Equilibrium electrochemistry

Transcription

1 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.

2 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.

3 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.

4 Any redox reaction may be expressed as the difference of two reduction half-reactions: Reduction of Cu 2+ : Cu 2+ (aq) + 2 e - Cu(s) Reduction of Zn 2+ : Zn 2+ (aq) + 2 e - Zn(s) Difference: Cu 2+ (aq) + Zn(s) Zn 2+ (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) + 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: Cu 2+ /Cu, Zn 2+ /Zn. In general, we write a couple as Ox/Red: Ox + ν e - Red Reaction quotient Q for a half-reaction (electrons are ignored): Cu 2+ (aq) + 2 e - Cu(s) Q = 1/a(Cu 2+ ) O 2 (g) + 4 H + (aq) + 4 e - 2 H 2 O(l) The reaction quotient for the reduction of O 2 to H 2 O in acid: Q = 4 a H + 2 a H2 O ( f O2 p Θ ) = p Θ a 4 H + p O2

5 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 Red 1 Ox 1 + ν e - at one electrode travel through the external circuit and reenter the cell through the other electrode Ox 2 + ν e - Red 2 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, Ox 2, 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.

6 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 + /H 2. This electrode is denoted as Pt(s) H 2 (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) H 2 (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) ZnSO 4 (aq) CuSO 4 (aq) Cu(s) Right-hand electrode: Cu 2+ (aq) + 2 e - Cu(s) Left-hand electrode: Zn 2+ (aq) + 2 e - Zn(s) The overall cell reaction: Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq)

7 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 w e,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 Δ r G 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 Δ r G The relation between the reaction Gibbs energy and the emf: -νfe = Δ r G F the Faraday constant, the magnitude of electric charge per mole of electrons: F = en A = kc mol -1. This equation the key connection between electrical measurements and thermodynamic properties.

8 Justification. We consider the change in G when the cell reaction advances by an infinitesimal amount dξ at some composition: dg = µ J dn J = ν J µ J dξ J J Δ r G = G = ν J µ J dg = Δ r Gdξ ξ p,t J The maximum non-expansion (electrical) work that the reaction can do as it advances by dξ at constant T and p: dw e = Δ r Gdξ 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 -νen A dξ (because νdξ is the amount of the electrons and the charge per mole of electrons is -en A ). Hence, the total charge transported is - νfdξ because en A = F. The work done when an infinitesimal charge -νfdξ travels from the anode to the cathode the product of the charge and the potential difference E: dw e = νfedξ = Δ r Gdξ -νfe = Δ r G 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.

9 The Nernst equation Δ r G = Δ r G + RT ln Q E = -Δ r G/νF E = -Δ r G /νf (RT/νF) ln Q E = E (RT/νF) ln Q E = -Δ r G /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 Δ r G 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 M + (aq, R) M + (aq, L) Q = a L /a R, ν = 1 E = RT F ln a L RT a R F ln b L b R 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.

10 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.

11 The standard potential of the Daniell cell is V. Then we can calculate the equilibrium constant for the cell reaction as Cu 2+ (aq) + Zn(s) Zn 2+ (aq) + Cu(s) ln K = 2 ( C mol -1 ) (1.10 V)/{( J K -1 mol -1 ) ( K)} = K = 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) H 2 (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) H 2 (g) H + (aq) Ag + (aq) Ag(s) V AgCl/Ag,Cl - couple: the standard emf of the following cell: Pt(s) H 2 (g) H + (aq) Cl - (aq) AgCl(s) Ag(s) E ( AgCl/Ag,Cl - ) = E = 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 - ) = V it should be understood as the potential for the overall reaction: AgCl(s) + 1/2 H 2 (g) Ag(s) + H + (aq) + Cl - (aq) E = V

12 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 Δ r G 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 = E R E L E R the standard potential of the right-hand electrode. E L 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) = = V Because Δ r G = -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) + Cu 2+ (aq) at 298 K Right-hand electrode: Cu(s) Cu + (aq) Cu + (aq) + e - Cu(s) E = V Left-hand electrode: Pt(s) Cu 2+ (aq),cu + (aq) Cu 2+ (aq) + e - Cu + (aq) The standard emf of the cell: E = V 0.16 V = V ln K = 0.36 V / V K = E = V

13 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 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 = V Left Ag + (aq) + e - Ag(s) E = V The standard emf is 0.58 V, ν = 1 ln K = νfe /RT = νe /(RT/F) = V / V K = K = a Ag+ a Cl- S 2 S = 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) H 2 (g) HCl(aq) AgCl(s) Ag(s) 1/2 H 2 (g) + AgCl(s) HCl(aq) + Ag(s) The Nernst equation: E = E Θ AgCl / Ag,Cl ( ) RT F ln a H +a Cl ( f H2 p Θ ) 1/2 For simplicity, we set f = p and express the activities in terms of the molality b and the mean activity coefficient γ + : E = E Θ RT F lnb2 RT F lnγ 2 ± E + 2RT F lnb = E Θ 2RT F lnγ ± From the Debye-Hückel limiting law: γ + -b 1/2 E + 2RT F lnb = E Θ +Cb 1/2 The expression on the left is plotted against b 1/2 and extrapolated to b = 0. The intercept at b 1/2 = 0 is the value of E for the silver/silver chloride electrode.

