Question

Calculate the activation energy for the following electrochemical reaction: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ if the half-cell potentials are: E°(O₂/H₂O) = 1.23 V E°(H⁺/H₂) = 0 V Assuming the reaction follows Arrhenius equation, where the rate constant k is related to the activation energy Ea, temperature T and pre-exponential factor A by: k = A exp(-Ea/RT) where R is the gas constant equal to 8.314 J/molK.

Answer 1

To calculate the activation energy for the given electrochemical reaction, we first need to determine the overall cell potential for the reaction. The cell potential can be calculated using the Nernst equation:E_cell = E cathode  - E anode In this case, the half-cell potentials are given as:E O/HO  = 1.23 V  cathode E H/H  = 0 V  anode So, the overall cell potential is:E_cell = 1.23 V - 0 V = 1.23 VNow, we can use the relationship between the cell potential, Gibbs free energy change  G , and the number of electrons transferred  n  in the reaction:G = -nFE_cellwhere F is the Faraday constant  96,485 C/mol .For the given reaction, 4 electrons are transferred:2HO l   O g  + 4H aq  + 4eSo, n = 4. Now we can calculate G:G = - 4  96,485 C/mol  1.23 V  = -475,676 J/molThe relationship between Gibbs free energy change  G  and activation energy  Ea  is given by the Eyring equation:G = Ea - RT ln k/A We are given the Arrhenius equation:k = A exp -Ea/RT Taking the natural logarithm of both sides:ln k  = ln A  - Ea/RTRearranging for Ea:Ea = -RT ln k/A Now, we can substitute the Eyring equation into this expression:Ea = -RT ln k/A  = G + RT ln k/A Since we don't have the values for k, A, and T, we cannot calculate the exact value of Ea. However, we can see that the activation energy is related to the Gibbs free energy change and the rate constant, pre-exponential factor, and temperature through these equations.