Question

Calculate the standard cell potential of the following electrochemical cell: Ag(s) │ AgCl(s) │ Cl^-(aq, 0.040 M) ‖ Cu^2+(aq, 0.020 M) │ Cu(s) at 298 K. The reduction potential of AgCl/Ag couple is +0.22 V and the reduction potential of Cu^2+/Cu couple is +0.34 V.

Answer 1

To calculate the standard cell potential, we can use the Nernst equation:E_cell = E_cell -  RT/nF  * ln Q where E_cell is the cell potential, E_cell is the standard cell potential, R is the gas constant  8.314 J/molK , T is the temperature in Kelvin  298 K , n is the number of electrons transferred, F is the Faraday constant  96,485 C/mol , and Q is the reaction quotient.First, we need to write the half-reactions for the given cell:AgCl s  + e  Ag s  + Cl aq   E = +0.22 VCu aq  + 2e  Cu s           E = +0.34 VNext, we need to balance the half-reactions to obtain the overall cell reaction:2 AgCl s  + e  Ag s  + Cl aq    2E = 2 +0.22 V Cu aq  + 2e  Cu s             E = +0.34 V----------------------------------------------2AgCl s  + Cu aq   2Ag s  + 2Cl aq  + Cu s Now we can calculate the standard cell potential  E_cell :E_cell = E_cathode - E_anodeE_cell =  +0.34 V  -  2 * +0.22 V E_cell = +0.34 V - 0.44 VE_cell = -0.10 VNow we can calculate the reaction quotient  Q :Q = [Cl]^2 / [Cu]Q =  0.040 ^2 /  0.020 Q = 0.0016 / 0.020Q = 0.08Now we can use the Nernst equation to calculate the cell potential  E_cell :E_cell = E_cell -  RT/nF  * ln Q E_cell = -0.10 V -   8.314 J/molK * 298 K  /  2 * 96,485 C/mol   * ln 0.08 E_cell = -0.10 V -   2.479 * 10 V  * ln 0.08  E_cell = -0.10 V + 0.036 VE_cell = -0.064 VThe standard cell potential of the given electrochemical cell is -0.064 V.