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Chapter 10 Equilibrium Electrochemistry

Chapter 10 Equilibrium Electrochemistry. 10.1 Thermodynamic functions of ions in solution. The standard enthalpy and Gibbs energy of ions are used in the same way as those for neutral compounds.

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Chapter 10 Equilibrium Electrochemistry

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  1. Chapter 10Equilibrium Electrochemistry

  2. 10.1 Thermodynamic functions of ions in solution • The standard enthalpy and Gibbs energy of ions are used in the same way as those for neutral compounds. • Cations cannot be prepared without their accompanying anions. Thus the individual formation reactions are not measurable. • Defining that hydrogen ion has zero standard enthalpy and Gibbs energy of formation at ALL Temperature. ΔfHθ(H+, aq) = 0; ΔfGθ(H+, aq) = 0 • The standard Gibbs energy and enthalpy of formation for other ions can be calculated in relative to the value of hydrogen ion.

  3. Consider: Ag(s) + ½ Cl2(g)  Ag+(aq) + Cl-(aq) ΔrHθ = ΔfHθ(Ag+, aq) + ΔfHθ(Cl-, aq) ΔrGθ = ΔfGθ(Ag+, aq) + ΔfGθ(Cl-, aq) (Once the standard reaction Gibbs energy is calculated, the calculation of the equilibrium constant will be the same as discussed for neutral solutions) • Consider: ½ H2(g) + ½ Cl2(g)  H+(aq) + Cl-(aq) ΔrGθ = -131.23kJ mol-1 ΔrGθ = ΔfGθ(H+, aq) + ΔfGθ(Cl-, aq) – ½ ΔfGθ(H2, g) + ½ ΔfGθ(Cl2, g) = 0 + ΔfGθ(Cl-, aq) – 0 – 0 = ΔfGθ(Cl-, aq) therefore the standard Gibbs energy of formation for Cl- ion can be obtained from the standard reaction Gibbs energy. Standard Gibbs energy and enthalpy of formation of other ions could be achieved through the same approach.

  4. Thermodynamic cycles

  5. The sum of the Gibbs energy for all steps around a circle is ZERO! • The Gibbs energy of formation of an ion includes contributions from the dissociation, ionization, and hydration of hydrogen. • Gibbs energies of solvation can be estimated from Max Born equation. where zi is the charge number, e is the elementary charge, NA is the Avogadro’s constant, ε0 is the vacuum permittivity, εr is the relative permittivity, ri is ion’s radius. • ΔsolvGθ is strongly negative for small, highly charged ions in media of high relative permittivity. • For water at 25oC:

  6. Example: Estimate the value of ΔsolvGө(Br-, aq) - ΔsolvGө (Cl-, aq) from the experimental data and from the Max Born equation. Solution: To calculate the difference of their experimental measurement, use the data provided in Table 2.6: ΔsolvGө(Br-, aq) = -103.96 kJ mol-1; ΔsolvGө(Cl-, aq) = -131.23 kJ mol-1; So ΔsolvGө(Br-, aq) - ΔsolvGө (Cl-, aq) = -103.96 – ( 131.23) = 27.27 kJ mol-1; In order to apply the Born equation (10.2), we need to know the radius of the corresponding ions. These numbers can be obtained from Table 23.3 r(Br-) = 196 pm; r(Cl-) = 181 pm; thus ΔsolvGө(Br-, aq) - ΔsolvGө (Cl-, aq) = - (1/196 – 1/181)*6.86*104 kJ mol-1 = 29.00 kJ mol-1 (The calculated result is slightly larger than the experimental value).

  7. Entropies of Ions • The partial molar entropy of the solute can be measured. However, The entropies of ions in solution are reported on a scale in which the standard entropy of H+ ions in water is taken as ZERO at all temperatures. • The entropies of ions can be either positive or negative. • Partial molar ion entropies vary on the basis that they are related to the degree to which the ions order the water molecules around them in solution.

  8. 10.2 Ion activities • The activity relates to the molality b via α = γ * b/bө where γ is called the activity coefficient and bө equal 1mol kg-1. • Now the chemical potential will be expressed by the following equation: μ =μө + RT ln(b/bө) + RTln(γ) = μideal + RTln(γ) • Consider an electrically neutral real solution of M+ X-, G = μ+ + μ- = μ+ideal + μ-ideal + RTln(γ+) + RTln(γ-) =Gideal + RT(γ+γ-) • Since there is no experimental way to separate the product (γ+γ-) into contributions from the cations and anions, mean activity coefficient γ is introduced here to assign equal responsibility for nonideality to both kind of ions.

