Final answer:
The standard cell potential (E°) for the reaction with an equilibrium constant (K) of 9.40×10⁻⁴ is approximately 0.089 V, assuming n = 2.
Step-by-step explanation:
To calculate the standard cell potential (E°) for the given reaction X(s) + Y+(aq) → X+(aq) + Y(s), we use the Nernst equation modified to determine E° using the equilibrium constant (K):
E° = (RT/nF) × ln(K)
Since we are dealing with standard conditions, T is 298 K (25°C), R is the gas constant (8.314 J/(mol·K)), n is the number of moles of electrons transferred in the balanced equation (which is not given but usually can be determined from the balanced equation), and F is the Faraday constant (96485 C/mol). However, the equation first needs to be converted to base 10 logarithm:
E° = - (0.0591/n) × log(K)
Since K is given as 9.40×10⁻⁴, we can substitute directly into the equation, once we determine the value of n. If n is assumed to be 2 (common for many reactions with a single electron transfer for each reactant), the equation would become:
E° = - (0.0591/2) × log(9.40×10⁻⁴)
Assuming n = 2, we can now calculate:
E° = -0.02955 × (-3.027) ≈ 0.089 V
The calculated standard cell potential for this reaction is approximately 0.089 V. Note that the sign of the log value is negative because K < 1 which corresponds to a non-spontaneous reaction under standard conditions, but the double negative signs in the equation make the E° a positive value.