Description of a Galvanic Cell. Describe completely the galvanic cell based on the following half-reactions under standard conditions:   Ag+ + e-  Ag Eocell = 0.80 V (1) Fe3+ + e-  Fe2+ Eocell = 0.77 V (2) In addition, draw the cell and write the line notation.

Cell Potential, Electrical Work The work that can be accomplished when electrons are transferred (emf) is defined in terms of a potential difference (in volts) between two circuits. emf = potential difference (V) = work (J) charge (C) or 1 V = 1 J/C Work is viewed from the point of view of the system. Therefore, E = -w/q or -w = q E

Maximum work The maximum work is defined as wmax = -q Emax *Achieving maximum work is impossible. In any real, spontaneous process, some energy is always wasted. The actual work realized is always less than the calculated maximum.

Free Energy The Faraday (F) is defined as the charge of 1 mole of electrons. F = 96,485 C/mol e- The purpose of a voltaic cell is to convert the free energy change of a spontaneous reaction into the KE of electrons moving through an external circuit. wmax = DG

For a galvanic cell DG = -n FEmax or DGo = -n FEomax Where n is the number of moles of electrons transferred in the redox reaction. Note the units of F is C/mol, and the units of E, volts, is J/C -nFE = (mol)(C/mol )(J/C) = J This equation is in Joules, not Kilojoules!!

Spontaneity If Ecell > 0, then DG < 0 and the process is spontaneous. If Ecell < 0, then DG > 0 and the process is nonspontaneous. If Ecell = 0, then DG = 0 and the process is at equilibrium.

Calculating DGo for a Cell Reaction. Using the Standard Reduction Potential Chart, calculate DGo for the reaction Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq)   Is the reaction spontaneous? Predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au3+ solution. Below is the half reaction of nitric acid acting as an oxidizing agent 4 H+ + NO3- + 3 e-  NO + 2 H2O Eo = .96 V

The Relationship to the Equilibrium Constant DGo = -R T ln K and DGo = -n FEocell  therefore -R T ln K = -n F Eocell When cadmium metal reduces Cu2+ in solution, Cd2+ forms in addition to copper metal. If DGo = -143 kJ, calculate K at 25o C. What would be Eocell in a voltaic cell that used this reaction?

The Effects of Concentration on E For the cell reaction 2Al(s) + 3Mn2+(aq) ⇌ 2Al3+(aq) + 3Mn(s) Eocell = 0.48 V   predict whether Ecell is larger or smaller than Eocell for the following cases. a) [Al3+] = 2.0 M, [Mn2+] = 1.0 M b) [Al3+] = 1.0 M, [Mn2+] = 3.0 M

Nernst Equation The Nernst equation gives the relationship between the cell potential and the concentrations of cell components. E = Eo – RT ln Q nF Calculations with this equation have been removed from the AP test, however, qualitative understanding of how concentration affects a cell are still on the test.

The Effect of Q A cell in which the concentrations are not in their standard states will continue to discharge until equilibrium is reached. Q values are never negative. For Q values less than 1, the concentration of reactant are higher than product initially, and the reaction will shift to the right to reach equilibrium, increasing the Ecell compared to Eocell. When Q < 1, [reactant] > [product], and therefore Ecell > Eocell.

Continued effect of Q For Q values higher than 1, the concentration of product are higher than reactant, and the reaction will shift to the left, decreasing the Ecell. When Q > 1, [reactant] < [product], and therefore Ecell < Eocell. When Q = 1, [reactant] = [product], Ecell = Eocell.

K and Q At equilibrium, K = Q and Ecell = 0. At equilibrium, the components in the two cell compartments have the same free energy, and DG = 0. At this point, the cell can no longer do work!

Problems Consider a cell based on the reaction   Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu (s) If [Cu2+] = .30 M, what [Fe2+] is needed to increase Ecell by 0.25 V above Eocell at 25o C? [Fe2+] would have to be 4.3x10^-9 M, which shows to make large voltage changes you need exceptionally large changes in concentration. Nearst equations problems generally have small changes in electric potential.

Describe the cell based on the following half-reactions: VO2+ + 2H+ + e-  VO2+ + H2O Eo = 1.00 V Zn2+ + 2e-  Zn Eo = -0.76 V   where T = 25o C [VO2+ ] = 2.0 M [H+ ] = 0.50 M [VO2+ ] = 0.01 M [Zn2+ ] = 0.1 M

Concentration Cells Concentration cells are constructed with the exact same half-reactions, with the exception of a difference in concentrations. Voltages are typically small as electrons are transferred from the cell of higher concentration to the cell of lower concentration. Eocell = 0.00 V, but these never deal with standard conditions because concentrations are not standard (not 1 M).

Calculating the Potential of a Concentration Cell. A concentration cell is built using two Au/Au3+ half-cells. In half-cell A, [Au3+] = 7.0 x 10-4 M, and in half-cell B, [Au3+] = 2.5 x10-2 M. What is Ecell, and which electrode is the anode?