To determine the activation energy Ea for the electrochemical reaction between copper and nitric acid, we can use the Arrhenius equation:k = A * exp -Ea / R * T where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant 8.314 J mol^-1 K^-1 , and T is the temperature in Kelvin.We have two sets of data: k1 = 2.31 x 10^-6 mol L^-1 s^-1 at T1 = 25C 298 K and k2 = 3.81 x 10^-5 mol L^-1 s^-1 at T2 = 45C 318 K . We can set up two equations using the Arrhenius equation:k1 = A * exp -Ea / R * T1 k2 = A * exp -Ea / R * T2 Now, we can divide the first equation by the second equation to eliminate the pre-exponential factor A :k1 / k2 = exp -Ea / R * T1 / exp -Ea / R * T2 Simplifying the equation:k1 / k2 = exp Ea / R * T2 - Ea / R * T1 Now, we can take the natural logarithm of both sides:ln k1 / k2 = Ea / R * T2 - Ea / R * T1 Rearranging the equation to solve for Ea:Ea = R * T1 * T2 * ln k1 / k2 / T2 - T1 Plugging in the given values:Ea = 8.314 J mol^-1 K^-1 * 298 K * 318 K * ln 2.31 x 10^-6 / 3.81 x 10^-5 / 318 K - 298 K Ea 8.314 * 94524 * -2.72 / 20Ea -214,799 J mol^-1Since the activation energy cannot be negative, there must be an error in the calculation. Let's check the equation again:ln k2 / k1 = Ea / R * T1 - Ea / R * T2 Rearranging the equation to solve for Ea:Ea = R * T1 * T2 * ln k2 / k1 / T1 - T2 Plugging in the given values:Ea = 8.314 J mol^-1 K^-1 * 298 K * 318 K * ln 3.81 x 10^-5 / 2.31 x 10^-6 / 298 K - 318 K Ea 8.314 * 94524 * 2.72 / -20Ea 107,399 J mol^-1So, the activation energy for the electrochemical reaction between copper and nitric acid is approximately 107,399 J mol^-1 or 107.4 kJ mol^-1.