1. Applications of standard potentials

2. The electrochemical series
For two redox couples Ox1/Red1 and Ox2/Red2,
Red1, Ox1 || Red2, Ox2
Eθ = Eθ2 – Eθ1
The cell reaction: Red1 + Ox2 → Ox1 + Red2
If Eθ > 0 then ΔrGθ < 0 (Nernst equation), the reaction will take place spontaneously. In other words, if Eθ2 > Eθ1, the Ox2 has the thermodynamic tendency to oxidize Red1.

4. The determination of activity coefficients
Once the standard potential of an electrode (Eθ) is known, one can use the following equation
E + (2RT/vF)ln(b) = E θ - (2RT/vF)ln(γ±)
to determine the mean activity coefficient of the ions at the concentration of interest via measuring the cell emf (E).

5. The determination of equilibrium constants
Self-test 7.11 Calculate the solubility constant (the equilibrium constant for reaction Hg2Cl2(s) ↔ Hg22+(aq) + 2Cl-(aq)) and the solubility of mercury(I) chloride at K. The mercury(I) ion is the diatomic species Hg22+.
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
Hg22+(aq) + 2e → 2Hg(l)
from reference table 7.2, Eθ = 0.79 V
Therefore the standard cell potential = 0.27 – 0.79 = V

6. K = (aHg(I) * a2cl-)/aHg2cl2 = b*(2b)2/1 = 4*b3 = 2.62 x 10-18
Using equation lnK = Eө*v/25.7mV,
here v = 2
lnK =
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

7. 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) = -

8. Two electrodes are required to build up an electrochemical cell
Two electrodes are required to build up an electrochemical cell. This is also 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?

9. Thermodynamic function
Nernst equation is the bridge to the connect 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 cell reaction. To do that, write down the two half reactions first 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) /2 H2(g) → Ag(s) + H+(aq)

10. Continued
ΔrGө = ΔfGө(Ag(s)) + ΔfGө (H+) - ΔfGө (Ag+) - (1/2)ΔrGө(H2(g))
ΔrGө = ΔfGө (Ag+) - 0
Since
ΔrGө = -νFEө
ΔfGө(Ag+) = vFEө = x 104 C mol-1 * 0.80V
= x 104 CV mol-1
= x 104 J mol-1 ( 1CV = 1J)

11. 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.

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

13. to identify the value of v, we need to write down the cell reaction
Ag /2H2(g) → H+(aq) + Ag(s)
ΔrGө = -vFEө = - (1) x x104 C mol-1 x (0.07V)
= x 103 CVmol-1
= 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 = x10-4 K-1V (at 298K)
then
ΔrSө = vF = 1 x x 104 C mol-1 x (- 4.11x10-4 K-1V )
= CV K-1 mol-1
= J K-1 mol-1
to calculate the standard reaction enthalpy:
ΔrHө = ΔrGө + TΔrSө = kJ mol K ( J K-1 mol-1)
= kJ mol-1

14. Example: The standard cell potential of
Pt(s)|H2(g)|HCl(aq)| Hg2Cl2(s)|Hg(l)
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
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 = V
ΔrGө = -2F Eө = kJ mol-1
The standard reaction entropy can then be calculated from
= (0.266V V)/10K = -3.0x10-4 VK-1
ΔrSө = 2F = - 58 JK-1 mol-1
then
ΔrHө = ΔrGө + TΔrSө = -69 kJ mol-1

15. Evaluate the reaction potential from two others
Example 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) Fe e- → Fe2+
2) Fe e- → Fe
3) Fe e- → 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 = V

16. Example: Given that the standard potentials of the Cu2+/Cu and Cu+/Cu couples are V and V, 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) e → 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) = F V
-1*F*Eө(Cu2+/Cu+) = F V
Eө(Cu2+/Cu+) = V