Thermodynamics in Electrochemistry PDF

Chemical Thermodynamics Consider a chemical reaction: aA + bB ⇆ cC + dD What criteria do we use to determine if the reaction goes forward or backwards? ∆G0rxn = ∆Gf0, products - ∆Gf0, reactants - Gibbs free energy can be used to calculate the maximum revers...

Chemical Thermodynamics Consider a chemical reaction: aA + bB ⇆ cC + dD What criteria do we use to determine if the reaction goes forward or backwards? ∆G0rxn = ∆Gf0, products - ∆Gf0, reactants - Gibbs free energy can be used to calculate the maximum reversible work that may be performed by system at a constant temperature and pressure. - - Gibbs energy is minimized when a system reaches chemical equilibrium at constant pressure and temperature. Electrochemical thermodynamics Gibbs energy change is the maximum reversible work the system can do. ∆𝐺 = -nF Ecell nF is the number of charges per mole of reaction Ecell is the voltage… charge × voltage = energy ∆𝐺 = ∆𝐺0 + 𝑅T ln𝑄 → −nFE = −nFE0 + 𝑅T ln𝑄 𝑹𝑻 E = E0 − ln𝑄 𝒏𝑭 𝑹𝑻 𝐥𝐨𝐠 𝑸 E = E0 − 𝒏𝑭 𝐥𝐨𝐠 𝒆 𝟐.𝟑𝟎𝟐 𝑹𝑻 E = E0 − 𝐥𝐨𝐠 𝑸 𝒏𝑭 𝟎.𝟎𝟓𝟗𝟐 𝑽 and at 298.15 K, E = E0 − log 𝑄 𝒏 Remember Q includes all species in the balanced reaction , including ions, solids, etc. The Relationship between Cell Potential and Free Energy Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C), an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A); 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s): 𝟏𝑱 = 1C = 1A⋅ s (1) 𝟏𝑽 In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F), named after the English physicist and chemist Michael Faraday (1791–1867): 𝟔.𝟎𝟐𝟐𝟏𝟒×𝟏𝟎𝟐𝟑 F = (1.60218×10−19C) (2) 𝟏𝐦𝐨𝐥𝐞 F = 9.64855×104 C / mole ≃ 96,486C / (mol e-) The total charge transferred from the reductant to the oxidant is therefore nF, where n is the number of moles of electrons The maximum amount of work that can be produced by an electrochemical cell (wmax) is equal to the product of the cell potential (Ecell) and the total charge transferred during the reaction (nF): wmax = nFEcell (3) Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings. The change in free energy (ΔG) is also a measure of the maximum amount of work that can be performed during a chemical process (ΔG = wmax). Consequently, there must be a relationship between the potential of an electrochemical cell and ΔG, the most important thermodynamic quantity. This relationship is as follows: ΔG = −nFEcell (4) A spontaneous redox reaction is therefore characterized by a negative value of ΔG and a positive value of Ecell, consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and E°cell is as follows: ΔGo = −nFEocell (5) Note the Pattern A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell. Example 1. Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table 1, calculate the free- energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous? Given: redox reaction Asked for: ΔG° for the reaction and spontaneity Strategy: A From the relevant half-reactions and the corresponding values of E°, write the overall reaction and calculate E°cell using Equation 2. B Determine the number of electrons transferred in the overall reaction. Then use Equation 5 to calculate ΔG°. If ΔG° is negative, then the reaction is spontaneous. Solution: A As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of E°. From Table 1, we can find the reduction and oxidation half-reactions and corresponding E° values: Cathode: Cr2O72- (aq) + 14H+ (aq) + 6 e− → 2Cr3+ (aq) + 7H2O (l) anode: 2Br− (aq) → Br2 (aq) +2e− Eocathode = 1.23 V Eoanode = 1.09 V To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of E° is not affected: cathode: Cr2O72- (aq)+14H+(aq) + 6e− → 2Cr3+ (aq) + 7H2O(l) anode: 6Br− (aq) → 3Br2 (aq) + 6e− overall: Cr2O72- (aq)+14H+ (aq) + 6BrI− →2Cr3+ (aq) + 7H2O (l) + 3Br2(aq) Eocathode=1.23V Eoanode=1.09V Eocell=0.14V B We can now calculate ΔG° using EQUATION.10 Because six electrons are transferred in the overall