Question

A concentration cell is built based on the following half reactions by using two pieces of copper as electrodes, two Cu2+ solutions, 0.125 M and 0.441 M, and all other materials needed for a galvanic cell. What will the potential of this cell be when the cathode concentration of Cu2+ has changed by 0.019 M at 303 K? Cu2+ + 2 e- → Cu Eo = 0.341 V

Answer 1

In concentration cell E std is zero

Reduction reaction

Cu2+ + 2e- --- > Cu(s)

Number of moles of electrons =2

Use Nernst equation

Ecell = Estd – RT/nF ln ([Cu2+]dil/ [Cu2+]conc.)

R = 8.314 J per (K mol) , T = 303 K

F= 96484 C/mol

Calculate equilibrium concentrations

[cu2+] cathode = 0.441 – change = 0.441 – 0.019 = 0.422 M

Anode = 0.125 + 0.019 = 0.144

Plug in and calculate Ecell

Ecell = 0 V – (8.314 J per ( K mol) x 303 K ) / (2 mol x 96484 C/mol) x ln (0.144/0.422)

Ecell= -0.014 V

Ecell would be -0.014 V