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Concentration Cells

Concentration Cells. M | M + (aq, L) || M + (aq, R) | M Cell reaction: M + (aq, R) → M + (aq, L) since Δ r G θ = 0 (why?). The cell emf. The emf of a cell can be calculated by the difference of the potentials of the two electrodes, E cell = E right - E left

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Concentration Cells

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

  2. The cell emf • The emf of a cell can be calculated by the difference of the potentials of the two electrodes, Ecell = Eright - Eleft • The potential of an electrode can be calculated from its standard potential. For example Fe+3(aq) + e- → Fe+2(aq) Cu+2(aq) + 2e- → Cu(s) • Consider: Ag(s)|Ag+(aq) || Cl-(aq) |AgCl(s)| Ag(s) Ecell = E(AgCl/Ag, Cl-) – E(Ag+/Ag) = Eθ - ??

  3. In the lead storage battery (used in automobiles), Pb | PbSO4 | H2SO4 | PbSO4|PbO2 | Pb would the voltage change if you changed the concentration of H2SO4? (yes/no) • Answer ... • Yes, because • The net cell reaction is Pb + PbO2 + 2HSO4- + 2H+→ 2 PbSO4 + 2 H2O • The Nernst equation E = E° - (0.0594/2)log{1/{[HSO4-]2[H+]2}}.

  4. Previous two examples involve the same number of electron transfer at the cathode and anode, how to calculate emf if the number of electrons transferred at the two electrodes are different? Example:What is the emf for the cell : Mn(s)|Mn+2||Fe+3|Fe+2|Pt(s) Solution: The two reduction half reactions Right: Fe+3(aq) + e-→ Fe+2(aq) Left: Mn+2(aq) + 2e- → Mn(s) It shows that the above two half reactions have different number of electrons being transferred! The 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) ? Answer: NO! it is still calculated with Ecell = Eright - Eleft 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

  5. Standard Cell emf • Eθcell = Eθ (right) – Eθ(Left) • Calculating equilibrium constant from the standard emf : 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 cell emf is : Eθ (right) – Eθ(Left) = - 0.58V K = 1.6x10-10 • The above example demonstrates the usefulness of using two half reactions to represent a non redox process. What would be the two half reactions for the autoprotolysis of H2O?

  6. 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)

  7. The Nernst equation of the above cell reaction is 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

  8. Example plot from the text book(the interception corresponds to E θ)

  9. 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) (Eө = ? ) The cell which corresponds to the above two half-reactions 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

  10. 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 So E2 – E1 = - ln( ) = - 25.7(mV)x ln( ) = 56.3 mV

  11. Choose the correct Nernst equation for the cell Zn(s) | Zn2+ || Cu2+ | Cu(s). A: ΔE = ΔE° - 0.0296 log([Zn2+ ]/[Cu2+]) B: ΔE = ΔE° - 0.0296 log([Cu2+] / [Zn2+]) C: ΔE = ΔE° - 0.0296 log(Zn / Cu) D: ΔE = ΔE° - 0.0296 log(Cu / Zn) • Answer ... Hint... • The cell as written hasReduction on the Right: Cu2+ + 2e = Cuoxidation on the left: Zn = Zn2+ + 2e • Net reaction of cell is Zn(s) + Cu2+ = Cu(s) + Zn2+

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