I assume you mean the reaction in which electrons flow from the anode to the cathode spontaneously.
Anode is on the left, it is oxidized; therefore, Eo oxidation = -0.37
Cathode is on the right so it receives the electrons and it is reduced so Eo redn = 0.67.
EoCell = Eox + Ered = ?
Then nFEo = RTlnK.
You know n, Eo and F (96,485), R(8.314) and T. Solve for K.
Under standard conditions of 1 atm and 298.15K, the half-cell reduction potential E,zero for the anode in a voltaic cell is 0.37V. The half-cell reduction potential E,zero for the cathode of the cell is 0.67V. The number of moles, n, of electrons transferred in the redox reaction is 3. Calculate the equilibrium constant K for this reaction at 25degC. Enter your answer in exponential notation with two decimal places.
1 answer
1 answer