Equilibrium electrochemistry

The principles of thermodynamics can be applied to solutions of electrolytes. For that weneed to take into account activity coefficients: they differ significantly from 1 on account of thestrong ionic interactions in electrolyte solutions. These coefficients are best treated as empiricalquantities, but it is possible to estimate them in very dilute solutions. This chapter describesthermodynamic properties of reactions in electrochemical cells, in which, as the reactionproceeds, it drives electrons through an external circuit. Thermodynamic arguments can be usedto derive an expression for the electric potential of such cells and the potential can be related totheir 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 thermodynamicproperties of chemical reactions.

Electrochemical cells An electrochemical cell consists of two electronic conductors (metal or graphite) dipping intoan electrolyte (an ionic conductor), which may be a solution, a liquid, or a solid. The electronicconductor and its surrounding electrolyte – electrode. The physical structure containing them –an electrode compartment. The two electrodes may share the same compartment (left). If theelectrolytes are different, then the two compartments may be joined by a salt bridge – andelectrolyte solution that completes the electrical circuit by permitting ions to move between thecompartments (right). Alternatively, the two solutions may be in direct physical contact (througha porous membrane) and form a liquid junction.

A galvanic cell – an electrochemical cell that produces electricity as a result of thespontaneous reaction occurring inside it. An electrolytic cell – and electrochemical cell inwhich a non-spontaneous reaction is driven by an external source of direct current. Thecommercially available dry cells, nickel-cadmium cells, mercury cells, and lithium ion cellsused to power electrical equipment are all galvanic cells and produce electricity as a results ofthe spontaneous chemical reaction between the substances built into them at manifacture. Afuel cell – a galvanic cell in which the reagents, such as hydrogen and oxygen, are suppliedfrom outside. They are used in manned spacecraft. Electric eels and electric catfish –biological versions of fuel cells. Electrolytic cells include the arrangement used to electrolyzewater 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 electrodesA 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 hasundergone an increase in oxidation number. Gain of electrons – reduction – an element hasundergone a decrease in oxidation number. The requirement to break and form covalent bondsin 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.

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-reactionin which the reactant is reduced to one of the products. Next, subtract that half-reaction fromthe overall reaction to identify the second half-reaction. In practice, it is easier to reverse thehalf-reaction and add it to the overall reaction. Write the resulting half-reaction as a reductionby reversing it.

AgCl(s) → Ag+(aq) + Cl-(aq) overallThe 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) + e-

Adding to the overall reaction gives:AgCl(s) + Ag(s) + Cl-(aq) → Ag+(aq) + Cl-(aq) + AgCl(s) + e-

Ag(s) → Ag+(aq) + e-

Reverse 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- → RedReaction quotient Q for a half-reaction (electrons are ignored):

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)

Cu2+(aq) + 2 e- → Cu(s) Q = 1/a(Cu2+) O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l)

The reaction quotient for the reduction of O2 to H2O in acid:

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Q =aH2O2

aH+4 fO2 pΘ( )

=pΘ

aH+4 pO2

Reactions at electrodesThe reduction and oxidation processes responsible for the

overall reaction in a cell are separated in space: oxidationtakes place in one electrode compartment and reductiontakes place in the other compartment. As the reactionproceeds, the electrons released in the oxidation

Red1 → Ox1 + ν e-

at one electrode travel through the external circuit and re-enter the cell through the other electrode

Ox2 + ν e- → Red2

The electrode at which oxidation occurs – anode; the electrode at which reduction occurs – thecathode. In a galvanic cell – the cathode has a higher potential than the anode: the speciesundergoing reduction, Ox2, withdraws electrons from its electrode (the cathode), so leaving arelative positive charge on it (a high potential). At the anode, oxidation results in the transfer ofelectrons to the electrode, so giving it a relative negative charge (a low potential). In anelectrolytic cell, the anode is still the location of oxidation (by definition), but now the electronsmust be withdrawn from the species in that compartment because the oxidation does not occurspontaneously, 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 itsions in the presence of an inert metal. The inert metal (Pt) acts as asource or sink of electrons but takes no other part in the reactionexcept acting as a catalyst.

One important example – the hydrogen electrode, in whichhydrogen is bubbled through an aqueous solution of hydrogen ionsand the redox couple is H+/H2. This electrode is denoted asPt(s)|H2(g)|H+(aq). In this electrode, the junctions are between theplatinum and the gas and between the gas and theliquid 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 placeinside it. The cell reaction – the reaction in the cell written on the assumption that the right-handelectrode is the cathode and the reduction is taking place in the right-hand compartment. We’ll seelater how to predict if the right-hand electrode is in fact the cathode; if it is, then the cell reaction isspontaneous as written. If the left-hand electrode turns out to be the cathode, then the reverse of thecell reaction is spontaneous.

