Calculate the standard cell potential at 25 ∘C for the reaction X(s)+2Y+(aq)�¨X2+(aq)+2Y(s) where ƒ¢H∘ = -653 kJ and ƒ¢S∘ = -263 J/K .
I thought I could find the answer on this, I'm sorry. I am as confused as you are.
To calculate the standard cell potential (E°) for the given reaction, you need to use the thermodynamic equation:
ΔG° = -nFE°
where:
ΔG° is the standard Gibbs free energy change of the reaction,
n is the number of electrons transferred in the reaction,
F is the Faraday constant (96,485 C/mol), and
E° is the standard cell potential.
First, you need to determine ΔG° using the equation:
ΔG° = ΔH° - TΔS°
where:
ΔH° is the standard enthalpy change of the reaction,
ΔS° is the standard entropy change of the reaction, and
T is the temperature in Kelvin.
Given:
ΔH° = -653 kJ
ΔS° = -263 J/K
T = 25 °C = 298 K (as the temperature needs to be in Kelvin)
Converting ΔH° to J:
ΔH° = -653 kJ = -653,000 J
Now, substitute the given values into the equation to calculate ΔG°:
ΔG° = ΔH° - TΔS°
ΔG° = -653,000 J - (298 K * -263 J/K)
ΔG° = -653,000 J + 78,274 J
ΔG° = -574,726 J
Next, calculate the value of n, the number of electrons transferred in the reaction. Looking at the balanced equation:
X(s) + 2Y+(aq) → X2+(aq) + 2Y(s)
We see that 2 moles of electrons are transferred in the reaction. Therefore, n = 2.
Now, substitute the calculated ΔG° and the value of n into the equation ΔG° = -nFE° to solve for E°:
-574,726 J = -2 * 96,485 C/mol * E°
E° = -574,726 J / (-2 * 96,485 C/mol)
E° = 2.98 V
So, the standard cell potential (E°) at 25 °C for the given reaction is 2.98 volts.