When enough NaCl(s) is added to the compartment containing gold to make the [Cl -] , the cell potential is observed to be 0.31 V. Assume that Au3+ is reduced and assume that the reaction in the compartment containing gold is
Au3+(aq) + 4Cl- (aq) ↔ AuCl4- . Calculate the k for this reaction at 250C.
In the reaction you have written, Au^+3 is not reduced. It is +3 on the left and +3 on the right. Cl^- is not reduced or oxidized either.
so how do you solve this problem?
To calculate the equilibrium constant (K) for the given reaction at 25°C, we can use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the equilibrium constant (K) and the concentrations of the reactants and products.
The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
- Ecell is the observed cell potential (0.31 V in this case)
- E°cell is the standard cell potential, which can be looked up in a table (we assume it to be 0.00 V in this case)
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in kelvin (25°C = 298 K)
- n is the number of electrons transferred in the reaction (in this case, 3)
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, which is the ratio of the product of the concentrations of the products (AuCl4-) to the product of the concentrations of the reactants (Au3+ and Cl-).
Since we are given the cell potential (Ecell) and we assume the standard cell potential (E°cell) to be 0.00 V, we can rearrange the Nernst equation and solve for ln(Q):
ln(Q) = (Ecell - E°cell) * (nF/RT)
Now, let's calculate ln(Q):
ln(Q) = (0.31 V - 0.00 V) * (3 * 96485 C/mol) / (8.314 J/mol·K * 298 K)
ln(Q) ≈ 5.57
Finally, we can calculate K using the relationship between Q and K:
K = e^(ln(Q))
K = e^(5.57)
K ≈ 262.60
Therefore, the equilibrium constant (K) for the given reaction at 25°C is approximately 262.60.