Consider an electrochemical cell containing a copper electrode in a 0.5 M CuSO4 solution and a silver electrode in a 1.0 M AgNO3 solution. Calculate the current density at a temperature of 25°C given that the copper electrode has a surface area of 5 cm² and the silver electrode has a surface area of 10 cm². The standard reduction potentials for Cu²⁺ and Ag⁺ are -0.34 V and 0.80 V, respectively. Use the Nernst equation to calculate the cell potential and then use Ohm's law to calculate the current density.

Question

Consider an electrochemical cell containing a copper electrode in a 0.5 M CuSO4 solution and a silver electrode in a 1.0 M AgNO3 solution. Calculate the current density at a temperature of 25°C given that the copper electrode has a surface area of 5 cm² and the silver electrode has a surface area of 10 cm². The standard reduct

Answer 1

First, we need to calculate the cell potential using the Nernst equation. The Nernst equation is:E_cell = E_cell -  RT/nF  * ln Q where E_cell is the standard cell potential, R is the gas constant  8.314 J/molK , T is the temperature in Kelvin  25C = 298 K , n is the number of electrons transferred in the redox reaction, F is the Faraday constant  96485 C/mol , and Q is the reaction quotient.The standard cell potential, E_cell, can be calculated using the standard reduction potentials:E_cell = E_cathode - E_anodeSince Ag has a higher reduction potential, it will act as the cathode, and Cu will act as the anode:E_cell = 0.80 V -  -0.34 V  = 1.14 VThe balanced redox reaction is:Cu s  + 2Ag aq  -> Cu aq  + 2Ag s The reaction quotient, Q, is:Q = [Cu]/[Ag]^2Since the concentrations are given, we can plug them in:Q =  0.5 M / 1.0 M ^2 = 0.5Now we can calculate the cell potential using the Nernst equation. The number of electrons transferred, n, is 2:E_cell = 1.14 V -  8.314 J/molK * 298 K /  2 * 96485 C/mol   * ln 0.5 E_cell  1.14 V - 0.0295 V = 1.1105 VNow we can use Ohm's law to calculate the current density. Ohm's law is:I = V/RHowever, we need to find the current density  I/A , where A is the surface area. Since the surface area of the copper electrode is smaller, it will limit the current density. Therefore, we will use the copper electrode's surface area  5 cm  to calculate the current density. We also need the resistance  R  of the cell, which is not given in the problem. Without the resistance, we cannot calculate the current density using Ohm's law.If the resistance were provided, the current density could be calculated as:Current density =  I/A  =  V/R  / A