Question

Considering the reaction: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+(aq) What is the cell potential (in volts) at 25°C when the concentrations of [Cu2+] and [Zn2+] are 0.5 M and 1.0 M, respectively, and the zinc electrode is the anode? The standard reduction potentials for Cu2+/Cu and Zn2+/Zn are 0.34 V and -0.76 V, respectively.

Answer 1

To calculate the cell potential, we can use the Nernst equation:E_cell = E_cell -  RT/nF  * ln Q Where:E_cell = cell potentialE_cell = standard cell potentialR = gas constant  8.314 J/molK T = temperature in Kelvin  25C = 298 K n = number of electrons transferred  2 in this case, since both Cu2+ and Zn2+ have a charge of 2+ F = Faraday's constant  96,485 C/mol Q = reaction quotientFirst, we need to find the standard cell potential  E_cell . Since the zinc electrode is the anode, the reaction at the anode is an oxidation process:Zn s  -> Zn2+ aq  + 2e-  E = -0.76 V At the cathode, the reaction is a reduction process:Cu2+ aq  + 2e- -> Cu s   E = 0.34 V The standard cell potential is the difference between the standard reduction potentials of the cathode and the anode:E_cell = E_cathode - E_anode = 0.34 V -  -0.76 V  = 1.10 VNext, we need to find the reaction quotient  Q . The reaction quotient is given by:Q = [Zn2+]/[Cu2+]Plugging in the given concentrations:Q =  1.0 M / 0.5 M  = 2Now we can plug all the values into the Nernst equation:E_cell = 1.10 V -  8.314 J/molK * 298 K  /  2 * 96,485 C/mol  * ln 2 E_cell = 1.10 V -  0.0257 V  * ln 2 E_cell  1.10 V - 0.018 V = 1.082 VThe cell potential at 25C when the concentrations of [Cu2+] and [Zn2+] are 0.5 M and 1.0 M, respectively, and the zinc electrode is the anode is approximately 1.082 V.