Answer:
The equilibrium constant is [tex]K = 3.565*10^{-31}[/tex]
Explanation:
From the question we are told that
The reduction potential of [tex]Cr^{3-}[/tex] = [tex]-0.74V[/tex]
The reduction potential of [tex]Fe^{2+}[/tex] = [tex]-0.0440[/tex]
In order for us to obtain the equilibrium constant we need to understand the process of the reaction
At the cathode [tex]Cr^{3-}[/tex] is reduced to Cr the reaction is
[tex]2Cr^{3-} + 6e^ - ------> 2Cr[/tex]
At the anode [tex]Fe[/tex] is oxidized to [tex]Fe^{2+}[/tex] the reaction is
[tex]3Fe -----> 3Fe^{2+} + 6e^-[/tex]
The net voltage of the cell is mathematically given as
[tex]E_{cell} = E_{cathode} - E_{anode}[/tex]
substituting value
[tex]E_{cell} = -0.74 - (-0.440)[/tex]
[tex]= -0.30V[/tex]
Next we need to mathematically evaluate the free energy of the cell as
[tex]\Delta G = -n_e F E_{cell}[/tex]
where [tex]n_e[/tex] is the number of moles of electron which is 6
F is the farad constant with a value [tex]F = 96,500 C/mole[/tex]
Substituting values
[tex]\Delta G = - ( 6 * 96500 *-0.3)[/tex]
[tex]= 17,3700 J/mol[/tex]
The equilibrium constant is mathematically represented as
[tex]K = e ^{\frac{\Delta G}{-(RT)} }[/tex]
Where R is the gas constant with value 8.314 J/mol
T is the temperature with a given value of [tex]T =298K[/tex]
Substituting values
[tex]K = e ^{\frac{173700}{- (8.314 * 298)} }[/tex]
[tex]K = 3.565*10^{-31}[/tex]
