Question

Calculate the standard cell potential of the following electrochemical cell at 298 K, given the half-reactions and their standard reduction potentials in brackets:Cu(s) | Cu^2+(0.001 M) || Ag+(0.1 M) | Ag(s)

Answer 1

To calculate the standard cell potential, we will 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 moles of electrons transferred, F is the Faraday constant  96,485 C/mol , and Q is the reaction quotient.First, we need to determine the half-reactions and their standard reduction potentials:Cu aq  + 2e  Cu s   E = +0.34 VAg aq  + e  Ag s   E = +0.80 VSince we want the Cu to be oxidized and Ag to be reduced, we will reverse the first half-reaction and subtract its standard reduction potential:Cu s   Cu aq  + 2e  E = -0.34 VNow, we can add the two half-reactions to get the overall reaction:Cu s  + 2Ag aq   Cu aq  + 2Ag s Next, we can calculate the standard cell potential  E_cell  by adding the standard reduction potentials of the two half-reactions:E_cell = E_red Ag  + E_ox Cu  = 0.80 V - 0.34 V = 0.46 VNow we can calculate the reaction quotient  Q  for the reaction:Q = [Cu]/[Ag] =  0.001 M / 0.1 M  = 0.001/0.01 = 0.1Now we can plug all the values into the Nernst equation:E_cell = E_cell -  RT/nF  * ln Q E_cell = 0.46 V -   8.314 J/molK  *  298 K  /  2 * 96,485 C/mol   * ln 0.1 E_cell = 0.46 V -  0.0129 V  * ln 0.1 E_cell  0.46 V + 0.0299 VE_cell  0.49 VThe standard cell potential of the electrochemical cell at 298 K is approximately 0.49 V.