Question

A zinc electrode is connected to a copper electrode by a wire in a beaker that has a solution of 0.1 M CuSO4 and 0.1 M ZnSO4. The initial reduction potential of the copper electrode is +0.34 V, and the initial oxidation potential of zinc is –0.76 V. What is the potential difference between the two electrodes at 298 K, and what is the rate of the electrochemical reaction when the cell operates at a current of 2.50 A?

Answer 1

To find the potential difference between the two electrodes, we can simply subtract the oxidation potential of zinc from the reduction potential of copper:Potential difference  E_cell  = E_reduction  Cu  - E_oxidation  Zn E_cell =  +0.34 V  -  -0.76 V E_cell = 1.10 VSo, the potential difference between the two electrodes at 298 K is 1.10 V.Now, to find the rate of the electrochemical reaction when the cell operates at a current of 2.50 A, we can use the following equation:Rate = Current  I  / nFWhere:- Rate is the rate of the electrochemical reaction  in mol/s - Current  I  is the current flowing through the cell  in A - n is the number of electrons transferred in the redox reaction  for the Cu2+/Cu and Zn/Zn2+ redox couple, n = 2 - F is Faraday's constant  approximately 96485 C/mol Rate = 2.50 A /  2 * 96485 C/mol Rate = 2.50 A / 192970 C/molRate = 1.29 x 10^-5 mol/sTherefore, the rate of the electrochemical reaction when the cell operates at a current of 2.50 A is 1.29 x 10^-5 mol/s.