  9. The mean activity coefficient γ is calculated as (γ+γ-)1/2 • The chemical potential for individual ions M+ and X- then becomes: μ+ = μideal + RTln(γ) μ- = μideal + RTln(γ) • For a general compound of the following form: MpXq, the mean activity coefficient is expressed as: γ = [(γ+)p(γ-)q]1/s with s = p+q • Debye-Hückel limiting law is employed to calculate the mean activity coefficient: log(γ) = -|z+z-|AI1/210.3 A = 0.509 for aqueous solution at 25oC. I is the ionic strength, which is calculated as the following: I = ½ zi2(bi/bө) 10.4 where zi is the charge number of the ion I and bi is the molarlity of ion I.

  10. Example: Relate the ionic strength of (a) MgCl2, (b) Fe2(SO4)3 solutions to their molalities, b. Solution: To use the equation 10.4, we need to know the charge numbers and the molalities of the ions: MgCl2: From its molecular formula, we can get b(Mg2+) = b(solution); b(Cl-) = 2*b(solution); Z(Mg2+) = +2; Z(Cl-) = -1; So I = ½((2)2*b +(-1)2*(2b)) = ½(4b + 2b) = 3b; Fe2(SO4)3: From the molecular formula, we can get b(Fe3+) = 2*b(solution); b(SO42-) = 3*b(solution); Z(Fe3+) = +3; Z(SO42-) = -2; So I = ½((3)2*(2b) +(-2)2*(3b)) = ½(18b + 12b) = 15b

  11. Solutions contain more than one types of electrolytes • The ionic strength of the solution equals the sum of the ionic strength of each individual compound. Example: Calculate the ionic strength of a solution that contains 0.050 mol kg-1 K3[Fe(CN)6](aq), 0.040 mol kg-1 NaCl(aq), and 0.03 mol kg-1 Ce(SO4)2 (aq). Solution: I(K3[Fe(CN)6]) = ½( 12*(0.05*3) + (3)2*0.05 + (-1)2*(0.05*6)) = 0.45; I(NaCl) = ½(12*0.04 + (-1)2*0.04) = 0.04; I(Ce(SO4)2) = ½(42*0.03 + (-2)2*(2*0.03)) =0.36; So, I = I(K3(Fe(CN)6]) + I(NaCl) + I(Ce(SO4)2) = 0.45 + 0.04 + 0.36 = 0.85

  12. Calculating the mean activity coefficient Example: Calculate the ionic strength and the mean activity coefficient of 2.0m mol kg-1 Ca(NO3)2 at 25 oC. Solution: In order to calculate the mean activity coefficient with the eq. 10.3, one needs to know the ionic strength of the solution. Thus, the right approach is first to get I and then plug I into the equation 10.3. I = ½(22*0.002 + (-1)2*(2*0.002)) = 3*0.002 = 0.006; From equation 10.3, log(γ±) = - |2*1|*A*(0.006)1/2; = - 2*0.509*0.0775; = -0.0789; γ± = 0.834;

  13. Experimental test of the Debye-Hückel limiting law

  14. Accuracy of the Debye-Hückel limiting law Example: The mean activity coefficient in a 0.100 mol kg-1 MnCl2(aq) solution is 0.47 at 25oC. What is the percentage error in the value predicted by the Debye-Huckel limiting law? Solution: First use equation 10.4 to calculate the ionic strength and then use eq. 10.3 to calculate the mean activity coefficient. From eq. 10.4, I = ½(22*0.1 + 12*(2*0.1)) = 0.3 From eq. 10.3 log(γ) = -|2*1|A*(0.3)1/2; = - 2*0.509*0.5477 = - 0.5576 so γ = 0.277 Error = (0.47-0.277)/0.47 * 100% = 41%

  15. Extended Debye-Hückel law • (10.5) B is an adjustable empirical parameter.

  16. Calculating parameter B Example : The mean activity coefficient of NaCl in a diluted aqueous solution at 25oC is 0.907 (at 10.0mmol kg-1). Estimate the value of B in the extended Debye-Huckel law. Solution: First calculate the ionic strength I = ½(12*0.01 + 12*0.01) = 0.01 From equation 10.5 log(0.907) = - (0.509|1*1|*0.011/2)/(1+ B*0.011/2) B = - 1.67

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