Determine the equilibrium constant for the following reaction at 25 °C. Cl₂(aq) + 2I⁻(aq) → 2 Cl⁻(aq) + I₂(aq) E° = +0.825 V
To determine the equilibrium constant for the given reaction, you need to use the Nernst equation and the standard cell potential (E°).
The Nernst equation relates the reaction quotient (Q) to the standard cell potential (E°) and the equilibrium constant (K). It is given by:
E = E° - (RT/nF) * ln(Q)
where:
E = Cell potential under non-standard conditions
E° = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
n = Number of electrons transferred in the reaction
F = Faraday's constant (96,485 C/mol)
ln = Natural logarithm
Q = Reaction quotient
In this case, n = 2 because two electrons are transferred in the reaction. We need to find the equilibrium constant, K, at 25 °C (which is 298 K).
First, let's calculate Q using the concentrations of the species involved in the reaction. However, we need the concentrations, not activities, because the reaction is taking place in an aqueous solution.
Now, given that Cl₂ is in the aqueous state, it will exist as Cl₂(aq), and the same applies for the other species in the equation. Furthermore, Cl⁻ and I⁻ are both spectator ions because they do not participate in the electron transfer process. So, we can assume that their concentrations remain constant, and we can treat them as 1 M (molarity) for simplicity.
As a result, we have:
[Cl₂(aq)] = x M
[I⁻(aq)] = 1 M
[Cl⁻(aq)] = 1 M
[I₂(aq)] = y M
The initial concentrations of Cl₂(aq) and I⁻(aq) are not given, which implies that we start with 0 M concentrations. Thus, at the beginning, [Cl₂(aq)] = 0 M and [I⁻(aq)] = 1 M.
Now, use the Nernst equation to calculate the cell potential (E):
E = E° - (RT/nF) * ln(Q)
E = +0.825 V
E° = +0.825 V
R = 8.314 J/(mol·K)
T = 298 K
n = 2
F = 96,485 C/mol
Since we have no information about the concentration of Cl₂(aq), Q is initially 0, and the natural logarithm of 0 is undefined. Thus, we cannot directly solve for K at this moment.
To achieve equilibrium, the reaction will proceed until Q reaches a certain value. At equilibrium, Q equals K, where K is the equilibrium constant. Therefore, at equilibrium, Q = K.
Now consider the stoichiometry of the reaction: Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq)
At equilibrium, the concentration of Cl₂(aq) can be expressed in terms of [I₂(aq)]. According to the stoichiometry, the concentration of Cl₂(aq) is twice the concentration of I₂(aq).
So, [Cl₂(aq)] = 2y
Now, substitute these values into the expression for Q:
Q = ([Cl⁻(aq)]^2 * [I₂(aq)]) / ([Cl₂(aq)] * [I⁻(aq)]^2)
Q = (1^2 * y) / (2y * 1^2)
Q = y / (2y)
Q = 1/2
Since Q = K at equilibrium, we have K = 1/2.
Therefore, the equilibrium constant for the given reaction at 25 °C is 1/2 (or 0.5).