reaction, the value of n is 6: ΔGo = −nFEocell = (6 mol) (96,468 J/(V⋅mol)) (0.14V) − 8.1×104J −81kJ/molCr2O7 Thus ΔG° is −81 kJ for the reaction as written, and the reaction is spontaneous. Exercise Use the data in Table 1 to calculate ΔG° for the reduction of ferric ion by iodide: 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(s) Is the reaction spontaneous? Answer: −44 kJ/mol I2; yes The Relationship between Cell Potential and the Equilibrium Constant We can use the relationship between ΔG° and the equilibrium constant K to obtain a relationship between E°cell and K. Recall that for a general reaction of the type aA + bB → cC + dD, the standard free-energy change and the equilibrium constant are related by the following equation: ΔGo = −RTln K (10) Given the relationship between the standard free-energy change and the standard cell potential (Equation 9), we can write −nFEocell = −RTln K (11) Rearranging this equation, 𝐑𝐓 Eocell = ( ) ln K (12) 𝐧𝐅 For T = 298 K, Equation 12 can be simplified as follows:  𝐑𝐓 Eocell = ( ) ln K 𝐧𝐅 𝟖.𝟑𝟏𝟒𝐉/(𝐦𝐨𝐥⋅𝐊)(𝟐𝟗𝟖𝐊) =[ ] 2.303 𝐧[𝟗𝟔,𝟒𝟖𝟔𝐉/(𝐕⋅𝐦𝐨𝐥)] 𝟎.𝟎𝟓𝟗𝟏 log K =( ) log K (13) 𝒏 Thus E°cell is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of E°cell and vice versa. Example 2 Use the data in Table 1 to calculate the equilibrium constant for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures. Given: redox reaction Asked for: K Strategy: A Write the relevant half-reactions and potentials. From these, obtain the overall reaction and E°cell. B Determine the number of electrons transferred in the overall reaction. Use Equation 13 to solve for log K and then K. Solution: A The relevant half-reactions and potentials from Table 1 are as follows: cathode: PbO2(s)+SO42- (aq)+4H+(aq)+2e−→PbSO4(s)+2H2O(l) anode: Pb(s)+ SO42- (aq)→PbSO4(s)+2e− overall: Pb(s)PbO2(s)+2 SO42- (aq)+4H+(aq)→2PbSO4(s)+2H2O(l) Eocathode=1.69V Eoanode=0.36V Eocell=2.05V B Two electrons are transferred in the overall reaction, so n = 2. Solving Equation 13 for log K and inserting the values of n and E°, nEo 2(2.05V) ln K =( ) =( 0.0591V ) = 69.37 0.0591V K=2.3×1069 Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know). Exercise Use the data in Table 1 to calculate the equilibrium constant for the reaction of Sn2+(aq) with oxygen to produce Sn4+(aq) and water under standard conditions. Report your answer to two significant figures. The reaction is as follows: 2Sn2+(aq)+O2(g)+4H+(aq)⇌2Sn4+ (aq) + 2H2O(l) Answer: 1.2 × 1073 Figure 1 summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between ΔG and the reaction quotient Q. Figure 1: The Relationships among Criteria for Thermodynamic Spontaneity. The three properties of a system that can be used to predict the spontaneity of a redox reaction under standard conditions are K, ΔG°, and E°cell. If we know the value of one of these quantities, then these relationships enable us to calculate the value of the other two. The signs of ΔG° and E°cell and the magnitude of K determine the direction of spontaneous reaction under standard conditions. The Effect of Concentration on Cell Potential: The Nernst Equation The actual free-energy change for a reaction under nonstandard conditions, ΔG, is given as follows: ΔG=ΔGo+RT ln Q (14) We also know that ΔG = −nFEcell and ΔG° = −nFE°cell. Substituting these expressions into Equation 14, we obtain −nFEcell=−nFEocell+RTln Q (15) Dividing both sides of this equation by −nF, 𝐑𝐓 Ecell=Eocell−( )ln Q (16) 𝐧𝐅 Equation 16 is called the Nernst equation, after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium (ΔG = 0), Equation 16 reduces to Equation 12 because Q = K, and there is no net transfer of electrons (i.e., Ecell = 0). Substituting the values of the constants into Equation 16 with T = 298 K and converting to base-10 logarithms give the relationship of the actual cell potential (Ecell), the standard cell potential (E°cell), and the reactant and product concentrations at room temperature (contained in Q): 𝟎.