Comments re Chapter 13 Problems:


An exercise in converting charge per mole of ions to charge in Coulombs.  The needed constant is Faraday’s constant, 96485 Coulombs per mole of electrons.


Exercise in working out the number of electrons transferred in a redox reaction or in a half-reaction.  Levine calls this number the “charge number”.  (Review of material from General Chemistry).


Use of DfGo data to calculate DrxnGo, which is then converted to an Eo value by using DrxnGo = -nFEo. An important point of this problem is the relationship between the magnitudes of the DrxnGo and Eo values. An Eo value of 0.5 V corresponds to a large, negative value of DrxnGo.
Note that DfGo values for ions are needed to calculate DrxnGo.  There are entries for the necessary ions in Levine's table of thermodynamic properties.




The following problems depend on the Nernst equation.  It is derived by starting with DrxnG = DrxnGo + RT lnQ (an equation that was covered earlier in the semester as well as in General Chemistry),
and then using
DrxnG - -nFE (and DrxnG° - -nFE°).
Remember that Q has the same form as the equilibrium constant, but that the solution activities and/or gas pressures om Q do not necessarily correspond to equilibrium values that would be in K.


Another T/F exercise.  It can be argued that part (b) is false because the units of n are mol e- for the reaction as written but Levine disagrees.


A small thought problem involving E, Eo, ionic strength, and activity coefficients.


Another small thought problem related to the Nernst equation


An exercise based on the relationship E = Eo - (RT/nF) lnQ, which is the Nernst equation.  The half reactions and overall reaction are shown in equation (13.33);  the Eo value for the cell is 0 + 0.2222 V = 0.2222 V.  The point of the problem is that the term (RT/nF) lnQ is usually small relative to Eo unless the activities are very different from 1 or unless E° = 0 (as in the case of a concentration cell).  Note that Q contains terms for HCl(aq) and H2(g) only because both the reduced (Ag) and oxidized (AgCl) forms of silver are solids and so have activities of 1.


Another problem based on the Nernst equation, but this time the activity coefficients are included.  The oxidation half reaction takes Cu(s) to Cu2+(aq) and the reduction half reaction takes Hg2SO4(s) to 2Hg(l) and SO42-(aq), so that the net reaction is:  Cu(s) + Hg2SO4(s) ® Cu2+(aq) + 2Hg(l).  The E° values for the two half reactions can be found in Table 13.1 on pg 419.


An exercise in adding two half-cell Eo values together to get an Eo value for a third half-reaction. The the answer obtained by simply summing the two Eo values for the two half-reactions given is wrong.


Estimation of the voltage for a simple cell from Eo values and from activity coefficients estimated using the Davies equation.




A short T/F question that points out (a) that is an intensive variable, and (b) that the overall reaction in an electrochemical cell does not have to be a redox reaction (see “Concentration Cells” on pg 421 and “Example 13.7” on pg 423).


Calculation of an interesting equilibrium constant from Eo values.  Remember that DrxnGo = -RT lnK° = -nFE°. 


Exercise that demonstrates how DfGo values can be determined for ions. Remember that Eo for the reduction of H+(aq), and DfGo and DfHo for H+(aq) are all zero.


Problem in which the Ksp for PbI2 is determined from Eo values.  (See pg 423 for a similar example).




Problem that reviews the idea that knowing the equilibrium constant or (here) as a function of T allows calculation of DG°, DH°, DS°, and DCp°, for the reaction.


A problem that shows (1) that the pH of a 0.100 m solution of HCl in water is not 1.00, and (2) that the value calculated from the Davies-equation approximation of the activity coefficient is very close to the measured value.  Remember that pH is defined as –log10aH+ rather than as –log10[H+] (or even –log10mH+).  This problem is given in this chapter because pH meters are electrochemical cells and the quantity that is measured by a pH meter is a voltage.  The voltage is then converted to pH.




A straightforward problem in which E is to be calculated given the cell and the ion activities.  The only complication is that E for the cell is negative, which means that the anode on the left is really the cathode if the cell is run as a battery.  That switch reverses the direction of electron flow.  In answering the question remember that electrons flow from regions of lower electrical potential to higher electrical potential because the sign of the charge on the electron was chosen to be negative.