Question

Calculate the polarization of a cell having an anode made of copper metal immersed in a 1.0 M Cu(NO3)2 solution and a cathode made of silver metal in a 1.0 M AgNO3 solution. Given that the potential of the copper electrode is -0.34 V and that of the silver electrode is 0.80 V. Also, assume that the current density is 0.1 A/cm² and that the temperature is 298 K.

Answer 1

To calculate the polarization of the cell, we need to determine the difference between the equilibrium potential and the actual potential of the cell.First, let's find the equilibrium potential  E  of the cell. The equilibrium potential is the difference between the standard reduction potentials of the two half-cells.E = E cathode  - E anode E = 0.80 V -  -0.34 V E = 1.14 VNow, let's find the actual potential  E  of the cell. To do this, we need to use the Nernst equation:E = E -  RT/nF  * ln Q where R is the gas constant  8.314 J/molK , T is the temperature  298 K , n is the number of electrons transferred in the redox reaction, F is the Faraday constant  96485 C/mol , and Q is the reaction quotient.For the redox reaction between copper and silver, the balanced equation is:Cu + 2Ag  Cu + 2AgThe number of electrons transferred, n, is 2.Since both Cu NO3 2 and AgNO3 solutions have a concentration of 1.0 M, the reaction quotient Q is:Q = [Cu][Ag] / [Cu][Ag]Q =  1.0  1.0  /  1.0  1.0 Q = 1Now, we can plug in the values into the Nernst equation:E = 1.14 V -  8.314 J/molK  298 K  /  2  96485 C/mol  * ln 1 E = 1.14 V - 0E = 1.14 VThe actual potential of the cell is 1.14 V.Finally, we can calculate the polarization  P  of the cell:P = E - EP = 1.14 V - 1.14 VP = 0 VThe polarization of the cell is 0 V.