14 Example 3. Determining the standard emf of a cell The emf of the cell Pt(s) H 2 (g,p ) HCl(aq,b) AgCl(s) Ag(s) at 25 C is: b/(10-3 b ) E/V E + 2RT F lnb = E Θ +Cb 1/2 (b/(10-3 b )) 1/ E/V + (2RT/F)lnb The intercept of the plot gives E = 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 molality b can be obtained from: lnγ ± = E Θ E 2RT F lnb 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 Red 1,Ox 1 Red 2,Ox 2 E = E 2 E 1 the cell reaction Red 1 + Ox 2 Ox 1 + Red 2 The reaction is spontaneous as written if E > 0 E 2 > E 1 Because in the cell reaction Red 1 reduces Ox 2, we conclude that Red 1 has a thermodynamic tendency to reduce Ox 2 if E 1 < E 2

15 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 (Zn 2+,Zn) = V < E (Cu 2+,Cu) = V Zn(s) has a thermodynamic tendency to reduce Cu 2+ (aq) under standard conditions. Hence, the reaction Zn(s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu(s) can be expected to have K > 1 (in fact, K = 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 pk a The half-reaction at a hydrogen electrode: ( ) 1/2 H + (aq) + e - 1/2 H 2 (g) Q = f H 2 / p Θ a H + ν = 1 We let f H2 = p Θ and E (H + /H 2 ) = 0. Then the electrode potential is E( H +,H 2 ) = RT F lna RT ln10 = ph H + F At 25 C E(H + /H 2 ) = mv ph Each unit increase in ph decreases the electrode potential by 59 mv.

16 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 (Hg 2 Cl 2 (s)) reference electrode with potential E(cal). The right-hand electrode the hydrogen electrode. ph = {E E(cal)} / ( 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) H 2 (g) S(aq) 3.5 M KCL(aq) X(aq) H 2 (g) Pt(s) S a solution of standard ph. The currently recommended primary standards: a saturated aqueous solution of potassium hydrogen tartrate (ph = at 25 C) and 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 pk a : the pk a of an acid is equal to the ph of a solution containing equal amounts of the acid and its conjugate base.

17 Thermodynamic functions The cell potential is related to the reaction Gibbs energy: Δ r G = -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) H 2 (g) H + (aq) Ag + (aq) Ag(s) E = V Ag + (aq) + 1/2 H 2 (g) H + (aq) + Ag(s) Δ r G = Δ f G (Ag +,aq) Δ f G (Ag +,aq) = -(-E F) 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 (Cu 2+,Cu) = V and E (Cu +,Cu) = V, calculate E (Cu 2+,Cu + ). We need to convert the E values to Δ r G, add them appropriately and convert the overall Δ r G to the required E. (a) Cu 2+ (aq) + 2 e - Cu(s) E = V Δ r G (a) = -2F (0.340 V) = ( V) F (b) Cu + (aq) + e - Cu(s) E = V Δ r G (b) = -F (0.522 V) = ( V) F The required reaction is (c) Cu 2+ (aq) + e - Cu + (aq) Δ r G = -FE Because (c) = (a) (b), Δ r G (c) = Δ r G (a) Δ r G (b) FE (c) = -( V)F + ( V)F E (c) = V In a general case: ν c E (c) = ν a E (a) ν b E (b)

18 The temperature coefficient of the standard cell emf, de /dt gives the standard entropy of the cell reaction: E = -Δ r G /νf ( G/ T) p = -S de /dt = Δ r S /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: Δ r H = Δ r G + TΔ r S = -νf{e T(dE /dt)} This provides a noncalorimetric method for measuring Δ r H and, through the convention Δ f H (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) H 2 (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 = (T/K 298) (T/K 298) 2 Evaluate the standard reaction Gibbs energy, enthalpy, and entropy at 298 K. At 298 K, E = V Δ r G = -νfe = -1( C mol -1 )( V) = kj mol -1 The temperature coefficient of the cell potential: de /dt = V K -1 2( )(T/K 298) V K -1 = V K -1 at 298 K Δ r S = 1( C mol -1 ) ( V K -1 ) = J K -1 mol -1 Δ r H = Δ r G + TΔ r S = ( kj mol -1 ) + (298 K) ( kj K -1 mol -1 ) = kj mol -1