zn + fe2+ gives zn2+ + fe. the value of kc for this reaction is 10^23 calculate the standard free energy change

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Solution

The free energy change of the reaction in any state, ΔG (when equilibrium has not been attained) is related to the standard free energy change of the reaction, ΔG0 (which is equal to the difference in free energies of formation of the products and reactants both in their standard states) according to the equation.
ΔG = ΔGo + RT InQ
Where Q is the reaction quotient
When equilibrium is attained, there is no further free energy change i.e. ΔG = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes.
ΔGo = –RT In K(eq)
or ΔGo = –2.303 RT log K(eq)
In case of galvanic cells. Gibbs energy change ΔG is related to the electrical work done by the cell.
ΔG = -nFE(cell) where n = no. of moles of electrons involved
F = the Faraday constant
E = emf of the cell
If reactants and products are in their standard states ΔGo = –nFEocell.
Here ,
ΔGo = –2.303 RT log K , Kc = 10^23. R and T unknown.
Log10^23 = 23.
∆Go = -52.969RT

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$K_{c}$ for the reaction, $F e^{2 +} + Z n \to Z n^{2 +} + F e$ is $10^{23}$ , The standard free energy change is

{A g}^{+} (a q .) + {F e}^{2 +} (a q .) ⟶ {F e}^{3 +} (a q .) + A g (s)

0.028 V

E^{o}

Z n \to Z n^{2 +} + 2 e^{-}; E^{o} = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E^{o} = + 0.41 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

E^{o}

Z n ⟶ {Z n}^{2 +} + 2 e^{-}

E^{o} = + 0.76 V

F e ⟶ {F e}^{2 +} + 2 e^{-}

E^{o} = + 0.41 V

{F e}^{2 +} + Z n ⟶ {Z n}^{2 +} + F e

E^{⊝}

Z n \to Z n^{2 +} + 2 e^{-}; E^{⊝} = + 0.76 V

F e \to F e^{2 +} + 2 e^{-}; E^{⊝} = 0.41 V

F e^{2 +} Z n \to Z n^{2 +} + F e

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