To calculate the polarization of the copper electrode, we need to determine the cell potential E_cell of the electrochemical cell formed by the Cu2+|Cu electrode and the H+|H2 electrode. We can use the Nernst equation to find the cell potential:E_cell = E_cell - RT/nF * ln Q Where:E_cell is the standard cell potentialR 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 2 for Cu2+ + 2e- -> Cu F is the Faraday constant 96485 C/mol Q is the reaction quotient, which is equal to [Cu2+]/[H+]^2 for this reactionFirst, we need to find the standard cell potential E_cell . Since the standard reduction potential of Cu2+|Cu electrode is 0.34 V and the standard reduction potential of H+|H2 electrode is 0 V, the standard cell potential is:E_cell = E_Cu2+|Cu - E_H+|H2 = 0.34 V - 0 V = 0.34 VNext, we need to find the reaction quotient Q . Since the CuSO4 solution is 0.1 M, the concentration of Cu2+ ions is 0.1 M. The hydrogen electrode has a hydrogen pressure of 1 atm, which means the concentration of H+ ions is also 1 M assuming 1 atm pressure corresponds to 1 M concentration . Therefore, Q = [Cu2+]/[H+]^2 = 0.1/1^2 = 0.1.Now we can plug these values into the Nernst equation:E_cell = 0.34 V - 8.314 J/molK * 298 K / 2 * 96485 C/mol * ln 0.1 E_cell = 0.34 V - 0.0129 V * ln 0.1 E_cell 0.34 V + 0.029 VE_cell 0.369 VThe polarization of the copper electrode is the difference between the standard reduction potential and the cell potential:Polarization = E_Cu2+|Cu - E_cell = 0.34 V - 0.369 V = -0.029 VSo, the polarization of the copper electrode is -0.029 V.