c14 JWBS043-Rogers September 13, 2010 11:27 Printer Name: Yet to Come
PROBLEMS AND EXAMPLES 231
Ag;AgCl electrode in the cell
Pt(s); H 2 (g,1 atm); HCl(aq=?); AgCl(s); Ag(s)
to determine the concentration of HCl(aq=?). If the cell potential for HCl(aq=?)
is 493 mV, what is the pH of the solution in which the electrodes are immersed?
Solution 14.1 The cell reaction is
AgCl(s)+^12 H 2 (g)←→H+(aq)+Cl−(aq)+Ag(s)
The Nernst equation is
E=E◦−
0. 0592
1
logmHCl= 0 .493 volts
because of the unit activities of Ag(s) and AgCl(s) andpH 2 =1 atm. We knowE◦
and we takemH+=mCl−because each HCl breaks up to give one of each kind of ion.
We t a kemas the geometric mean of the ion concentrationsmHCl=mH+mCl−=m^2 H+:
0. 493 = 0. 222 − 0 .0592 logm^2 H+
0. 493 − 0. 222 = 2 ( 0. 0592 )(−logmH+)
0. 271 =( 0. 1184 )(−logmH+)=( 0. 1184 )(pH)
pH= 0. 271 / 0. 1184 = 2. 29
Comment:The pH is not carried out to the same five-digit accuracy that the Ag;AgCl
electrode presents because a number of approximations appear in the development of
the pH equation. One is that the ionic interference between the dissociated HCl ions
is not taken into account because the ions are very dilute and, second, dissolution of
AgCl causes a slight imbalance between the H+and Cl−ions.
Example 14.2 A Mean Activity Coefficient
If a silver chloride–hydrogen cell withmHCl= 0 .1000 produces an electrical potential
ofE= 0 .353, what is the mean activity coefficient of H+at this molarity?
Solution 14.2 Notice that logmHCl=log(0.1000)=−1, so
logγ±=
−E+E◦− 0 .1184 logmHCl
0. 1184