Mistake in Pub. Soln on Carmen - 20.40(b) in 12th-14th ed.

[robert zellmer]

There is a mistake in the solutions manual on Carmen (12th-14th ed.) for 
20.40(b).

20.40(b)  The solutions use an electrode not given in the problem.
                 To get the smallest (+) emf from the 1/2-cells given would
                 be using the first two half-cells listed,

     Anode: AuBr4^- (aq)  +  3 e-  ---->  Au(s)  +  4 Br- (aq)

     Cat:    Eu^3+ (aq)  +  1 e-  ----> Eu^2+ (aq)

     To balance the electrons you need to multiply the cathode rxn by 3

      Cat:    3 (Eu^3+ (aq)  +  1 e-  ----> Eu^2+ (aq))

     Then you get:

     Anode: AuBr4^- (aq)  +  3 e-  ---->  Au(s)  +  4 Br- (aq)

     Cat:    3 Eu^3+ (aq)  +  3 e-  ---->  3 Eu^2+ (aq)

     Add them to get the overall rxn.

     AuBr4^- (aq)  +  Eu^3+ (aq) ---->  Au(s)  +  4 Br- (aq)  +  3 Eu^2+ 
(aq)

             E^o = E^o (cathode) - E^o (anode)

                     = (-.43 V) - (-.858 V)

                     = + 0.43 V

Dr. Zellmer
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