The determination of equilibrium constants Self-test 7.11 Calculate the solubility constant (the equilibrium constant for reaction Hg 2 Cl 2 (s) ↔ Hg 2 2+ (aq) + 2Cl - (aq)) and the solubility of mercury(I) chloride at K. The mercury(I) ion is the diatomic species Hg Answer: This chemical process does not involve electron transfer, i.e. is not a redox reaction. Choosing cathode reaction as: Hg 2 Cl 2 (s) + 2e → 2Hg( l ) + 2Cl - (aq) from reference table 7.2, E θ = 0.27 V the anode reaction can be obtained through R – Cell Hg 2 2+ (aq) + 2e → 2Hg( l ) from reference table 7.2, E θ = 0.79 V Therefore the standard cell potential = 0.27 – 0.79 = V

Using equation lnK = E ө *v/25.7mV, here v = 2 lnK = K = 2.62 x K = ( a Hg(I) * a 2 cl- )/ a Hg2cl2 = b*(2b) 2 /1 = 4*b 3 = 2.62 x therefore b = 8.68 x mol/kg

Species-selective electrodes: Measuring pH The half reaction at a hydrogen electrode: H + (aq) + e - → ½ H 2 (g) Here v= 1, Q = The potential E(H + /H 2 ) = E ө - E(H + /H 2 ) = - Assuming that the pressure of H 2 gas equals 1 bar E(H + /H 2 ) = - = = E(H + /H 2 ) = -

Two electrodes are required to build up an electrochemical cell. This is why we need a reference electrode when measuring the pH of a solution. A regular reference electrode is calomel (Hg 2 Cl 2 (s)). The hydrogen electrode is used as the right hand electrode, i.e. cathode. E(cell) = E(H + /H 2 ) - E(ref.) E(cell) + E(ref) = - Why do we need to calibrate the pH electrode before its usage?

Thermodynamic function Nernst equation is a bridge connecting the thermodynamic quantify, Gibbs energy, and the electromotive force. Consider: Pt(s)|H 2 (g)|H + (aq)|Ag + (aq)|Ag(s) E ө = 0.80V. Calculate the Δ f G ө (Ag + (aq)). Solution: First, write down the two reduction half reactions and then do a simple subtraction (R-L) to get the cell reaction. R: Ag + (aq) + e - → Ag(s) L: H + (aq) + e - → 1/2 H 2 (g) Cell : Ag + (aq) + 1/2 H 2 (g) → Ag(s) + H + (aq)

Continued Δ r G ө = Δ f G ө (Ag(s)) + Δ f G ө (H + ) - Δ f G ө (Ag + ) - (1/2)Δ f G ө (H 2 (g)) Δ r G ө = Δ f G ө (Ag + ) - 0 Since Δ r G ө = -νFE ө Δ f G ө (Ag + ) = vFE ө = x 10 4 C mol -1 * 0.80V = x 10 4 CV mol -1 = x 10 4 J mol -1 ( 1CV = 1J)

Temperature dependence of emf Δ r G ө = -νFE ө take the derivate of temperature for both sides is called temperature coefficient of standard cell emf. Because one gets therefore, one can use electrochemical method to obtain reaction entropy and relate them to entropies of ions in solution.

noncalorimetric method of measuring Δ r H ө Δ r H ө = Δ r G ө + TΔ r S ө = -vFE ө + T(vF ) = -vF(E ө - T ) Example: The standard electromotive force of the cell Pt(s)|H 2 (g)|H + (aq)||Ag + (aq)|Ag(s) was measured over a broad range of temperatures, and the data were fitted to the following polynomial: E ө /V = 0.07 – 4.11x10 -4 (T/K – 298) – 3.2x10 -6 (T/K -298) 2 Evaluate the standard reaction Gibbs energy, enthalpy, and entropy at 298K. Solution: The standard reaction Gibbs energy can be calculated once we know the standard emf of the above cell: At 298K, E ө /V = x10 -4 (298K/K – 298) – 3.2x10 -6 (298K/K - 298) 2 E ө /V = 0.07 E ө = 0.07 V

