Find the value of the equilibrium constant and standard Gibbs free energy change at 25 °C for the following cell; Cu|Cu+2(1 M)||Ag+(1 M)|Ag. (Given: E°cell(Ag+|Ag)=0.799 V, E°cell(... Find the value of the equilibrium constant and standard Gibbs free energy change at 25 °C for the following cell; Cu|Cu+2(1 M)||Ag+(1 M)|Ag. (Given: E°cell(Ag+|Ag)=0.799 V, E°cell(Cu2+|Cu)=0.518 V). Read more

Steps to Solve

1. Calculate the Standard Cell Potential

The standard cell potential ($E^\circ_{cell}$) can be calculated using the formula:

$$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $$

From the problem, we know:

• $E^\circ_{cell}(Ag^+|Ag) = 0.799 , \text{V}$ (cathode)
• $E^\circ_{cell}(Cu^{2+}|Cu) = 0.518 , \text{V}$ (anode)

So,

$$ E^\circ_{cell} = 0.799 , \text{V} - 0.518 , \text{V} = 0.281 , \text{V} $$

2. Calculate Gibbs Free Energy Change

The Gibbs free energy change ($\Delta G^\circ$) can be calculated with the equation:

$$ \Delta G^\circ = -nFE^\circ_{cell} $$

Where:

• $n = 2$ (the number of moles of electrons transferred per reaction)
• $F = 96485 , \text{C/mol}$ (Faraday's constant)
• $E^\circ_{cell} = 0.281 , \text{V}$

Thus,

$$ \Delta G^\circ = -2 \times 96485 , \text{C/mol} \times 0.281 , \text{V} $$

Calculating this gives:

$$ \Delta G^\circ = -2 \times 96485 \times 0.281 = -54362.37 , \text{J/mol} $$

To convert to kJ/mol:

$$ \Delta G^\circ = -54.36 , \text{kJ/mol} $$

3. Calculate the Equilibrium Constant

The equilibrium constant ($K$) is related to Gibbs free energy by the equation:

$$ \Delta G^\circ = -RT \ln K $$

Where:

• $R = 8.314 , \text{J/(mol K)}$
• $T = 298 , \text{K}$ (25 °C)

Rearranging gives:

$$ K = e^{-\frac{\Delta G^\circ}{RT}} $$

Substituting the values:

$$ K = e^{-\frac{-54362.37 , \text{J/mol}}{8.314 , \text{J/(mol K)} \times 298 , \text{K}}} $$

Calculate:

$$ K = e^{21.7663} \approx 2.85 \times 10^9 $$