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REDOX EQM Nernst Equation &

p.01. C. Y. Yeung (CHW, 2009). reduced form. [product]. 0.0592. log. [reactant]. oxidized form. n. no. of e - involved in rxn. E = E –. 05. REDOX EQM Nernst Equation &. Factors affecting E and E cell. Nernst Equation:.

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REDOX EQM Nernst Equation &

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

  2. p.02 EXAMPLE (1): Predict whether the following reaction would proceed spontaneously at 298K: Co(s) + Fe2+(aq)  Co2+(aq) + Fe(s) Given: [Co2+(aq)] = 0.15M and [Fe2+(aq)] = 0.68M E = -0.44 V = -0.28 V E Co2+| Co Fe2+| Fe (anode) (cathode) E = -0.44 – (-0.28) = -0.16 V cell (0.15) 0.0592 Ecell = -0.16 – = -0.141 V log (0.68) 2 Co(s)  Co2+(aq) + 2e- Fe2+(aq) + 2e- Fe(s) Ecell < 0,  the reaction is NOT spontaneous in the direction written.

  3. p.03 EXAMPLE (2): In the following reaction, the e.m.f. of the cell is found to be +0.54V at 250C. Suppose that [Zn2+(aq)] = 1.0M and PH2(g)= 1.0 atm. Calculate the molarity of H+(aq). Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g) = -0.76 V E Zn2+| Zn (anode) (cathode) (1.0)(1.0) 0.0592 E = 0 – (-0.76) = +0.76 V log +0.54 = +0.76 – cell 2 [H+]2 Zn(s)  Zn2+(aq) + 2e- 2H+(aq) + 2e- H2(g)  [H+] = 1.92  10-4 M

  4. p.04 EXAMPLE (3): Consider an electrochemical cell in which zinc electrodes are put into two aq. solutions of ZnSO4(aq) at 0.10M and 1.0M concentrations. Write the cell diagram and calculate the cell e.m.f. = -0.76 V E Zn2+| Zn Given: Cell Diagram: Zn(s) | Zn2+(aq, 0.10M) Zn2+(aq, 1.0M) | Zn(s) 0.10 0.0592 log = 0.0296 V E = 0 – 2 1.0 Concentration Cell: an electrochemical cell from 2 half cells composed of the same material but differing in ion concentrations. Reduction should occur in the more concentrated compartment, therefore: ** 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 EXAMPLE (4): Concentration of K+ ions in the interior and exterior of a nerve cell are 400mM and 15mM respectively. Calculate the membrane potential. 15 0.0592 log E = 0 – 1 400 = 0.0844 V = 84.4 mV 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)

  6. p.06 e- released from electrode 1 (anode)!  [H+] is higher in electrode 2 (cathode)! x2(1) 0.0592 0.211 = 0 – log 2 (1)2(1) x = 2.73  10-4 M pH = 3.56 Exercise A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2 = 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? Electrode 1 (anode): H2(g, 1atm)  2H+(aq, xM) + 2e- Electrode 2 (cathode): 2H+(aq, 1M ) + 2e- H2(g, 1atm)

  7. p.07 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 = Keq. Given = +0.154 V, = +0.769 V Find the Keq for the following rxn at 298K: E Sn4+|Sn2+ Sn2+ + 2Fe3+ Sn4+ + 2Fe2+ [Sn4+][Fe2+]2 E log  0 = (+0.769 – 0.154) – Fe3+|Fe2+ [Sn2+][Fe3+]2 [product] 0.0592 log [reactant] n Keq log 0.0296 0 = 0.615 – 0.0592 E = E – 2 Keq = 5.98  1020 Understand more about the Nernst Equation: At eqm, E = 0 V

  8. p.08 Calculate the concentration of Sn4+ ion in solution with 1.00M Sn2+ ion in a half-cell which would have a zero potential when connected to a S.H.E. Would Sn2+ ion tend to be oxidized or would Sn4+ tend to be reduced under these conditions? Given: E (1)(1)2 log  0 = (+0.13 – 0) – [Sn4+](1) 0.0592 2 Exercise = +0.13 V Sn4+|Sn2+ Sn4+ + H2 Sn2+ + 2H+ [Sn4+] = 4.06  10-5 M Sn2+ would NOT tend to be oxidized, and Sn4+ would NOT tend to be reduced, because E = 0 V. Eqm is reached.

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

  10. p.10 Summary By using the Nernst Equation: ** we can calculate the values of E or Ecell under non-standard conditions (conc.  1M) at 298K. ** calculate the concentration ratio of R.A. and O.A. with a given E or Ecell at 298K. ** calculate the value of Keq 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 Ecell are affected by [electrolyte], temperature and pressure.

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

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