Nerst 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

Concentration Cells Cell reaction: M+(aq, R) → M+(aq, L) E = M | M+(aq, L) || M+(aq, R) | M Cell reaction: M+(aq, R) → M+(aq, L) E = since ΔrGθ = 0 (why?)

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 red - E θox = 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

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 the electrode is the right-hand electrode (e.g. Cathode electrode) 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) its 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

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 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 can be obtained via 2*R – L, 2Fe+3(aq) + Mn(s) → 2Fe+2(aq) + Mn+2(aq) therefore, ΔrGθ = 2ΔrGθ (R) - ΔrGθ(L) 2FE θ = 2(1*F* E θ (R) – 2*F* E θ (L) At the end: 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

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 = Cu oxidation on the left: Zn = Zn2+ + 2e Net reaction of cell is Zn(s) + Cu2+ = Cu(s) + Zn2+