To write the cell reaction corresponding to the cell diagram, we first write the half-reactionsat both electrodes as reductions, and then subtract the left-hand equation from the right-handequation. 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 potentialA 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 – thepotential difference between the two electrodes (measured in volts, 1 V = 1 J C-1). Large cellpotential – large amount of electrical work can be done by given number of electrons travelingbetween the electrodes.

The maximum electrical work that a system (the cell) can do is given by ΔG. For aspontaneous 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 thereaction mixture. Therefore, to measure ΔrG we must ensure that the cell is operating reversiblyat a specific constant composition – measuring the cell potential when it is balanced by anexactly opposing sourse of potential so that the cell reaction occurs reversibly, the composition isconstant, and no current flows. In effect, the cell reaction is poised for change, but not actuallychanging. The resulting potential difference – the electromotive force (emf), E, of the cell.

The relation between E and ΔrGThe 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 andthermodynamic properties.

Justification. We consider the change in G when the cell reaction advances by an infinitesimalamount dξ at some composition:

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dG = µJJ∑ dnJ = ν JµJ

J∑ dξ

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ΔrG =∂G∂ξ

p,T

= ν JµJJ∑

€

dG = ΔrGdξ

The maximum non-expansion (electrical) work that the reaction can do as it advances by dξ atconstant T and p:

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dwe = ΔrGdξThis work is infinitesimal and the composition of the system is virtually constant when itoccurs. When the reaction advances by dξ, then νdξ electrons must travel from the anode tothe cathode. The total charge transferred is -νeNAdξ (because νdξ is the amount of theelectrons 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 theanode to the cathode – the product of the charge and the potentialdifference E:

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dwe = −νFEdξ = ΔrGdξ -νFE = ΔrGBy knowing the reaction Gibbs energy at a specified composition, we canstate the cell emf at that composition. A negative Gibbs energy corres-ponding to a spontaneous cell reaction, corresponds to a positive cellemf. The driving power of a cell (emf) is proportional to the slope of Gibbsenergy with respect to the extent of reaction. A reaction far from equi-librium (a steep slope) has a large emf. Close to equilibrium – small emf.

The Nernst equationΔrG = ΔrG

∅ + RT ln Q E = -ΔrG/νFE = -ΔrG

∅/νF – (RT/νF) ln QE = E∅– (RT/νF) ln Q E∅ = -ΔrG

∅/νFNernst equation Standard emf of the cellAccording to the Nernst equation, there is a linear dependence of cell potential onlog Q. The standard emf can be interpreted as the emf when all the reactants andproducts are in their standard states: Q = 1 ln Q = 0E∅ 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 QFor a reaction with ν = 1, if Q is decreased by a factor of 10, the emf becomes more positive by59.2 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 electrolyteconcentration 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 LeftM+(aq, R) → M+(aq, L) Q = aL/aR, ν = 1

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E = −RTFln aLaR

≈ −RTFln bLbR

If R is the more concentrated solution, E > 0. The positive potential arises because positive ions tendto 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 morepermeable to K+ ions than either Na+ or Cl- ions. In the enzyme (Na+-K+)-ATP, the hydrolysisof ATP drives the pumping of ions across the membrane. The concentration of K+ inside aninactive 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 ionsresults in a transmembrane potential difference of about –62 mV. This negative potentialdifference – the resting potential of the cell membrane.

The transmembrane potential difference – important for the transmission of nerveimpulses. Upon receiving an impulse (called an action potential), a site in the nerve cellmembrane becomes transiently permeable to Na+ and the transmembrane potential differencechanges. To propagate along a nerve cell, the action potential must change the transmembranepotential by at least 20 mV, to values less negative than –40 mV. Propagation occurs when anaction 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 equilibriumconstant of the cell reaction. Because chemical reaction at equilibrium cannot do work, it generateszero potential difference between the electrodes. Q = K and E = 0 gives

ln K = νFE∅/RTUsing this equation we can predict equilibrium constants from standard cell potentials.

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)} = 42.81K = 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 potentialsEach 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 electrodecan be assigned a value zero and the others assigned relative values on that basis. Thespecially 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 onthe left. For example, the standard potential of the Ag+/Ag couple is the standard potential ofthe 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 reactionAgCl(s) + e- → Ag(s) + Cl-(aq) E∅(AgCl/Ag,Cl-) = +0.22 Vit 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 of the Daniell cell is +1.10 V. Then we can calculate theequilibrium constant for the cell reaction as

The standard potential is thus determined by properties of both the hydrogen electrode and thecouple to which the potential refers.

An important feature of standard emf of cells and standard potentials of electrodes – theyare unchanged if the chemical equation for the cell reaction or a half-reaction is multiplied bya numerical factor. A numerical factor increases ΔrG∅ but it also increases the number ofelectrons transferred by the same factor, so the value of E∅ remains unchanged.

