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Multiple Choice
Calculate the equilibrium constant (K) for an oxidation-reduction reaction (n = 2) with a cell potential of -0.51 V at 298 K.
A
K = 3.5 x 10^5
B
K = 1.2 x 10^-17
C
K = 2.1 x 10^3
D
K = 4.8 x 10^-9
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Verified step by step guidance
1
Identify the relationship between the cell potential (E°), the equilibrium constant (K), and the number of moles of electrons transferred (n) using the Nernst equation: E° = (RT/nF) * ln(K), where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and F is the Faraday constant (96485 C/mol).
Rearrange the Nernst equation to solve for the equilibrium constant (K): ln(K) = (nF * E°) / (RT).
Substitute the given values into the equation: n = 2, E° = -0.51 V, R = 8.314 J/mol·K, T = 298 K, and F = 96485 C/mol.
Calculate the value of ln(K) using the substituted values. Remember to convert the cell potential from volts to joules by multiplying by the Faraday constant.
Exponentiate the result from the previous step to solve for K: K = e^(ln(K)). This will give you the equilibrium constant for the reaction.