Electrochemistry Part IV: Spontaneity, Nernst Equation & Electrolysis Jespersen Chap. 20 Sec 4 to 7 Skipping Sec 6 & 8 Dr. C. Yau Spring 2013 1 1

Spontaneity of Reaction We know to have a spontaneous rxn… E > 0 ΔG < 0 How are these two related? ΔG = - n FEcell where n = moles of e- F = Faraday's constant 9.65x104 C/mol e- (remember 1 V = 1J/C) ΔGo = - n FEocell (under standard conditions) 2

NiO2 (s) + 2Cl(aq) + 4H+ (aq)  Cl2 (g) + Ni2+ (aq) + 2H2O (l) Example 20.7 p. 937 Calculate ΔGo for the reaction, given that its standard cell potential is 0.320 V at 25oC. NiO2 (s) + 2Cl(aq) + 4H+ (aq)  Cl2 (g) + Ni2+ (aq) + 2H2O (l) ΔGo = - n FEocell F = 9.65x104 C/mol e- (1 V = 1J/C, so Eo = 0.320 J/C) How do we figure out what n is? Ans. -61.8 kJ Do Pract Exer 13 & 14 p. 938

Calculating K from Cell Potential We know ΔGo = - n FEocell We also know ΔGo = - RT ln K (Chap. 19) so - n FEo = - RT ln K Example 20.8 p. 789 Calculate K for the reaction in Example 20.8. NiO2 (s) + 2Cl(aq) + 4H+ (aq)  2Cl2 (g) + Ni2+ (aq) + 2H2O (l) Collect all the constants we need. Do Pract Exer 15 & 16 p. 939

Derivation of the Nernst Eqn What happens when it is not under standard conditions? Divide both sides of eqn by (-nF) we get... ΔGo = - n FEocell ΔG = - n FEcell Nernst Equation

Common simplified version of the Nernst Equation for 25.0 oC: This version of Nernst Eqn will be given also, but remember it’s only for 25.0oC

Nernst Equation Eo is the cell potential under standard conditions (for aqueous soln, 1M) What if it is not 1 M? Example 20.9 p. 940 Suppose a galvanic cell employs the following: Ni2+ + 2e-  Ni Eo = - 0.25 V Cr3+ + 3e-  Cr Eo = - 0.74 V Calculate the cell potential when [Ni2+] = 4.87x10-4M and [Cr3+] = 2.48x10-3 M This type of quest will be on your final exam. Ans. +0.44 V

The rxn of tin metal with acid can be written as Example 20.10 p. 941 The rxn of tin metal with acid can be written as Sn (s) + 2H+ (aq)  Sn2+ (aq) + H2 (g) Calculate the cell potential (a) when the system is at standard state. (b) when the pH = 2.00 (c) when the pH is 5.00. Assume that [Sn2+] = 1.00 M and the partial pressure of H2 is also 1.00 atm. Do Pract Exer 17, 18, 20 p. 942 Ans. +0.02 V Ans. -0.16V

What we are skipping in Chap. 20: Concentration from E Measurements Sec 20.6 Electricity Batteries: Lead Storage Batteries Zinc-Manganese Dioxide Cells (LeClanche cell) Nickel-Cadmium Rechargeable Batteries Nickel-Metal Hydride Batteries Lithium Batteries Lithium Ion Cells Fuel Cells Photovoltaic Cells

Electrolytic Cell (Sec. 20.7) We had been considering 2 half-reactions, one of which is the "drive" for the forward reaction. In the electrolytic cell (or electrolysis cell), the driving force is the electric current. "Lysis" means to break apart. "Electrolysis" means to break with an electric current. We know 2Na + Cl2 2NaCl is spontaneous. We can do the reverse reaction by applying an electric current. 2NaCl  2Na + Cl2 What are the physical states? Usually the electrolytic cell has only one chamber.

Electrolysis in Aqueous Solutions Can we "break apart" potassium sulfate? It is not easy to melt an ionic compound. What if we did this in an aqueous solution? Water can be oxidized as well as reduced. Why? How? H2O  H2 reduction at the … H2O  O2 oxidation at the… Overall: 2 H2O (l)  2H2 (g) + O2 (g) This cannot take place without an electrolyte present (acting as a salt bridge). Do Pract Exer 23, 24 p. 958

Example 20.12 p. 957 Electrolysis is planned for an aq soln that contains a mixture of 0.50 M ZnSO4 and 0.50 M NiSO4. On the basis of standard reduction potentials, what products are expected to be observed at the electrodes? What is the expected net cell reaction? First determine which species are present. Next determine which might be reduced, which might be oxidized. (Remember those to be reduced are on the left side of Table 19.1, those to be oxidized are on the right side. Which is reduced the easiest on the table? Which is oxidized the easiest?)

Which is most easily reduced? Example 20.12 p. 957 (continued) Ni2+ + 2e-  Ni Eo = - 0.25 V Zn2+ + 2e-  Zn Eo = - 0.76 V 2H2O + 2e-  H2 + 2OH- Eo = - 0.83 V Which is most easily reduced? Now examine what is most easily oxidized. S2O82- + 2e-  2SO42- Eo = +2.01 V O2 + 4H+ + 4e-  2H2O Eo =+1.23 V What is the net cell rxn? What is the Eocell? Ans. Ni2+ Ans. H2O

Ni2+ + 2e-  Ni Eo = - 0.25 V O2 + 4H+ + 4e-  2H2O Eo =+1.23 V Example 20.12 p. 957 (continued) Ni2+ + 2e-  Ni Eo = - 0.25 V O2 + 4H+ + 4e-  2H2O Eo =+1.23 V 2H2O O2 + 4H+ + 4e- Eo =-1.23 V What is the net cell rxn? 2Ni2+ + 4e-  2Ni Eo = - 0.25 V 2H2O O2 + 4H+ + 4e- Eo = -1.23 V 2Ni2+ + 2H2O  2Ni + O2 + 4H+ Eo = -1.48 V Do Pract Exer 23, 24 p. 958 How can it be negative?

Easiest to be reduced Easiest to be oxidized

Periodic Table That’s what nonmetals do in redox rxns: X X- (reduction) Easiest to be reduced Periodic Table Easiest to be oxidized That’s what metals do (oxidation): M M+