Respuesta :
Answer : The concentration of [tex]X^+[/tex] is 0.0861 M
Explanation :
First we have to calculate the standard electrode potential of the cell.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{1}{[X]}[/tex]
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = [tex]25^oC=273+25=298K[/tex]
n = number of electrons in oxidation-reduction reaction = 1
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = ?
[tex]E_{cell}[/tex] = emf of the cell = 0.0460 V
[X] = concentration = 0.0482 M
Now put all the given values in the above equation, we get:
[tex]0.0460=E^o_{cell}-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{0.0482}[/tex]
[tex]E^o_{cell}=0.124V[/tex]
Now we have to calculate the concentration of [tex]X^+[/tex] when the potential is 0.0610 V.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{1}{[X]}[/tex]
[tex]0.0610=0.124-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{[X]}[/tex]
[tex][X]=0.0861M[/tex]
Therefore, the concentration of [tex]X^+[/tex] is 0.0861 M
