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[Dudesons123]

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 E^o_cell 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^-3 M 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

0.8803+(8.313x298/96490)ln{2(0.2144x10^-3)^3/2}=0.559-{3 ((8.313)(298)/ 2(96490)}ln{f_±}

 

Or by using the graph, knowing that c^1/2=0.2144x10^-3, we can find E + {RT/F}ln{2c^3/2}=0.57277). From here, the Nernst equation can be used to find the mean ionic activity coefficient, f_±.

 

^Hence,

f_±=0.700

 

Thanks.

Edited by Dudesons123, 11 June 2018 - 07:07 AM.

[Dudesons123]

Help will be greatly appreciated.

[Dudesons123]

Help me