137 
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. 
1313 
Exercise in working out the number of electrons transferred in a redox reaction or in a halfreaction. Levine calls this number the “charge number”. (Review of material from General Chemistry). 
1314 
Use of D_{f}G^{o}
data to calculate D_{rxn}G^{o}, which is then converted to
an E^{o} value by using D_{rxn}G^{o}
= nFE^{o}. An important point of
this problem is the relationship between the magnitudes of the D_{rxn}G^{o}
and E^{o} values. An E^{o} value of 0.5 V corresponds to a large,
negative value of D_{rxn}G^{o}. 


1312 
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. 
1315 
A small thought problem involving E, E^{o}, ionic strength, and activity coefficients. 
1320 
Another small thought problem related to the Nernst equation 
1327 
An exercise based on the relationship E = E^{o}  (RT/nF) lnQ, which is the Nernst equation. The half reactions and overall reaction are shown in equation (13.33); the E^{o} 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 E^{o} 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 H_{2}(g) only because both the reduced (Ag) and oxidized (AgCl) forms of silver are solids and so have activities of 1. 
1329 
Another problem based on the Nernst equation, but this time the activity coefficients are included. The oxidation half reaction takes Cu(s) to Cu^{2+}(aq) and the reduction half reaction takes Hg_{2}SO_{4}(s) to 2Hg(l) and SO_{4}^{2}(aq), so that the net reaction is: Cu(s) + Hg_{2}SO_{4}(s) ® Cu^{2+}(aq) + 2Hg(l). The E° values for the two half reactions can be found in Table 13.1 on pg 419. 
1331 
An exercise in adding two halfcell E^{o} values together to get an E^{o} value for a third halfreaction. The the answer obtained by simply summing the two E^{o} values for the two halfreactions given is wrong. 
1332 
Estimation of the voltage for a simple cell from E^{o} values and from activity coefficients estimated using the Davies equation. 


1337 
A short T/F question that points out (a) that E° 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). 
1339 
Calculation of an interesting equilibrium constant from E^{o} values. Remember that D_{rxn}G^{o} = RT lnK° = nFE°. 
1340 
Exercise that demonstrates how D_{f}G^{o} values can be determined for ions. Remember that E^{o} for the reduction of H^{+}(aq), and D_{f}G^{o} and D_{f}H^{o} for H^{+}(aq) are all zero. 
13.41 
Problem in which the K_{sp} for PbI_{2} is determined from E^{o} values. (See pg 423 for a similar example). 


1346 
Problem that reviews the idea that knowing the equilibrium constant or (here) E° as a function of T allows calculation of DG°, DH°, DS°, and DC_{p}°, for the reaction. 
1368 
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 Daviesequation approximation of the activity coefficient is very close to the measured value. Remember that pH is defined as –log_{10}a_{H}+ rather than as –log_{10}[H^{+}] (or even –log_{10}m_{H}+). 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. 