Nernst equation• E = Eө - lnQ

• Eө = -

• At 25 oC, the Nernst equation can be simplified further: E = Eө - lnQ

• Under equilibrium condition: Q = K, and E = 0

so lnK = Eө*v/25.7mV

vFRT

vmV7.25

vFGr

Concentration Cells• M | M+(aq, L) || M+(aq, R) | M

• Cell reaction: M+(aq, R) → M+(aq, L)

• E =

since ΔrGθ = 0 (why?)

E =

QvFRT

vFGr ln

R

Lbb

vFRT ln

The cell emf • The standard emf of a cell can be calculated by the difference of

the standard potentials of the two electrodes.

• Consider: Ag(s)|Ag+(aq) || Cl-(aq) |AgCl(s)| Ag(s) The overall potential of the cell is: Eθ = Eθ(AgCl/Ag, Cl-) – Eθ(Ag+/Ag)

• In a general form: Eθ = Eθ(right) – Eθ(left) • Example: Calculate the standard voltage for the cell: Mn+2|Mn+3||Fe+3|Fe+2

Solution: First, separate the cell into two half reactions Reduction: Fe+3(aq) + e- → Fe+2(aq) Reduction: Mn+3(aq) + e- → Mn+2(s)

To compute the cell voltage, simply use the right-hand electrode subtract the left-hand electrode.

E θ = E θ right - E θleft = 0.769 - (1.51) = - 0.741 V (???)

Calculating equilibrium constant from the standard emf

• Example: Evaluate the solubility constant of silver chloride, AgCl, from cell potential data at 298.15K.

• Solution: AgCl(s) → Ag+(aq) + Cl-(aq) Establish the electrode combination: Right: AgCl + e- → Ag(s) + Cl-(aq) Eθ = 0.22V Left: Ag+(aq) + e- → Ag(s) Eθ = 0.80V The standard emf is : Eθ (right) – Eθ(Left) = - 0.58V

lnK = = 0.58/0.025693

K = 1.6x10-10

RTvFE

The measurement of standard potentials• The potential of standard hydrogen electrode: Pt(s)|H2(g)|H+(aq)

is defined as 0 at all temperatures.• The standard potential of other electrodes can be obtained by

constructing an electrochemical cell, in which hydrogen electrode is employed as the left-hand electrode (i.e. anode)

• Example: the standard potential of the AgCl/Ag couple is the standard emf of the following cell:

Pt(s)|H2(g)|H+(aq), Cl-(aq)|AgCl(s)|Ag(s)or

Pt(s)|H2(g)|H+(aq) || Cl-(aq)|AgCl(s)|Ag(s)

with the cell reaction is: ½ H2(g) + AgCl(s) → H+(aq) + Cl-(aq) + Ag(s)

ΔrG = ΔrGθ + RTln

According to Nernst equation: –vFE = -vFE θ + RTln

Using the molality and the activity coefficient to represent the activity:

E = E θ – (RT/vF)ln(b2) - (RT/vF)ln(γ±2)

Reorganized the above equation: E + (2RT/vF)ln(b) = E θ - (2RT/vF)ln(γ±)

Since ln(γ±) is proportional to b1/2, one gets

E + (2RT/vF)ln(b) = E θ - C* b1/2, C is a constant

Therefore the plot of E + (2RT/vF)ln(b) against b1/2 will yield a straight line with the interception corresponds to E θ

212

/)/( pf

aa

H

ClH

212

/)/( pf

aa

H

ClH

Example plot from the text book

Example: Devise a cell in which the cell reaction is Mg(s) + Cl2(g) → MgCl2(aq)

Give the half-reactions for the electrodes and from the standard cell emf of 3.00V deduce the standard potential of the Mg2+/Mg couple.

Solution: the above reaction indicates that Cl2 gas is reduced and Mg is oxidized. Therefore,

R: Cl2(g) + 2e- → 2Cl-(aq) (Eө = + 1.36 from Table 10.7)

L: Mg2+(aq) + 2e- → Mg(s) The cell which corresponds to the above two half-reaction is :

Mg(s)|MgCl2(aq)|Cl2(g)|Pt

Eөcell = Eө(R) - Eө(L) = 1.36 – Eө(Mg2+, Mg)

Eө(Mg2+, Mg) = 1.36V – 3.00V = - 1.64V

Example: What is the standard voltage for the cell Mn|Mn+2||Fe+3|Fe+2|Pt

Solution: First, separate the cell into two reduction half reactions Right: Fe+3(aq) + e- → Fe+2(aq) Left: Mn+2(aq) + 2e- → Mn(s)

Note that the above two half reactions have different number of electrons being transferred!

To calculate the cell emf, write done the cell reaction and go through the standard reaction Gibbs energy.

The following cell reaction is obtained via 2*R – L, 2Fe+3(aq) + Mn(s) → 2Fe+2(aq) + Mn+2(aq)

should the standard cell potential be calculated as 2*Eө(R) - Eө(L) ?

Consider: ΔrGθ = 2ΔrGθ (R) - ΔrGθ(L) 2FE θ = 2(1*F* E θ (R) – 2*F* E θ (L)

it leads to Eөcell = Eө(R) - Eө(L)

= 0.769 - (- 1.182) = 1.951 V

Example: Consider a hydrogen electrode in aqueous HCl solution at 25oC operating at 105kPa. Calculate the change in the electrode potential when the molality of the acid is changed from 5.0 mmol kg-1 to 50 mmol kg-1. Activity coefficient can be found from Atkin’s textbook (Table 10.5 ).

Solution: first write down the half reaction equation: H+(aq) + e- → ½ H2(g) Based on Nernst equation

E = Eө - ln(Q)

Q = So E = - ln( ) E2 – E1 = - ln( ) = - 25.7(mV)x ln( )

= 56.3 mV

FvRT

212

/)/( PPH

212

/)/( PPHFvRT

FvRT

2

1

05.0*830.0005.0*929.0

• In the lead storage battery, Pb | PbSO4 | H2SO4 | PbSO4, PbO2 | Pb

would the voltage change if you changed the concentration of H2SO4? (yes/no)

• Answer ... • Hint...

The net cell reaction is • Pb + PbO2 + 2HSO4

- + 2H+ → 2 PbSO4 + 2 H2O • The Nernst equation • ΔE = Δ E° - (0.0592/2)log{1/{[HSO4

-]2[H+]2}}.

• Choose the correct Nernst equation for the cell Zn(s) | Zn2+ || Cu2+ | Cu(s).

A: Δ E = Δ E° - 0.0296 log([Zn2+ ]/[Cu2+]) B: Δ E = Δ E° - 0.0296 log([Cu2+] / [Zn2+]) C: Δ E = Δ E° - 0.0296 log(Zn / Cu) D: Δ E = Δ E° - 0.0296 log(Cu / Zn) • Answer ...

Hint... • The cell as written has

Reduction on the Right: Cu2+ + 2e = Cuoxidation on the left: Zn = Zn2+ + 2e

• Net reaction of cell is Zn(s) + Cu2+ = Cu(s) + Zn2+