Question

A student needs to determine the activation energy for an electrochemical reaction involving the oxidation of Zinc using a voltage of 1.85V, a current of 0.25A and a temperature of 25°C. Given that the standard potential for this reaction at 25°C is -0.76V, what is the activation energy of the reaction?

Answer 1

To determine the activation energy for the electrochemical reaction, we can use the Nernst equation:E = E -  RT/nF  * ln Q Where:E = cell potential  1.85 V E = standard cell potential  -0.76 V R = gas constant  8.314 J/molK T = temperature in Kelvin  25C + 273.15 = 298.15 K n = number of electrons transferred in the reaction  for Zinc, n = 2 F = Faraday's constant  96485 C/mol Q = reaction quotient  for a simple oxidation reaction, Q = 1 First, we need to solve for the natural logarithm  ln  term:1.85 V = -0.76 V -  RT/nF  * ln 1 1.85 V + 0.76 V =  RT/nF  * ln 1 2.61 V =  RT/nF  * ln 1 Since ln 1  = 0, the equation simplifies to:2.61 V =  RT/nF  * 02.61 V = 0This result indicates that there is an inconsistency in the given data. The cell potential  1.85 V  and the standard cell potential  -0.76 V  do not align with the Nernst equation for the given reaction conditions. Please check the input data for any errors. If the data is correct, it is not possible to determine the activation energy for this reaction using the Nernst equation.