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4-5-2011

4-5-2011. Calculation of the standard emf of an electrochemical cell. The procedure is simple: Arrange the two half reactions placing the one with the larger E o up (the cathode). The half reaction with the lower E o is placed down (the anode). E o cell = E o cathode - E o anode.

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4-5-2011

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  1. 4-5-2011

  2. Calculation of the standard emf of an electrochemical cell The procedure is simple: • Arrange the two half reactions placing the one with the larger Eo up (the cathode). • The half reaction with the lower Eo is placed down (the anode). • Eocell = Eocathode - Eoanode

  3. A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 and a Ag electrode in a 1.0 M AgNO3 solution. Find the overall reaction and the standard cell emf. From table of standard reduction potentials, we have: Ag+ + e g Ag(s) Eo = 0.80V Mg2+ + 2e g Mg(s) Eo = -2.37V From diagonal rule Ag+ will react with Mg(s), therefore the reaction will be as follows:

  4. Multiply the first half reaction by 2 to account for the number of electrons in the second one and reverse the second half reaction: 2Ag+ + 2eg2Ag(s) Eo = 0.80V Mg(s) g Mg2+ + 2eEo = +2.37V 2Ag+ + Mg(s) g Mg2+ + 2Ag(s) Eo = 3.17V Eorxn = Eocathode - Eoanode Eorxn = 0.80 – (- 2.37) = 3.17V Remember to use Eo values with sign as it appears in the table (reduction potential).

  5. Cd2+(aq) + 2e- Cd (s)E0 = -0.40 V Cr3+(aq) + 3e- Cr (s)E0 = -0.74 V Cr (s) Cr3+ (1 M) + 3e- 2e- + Cd2+ (1 M) Cd (s) What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? x 2 Anode (oxidation): x 3 Cathode (reduction): 2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)

  6. E0 = Ecathode - Eanode E0 = -0.40 – (-0.74) E0 = 0.34 V cell cell cell 0 0 Remember that multiplying a half reaction with any coefficient will not change the value of Eo.

  7. DG0 = -nFEcell 0 0 0 0 0 = -nFEcell Ecell Ecell Ecell F = 96,500 J RT V • mol ln K nF (8.314 J/K•mol)(298 K) ln K = 0.0257 V 0.0592 V log K ln K n (96,500 J/V•mol) = n n = = Spontaneity of Redox Reactions DG = -nFEcell n = number of moles of electrons in reaction = 96,500 C/mol DG0 = -RT ln K

  8. 0 Ecell 0 0 2+ + E0 = -0.44 – (0.80) E0 = EFe /Fe – EAg /Ag E0 E0 = -1.24 V cell e x n = e K = 0.0257 V 0.0257 V -1.24 V x2 0.0257 V ln K n = What is the equilibrium constant for the following reaction at 250C? Fe2+(aq) + 2Ag (s)D Fe (s) + 2Ag+(aq) From the reaction as written, Fe half cell is the cathode, and Ag half cell is the anode: 2Ag+ + 2e g Ag(s) Eo = 0.80V Fe2+ + 2e g Fe(s) Eo = -0.44V From diagonal rule, the rxn is non spontaneous n = 2 K = 1.23 x 10-42

  9. Calculate the equilibrium constant for the following reaction at 25 oC: Sn(s) + 2Cu2+D Sn2+ + 2Cu+ From the reaction as written, Cu half cell is the cathode, and Sn half cell is the anode: 2Cu2+ + 2e g 2Cu+ Eo = 0.15V Sn2+ + 2e g Sn(s) Eo = -0.14V According to the diagonal rule the reaction occurs spontaneously as written: Eocell = Eocathode (Cu2+, Cu+) – Eoanode(Sn2+, Sn(s)) Eocell = 0.15 – (-0.14) = 0.29 V

  10. e K = K = e 0.0257 E0 0.29 cell x n x 2 0.0257 V K = 6.3*109 K = 6.3*109

  11. Calculate the standard free energy for the following reaction at 25 oC: 2Au(s) + 3Ca2+D2Au3+ + 3Ca(s) First, we write the two half cells and calculate the cell potential as they appear in the reaction. From the reaction as written, Ca half cell is the cathode, and Au half cell is the anode: 2Au3+ + 6e g 2Au(s) Eo = +1.50V anode 3Ca2+ + 6e g 3Ca(s) Eo = -2.87V cathode According to the diagonal rule the reaction will not occur spontaneously as written: Eocell = Eocathode– Eoanode Eocell = -2.87 – 1.50 = -4.37 V DGo = - nFEocell DGo = - (6)(96500)(-4.37) = 2.53*106 J DGo = 2.53*103 kJ The large +ve DGo suggests a non-spontaneous reaction at the standard states.

  12. DG0 = -nFE 0 RT nF E = E0 - ln Q 0 0 E = E = E E 0.0257 V 0.0592 V log Q ln Q n n - - The Effect of Concentration on Cell Emf DG = DG0 + RT ln Q DG = -nFE -nFE = -nFE0+ RT ln Q Nernst equation At 298 K

  13. Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+(aq) + Cd (s) Fe (s) + Cd2+(aq) 0 0 2+ 2+ E0 = -0.44 – (-0.40) E0 = EFe /Fe – ECd /Cd E = -0.04 V E0 = -0.04 V 0.010 0.60 0 E = E 0.0257 V 0.0257 V ln ln Q 2 n - - From the reaction as written, Fe half cell is the cathode, and Cd half cell is the anode, use these to calculate the standard electrode potential: We cannot use Eo to predict whether the rxn is spontaneous or not since we do not have standard conditions E = 0.013 E > 0 Spontaneous

  14. Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.68 M and [Co2+] = 0.15 M? Fe2+(aq) + Co (s) Fe (s) + Co2+(aq) 0 0 2+ 2+ E0 = -0.44 – (-0.28) E0 = EFe /Fe – ECo /Co E0 = -0.16 V 0 E = E 0.0257 V ln ([Co2+]/[Fe2+]) n - From the reaction as written, Fe half cell is the cathode, and Co half cell is the anode, use these to calculate the standard electrode potential:

  15. E = -0.16 V 0.15 0.68 0.0257 V 0.0257 V 0 = - 0.16 ln ([Co2+]/[Fe2+]) ln 2 n - - E = - 0.14 E < 0 a Non-spontaneous Now think at what ratio of [Co2+]/[Fe2+] will the reaction be spontaneous? To find such a ratio we must pass by the point where Eocell = 0 [Co2+]/[Fe2+] = 4*10-6, the ratio must be less than 4*10-6

  16. Concentration Cells Galvanic cell from two half-cells composed of the same material but differing in ion concentrations.

  17. Selected Problems 2, 3, 6, 11-13, 15, 16, 17, 18, 21, 22, 24, 25, 29, 32, 34.

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