The determination of equilibrium constants

• Self-test 7.11 Calculate the solubility constant (the equilibrium constant for reaction Hg2Cl2(s) ↔ Hg2

2+(aq) + 2Cl-(aq)) and the solubility of mercury(I) chloride at 298.15K. The mercury(I) ion is the diatomic species Hg2

2+.

• Answer: This chemical process does not involve electron transfer, i.e. is not a redox reaction.

Choosing cathode reaction as: Hg2Cl2(s) + 2e → 2Hg(l) + 2Cl-(aq) from reference table 7.2, Eθ = 0.27 V the anode reaction can be obtained through R – Cell Hg2

2+(aq) + 2e → 2Hg(l) from reference table 7.2, Eθ = 0.79 V

Therefore the standard cell potential = 0.27 – 0.79 = -0.52 V

• Using equation lnK = Eө*v/25.7mV, here v = 2 lnK = - 40.467 K = 2.62 x 10-18

• K = (aHg(I) * a2cl-)/aHg2cl2

= b*(2b)2/1 = 4*b3 = 2.62 x 10-18

therefore b = 8.68 x 10-7 mol/kg

Species-selective electrodes: Measuring pH

• The half reaction at a hydrogen electrode: H+(aq) + e- → ½ H2(g) Here v= 1, Q =

The potential E(H+/H2) = Eө -

E(H+/H2) = -

Assuming that the pressure of H2 gas equals 1 bar

E(H+/H2) = - = =

E(H+/H2) = -

Ha

PHf 2/12 )/)((

))/)((

ln(2/1

2

Ha

PHf

vF

RT

))/)((

ln(2/1

2

Ha

PHf

vF

RT

)1

ln(H

avF

RT )ln( Ha

F

RTH

aF

RTlog)10ln(

pHF

RT)10ln(

• 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 (Hg2Cl2(s)).

• The hydrogen electrode is used as the right hand electrode, i.e. cathode.

• E(cell) = E(H+/H2) - E(ref.)

• E(cell) + E(ref) = -

• Why do we need to calibrate the pH electrode before its usage?

pHF

RT)10ln(

Thermodynamic function

• Nernst equation is a bridge connecting the thermodynamic quantify, Gibbs energy, and the electromotive force.

• Consider: Pt(s)|H2(g)|H+(aq)|Ag+(aq)|Ag(s) Eө = 0.80V. Calculate the ΔfGө(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 H2(g)

Cell : Ag+(aq) + 1/2 H2(g) → Ag(s) + H+(aq)

Continued

ΔrGө = ΔfGө(Ag(s)) + ΔfGө (H+) - ΔfGө (Ag+) - (1/2)ΔfGө(H2(g))

ΔrGө = 0 + 0 - ΔfGө (Ag+) - 0

Since

ΔrGө = -νFEө

ΔfGө(Ag+) = vFEө = 9.648 x 104 C mol-1 * 0.80V

= 7.715 x 104 CV mol-1

= 7.715 x 104 J mol-1 ( 1CV = 1J)

Temperature dependence of emf• ΔrGө = -νFEө

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.

dT

vFdE

dT

Gd r

dTvF

Gd

dT

dE r

dT

dE

SdT

Gdr

r

vF

S

dT

dE r

• noncalorimetric method of measuring ΔrHө

ΔrHө = ΔrGө + TΔrSө = -vFEө + T(vF ) = -vF(Eө - T )

• Example: The standard electromotive force of the cell Pt(s)|H2(g)|H+(aq)||Ag+(aq)|Ag(s)

was measured over a broad range of temperatures, and the data were fitted to the following polynomial:

Eө/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, Eө/V = 0.07- 4.11x10-4(298K/K – 298) – 3.2x10-6(298K/K -298)2

Eө/V = 0.07

Eө = 0.07 V

dT

dE

dT

dE

• to identify the value of v, we need to write down the cell reaction

Ag+ + 1/2H2(g) → H+(aq) + Ag(s)

ΔrGө = -vFEө = - (1) x 9.6485 x104 C mol-1 x (0.07V) = - 6.754 x 103 CVmol-1

= - 6.754 x 103 J mol-1

Calculate the temperature coefficient of the cell electromotive force

= – 4.11x10-4 K-1V – 3.2x10-6x2x(T/K -298) K-1V = - 4.11x10-4 K-1V (at 298K)

then

ΔrSө = vF = 1 x 9.6485 x 104 C mol-1 x (- 4.11x10-4 K-1V ) = - 39.66 CV K-1 mol-1

= - 39.66 J K-1 mol-1

to calculate the standard reaction enthalpy:

ΔrHө = ΔrGө + TΔrSө = -6.754kJ mol-1 + 298K (- 39.66 J K-

1 mol-1) = -18.572 kJ mol-1

dT

dE

dT

dE

Example: The standard cell potential of Pt(s)|H2(g)|HCl(aq)| Hg2Cl2(s)|Hg(l)|Ag(s)

was found to be +0.269 V at 293 K and + 0.266 V at 303K. Evaluate the standard reaction Gibbs function, enthalpy, and entropy at 298K.

Solution: Write the cell reaction Hg2Cl2(s) + H2(g) → 2Hg(l) + 2HCl(aq) So v = 2, To find the ΔrGө at 298 K, one needs to know the standard emf at 298K,

which can be obtained by linear interpolation between the two temperatures.

Eө = 0.2675 V ΔrGө = -2F Eө = -51.8 kJ mol-1

The standard reaction entropy can then be calculated from = (0.266V- 0.269V)/10K = -3.0x10-4 VK-1 ΔrSө = 2F = - 58 JK-1 mol-1 then ΔrHө = ΔrGө + TΔrSө = -69 kJ mol-1

dT

dEdT

dE

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 Fe3+/Fe from the values for the Fe3+/Fe2+ (+0.77V) and Fe2+/Fe( -0.44V).

• Solution: : first write down the half reactions for these three couples: 1) Fe3+ + e- → Fe2+

2) Fe2+ + 2e- → Fe 3) Fe3+ + 3e- → Fe

Reaction 3 is the sum of 1 and 2, yet one cannot use E3 = E1+ E2

ΔrGө(1) = - 1x F x 0.77V

ΔrGө(2) = - 2x F x (-0.44)V

ΔrGө(3) = ΔrGө(1) + ΔrGө(2) = 0.11F V

ΔrGө(3) = - 3*F*E3 = 0.11 F V

E3 = -0.033VA 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 Cu2+/Cu and Cu+/Cu couples are +0.340V and + 0.522V, respectively, evaluate Eө(Cu2+, Cu+).

Solution: Again, we should go through the standard Gibbs energy to calculate it.

First write the half-reactions: (1) Cu2+(aq) + 2e- → Cu(s) (2) Cu+(aq) + e- → Cu(s) (3) Cu2+(aq) + e- → Cu+

it can be identified easily that reaction (3) equals (1) - (2) thus: ΔrGө(3) = ΔrGө(1) - ΔrGө(2) = (-2*F*0.340V) – (-1*F*0.522V) ΔrGө(3) = -0.158F V

-1*F*Eө(Cu2+/Cu+) = -0.158F V Eө(Cu2+/Cu+) = 0.158 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.