Question

A student needs to calculate the current density (in Amperes per square meter) of a copper electrode (Area = 10 cm²) in an electrochemical cell where the half-cell reaction is Cu²⁺ + 2e⁻ → Cu and the cell voltage is 1.1 V. The concentration of Cu²⁺ in the cell is 0.1 M.

Answer 1

To calculate the current density, we first need to determine the current  I  flowing through the electrode. We can use the Nernst equation to find the standard cell potential  E  for the half-cell reaction, and then use Ohm's law to find the current.The Nernst equation is:E = E -  RT/nF  * ln Q where E is the cell potential, E is the standard cell potential, R is the gas constant  8.314 J/molK , T is the temperature in Kelvin  assume 298 K for room temperature , n is the number of electrons transferred in the reaction  2 for Cu + 2e  Cu , F is the Faraday constant  96485 C/mol , and Q is the reaction quotient.For the Cu/Cu half-cell, the standard cell potential  E  is 0.34 V. Since the cell is at equilibrium, Q = 1. Therefore, the Nernst equation becomes:E = 0.34 V -  8.314 J/molK * 298 K /  2 * 96485 C/mol   * ln 1 E = 0.34 VNow we can use Ohm's law to find the current:I = E / Rwhere I is the current, E is the cell potential, and R is the resistance. The cell voltage is given as 1.1 V, so the resistance is:R = E / IR = 1.1 V / 0.34 VR  3.24Now we can find the current:I = E / RI = 0.34 V / 3.24I  0.105 AFinally, we can calculate the current density  J :J = I / Awhere J is the current density, I is the current, and A is the area of the electrode. The area of the electrode is given as 10 cm, which is equal to 0.001 m.J = 0.105 A / 0.001 mJ = 105 A/mSo, the current density of the copper electrode is approximately 105 Amperes per square meter.