Given the following two half reactions: Al³⁺(aq) + 3e⁻ → Al(s) E° = -1.66 V; Cl₂(aq) + 2e⁻ → 2 Cl⁻(aq) E° = +1.36 V. Determine the equilibrium constant for 2 Al³⁺(aq) + 6 Cl⁻(aq) → 2 Al(s) + 3 Cl₂(aq)
Oh, I see what you're doing here with all those half reactions and equilibrium constants! You know, chemistry is all about finding the right balance, just like in real life. So, let's find the equilibrium constant for this reaction and strike a balance!
To do that, we can use the Nernst equation, which relates the equilibrium constant (K) to the standard cell potential (E°) and the reaction quotient (Q).
The reaction quotient (Q) is calculated by taking the products raised to their stoichiometric coefficients divided by the reactants raised to their stoichiometric coefficients. So, for this reaction, Q is:
Q = [Al]²/[Cl]³
Now, we can substitute in the values from the given half reactions:
Q = ([Al³⁺]²/[Cl⁻]³)²/([Al³⁺][Cl₂]²)³
Since we know that the standard electrode potentials (E°) for the given reactions are -1.66 V and +1.36 V, respectively, we can plug those in too.
E° = E°(cathode) - E°(anode)
E° = (+1.36 V) - (-1.66 V)
E° = 2.20 V
Finally, we can use the Nernst equation to find the equilibrium constant (K):
E = E° - (0.0592 V/n) * log(K)
2.20 V = -0.0592 V/n * log(K)
Now, n represents the number of electrons transferred in the reaction. Looking at the given reaction, we can see that 6 electrons are transferred. So, plugging in n = 6 and solving for K:
2.20 V = (-0.0592 V/6) * log(K)
2.20 V = -0.00987 V * log(K)
log(K) = 2.20 V / (-0.00987 V)
log(K) = -222.92
K ≈ 1.58 x 10^(-223)
Well, that's one tiny equilibrium constant! It's probably too small to even fit on your calculator. So, I guess you could say this reaction really doesn't like to reach equilibrium. It prefers to stay far, far away from it!
To determine the equilibrium constant for the reaction:
2 Al³⁺(aq) + 6 Cl⁻(aq) → 2 Al(s) + 3 Cl₂(aq)
We need to use the Nernst equation, which relates the standard reduction potentials (E°) of the half reactions to the equilibrium constant (K) of the overall reaction.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(K)
Where:
E = cell potential at non-standard conditions
E° = standard cell potential
R = gas constant (8.314 J/K*mol)
T = temperature in Kelvin
n = number of moles of electrons transferred in the balanced equation
F = Faraday's constant (96,485 C/mol)
First, let's calculate the cell potential for the reaction using the given half-reactions:
For the reduction half-reaction:
Al³⁺(aq) + 3e⁻ → Al(s)
E° = -1.66 V
For the oxidation half-reaction:
Cl₂(aq) + 2e⁻ → 2 Cl⁻(aq)
E° = +1.36 V
Now, we need to multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3 to ensure the electrons cancel out:
2 Al³⁺(aq) + 6e⁻ → 2 Al(s)
3 Cl₂(aq) + 6e⁻ → 6 Cl⁻(aq)
Next, we add the two half-reactions:
2 Al³⁺(aq) + 3 Cl₂(aq) → 2 Al(s) + 6 Cl⁻(aq)
The total cell potential (E) for the reaction is the sum of the reduction and oxidation potentials:
E = (-1.66 V) + (+1.36 V)
E = -0.30 V
Now, substituting the values into the Nernst equation to find the equilibrium constant (K):
-0.30 V = -1.66 V - (8.314 J/K*mol) * (T/K) * ln(K)
Rearranging the equation:
ln(K) = (E° - E) / [(8.314 J/K*mol) * (T/K)]
Substituting the values and solving for ln(K):
ln(K) = (-1.66 V - (-0.30 V)) / [(8.314 J/K*mol) * (T/K)]
ln(K) = -1.36 V / [(8.314 J/K*mol) * (T/K)]
Finally, taking the exponential of both sides to find K:
K = e^(ln(K))
K = e^(-1.36 V / [(8.314 J/K*mol) * (T/K)])
Thus, the equilibrium constant (K) for the reaction 2 Al³⁺(aq) + 6 Cl⁻(aq) → 2 Al(s) + 3 Cl₂(aq) is given by the expression e^(-1.36 V / [(8.314 J/K*mol) * (T/K)]).
To determine the equilibrium constant for the given reaction, we can use the Nernst equation and the relationship between standard cell potentials and equilibrium constants.
The Nernst equation relates the standard cell potential (E°) to the reaction quotient (Q) 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
First, let's write the balanced equation for the reaction:
2 Al³⁺(aq) + 6 Cl⁻(aq) → 2 Al(s) + 3 Cl₂(aq)
From the given half-reactions, we can find the standard potentials for the reactants and products:
E° for 2 Al³⁺(aq) + 6 Cl⁻(aq) → 2 Al(s) + 3 Cl₂(aq) = E°(Al³⁺ → Al) + 3E°(Cl₂ → 2Cl⁻)
Next, we substitute the values into the Nernst equation to find the cell potential (E):
E = E° - (RT/nF) * ln(Q)
Since we are interested in the equilibrium constant (K), we can relate it to the reaction quotient (Q) using the expression:
K = e^(nF(E°-E)/RT)
Now, let's calculate the equilibrium constant using the given values:
E°(Al³⁺ → Al) = -1.66 V
E°(Cl₂ → 2Cl⁻) = +1.36 V
Substituting these values into the equation:
E = E° - (RT/nF) * ln(Q)
E = (-1.66 V + 3 * 1.36 V) - (8.314 J/(mol*K) * T / (6 * 96,485 C/mol) * ln(Q)
Once you have the value of E, you can substitute it into the expression for K to find the equilibrium constant:
K = e^(nF(E°-E)/RT)
By following these steps, you can determine the equilibrium constant for the given reaction.