Respuesta :
Answer:
Equilibrium Constant = (3.60 × 10⁶²)
Explanation:
The change in Gibb's free energy for a galvanic cell is given as
Δ = -nFE°
where n = number of electrons transferred = 5 for this reaction
F = Faraday's constant = 96500 C
E° = cell potential = +0.74 V
But the change in Gibb's free energy for galvanic cell reaction is also given as
ΔG = -RT In K
R = molar gas constant = 8.314 J/mol.K
T = absolute temperature in Kelvin = 25 + 273.15 = 298.15 K
K = Equilibrium constant.
Equating these two expressions
-nFE° = - RT In K
RT In K = nFE°
In K = (nFE°) ÷ (RT)
In K = (5 × 96500 × 0.74) ÷ (8.314 × 298.15)
In K = 144.04
K = e^(144.04)
K = (3.60 × 10⁶²)
Hope this Helps!!!
The equilibrium constant for the galvanic cell has been [tex]\rm \bold{e^1^4^4^.^0^4}[/tex].
Gibb's free energy has been defined as the reversible work performed by the thermodynamic system.
The Gibbs free energy can be given as:
[tex]\Delta[/tex]G = -nF[tex]\rm E^\circ[/tex]
The [tex]\Delta[/tex]G for the galvanic cell can also be given as:
[tex]\Delta[/tex]G = -RT ln K
Thus, for the galvanic cell:
-nF[tex]\rm E^\circ[/tex] = -RT ln K
There has been the transfer of 5 moles of electrons, thus:
- 5 mol × 96500 C × 0.74 V = - 8.314 J/mol.K × 298.15 K × In Equilibrium constant (K)
357,050 = 2,478.8191 K
In K = 144.04
K = [tex]\rm e^1^4^4^.^0^4[/tex]
The equilibrium constant for the galvanic cell has been [tex]\rm e^1^4^4^.^0^4[/tex].
For more information about the galvanic cell, refer to the link:
https://brainly.com/question/25606438
