Calculate the current density in an electrochemical cell if a current of 2.5 A is passed through a copper electrode with a surface area of 25 cm² and a silver electrode with a surface area of 10 cm² for 3 hours. The half-cell potentials of copper and silver electrodes are +0.34 V and +0.80 V, respectively, and the molar concentrations of copper and silver ions in the electrolyte are 0.5 M and 1 M, respectively.

Question

Calculate the current density in an electrochemical cell if a current of 2.5 A is passed through a copper electrode with a surface area of 25 cm² and a silver electrode with a surface area of 10 cm² for 3 hours. The half-cell potentials of copper and silver electrodes are +0.34 V and +0.80 V, respectively, and the m

Answer 1

To calculate the current density in the electrochemical cell, we need to consider the total surface area of both electrodes and the current passing through them. The current density  J  is defined as the current  I  divided by the surface area  A  of the electrode:J = I / AFirst, we need to convert the surface areas of the copper and silver electrodes from cm to m:Copper electrode surface area: 25 cm *  1 m / 10,000 cm  = 0.0025 mSilver electrode surface area: 10 cm *  1 m / 10,000 cm  = 0.0010 mNext, we need to find the total surface area of both electrodes:Total surface area  A_total  = 0.0025 m + 0.0010 m = 0.0035 mNow, we can calculate the current density  J  using the given current  I = 2.5 A  and the total surface area  A_total = 0.0035 m :J = I / A_total = 2.5 A / 0.0035 m = 714.29 A/mSo, the current density in the electrochemical cell is approximately 714.29 A/m.