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Hi Everyone,
 
I am not entirely sure about my solution to question 10 (part c). I have briefly explained how I came up with my solution and the online source which gave me the answer to the question. The question, however, did not specifically say what the values of E and Eocell are. Your help will be greatly appreciated.
 
The answer to the question is available at
http://compton.chem....wers&number=ten
 
(10) At 298 K and at one atmosphere pressure, the EMF, E, of the cell,
 
             Cd (amalgam)(4.6% Cd) | CdCl₂ (aq) (c M) | AgCl (s) | Ag
 
varies with the concentration, c, of CdCl₂ as follows
 
              c/10^-3 mol dm^-3    0.1087    0.1296    0.2144    0.3659
             E/V                        0.9023    0.8978    0.8803    0.8641                           
 
Write down the Nernst equation for this cell and, using a Debye-Huckel extrapolation procedure, deduce the cell EMF when both Cd²⁺ and Cl^- are at unit activity in the solution.
 
      Under the same conditions, the EMF of the cell
 
             Cd (amalgam)(4.6% Cd) | CdCl₂ (aq) (0.5 M) | Cd
 
is -0.0534 V. The standard electrode potential of the silver/silver chloride electrode is 0.2222 V. The standard electrode potential of the Cd/Cd2+ couple and the Gibbs free energy of formation, ∆G^Ø_f, of Cd²⁺​. The question
I am stuck on is the one where I have to determine the mean ionic activity coefficient of 0.2144 x 10^-3M CdCl² (part c). The answer should be f±=0.701.
My solution:
From the equation below
E + {RT/F}ln{2c^3/2} = E⁰_cell - {3 RT/ 2F}ln{f±}
 
I substituted in what I think the values for the parameters are
-0.0534+(8.313x298/96490)ln{2(0.2144x10-3)^3/2}=E^o_cell+(3.4545x8.313x298/96490)(√3)*(0.2144x10-3)^1/2
 
Hence,
E^o_cell=-0.3632
 
-0.0534+(8.313x298/96490)ln{2(0.2144x10-3)^3/2}=-0.3632-(3x8.313x298/(2x96490))xln{f±}
=>f±=0.943
 
Thanks.