Free Energy ∆G & Nernst Equation [ ]

Cell Potentials (emf) Zn  Zn e volts Cu e-  Cu volts Cu +2 + Zn  Cu + Zn volts Zn │Zn +2 ║Cu +2 │Cu line notation of eCells When E o >0 redox reaction is a spontaneous process Anode: dissolves/loses mass Salt bridge/porous disc Cathode: gains mass

Gibbs Free Energy & eCells The maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell. ∆G o = -n₣ Ɛ o # of moles transferred per mole of reactant & product Faraday = 96,485 coulombs of charge per mole of electrons Volts = work (J) / charge (C)

For example: Cu +2 + Mg ↔ Mg +2 + Cu Cu e- ↔ Cu 0.34 volts Mg ↔ Mg e volts Cu +2 + Mg ↔ Mg +2 + Cu volts ∆G = -(2 mole)(96485 C/mol e-)(2.71 J/C) ∆G = x 10 5 J ***some energy is always wasted, always less than theoretical value

Cell Potential & Concentration Ɛ o = 0.48 volts under standard conditions (1.0 M) 2 Al + Mn +2 → 2 Al Mn a.[Al +3 ] = 2.0 M; [Mn +2 ] = 1.0 M *reverse reaction favored; decreases Ɛ o < 0.48 volts b. [ Al +3 ] = 1.0 M; [Mn +2 ] = 3.0 M *forward reaction favored; increases Ɛ o > 0.48 volts

Concentration Cells *These cells typically produce small voltages

Nernst Equation Remember that  G =  G  + RT ln Q This means −nFE = −nFE  + RT ln Q

Nernst Equation Dividing both sides by −nF, we get the Nernst equation: E = E  − RT nF ln Q or, using base-10 logarithms, E = E  − RT nF log Q

Nernst Equation At room temperature (298 K), Thus the equation becomes E = E  − n log Q RT F = V

Concentration Cells Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell,would be 0, but Q would not. E cell  Therefore, as long as the concentrations are different, E will not be 0.

SAMPLE EXERCISE Determining  G ° and K Plan: We use the data in Table 20.1 and Equation to determine E° for the reaction and then use E° in Equation to calculate  G°. We will then use Equation 19.22,  G° = –RT in K, to calculate K. What are the values of E°,  G°, and K when the reaction is written in this way? (a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change,  G°, and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction (b) Suppose the reaction in part (a) was written Solve: (a) We first calculate E° by breaking the equation into two half- reactions, as we did in Sample Exercise 20.9, and then obtain values from Table 20.1 (or Appendix E): Even though the second half-reaction has 4 Ag, we use the value directly from Table 20.1 because emf is an intensive property. Using Equation 20.10, we have

SAMPLE EXERCISE K is indeed very large! This means that we expect silver metal to oxidize in acidic environments, in air, to Ag +. Notice that the voltage calculated for the reaction was 0.43 V, which is easy to measure. Directly measuring such a large equilibrium constant by measuring reactant and product concentrations at equilibrium, on the other hand, would be very difficult. The half-reactions show the transfer of four electrons. Thus, for this reaction n = 4. We now use Equation to calculate  G°: The positive value of E° leads to a negative value of  G°. Now we need to calculate the equilibrium constant, K, using  G° = –RT ln K. Because  G° is a large negative number, which means the reaction is thermodynamically very favorable, we expect K to be large.

SAMPLE EXERCISE continued Comment: E° is an intensive quantity, so multiplying a chemical equation by a certain factor will not affect the value of E°. Multiplying an equation will change the value of n, however, and hence the value of  G°. The change in free energy, in units of J/mol of reaction as written, is an extensive quantity. The equilibrium constant is also an extensive quantity. (b) The overall equation is the same as that in part (a), multiplied by The half-reactions are The values of are the same as they were in part (a); they are not changed by multiplying the half- reactions by Thus, E° has the same value as in part (a): Notice, though, that the value of n has changed to n = 2, which is the value in part (a). Thus,  G° is half as large as in part (a). Now we can calculate K as before:

SAMPLE EXERCISE Voltaic Cell EMF Under Nonstandard Conditions Solution Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 when [Cr 2 O 7 2– ] = 2.0 M, [H + ] = 1.0 M, [I – ] = 1.0 M, and [Cr 3+ ] = 1.0  10 –5 M. Solve: We first calculate E° for the cell from standard reduction potentials (Table 20.1 or Appendix E). The standard emf for this reaction was calculated in Sample Exercise 20.6: E° = 0.79 V. As you will see if you refer back to that exercise, the balanced equation shows six electrons transferred from reducing agent to oxidizing agent, so n = 6. The reaction quotient, Q, is Using Equation 20.16, we have

SAMPLE EXERCISE continued Check: This result is qualitatively what we expect: Because the concentration of Cr 2 O 7 2– (a reactant) is greater than 1 M and the concentration of Cr 3+ (a product) is less than 1 M, the emf is greater than E°. Q is about 10 –10, so log Q is about –10. Thus, the correction to E° is about 0.06  (10)/6, which is 0.1, in agreement with the more detailed calculation.

SAMPLE EXERCISE Concentrations in a Voltaic Cell If the voltage of a Zn–H + cell (like that in Figure 20.11) is 0.45 V at 25°C when [Zn 2+ ] = 1.0 M and atm, what is the concentration of H + ? Solution Plan: First, we write the equation for the cell reaction and use standard reduction potentials from Table 20.1 to calculate E° for the reaction. After determining the value of n from our reaction equation, we solve the Nernst equation for Q. Finally, we use the equation for the cell reaction to write an expression for Q that contains [H + ] to determine [H + ]. Solve: The cell reaction is The standard emf is Because each Zn atom loses two electrons,

SAMPLE EXERCISE continued Using Equation 20.16, we can solve for Q: Q has the form of the equilibrium constant for the reaction Solving for [H + ], we have Comment: A voltaic cell whose cell reaction involves H + can be used to measure [H + ] or pH. A pH meter is a specially designed voltaic cell with a voltmeter calibrated to read pH directly. (Section 16.4)

SAMPLE EXERCISE pH of a Concentration Cell A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has atm and an unknown concentration of H + (aq). Electrode 2 is a standard hydrogen electrode ([H + ] = 1.00 M, atm). At 298 K the measured cell voltage is V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate[H + ] for the solution at electrode 1. What is its pH? Solve: Using the Nernst equation, we have Because electrons flow from electrode 1 to electrode 2, electrode 1 is the anode of the cell and electrode 2 is the cathode. The electrode reactions are therefore as follows, with the concentration of H + (aq) in electrode 1 represented with the unknown x:

SAMPLE EXERCISE continued Comment: The concentration of H + at electrode 1 is lower than that in electrode 2, which is why electrode 1 is the anode of the cell: The oxidation of H 2 to H + (aq) increases [H + ] at electrode 1. Thus, At electrode 1, therefore, and the pH of the solution is