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Question
Calculate ΔrG° and log KC of the reaction:
\[\ce{Fe{^{2+}_{(aq)}} + Ag{^{+}_{(aq)}} -> Fe{^{3+}_{(aq)}} + Ag_{(s)}}\]
Given `E_(Ag^+//Ag)^° = 0.80 V, E_(Fe^(3+)//Fe^(2+))^° = 0.77 V`
[R = 8.314 J K−1, F = 96500 C mol−1]
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Solution
Given: `E_(Ag^+//Ag)^° = 0.80 V`
`E_(Fe^(3+)//Fe^(2+))^° = 0.77 V`
Formula: The standard Gibbs free energy change formula is:
ΔrG° = −nFE°cell
where:
n = number of moles of electrons transferred (in this case, n = 1, as one electron is transferred in the reaction).
F = Faraday constant (96500 C mol−1)
E°cell = standard cell potential
The standard cell potential (E°cell) is calculated using the standard reduction potentials.
`E°_"cell" = E°_"cathode" − E°_"anode"`
For this reaction:
\[\ce{Fe{^{2+}_{(aq)}} + Ag{^{+}_{(aq)}} -> Fe{^{3+}_{(aq)}} + Ag_{(s)}}\]
Ag+ is reduced (cathode).
Fe2+ is oxidised to Fe3+ (anode).
So,
= `E_(Ag^+//Ag)^° - E_(Fe^(3+)//Fe^(2+))^° `
= 0.80 V − 0.77 V
= 0.03 V
Now, substitute the respective values into the Gibbs free energy equation:
ΔrG° = −1 × 96500 C/mol × 0.03 V
= −2895 CV/mol
= −2895 J/mol
= −2.895 kJ/mol
The correlation between the standard cell potential and the equilibrium constant is described by the Nernst equation:
ΔrG° = −RT ln KC
Rearranging for log KC:
`log K_C = (nE_"cell"^°)/0.0591`
Substitute the known values:
`log K_C = (1 xx 0.03 V)/0.0591`
= 0.507
