity coefficients
Once the standard potential of an electrode in a cell is known, we can use it to determine the
activities of the ions. For example, the mean activity coefficient of the ions in HCl(aq) of
EΘ − E
ln γ ± =
− ln b
molality b can be obtained from:
2RT F
Applications of standard potentials
The measurement of cell emfs – a convenient source of data on the Gibbs energies,
enthalpies, and entropies of reactions. In practice the standard values are normally determined.
€
The electrochemical series
For two redox couples Red1,Ox1||Red2,Ox2 E∅ = E2∅ – E1∅
the cell reaction
Red1 + Ox2 → Ox1 + Red2
The reaction is spontaneous as written if E∅ > 0 E2∅ > E1∅
Because in the cell reaction Red1 reduces Ox2, we conclude that
Red1 has a thermodynamic tendency to reduce Ox2 if E1∅ < E2∅
A species with a low standard potential has a thermodynamic tendency to reduce a species
with a high standard potential.
Low reduces high and high oxidizes low.
E∅(Zn2+,Zn) = -0.76 V < E∅(Cu2+,Cu) = +0.34 V
Zn(s) has a thermodynamic tendency to reduce Cu2+(aq) under standard conditions. Hence, the
reaction
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
can be expected to have K > 1 (in fact, K = 1.5×1037 at 298 K).
In the tabulated electrochemical series (see Table 7.3), the metallic elements (and
hydrogen) are arranged in the order of their reducing power as measured by their standard
potentials in aqueous solution. A metal low in the series (with a lower standard potentials) can
reduce the ions of metals with higher standard potentials.
The measurement of pH and pKa
The half-reaction at a hydrogen electrode:
Θ 1/2
f
/
p
H2
H+(aq) + e- → 1/2 H2(g)
ν=1
Q=
aH +
We let fH 2 = pΘ and E∅(H+/H2) = 0. Then the electrode potential is
RT
RT ln10
E H + ,H 2 =
ln aH + = −
pH
F€
F
At 25°C
E(H+/H2) = -59.16 mV × pH
€
Each unit increase in pH decreases the electrode potential by 59 mV.
(
(
€
)
)
The measurement of the pH of a solution is simple – it is based on the measurement of
the potential of a hydrogen electrode immersed in the solution. The left-hand electrode is
typically a saturated calomel (Hg2Cl2(s)) reference electrode with potential E(cal).
The right-hand electrode – the hydrogen electrode.
pH = {E – E(cal)} / (-59.16 mV)
The practical definition of the pH of a solution X is
pH = pH(S) – (FE/RTln10)
E – the emf of the cell
Pt(s)|H2(g)|S(aq)||3.5 M KCL(aq)||X(aq)|H2(g)|Pt(s)
S – a solution of standard pH. The currently recommended primary standards: a saturated aqueous
solution of potassium hydrogen tartrate (pH = 3.557 at 25°C) and 0.0100 mol kg-1 disodium
tetraborate (pH = 9.180).
In practice, indirect methods are much more convenient and the hydrogen electrode is
replaced by the glass electrode. This electrode is sensitive to the hydrogen ion
activity and has a potential proportional to pH. It is filled with a phosphate buffer
containing Cl- ions and conveniently has E = 0 when the external media is at pH = 7.
The glass electrode is much more convenient to handle that the gas electrode and it is
calibrated using solutions of known pH.
The electrochemical determination of pH opens up a route to the electrochemical
determination of pKa: the pKa of an acid is equal to the pH of a solution containing
equal amounts of the acid and its conjugate base.
Thermodynamic functions
The cell potential is related to the reaction Gibbs energy:
ΔrG∅ = -νFE∅. By measuring the standard potential of a cell driven by the reaction of interest
we can obtain the standard Gibbs energy.
The cell reaction taking place in
Pt(s)|H2(g)|H+(aq)||Ag+(aq)|Ag(s)
E∅ = +0.7996 V
Ag+(aq) + 1/2 H2(g) → H+(aq) + Ag(s) ΔrG∅ = ΔfG∅(Ag+,aq)
ΔfG∅(Ag+,aq) = -(-E∅F) + +77.15 kJ mol-1
The relation between the standard potential of a cell and the standard Gibbs energy – a
convenient route for the calculation of the standard potential of a couple from two other
couples. G is a state function – the Gibbs energy of an overall reaction is the sum of the Gibbs
energies of the reactions into which it can be divided. In general, we cannot combine E∅
values directly because they depend on the value of ν, which may differ for the two couples.
Example 4. Calculating a standard potential from two others
Given the standard potentials E∅(Cu2+,Cu) = 0.340 V and E∅(Cu+,Cu) = 0.522 V, calculate
E∅(Cu2+,Cu+).
We need to convert the E∅ values to ΔrG∅, add them appropriately and convert the overall
ΔrG∅ to the required E∅.
(a) Cu2+(aq) + 2 e- → Cu(s) E∅ = +0.340 V ΔrG∅(a) = -2F×(0.340 V) = (-0.680 V)×F
(b) Cu+(aq) + e- → Cu(s)
E∅ = +0.522 V ΔrG∅(b) = -F×(0.522 V) = (-0.522 V)×F
The required reaction is
(c) Cu2+(aq) + e- → Cu+(aq) ΔrG∅ = -FE∅
Because (c) = (a) – (b), ΔrG∅(c) = ΔrG∅(a) – ΔrG∅(b)
FE∅(c) = -(-0.680 V)F + (-0.522 V)F
E∅(c) = +0.158 V
In a general case:
νcE∅(c) = νaE∅(a) – νbE∅(b)
The temperature coefficient of the standard cell emf, dE∅/dT gives the standard entropy
of the cell reaction:
E∅ = -ΔrG∅/νF
(∂G/∂T)p = -S
dE∅/dT = ΔrS∅/νF
Hence we have an electrochemical technique for obtaining standard reaction entropies and
through them the entropies of ions in solution. Finally, we can combine the results to obtain
the standard reaction enthalpy:
ΔrH∅ = ΔrG∅ + TΔrS∅ = -νF{E∅ – T(dE∅/dT)}
This provides a noncalorimetric method for measuring ΔrH∅ and, through the convention
ΔfH∅(H+,aq) = 0, the standard enthalpies of formation of ions in solution.
Example 5. Using the temperature coefficient of the cell potential
The standard potential of the cell Pt(s)|H2(g)|HBr(aq)|AgBr(Aq)|Ag(s) was measured over a
range of temperatures and the data were fitted to the following polynomial:
E∅/V = 0.07131 – 4.99×10-4(T/K – 298) – 3.45×10-6(T/K – 298)2
Evaluate the standard reaction Gibbs energy, enthalpy, and entropy at 298 K.
At 298 K, E∅ = 0.07131 V
ΔrG∅ = -νFE∅ = -1(9.6485×104 C mol-1)(0. 0.07131 V) =
–6.880 kJ mol-1
The temperature coefficient of the cell potential:
dE∅/dT = –4.99×10-4 V K-1 – 2(3.45×10-6)(T/K – 298) V K-1 = –4.99 ×10-4 V K-1 at 298 K
ΔrS∅ = 1(9.6485×104 C mol-1)×(–4.99 ×10-4 V K-1) = -48.2 J K-1 mol-1
ΔrH∅ = ΔrG∅ + TΔrS∅ = (-6.880 kJ mol-1) + (298 K)×(-0.0482 kJ K-1 mol-1) = -21.2 kJ mol-1