A student is given a concentration of 0.1 M of CuSO4 and 0.2 M of AgNO3, and is asked to calculate the total resistance of an electrochemical cell with electrodes made of these two solutions separated by a salt bridge. The standard reduction potentials for Cu and Ag are +0.34 V and +0.80 V, respectively. Calculate the total resistance of the cell at 25°C assuming that the cell has a thickness of 1 cm and a cross-sectional area of 10 cm².

Question

A student is given a concentration of 0.1 M of CuSO4 and 0.2 M of AgNO3, and is asked to calculate the total resistance of an electrochemical cell with electrodes made of these two solutions separated by a salt bridge. The standard reduction potentials for Cu and Ag are +0.34 V and +0.80 V, respectively. Calculate the total resistance of the cell at 25°C assuming that the cell has a thickness of 1 cm an

Answer 1

To calculate the total resistance of the electrochemical cell, we first need to determine the cell potential  E_cell  using the Nernst equation. Then, we can use Ohm's law to find the resistance.1. Calculate the cell potential  E_cell  using the Nernst equation:E_cell = E_cathode - E_anodeSince Ag has a higher standard reduction potential, it will act as the cathode, and Cu will act as the anode. Therefore,E_cell = E_Ag - E_CuE_cell =  +0.80 V  -  +0.34 V E_cell = 0.46 V2. Calculate the current  I  using the Faraday's law of electrolysis:I = n * F * A *  C_Ag - C_Cu  / dwhere n is the number of electrons transferred  which is 2 for both Cu and Ag , F is the Faraday constant  96485 C/mol , A is the cross-sectional area  10 cm , C_Ag and C_Cu are the concentrations of AgNO3 and CuSO4  0.2 M and 0.1 M, respectively , and d is the thickness  1 cm .I = 2 * 96485 C/mol * 10 cm *  0.2 M - 0.1 M  / 1 cmI = 1930 C/s3. Calculate the total resistance  R  using Ohm's law:R = V / IR = 0.46 V / 1930 C/sR = 2.38 x 10 So, the total resistance of the electrochemical cell is approximately 2.38 x 10  at 25C.