Calculate the rate of the electrochemical reaction that takes place in the following electrolytic cell:Cu(s) | Cu2+(aq, 0.050 M) || Ag+(aq, 0.010 M) | Ag(s)Given that the current density is 0.500 A/m2 and the Faraday constant is 96,485 C/mol. The reduction potential of Cu2+(aq) + 2e- → Cu(s) is +0.34 V, and the reduction potential of Ag+(aq) + e- → Ag(s) is +0.80 V. Assume the temperature is 25°C and one-unit charge is exchanged.

Question

Calculate the rate of the electrochemical reaction that takes place in the following electrolytic cell:Cu(s) | Cu2+(aq, 0.050 M) || Ag+(aq, 0.010 M) | Ag(s)Given that the current density is 0.500 A/m2 and the Faraday constant is 96,485 C/mol. The reduction potential of Cu2+(aq) + 2e- → Cu(s) is +0.34 V, and the reduction potential of Ag+(aq) + e- → Ag(s) is +0.80 V. Assume the temperature is 25°C and one-unit charge is

Answer 1

To calculate the rate of the electrochemical reaction, we first need to determine the overall cell potential  E_cell  and the number of electrons transferred  n  in the reaction.The overall cell potential can be calculated using the Nernst equation:E_cell = E_cathode - E_anodeSince Ag+ has a higher reduction potential  +0.80 V  than Cu2+  +0.34 V , Ag+ will be reduced at the cathode and Cu will be oxidized at the anode. Therefore:E_cell = 0.80 V - 0.34 V = 0.46 VNow, we need to determine the number of electrons transferred  n  in the reaction. The balanced half-reactions are:Cu s   Cu2+ aq  + 2e-  anode Ag+ aq  + e-  Ag s   cathode The overall balanced reaction is:Cu s  + 2Ag+ aq   Cu2+ aq  + 2Ag s From the balanced reaction, we can see that 2 moles of electrons are transferred for every mole of Cu oxidized and 1 mole of electrons is transferred for every mole of Ag reduced. Since one-unit charge is exchanged, we will consider n = 1 for this problem.Now, we can calculate the rate of the electrochemical reaction using the current density  0.500 A/m  and the Faraday constant  96,485 C/mol :rate =  current density  /  n * Faraday constant rate =  0.500 A/m  /  1 * 96,485 C/mol  = 5.18 x 10^-6 mol/msThe rate of the electrochemical reaction in the electrolytic cell is 5.18 x 10^-6 mol/ms.