Answers to problems in "Electrode Potentials"
Question 10
The reader is advised to re-read pp. 70-72 before considering this solution.Cell diagram:
Cd(amalgam, 4.6% Cd) | CdCl_{2}(aq, cM) | AgCl(s) | Ag(s) |
Potential determining equilibria:
RHE: |
AgCl(s) + e^{-} ⇌ Ag(s) + Cl^{-}(aq) |
LHE: |
½Cd^{2+} + e^{- ⇌ }½Cd(amal) |
Cell Reaction:
AgCl(s) + ½Cd(amal) ⇌ Ag(s) + ½Cd^{2}+(aq) + Cl-(aq)
The Nernst Equation for this cell reaction is: E = E^{0}_{cell} - {RT/F}ln {a_{Cl-} a_{Cd2+}^{½}}
(where the activity of cadmium in the amalgam is taken to be unity).
a_{Cl-} = f_{-}(2c) and a_{Cd2+} = f_{+}(c). Since c < 10^{-2}M, we can use the DHL. The ionic strength of the solution, I = ½[ 4c +2c] = 3c. Since this is the same for both Cd^{2+} and Cl^{-}, and since the ratio
z_{Cd2+}^{2}/z_{Cl-}^{2}
= 4, we can deduce that lg f_{+}/lg f_{-} = 4. ie.,
f_{+} = f_{-}^{4}. We can use this result
to determine E^{0}_{cell}. However, it is easier
and more convenient to use the mean ionic activity coefficient,
f_{±}, which is defined on p.48.
Hence, a_{Cl-} = f_{±}(2c)
= f_{±}[Cl^{-}] and a_{Cd2+}
= f_{±}(c) = f_{±}[Cd^{2+}]
Therefore, E = E^{0}_{cell} - {RT/F}ln{f_{±}^{3/2}
(2)(c)^{3/2}}
or,
E + {RT/F}ln{2c^{3/2}} = E^{0}_{cell} - {3 RT/ 2F}ln{f_{±}} | (*) |
or, using DHL, E + {RT/F}ln{2c^{3/2}} = E^{0}_{cell} - {6.909 RT A 3^{1/2} / 2F} c^{1/2}
We plot a graph of E + {RT/F}ln{2c^{3/2}} against c^{1/2}, and extrapolate to find the ordinate intercept (ie. a value for E^{0}_{cell}). The cell emf when a_{Cd2+} = a_{Cl-} = 1 corresponds to E_{cell}.
c / 10^{-3} mol dm^{-3} | E / V | c^{1/2} / mol^{1/2} dm^{-3/2} | E + {RT/F}ln{2c^{3/2}} / V |
0.1087 |
0.9023 | 0.01041 | 0.56861 |
0.1269 |
0.8978 | 0.01126 | 0.57007 |
0.2144 |
0.8803 | 0.01464 | 0.57277 |
0.3659 |
0.8641 | 0.01913 | 0.57715 |
From graph, ordinate-intercept = E^{0}_{cell} = +0.559V
(a). E^{0}_{cell} = E^{0}_{Ag|AgCl} - E^{0}_{Cd|Cd(amal)}
Therefore, E^{0}_{Cd|Cd(amal}) = 0.2222
- 0.559 = -0.3369V
Now consider the cell:
Cd(amal, 4.6% Cd) | CdCl_{2}(aq, 0.5M) | Cd | E = -0.0534V |
For this cell, E^{0}_{cell }= E^{0}_{Cd|Cd2+} - E^{0}_{Cd2+|Cd(amal) }
Cell Reaction is:
Cd^{2+}(aq) + Cd(amal) Cd(s)
+ Cd^{2+}(aq)
Thus, the Nernst Equation becomes: E = E^{0}_{cell} + {RT/ 2F} ln {a_{Cd2+}/ a_{Cd2+}}
E = E^{0}_{cell }
Hence, E^{0}_{Cd|Cd2+} = -0.0534 + (-0.3369) = -0. 39V
Therefore, the S.E.P. for the Cd|Cd^{2+} couple is -0.39V
This result is consistent with the data supplied in Table 1.1 on p. 21.
(b). The Gibbs free energy of the formation of the
hydrated ion Cd^{2+}(aq) refers to the process:
But this process corresponds to the inverse of the Cd|Cd^{2+} S.E.P.
Since DG^{0} = -nFE^{0},
and this process corresponds to a two electron transfer (n = 2),
DG^{0} = -(2)(96 490)(0.3903)
DG^{0} = -75.32 kJ mol^{-1}
(c). Using the equation labelled with an asterix (*), we can deduce that the mean ionic activity coefficient (f_{±})_{ }of 0.2144 x 10^{-3} M CdCl_{2} is 0.701.