𝟎𝟓𝟗𝟏𝐕 Ecell=Eocell−( )ln Q (17) 𝐅 Note the Pattern The Nernst equation can be used to determine the value of Ecell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions. Equation 17 allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation 17 that the cell potential changes by 0.0591/n V for each 10-fold change in the value of Q because log 10 = 1. Example 3 In the exercise in Example 6, you determined that the following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which you now know means that ΔG° < 0): 2Ce4+(aq)+2Cl−(aq)→2Ce3+(aq)+Cl2(g) Calculate E for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce4+] = 0.013 M, [Ce3+] = 0.60 M, [Cl−] = 0.0030 M, Cl2= 1.0 atm, and T = 25°C. Given: balanced redox reaction, standard cell potential, and nonstandard conditions Asked for: cell potential Strategy: Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions. Solution: We can use the information given and the Nernst equation to calculate Ecell. Moreover, because the temperature is 25°C (298 K), we can use Equation 17 instead of 19.46. The overall reaction involves the net transfer of two electrons: 2Ce4+(aq) + 2e− → 2Ce3+(aq) 2Cl−(aq) → Cl2(g) + 2e− so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl2, and the value of E°cell into Equation 17, 0.0591V Ecell=Eocell−( ) log Q F 0.0591V [Ce3+]2PCl2 =0.25V−( ) logn ([Ce4+]2[Cl−]2) = 0.25V−[(0.0296V)(8.37)] =0.00V 2 Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture. Exercise In the exercise in Example 6, you determined that molecular oxygen will not oxidize MnO2 to permanganate via the reaction 4MnO2(s)+3O2(g)+4OH−(aq)→4MnO−4(aq)+2H2O(l) E_{cell}^{o}=-0.20\;V \) Calculate Ecell for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, P(O2) = 0.20 atm, [MNO4−] = 1.0 × 10−4 M, and T = 25°C. Answer: Ecell = −0.22 V; the reaction will not occur spontaneously. Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s) E_{cell}^{o}=1.10\;V \tag{19.4.18}\) The reaction quotient is therefore Q = [Zn2+]/[Cu2+]. Suppose that the cell initially contains 1.0 M Cu2+ and 1.0 × 10−6 M Zn2+. The initial voltage measured when the cell is connected can then be calculated from Equation 17: 𝟎.𝟎𝟓𝟗𝟏𝐕 [𝐙𝐧𝟐+] Ecell=Eocell−( )log([𝐂𝐮𝟐+]) (19) 𝐧 𝟎.𝟎𝟓𝟗𝟏𝐕 𝟏.𝟎×𝟏𝟎−𝟔 Ecell=1.10V−( )log =1.28V 𝟐 𝟏.𝟎 Thus the initial voltage is greater than E° because Q < 1. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn2+]/[Cu2+] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn2+] = [Cu2+], so Q = 1 and Ecell = E°cell. Beyond this point, [Zn2+] will continue to increase in the anode compartment, and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in Ecell. When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn2+ and 1.0 × 10−6 M Cu2+), Q = 1.0 × 106, and the cell potential will be reduced to 0.92 V. The variation of Ecell with log Q over this range is linear with a slope of −0.0591/n, as illustrated in Figure 2. As the reaction proceeds still further, Q continues to increase, and Ecell continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of Q when Ecell = 0 is calculated as follows: 𝟎.𝟎𝟓𝟗𝟏𝐕 Ecell=Eocell− log Q = 0 (20) 𝐧 𝟎.𝟎𝟓𝟗𝟏𝐕 Eo= log Q 𝐧 𝐄𝐨 𝐧 (𝟏.𝟏𝟎𝐕)(𝟐) Log Q= = = 37.23 𝟎.𝟎𝟓𝟗𝟏𝐕 𝟎.𝟎𝟓𝟗𝟏𝐕 Q=1037.23 =1.7×1037 Figure 2: The Variation of Ecell with Log Q for a Zn/Cu Cell Initially, log Q < 0, and the voltage of the cell is greater than E°cell. As the reaction progresses, log Q increases, and Ecell decreases. When [Zn2+] = [Cu2+], log Q = 0 and Ecell = E°cell = 1.10 V. As long as the electrical circuit remains intact, the reaction will continue, and log Q will increase until Q = K and the cell voltage reaches zero. At this point, the system will have reached equilibrium. Recall that at equilibrium, Q = K. Thus the equilibrium constant for the reaction of Zn metal with Cu2+ to give Cu metal and Zn2+ is 1.7 × 1037 at 25°C. Concentration Cells A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO3 