The cell emf in terms of individual standard potentialsTo 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 VBecause ΔrG

∅ = -νFE∅, if E∅ > 0, then the corresponding cell reaction is spontaneous in thedirection written (K > 1). Once we have the value of E∅, we can use the Nernst equation forequilibrium 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) E∅ = +0.16 VThe 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

We need to find an electrode that reproduces the solubility equilibrium. Then we identifythe solubility constant with the equilibrium constant of the cell reaction. We calculate thestandard 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 Sand 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 VLeft Ag+(aq) + e- → Ag(s) E∅ = +0.80 VThe 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 potentialsConsider 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)

The Nernst equation:

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E = EΘ AgCl /Ag,Cl−( )− RTF lnaH+aCl−fH2 pΘ( )

1/2

For simplicity, we set f = p∅ and express the activities in terms of the molality b and the mean

activity coefficient γ+:

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E = EΘ −RTFlnb2 − RT

Flnγ±

2

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E +2RTFlnb = EΘ −

2RTFlnγ±

Example 2. Calculating an equilibrium constantEvaluate 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.

From the Debye-Hückel limiting law: γ+ ∝ -b1/2

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E +2RTFlnb = EΘ +Cb1/2

The expression on the left is plotted against b1/2 and extrapolated to b = 0. The intercept at b1/2

= 0 is the value of E∅ for the silver/silver chloride electrode.

Example 3. Determining the standard emf of a cellThe 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 25.63E/V 0.52053 0.49257 0.46860 0.41824

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E +2RTFlnb = EΘ +Cb1/2

(b/(10-3 b∅))1/2 1.793 2.370 3.023 5.063E/V + (2RT/F)lnb 0.2256 0.2263 0.2273 0.2299The intercept of the plot gives E∅ = 0.2232 V.

The measurement of activity coefficientsOnce 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

molality b can be obtained from:

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lnγ± =EΘ − E2RT F

− lnb

Applications of standard potentialsThe 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 thatRed1 has a thermodynamic tendency to reduce Ox2 if E1

∅ < E2∅

A species with a low standard potential has a thermodynamic tendency to reduce a specieswith 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, thereaction 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 (andhydrogen) are arranged in the order of their reducing power as measured by their standardpotentials in aqueous solution. A metal low in the series (with a lower standard potentials) canreduce the ions of metals with higher standard potentials.

The measurement of pH and pKa

The half-reaction at a hydrogen electrode:

H+(aq) + e- → 1/2 H2(g)

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Q =fH2 / p

Θ( )1/2

aH+

ν = 1

We let

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fH2 = pΘ and E∅(H+/H2) = 0. Then the electrode potential is

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E H + ,H2( ) =RTFlnaH+ = −

RT ln10F

pH

At 25°C E(H+/H2) = -59.16 mV × pHEach 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 ofthe potential of a hydrogen electrode immersed in the solution. The left-hand electrode istypically 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 cellPt(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 aqueoussolution of potassium hydrogen tartrate (pH = 3.557 at 25°C) and 0.0100 mol kg-1 disodiumtetraborate (pH = 9.180).

In practice, indirect methods are much more convenient and the hydrogen electrode isreplaced by the glass electrode. This electrode is sensitive to the hydrogen ionactivity and has a potential proportional to pH. It is filled with a phosphate buffercontaining 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 iscalibrated 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 containingequal amounts of the acid and its conjugate base.

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 VAg+(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 – aconvenient route for the calculation of the standard potential of a couple from two othercouples. G is a state function – the Gibbs energy of an overall reaction is the sum of the Gibbsenergies 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, calculateE∅(Cu2+,Cu+).

We need to convert the E∅ values to ΔrG∅, add them appropriately and convert the overallΔrG∅ 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)×FThe required reaction is(c) Cu2+(aq) + e- → Cu+(aq) ΔrG∅ = -FE∅

Thermodynamic functionsThe 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

Because (c) = (a) – (b), ΔrG∅(c) = ΔrG∅(a) – ΔrG∅(b)FE∅(c) = -(-0.680 V)F + (-0.522 V)F E∅(c) = +0.158 VIn a general case: νcE∅(c) = νaE∅(a) – νbE∅(b)

The temperature coefficient of the standard cell emf, dE∅/dT gives the standard entropyof the cell reaction:

E∅ = -ΔrG∅/νF (∂G/∂T)p = -S dE∅/dT = ΔrS

∅/νFHence we have an electrochemical technique for obtaining standard reaction entropies andthrough them the entropies of ions in solution. Finally, we can combine the results to obtainthe 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 potentialThe 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 VΔrG∅ = -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