chem quiz
Post on 02-Nov-2015
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2) Calculate the potential of the following cell
2) Calculate the potential of the following cell
a. Ag/Ag+ (0.0200 M) // Cu2+ (0.0200 M) / Cu
Ag+ + e ( Ag(s) Eo= 0.799V
Cu2+ + 2e ( Cu(s) Eo= 0.337 V
1. E= Eo 0.0592 log 1
1 [Ag+]
= 0.799 0.0592 log 1
1 [0.0200 M]
= 0.698 V
2. E= Eo= 0.0592 log 1
2 [Cu2+]
= 0.377 V 0.0592 log 1
2 [0.0200]
= 0.287 V
Ecell = E cathode E anode
= E Cu2+/ Cu E Ag+/ Ag
= 0.287 0.698 V
= -0.411 V (non-spontaneous)
b. Zn/ Zn2+ (0.0955 M) // Co2+ (6.78 x 10-3 M) / Co
Eo= Zn2+/ Zn = -0.763 V
Eo= Co2+/Co = -0.277 V
1. E= Eo= - 0.0592 log 1
2 0.0955 M
= -0.763 0.0592 log
1
2 0.0955 M
=-0.793
2. E= Eo 0.0592 log 1
2 6.78 x 10-3
= -0.277 V 0.0592 log
1
2 6.78 x 10-3
= -4.647
E cell= E cathode Eanode
= -4.647-(-0.793)
= -3.854
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