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What is the activation energy for the oxidation of copper in a 1 M copper sulfate solution with a standard reduction potential of +0.34 V at 298 K, given that the exchange current density is 5.6 x 10^-6 Acm^-2 and the reaction order is unity?

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To calculate the activation energy for the oxidation of copper, we can use the Tafel equation, which relates the overpotential    to the exchange current density  i0  and the current density  i  of the electrochemical reaction: =  RT/nF  * ln i/i0 where R is the gas constant  8.314 J/molK , T is the temperature  298 K , n is the number of electrons transferred  2 for the Cu/Cu2+ redox couple , and F is the Faraday constant  96485 C/mol .First, we need to find the current density  i  for the reaction. We can use the Nernst equation to find the equilibrium potential  E  for the reaction:E = E -  RT/nF  * ln [Cu2+]/[Cu] Since the reaction order is unity, the concentrations of Cu2+ and Cu are equal, so the ln term becomes ln 1  = 0. Therefore, the equilibrium potential is equal to the standard reduction potential:E = +0.34 VNow, we can use the Butler-Volmer equation to find the current density  i :i = i0 *  exp nF/RT  - exp -nF/RT  Since the reaction is at equilibrium, the net current density  i  is zero:0 = i0 *  exp nF/RT  - exp -nF/RT  We can rearrange the equation to solve for the overpotential   : =  RT/nF  * ln exp -nF/RT  Now, we can plug in the values for R, T, n, F, and i0: =  8.314 J/molK * 298 K  /  2 * 96485 C/mol  * ln 5.6 x 10^-6 Acm^-2   0.0429 VFinally, we can use the Arrhenius equation to find the activation energy  Ea :Ea =  * n * FEa = 0.0429 V * 2 * 96485 C/molEa  8260 J/molSo, the activation energy for the oxidation of copper in a 1 M copper sulfate solution is approximately 8260 J/mol.

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