Several cell voltages are given in the following table. The standard reduction p

Question

Several cell voltages are given in the following table. The standard reduction potential for the Cu/Cu2+ pair is:
Cu2+ + 2e- ----> Cu(s) E*red = +0.342V
a) Calcuate the standard reduction potentials for the other two half-cells.
b) Fill in the blanks in the table. Write "non-spontaneous" in the appropriate cell.
c) For the remaining pair, write the spontaneous reactions that occur at the cathode and the anode and the balanced overall spontaneous reaction.

Anode (right)

Cathode (below)

This is what I did...
a) E*cell = E*red(cathode) - E*red(anode)
   +0.342V = E*red(Pd/Pd2+) - (+0.609V)
   Pd/Pd2+ = +0.342V + (+0.609V) = +0.951V
E*cell = E*red(cathode) - E*red(anode)
+0.745V = (+0.342V) - E*red(Cd/Cd2+)
Cd/Cd2+ = +0.745V - (+0.342V) = -0.403V
E*cell = (+0.951V) - (-0.403V) = +1.354V ?? (Cd/Cd2+anode + Pd/Pd2+cathode)

b) Pd/Pd2+ (anode) and Cd/Cd2+ (cathode) = non-spontaneous ??
c) Cd2+ + 2e- --> Cd(s)
   Pd(s) --> Pd2+ + 2e-
= Cd2+ + Pd(s) --> Cd(s) + Pd2+ ??

     

Anode (right)

Cathode (below)

Pd/Pd2+ Cd/Cd2+ Cu/Cu2+ Pd/Pd2+ X +1.354V? +0.609V Cd/Cd2+ non-spontaneous? X non-spontaneous Cu/Cu2+ non-spontaneous +0.745V X

Explanation / Answer

a) E0cell = E0red(cathode) - E0red(anode)

   +0.342V = E0red(Pd/Pd2+) - (+0.609V)

   Pd/Pd2+ = +0.342V + (+0.609V) = +0.951V

E0cell = E0red(cathode) - E0red(anode)

+0.745V = (+0.342V) - E*red(Cd/Cd2+)

Cd/Cd2+ = +0.745V - (+0.342V) = -0.403V

E*cell = (+0.951V) - (-0.403V) = +1.354V (Cd/Cd2+anode + Pd/Pd2+cathode) b) Pd/Pd2+ (anode) and

Cd/Cd2+ (cathode) = spontaneous

c) Cd2+ + 2e- --> Cd(s)
   Pd(s) --> Pd2+ + 2e-
   Overall reaction = Cd2+ + Pd(s) ----> Cd(s) + Pd2+

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