First, let's calculate the equilibrium constant, K, using the relationship between ΔG° and K, which is given by the formula:
ΔG° = -RT * ln(K),
where ΔG° is the change in Gibbs free energy, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298K, since it's at 25°C), and K is the equilibrium constant.
Rearranging the formula to solve for K, we get:
K = e^(-ΔG° / (RT)).
Plugging the given values:
K = e^(-(143kJ * 1000 J/kJ) / (8.314 J/mol·K * 298 K))
K ≈ 6.24 x 10^10
Now, let's calculate the standard cell potential, E°, related to ΔG° using the formula:
ΔG° = -nFE°,
where n is the number of electrons transferred in the redox reaction, F is the Faraday constant (96,485 C/mol), and E° is the standard cell potential.
In this case, the balanced redox reaction is:
Cd (s) + Cu^2+ (aq) → Cd^2+ (aq) + Cu(s).
We can see that there are 2 moles of electrons transferred (one from Cd to Cd^2+ and one from Cu^2+ to Cu). Therefore, n = 2.
Rearranging the formula to solve for E°, we get:
E° = -ΔG° / (nF).
Plugging the given values:
E° = -(143kJ * 1000 J/kJ) / (2 * 96,485 C/mol)
E° ≈ 0.74 V
So, the standard cell potential, E°, is 0.74 V.
When cadmium metal is reduced Cu^2+ in solution Cd^2+ forms in addition to copper metal. If ∆G° =143kJ, Calculate K at 25°C.
What is E ^ alpha in a voltaic cell that uses the reaction above?
1 answer
1 answer