Question

Calculate the polarization potential in an electrochemical cell if the concentration of zinc ions in the anode compartment is 0.5 M, the concentration of copper(II) ions in the cathode compartment is 1.0 M, and the voltage of the cell is 1.1 V. The standard reduction potential of copper(II) ions is +0.34 V and the standard reduction potential of zinc ions is -0.76 V.

Answer 1

To calculate the polarization potential in an electrochemical cell, we need to use the Nernst equation:E_cell = E_cell -  RT/nF  * ln Q Where:E_cell = cell potential  voltage  under non-standard conditionsE_cell = cell potential  voltage  under standard conditionsR = gas constant  8.314 J/molK T = temperature  in Kelvin, assuming 298 K or 25C n = number of electrons transferred in the redox reaction  2 for the Cu/Zn redox couple F = Faraday's constant  96,485 C/mol Q = reaction quotient, which is the ratio of the concentrations of the products to the reactantsFirst, we need to find the standard cell potential  E_cell :E_cell = E_cathode - E_anodeE_cell =  +0.34 V  -  -0.76 V E_cell = 1.10 VNow, we can calculate the reaction quotient  Q :Q = [Zn]/[Cu]Q =  0.5 M / 1.0 M Q = 0.5Now, we can plug the values into the Nernst equation:E_cell = 1.10 V -  8.314 J/molK * 298 K  /  2 * 96,485 C/mol  * ln 0.5 E_cell = 1.10 V -  0.0257 V  * ln 0.5 E_cell = 1.10 V -  -0.0179 V E_cell = 1.1179 VThe polarization potential in the electrochemical cell is approximately 0.0179 V  1.1179 V - 1.10 V .