to identify the value of v, we need to write down the cell reaction Ag + + 1/2H 2 (g) → H + (aq) + Ag(s) Δ r G ө = -vFE ө = - (1) x x10 4 C mol -1 x (0.07V) = x 10 3 CVmol -1 = x 10 3 J mol -1 Calculate the temperature coefficient of the cell electromotive force = – 4.11x10 -4 K -1 V – 3.2x10 -6 x2x(T/K -298) K -1 V = x10 -4 K -1 V (at 298K) then Δ r S ө = vF = 1 x x 10 4 C mol -1 x (- 4.11x10 -4 K -1 V ) = CV K -1 mol -1 = J K -1 mol -1 to calculate the standard reaction enthalpy: Δ r H ө = Δ r G ө + TΔ r S ө = kJ mol K ( J K -1 mol -1 ) = kJ mol -1

Example: The standard cell potential of Pt(s)|H 2 (g)|HCl(aq)| Hg 2 Cl 2 (s)|Hg( l )|Ag(s) was found to be V at 293 K and V at 303K. Evaluate the standard reaction Gibbs function, enthalpy, and entropy at 298K. Solution: Write the cell reaction Hg 2 Cl 2 (s) + H 2 (g) → 2Hg( l ) + 2HCl(aq) So v = 2, To find the Δ r G ө at 298 K, one needs to know the standard emf at 298K, which can be obtained by linear interpolation between the two temperatures. E ө = V Δ r G ө = -2F E ө = kJ mol -1 The standard reaction entropy can then be calculated from = (0.266V V)/10K = -3.0x10 -4 VK-1 Δ r S ө = 2F = - 58 JK-1 mol-1 then Δ r H ө = Δ r G ө + TΔ r S ө = -69 kJ mol -1 What is the quotient, Q, of the above cell reaction?

Evaluate the reaction potential from two others Example 1. Calculate the standard potential of the Fe 3+ /Fe from the values for the Fe 3+ /Fe 2+ (+0.77V) and Fe 2+ /Fe( -0.44V). Solution: : first write down the half reactions for these three couples: 1) Fe 3+ + e - → Fe 2+ 2) Fe e - → Fe 3) Fe e - → Fe Reaction 3 is the sum of 1 and 2, yet one cannot use E 3 = E 1 + E 2 Δ r G ө (1) = - 1x F x 0.77V Δ r G ө (2) = - 2x F x (-0.44)V Δ r G ө (3) = Δ r G ө (1) + Δ r G ө (2) = 0.11F V Δ r G ө (3) = - 3*F*E3 = 0.11 F V E 3 = V A good practice of calculating the potential of a redox couple from other redox coupled is going through the reaction Gibbs energy!

Example 2: Given that the standard potentials of the Cu 2+ /Cu and Cu + /Cu couples are V and V, respectively, evaluate E ө (Cu 2+, Cu + ). Solution: Again, we should go through the standard Gibbs energy to calculate it. First write the half-reactions: (1) Cu 2+ (aq) + 2e - → Cu(s) (2) Cu + (aq) + e - → Cu(s) (3) Cu 2+ (aq) + e - → Cu + it can be identified easily that reaction (3) equals (1) - (2) thus: Δ r G ө (3) = Δ r G ө (1) - Δ r G ө (2) = (-2*F*0.340V) – (-1*F*0.522V) Δ r G ө (3) = F V -1*F*E ө (Cu2+/Cu + ) = F V E ө (Cu 2+ /Cu + ) = V

Summary of chapter 7 Equilibrium constant, Gibbs energy vs the extent of reaction Reaction Gibbs energy Standard reaction Gibbs energy and thermodynamic equilibrium constant. Le Chaterlier principle & van’t Hoff relationship. Reactions in charged environment: Nerst equation. Cell potential and standard cell potential. Connection between thermal and electrochemical quantities.