0 votes
31 views
in ElectroChemistry by (138k points)
What is the activation energy for the electrochemical reaction between copper and zinc ions using a copper-zinc galvanic cell at 25°C and a potential difference of 1.8V?

1 Answer

0 votes
by (138k points)
To calculate the activation energy  Ea  for the electrochemical reaction between copper and zinc ions in a copper-zinc galvanic cell, 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/molK , and T is the temperature in Kelvin  25C = 298.15 K .However, we need to relate the rate constant  k  to the potential difference  1.8 V  in the galvanic cell. We can do this using the Butler-Volmer equation:i = i0 *  exp   * F *   /  R * T   - exp  -  1 -   * F *   /  R * T   where i is the current, i0 is the exchange current density,  is the charge transfer coefficient, F is the Faraday constant  96485 C/mol ,  is the overpotential, and the other variables are as previously defined.In a galvanic cell, the potential difference  1.8 V  is equal to the difference between the standard reduction potentials of the two half-reactions:E_cell = E_cathode - E_anodeFor a copper-zinc galvanic cell, the standard reduction potentials are:E_cathode  Cu + 2e  Cu  = +0.34 VE_anode  Zn  Zn + 2e  = -0.76 VE_cell = 0.34 V -  -0.76 V  = 1.1 VSince the potential difference in the given problem is 1.8 V, which is higher than the standard potential difference  1.1 V , we have an overpotential    of: = 1.8 V - 1.1 V = 0.7 VNow, we need to find the relationship between the rate constant  k  and the current  i  or the exchange current density  i0 . This relationship is not straightforward and depends on the specific reaction kinetics and the electrode surface properties. Therefore, without further information, it is not possible to accurately calculate the activation energy  Ea  for this electrochemical reaction.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...