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What is the equilibrium constant expression for the following redox reaction at standard conditions: Fe2+(aq) + Cu(s) ⇌ Fe3+(aq) + Cu2+(aq)

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The equilibrium constant expression for a redox reaction can be determined using the Nernst equation. First, we need to identify the half-reactions:Oxidation half-reaction: Fe2+ aq   Fe3+ aq  + e-Reduction half-reaction: Cu2+ aq  + 2e-  Cu s Now, we can write the Nernst equation:E = E -  RT/nF  * ln Q where E is the cell potential, E is the standard cell potential, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.At equilibrium, E = 0, and the reaction quotient Q becomes the equilibrium constant K:0 = E -  RT/nF  * ln K Rearranging for K:K = e^nFE/RT For this reaction, n = 1  since one electron is transferred in each half-reaction . To find E, we need the standard reduction potentials for each half-reaction:E Fe3+/Fe2+  = +0.77 VE Cu2+/Cu  = +0.34 VSince the Cu2+/Cu half-reaction is reversed in the overall reaction, we need to change the sign of its standard reduction potential:E Cu/Cu2+  = -0.34 VNow, we can find the standard cell potential E for the overall reaction:E = E Fe3+/Fe2+  + E Cu/Cu2+  = 0.77 V +  -0.34 V  = 0.43 VFinally, we can plug the values into the equation for K:K = e^1 * F * 0.43 V /  R * 298 K  Using the values R = 8.314 J/ molK  and F = 96485 C/mol, we get:K = e^1 * 96485 C/mol * 0.43 V /  8.314 J/ molK  * 298 K  K  1.1 x 10^18So, the equilibrium constant expression for the redox reaction is:K = [Fe3+][Cu2+] / [Fe2+]And the value of K at standard conditions is approximately 1.1 x 10^18.

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