A student measures Ecell of a concentration cell created by having the saturated solution of AgBr on one side and 0.10 M Ag+(aq) on the other (along with silver electrodes). The experimental value is 0.603 V. The student is told to use the theoretical value of the Nernst plot (−0.0591 V). (The concentration of the halide is also 0.10 M.) (a) In which direction will electrons spontaneously flow? from the 'saturated side' to the '0.10 M side' from the '0.10 M side' to the 'saturated side' the electrons will not flow in this cell (b) What is the value of [Ag+]dilute? M (c) What is the experimental Ksp? (d) Calculate ΔG° in kJ/mol using ΔG° = −RTlnKsp. (R = 8.314 J/K · mol, T = 298 K.) kJ/mol

Complete Question

The complete question is shown on the first uploaded image  

a

  From  the saturated side to the '0.10 M ' side

b

 The diluted is  

c

 The experimental values of is    

d

Explanation:

   From the question we are told that

        The concentration of is  

        The experiment value is

         The value of the  Nernst plot is

         The concentration of the halide is

From the question we are told that one side has a saturated and the other side has a ion 0.10M  Ag+(aq) this will cause a difference in potential causing the electron to flow from the the saturated side to the side with Ag+(aq) in other to reduce  it

 The equation for the reaction is given as.

This values are constant potential  data for standard reduction of AgBr

      The overall equation of reaction in the cell is  

This potential for the cell can also be represented as

Where  is the number of moles of electron which 1 from the chemical equation

is the solubility product which is mathematically represented as

So  

=>  

=>    

  Now

So

The Free energy ΔG° is mathematically represented as

Where R is the gas constant with value  

        So

Converting to KJ