Question

An electrochemical cell is expressed as Cu(s) | Cu^2+ (0.20 M) || I^- (0.10 M) | I_3^- (0.20 M) | Pt. I_3^- + 2e^- rightarrow 3I^-, E^0 = 0.535 V Cu^2+ + 2e^- rightarrow Cu(s) E^0 = 0.339 V Please answer the following questions: a. Write half-cell reactions at anode and at cathode. b. Write whole-cell reaction. c. Calculate the potential/voltage of the cell. d. Calculate equilibrium constant for this whole-cell reaction.

Answer 1

a)

anode --> oxidtion

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

cathode

I3-(aq) + 2e- --> 3I-(aq)

b)

balanced

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

I3-(aq) + 2e- --> 3I-(aq)

add all

I3-(aq) + 2e- + Cu(s) --> Cu2+(aq) + 2e- +  I3-(aq)

I3-(aq) + Cu(s) --> Cu2+(aq) + 3I-(aq)

c)

E° = ERed - Eox = 0.535 - 0.339 = 0.196 V

d)

E = E° -0.0592/n * log(Q)

I3-(aq) + Cu(s) --> Cu2+(aq) + 3I-(aq)

Q = [Cu2+][I-]^3/[I3-]

Q = (0.2)(0.1^3)/(0.2) = 0.001

n = 2 electrons

E = 0.196 - 0.0592/2 * log(0.001) = 0.2848 V