Answers to problems in "Electrode Potentials"
Question 8
ERRATUM: K_{w} = a_{H+} a_{OH} and not as printed in the book.
Both cells A and B are Harned Cells
Pt  H_{2}(g)  HCl(aq)  AgCl(s)  Ag(s) 
RHE:  AgCl(s) + e^{} ⇌
Ag(s) + Cl^{}(aq)

LHE:  H^{+}(aq) + e^{ } ⇌
½H_{2}(g)

Cell Reaction:
AgCl(s) + ½H_{2}(g) ⇌ Ag(s) + H^{+}(aq) + Cl^{}(aq) 
Nernst Equation: E_{cell } = E^{0}_{cell}  {RT/F}ln{a_{H+} a_{Cl}},
for P_{H2} = 1 bar which approximates to 1 atm., and where E^{0}_{cell} = E^{0}_{AgAgCl} (by definition of E^{0}_{H+H2} = 0V).
Consider Cell A
a_{H+} = K_{w}/ a_{OH}
Hence, E_{cell} = E^{0}_{AgAgCl}
 {RT/F}ln{K_{w} a_{Cl}/ a_{OH}}
Now, a_{X} = f_{X}[X], where f_{X} is the activity coefficient of X (also given the lower case Greek symbol gamma) and, [Cl^{}] = [OH^{}] = c
Also, since c < 10^{2}M, the DebyeHückel Limiting Law (DHL) applies.
Since lgf_{i} = Az_{i}^{2}I^{½}, and I, z_{i}^{2} and A are all the same for both Cl^{} and OH^{}, f_{Cl} = f_{OH} and so,
E_{cell} = E^{0}_{AgAgCl}  {RT/F}lnK_{w}  (1) 
From the data given, we notice that E_{cell A }is not a function of concentration (as equation (1) predicts), and therefore,
1.0493 = E^{0}_{AgAgCl}  {RT/F}lnK_{w}  (2) 
We next need a value for E^{0}_{AgAgCl} which (if we are to find K_{w}must therefore be obtained from cell B.
Now we consider Cell B:
E_{cell} = E^{0}_{AgAgCl}
 {RT/F}ln{a_{H+} a_{Cl}}
a_{H+ }= f_{+}c and a_{Cl} = f_{}(2c)
Using the DHL, we see that f_{+} = f_{}
= f , since z_{i}^{2}, A(= 0.509 in water at 298K)
and I are the same for both H^{+} and Cl^{}, as
they coexist in a solution of ionic strength I.
Therefore, E = E^{0}_{AgAgCl}  {RT/F}ln{f^{2} 2c^{2}}
E + {RT/F}ln{2c^{2}} = E^{0}_{AgAgCl}  {4.606RT/F}lg(f)
E + {RT/F}ln{2c^{2}} = E^{0}_{AgAgCl}  4.606(0.509)RT(3/2)^{½} c^{½}/ F
Hence, a plot of E + {RT/F}ln{2c^{2}} (on the ordinate axis) against c^{½} (on the abscissa) will have an ordinate intercept equal to the Standard Electrode Potential for AgAgCl.
E/V  c/mol dm^{3}  c^{½}/ mol^{½} dm^{3/2}  E + {RT/F}ln{2c^{2}}/V 
0.578 
0.001  0.0316  0.2411 
04969 
0.005  0.0707  0.2426 
0.4624 
0.010  0.1000  0.2437 
From the graph, ordinate intercept = E^{0}_{AgAgCl} = +0.2395V
Thus, from (2), K_{w} = 2.00 x 10^{14} mol^{2} dm^{6}.
Of the course the true answer for K_{w} is only one half as big; students 'refining' their results beware!