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The reaction X₃⁺(aq) + Y(l) + Y₃⁺(aq) + X(s) has a standard state cell potential of 0.0500 V.

What is the equilibrium constant for the reaction?

1 Answer

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Final answer:

The equilibrium constant (K) for the reaction can be calculated using the relationship between the standard cell potential (E°cell) and K, applying the Nernst equation and the standard free energy change equation at standard conditions.

Step-by-step explanation:

The relationship between the standard cell potential (E°cell) and the equilibrium constant (K) is given by the Nernst equation at standard conditions (25°C, 1 atm, 1 M concentrations) which is combined with the standard free energy change (ΔG°) equation:

ΔG° = -nFE°cell

where ΔG° is the standard free energy change, n is the number of moles of electrons transferred, F is the Faraday's constant (96,485 C/mol e⁻), and E°cell is the standard cell potential. Using the standard free energy change equation, we can link ΔG° with the equilibrium constant (K) through the following relationship:

ΔG° = -RT ln(K)

Here, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (K), and K is the equilibrium constant. By setting the two expressions for ΔG° equal to each other and solving for K, we find:

K = e^{⁡(-nFE°cell / RT)}

For the reaction X₃⁺(aq) + Y(l) → Y₃⁺(aq) + X(s) with a standard cell potential (E°cell) of 0.0500 V and assuming the reaction involves the transfer of a single electron (n = 1), we can calculate the equilibrium constant (K) at room temperature (25°C or 298.15 K), using the given values:

K = e^{⁡(-(1)(96,485 C/mol)(0.0500 V) / (8.314 J/mol·K)(298.15 K))}

After calculating this expression, the equilibrium constant (K) for the reaction can be known.

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