1. 05 REDOX EQM Nernst Equation & C. Y. Yeung (CHW, 2009)p.01 The math. relationship between cell e.m.f. & [reactant] and [product] in a redox rxn under non-standard conditions. [conc.  1M] Nernst Equation: E = E – 0.0592 n log [product] [reactant] reduced form oxidized form no. of e - involved in rxn reaction quotient (Q) Factors affecting E and E cell

2. p.02 Co(s)  Co 2+ (aq) + 2e - Fe 2+ (aq) + 2e -  Fe(s) EXAMPLE (1): Predict whether the following reaction would proceed spontaneously at 298K: Co(s) + Fe 2+ (aq)  Co 2+ (aq) + Fe(s) Given:[Co 2+ (aq)] = 0.15M and [Fe 2+ (aq)] = 0.68M E Fe 2+ | Fe = -0.44 V E Co 2+ | Co = -0.28 V (anode) (cathode) E cell = -0.44 – (-0.28) = -0.16 V E cell = -0.16 – 0.0592 2 log (0.15) (0.68) = -0.141 V E cell < 0,  the reaction is NOT spontaneous in the direction written.

3. p.03 Zn(s)  Zn 2+ (aq) + 2e - 2H + (aq) + 2e -  H 2 (g) (anode) (cathode) E cell = 0 – (-0.76) = +0.76 V +0.54 = +0.76 – 0.0592 2 log (1.0)(1.0) [H + ] 2  [H + ] = 1.92  10 -4 M EXAMPLE (2): In the following reaction, the e.m.f. of the cell is found to be +0.54V at 25 0 C. Suppose that [Zn 2+ (aq)] = 1.0M and P H2(g) = 1.0 atm. Calculate the molarity of H + (aq). Zn(s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g) E Zn 2+ | Zn = -0.76 V

4. p.04 Concentration Cell : an electrochemical cell from 2 half cells composed of the same material but differing in ion concentrations. EXAMPLE (3): Consider an electrochemical cell in which zinc electrodes are put into two aq. solutions of ZnSO 4 (aq) at 0.10M and 1.0M concentrations. Write the cell diagram and calculate the cell e.m.f.E Zn 2+ | Zn = -0.76 V Given: Cell Diagram: Zn(s) | Zn 2+ (aq, 0.10M) Zn 2+ (aq, 1.0M) | Zn(s) Reduction should occur in the more concentrated compartment, therefore: E = 0 – 0.0592 2 log 0.10 1.0 = 0.0296 V **When the concentrations in the two compartments are the same, E becomes zero, and no further change occurs. i.e. eqm established.

5. p.05 Membrane potential : the electrical potential exists across the membrane of biological cells. It is developed when there are unequal concentrations of the same type of ion in the “interior” and “exterior” of a cell. (e.g. nerve cell) E = 0 – 0.0592 1 log 15 400 = 0.0844 V = 84.4 mV Concentration of K + ions in the interior and exterior of a nerve cell are 400mM and 15mM respectively. Calculate the membrane potential. EXAMPLE (4):

6. p.06 Ex ercise A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has P H2 = 1.00 atm and an unknown [H + ]. Electrode 2 is a S.H.E. At 298K, the measured voltage is 0.211 V, and the electrical current is observed to flow from electrode 2 through the external circuit to electrode 1. What is the pH of the solution at electrode 1? e - released from electrode 1 (anode)!  [H + ] is higher in electrode 2 (cathode)! Electrode 1 (anode): H 2 (g, 1atm)  2H + (aq, xM) + 2e - Electrode 2 (cathode): 2H + (aq, 1M ) + 2e -  H 2 (g, 1atm) 0.211 = 0 – 0.0592 2 log x 2 (1) x = 2.73  10 -4 M (1) 2 (1) pH = 3.56

7. p.07 Understand more about the Nernst Equation: E = E – 0.0592 n log [product] [reactant] As electrons flow from the anode to the cathode, [reactant] , [product] .  Value of Q  At eqm, no net transfer of e -.  E = 0, then Q = K eq. Given = +0.154 V, Sn 4+ |Sn 2+ E Fe 3+ |Fe 2+ E = +0.769 V Find the K eq for the following rxn at 298K: Sn 2+ + 2Fe 3+  Sn 4+ + 2Fe 2+ At eqm, E = 0 V  0 = (+0.769 – 0.154) – [Sn 4+ ][Fe 2+ ] 2 [Sn 2+ ][Fe 3+ ] 2 log 0.0592 2 0 = 0.615 – K eq log 0.0296 K eq = 5.98  10 20

8. p.08 Ex ercise Sn 4+ |Sn 2+ Calculate the concentration of Sn 4+ ion in solution with 1.00M Sn 2+ ion in a half-cell which would have a zero potential when connected to a S.H.E. Would Sn 2+ ion tend to be oxidized or would Sn 4+ tend to be reduced under these conditions? Given: E = +0.13 V Sn 4+ + H 2  Sn 2+ + 2H +  0 = (+0.13 – 0) – (1)(1) 2 [Sn 4+ ](1) log 0.0592 2 [Sn 4+ ] = 4.06  10 -5 M Sn 2+ would NOT tend to be oxidized, and Sn 4+ would NOT tend to be reduced, because E = 0 V. Eqm is reached.

9. p.09 Factors affecting the values of E and E cell Conc. of electrolyte: ** eqm position would be shifted in accordance to the Le Chatelier’s Principle. Temperature ** For redox rxns having E cell > 0, i.e.  H 0, i.e.  H < 0 If temp. , eqm position would be shifted BW to absorb excessive heat. (i.e. e.m.f.  ) Pressure ** affect the eqm position of systems involving gaseous species.

10. p.10Summary By using the Nernst Equation: ** we can calculate the values of E or E cell under non-standard conditions (conc.  1M) at 298K. ** calculate the concentration ratio of R.A. and O.A. with a given E or E cell at 298K. ** calculate the value of K eq when E = 0 V. (i.e. no net e- transfer occurs and an eqm is reached.) Concentration Cell and Membrane Potential The values of E and E cell are affected by [electrolyte], temperature and pressure.

11. p.11Assignment p.214 Check Point 20.6 [due date: 18/5 (Mon)] Next …. Primary Cells & Secondary Cells (p. 215-216) p.229 Q.3(a), 7, 8(c), 9(a), 16, 20(b), 25 [due date: 13/5 (Wed)]