Question

Calculate the power output of a Galvanic cell consisting of a standard hydrogen electrode and a standard copper electrode at 298 K when the concentration of copper ions in the cathode half-cell is 0.1 M and hydrogen gas pressure in the anode half-cell is at 1 bar. The standard reduction potentials for the hydrogen and copper half-reactions are +0.0 V and +0.34 V, respectively.

Answer 1

To calculate the power output of the Galvanic cell, we first need to determine the cell potential  E_cell  under non-standard conditions using the Nernst equation:E_cell = E_cell -  RT/nF  * ln Q where:E_cell = standard cell potentialR = gas constant  8.314 J/molK T = temperature  298 K n = number of electrons transferred in the redox reaction  2 for this case, as Cu + 2e  Cu and 2H + 2e  H F = Faraday's constant  96485 C/mol Q = reaction quotientThe standard cell potential  E_cell  can be calculated by subtracting the standard reduction potential of the anode  hydrogen  from that of the cathode  copper :E_cell = E_cathode - E_anode = 0.34 V - 0.0 V = 0.34 VThe reaction quotient  Q  can be calculated using the concentrations and pressures of the species involved in the reaction:Q = [Cu]/[H] * P H Since the hydrogen electrode is under standard conditions, the concentration of H ions is 1 M and the pressure of H gas is 1 bar:Q =  0.1 M / 1 M  * 1 bar = 0.1Now, 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 - 0.0296 V = 0.3104 VThe power output  P  of the Galvanic cell can be calculated using the formula:P = E_cell * IHowever, we do not have information about the current  I  flowing through the cell. Therefore, we cannot calculate the exact power output of the Galvanic cell. But we have determined the cell potential under non-standard conditions, which is 0.3104 V.