in one compartment and 1.0 M AgNO3 in the other. The cell diagram and corresponding half-reactions are as follows:  Ag(s) ∣ Ag+ (aq,0.010M) ∥ Ag+ (aq,1.0M) ∣ Ag (s) (21) cathode: Ag+(aq,1.0M)+e−→Ag(s) (22) anode: Ag(s)→Ag+(aq,0.010M)+e− (23) overall: Ag+(aq,1.0M)→Ag+(aq,0.010M) (24) As the reaction progresses, the concentration of Ag+ will increase in the left (oxidation) compartment as the silver electrode dissolves, while the Ag+ concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of Ag(s) in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode: 0.0591V Ecell=Eocell− logQ=0 n (0.0591V) 0.010 − log =0.12 (25) 1 1.0 An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cell. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (Ecell = 0). Example 4 Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO4 as the anode (T = 25°C). Given: galvanic cell, identities of the electrodes, and solution concentrations Asked for: voltage Strategy: A Write the overall reaction that occurs in the cell. B Determine the number of electrons transferr 0.0591V Ecell=Eocell− logQ=0 n 0.0591V 5.2×10−2 − log =0.047 2 2.0 ed. Substitute this value into the Nernst equation to calculate the voltage. Solution: A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl− and SO42−) do not participate in the reaction, so their identity is not important. The overall reaction is as follows: Mn2+(aq, 2.0 M) → Mn2+(aq, 5.2 × 10−2 M) B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into Equation 17: Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution. Exercise Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na 2SO4 at pH 7.00. Both cells are in contact with the atmosphere, with P(O2) = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four- electron reduction of oxygen to water, O2(g) + 4H+(aq) + 4e− → 2H2O(l), what will be the potential when the circuit is closed? Answer: 0.41 V Using Cell Potentials to Measure Solubility Products Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products (Ksp) of sparingly soluble substances. Solubility products can be very small, with values of less than or equal to 10−30. Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods. To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure 3, which is designed to measure the solubility product of silver chloride: Ksp = [Ag+][Cl−]. In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag+; the other compartment contains a silver wire inserted into a 1.0 M Cl− solution saturated with AgCl. In this system, the Ag+ ion concentration in the first compartment equals Ksp. We can see this by dividing both sides of the equation for Ksp by [Cl−] and substituting: [Ag+] = Ksp/[Cl−] = Ksp/1.0 = Ksp. The overall cell reaction is as follows: Ag+(aq, concentrated) → Ag+(aq, dilute) Thus the voltage of the concentration cell due to the difference in [Ag +] between the two cells is as follows: Ecell=0 0.0591V [Ag+]dilute − log( )= 1 [Ag+]concentrated Ksp −0.0591 V log ( )= −0.0591 V (26) 1.0 Figure 3: A Galvanic Cell for Measuring the Solubility Product of AgCl One compartment contains a silver wire inserted into a 1.0 M solution of Ag+, and the other compartment contains a silver wire inserted into a 1.0 M Cl− solution saturated with AgCl. The potential due to the difference in [Ag+] between the two cells can be used to determine Ksp. By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation 26 for Ksp, Ecell − 0.580V Log Ksp = − = = −9.81 (27) 0.0591V 0.0591V Ksp=1.5×10−10Ksp=1.5×10−10 Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt. Example 5 To measure the solubility product of lead(II) sulfate (PbSO4) at 25°C, you construct a galvanic cell like the one shown in Figure 3, which contains a 1.0 M solution of a very soluble Pb2+ salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na2SO4 saturated with PbSO4 in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is Ksp for PbSO4? Report your answer to two significant figures. Given: galvanic cell, solution concentrations, electrodes, and voltage Asked for: K sp Strategy: A From the information given, write the equation for Ksp. Express this equation in terms of the concentration of Pb2+. B Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into Equation 26 and solve for Ksp. Solution: A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of Pb2+ and the other containing a dilute solution of Pb2+ in 1.0 M Na2SO4. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb 2+ in the dilute solution to Ksp: [Pb2+][SO42-]=Ksp Ksp Ksp [Pb2+]= = =Ksp [SO42−] 1.0M B The reduction of Pb2+ to Pb is a two-electron process and proceeds according to the following reaction: Pb2+(aq, concentrated) → Pb2+(aq, dilute) so \( E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{n}\;log \;Q = 0.0591V [Pb2+]dilute Ksp 0.230V=0 − log( )=−0.0296 V log( 1.0 ) 2 [Pb2+]concentrated −7.77=logKsp 1.7×10−8=Ksp Exercise A concentration cell similar to the one described in Example 11 contains a 1.0 M solution of lanthanum nitrate [La(NO3)3] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF3 in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the Ksp for LaF3? Report your answer to two significant figures. Answer: 5.7 × 10−17 Using Cell Potentials to Measure Concentrations Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example 11, in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to Ksp. Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H+] using a pH meter, as illustrated in Example 12. Example 6 Suppose a galvanic cell is constructed with a standard Zn/Zn 2+ couple in one compartment and a modified hydrogen electrode in the second compartment (Figure 3). The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows: Zn(s)∣Zn2+(aq, 1.0 M) ∥ H+(aq, ? M)∣H2(g, 1.0 atm)∣Pt(s) What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C? Given: galvanic cell, cell diagram, and cell potential Asked for: pH of the solution Strategy: A Write the overall cell reaction. B Substitute appropriate values into the Nernst equation and solve for −log[H+] to obtain the pH. Solution: A Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H2 (note that Zn lies below H2 in Table 1): Zn(s)+2H+(aq)→Zn2+(aq)+H2(g) Eo=0.76 V B By substituting the given values into the simplified Nernst equation (Equation 64), we can calculate [H+] under nonstandard conditions: \( E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{n}\;log \left ( \dfrac{\left [ Zn^{2+} \right ]P_{H_{2}}}{\left [ H^{+} \right ]^{2}} \right ) 0.26V=0.76V 0.0591V 0.0591V −( )log( )=−0.0296 2 2 Ksp V log ( ) 1.0 1 16.9=log( [H+]2 )=log[H+]-2 =−2 log[H+] 8.46=−log[H+] 8.5=ph Thus the potential of a galvanic cell can be used to measure the pH of a solution. Exercise Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb2+ in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E°cathode = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows” Pb(s) ∣Pb2+(aq, ? M)∥H+(aq), 1.0 M∣O2(g, 1.0 atm)∣Pt(s) When the circuit is closed, the cell has a measured potential of 1.62 V. Use Table 1 and Table T1 to determine the concentration of Pb2+ in the groundwater. Answer: 1.2 × 10−9 M References: 1. Prof. Shannon Boettcher. Electrochemical Thermodynamics and Potentials: Equilibrium and the Driving Forces for Electrochemical Processes. Oregon Center for Electrochemistry. Department of Chemistry University of Oregon. https://bpb-us- e1.wpmucdn.com/blogs.uoregon.edu/dist/b/17323/files/2021/02/El ectrochemical-Thermodynamics-and-Potentials-LiSA-101- Boettcher.pdf 2. Howard University General Chemistry: An Atoms First Approach. Chapter 19.4: Electrochemical Cells and Thermodynamics. https://chem.libretexts.org/Courses/Howard_University/General_C hemistry%3A_An_Atoms_First_Approach/Unit_7%3A_Thermody namics_and_Electrochemistry/Chapter_19%3A_Electrochemistry/ Chapter_19.4%3A_Electrochemical_Cells_and_Thermodynamics