Standard cell potentials are determined with 1.0 M solutions of ions. In this ex
ID: 894395 • Letter: S
Question
Standard cell potentials are determined with 1.0 M solutions of ions. In this experiment, we will use 0.10 M solutions of Zn2+, Cu2+, Pb2+, and Ag1+ to minimize the amount of hazardous waste generated. In this series of questions, you will figure out what effect this will have on the voltages you observe.
Consider a cell consisting of a Cu2+/Cu couple and an Ag1+/Ag couple.
(a) Starting with 1.0 M solutions of ions (standard conditions), evaluate Q in the Nernst Equation.
Q =
(b) What is log Q for this cell?
Q =
(c) The cell potential in this case should be which of the following?
less than E°
greater than E°
equal to E°
(d) Starting with 0.1 M solutions of ions (at room conditions), evaluate Q in the Nernst Equation.
Q =
(e) What is log Q for this cell?
Q =
(f) The cell potential in this case should be which of the following?
equal to E°
less than E°
greater than E°
Explanation / Answer
Cu(s) + 2Ag+(aq) <----> Cu2+(aq) + 2Ag(s)
E0cell = E0cathode- E0anode
= 0.8 - 0.337 = 0.463 V
nernest equation :
E = E0 - 0.0591/n log[Cu2+]/[Ag+]^2
Q = [Cu2+]/[Ag+]^2
if the [Cu2+] = 1 M [Ag+] = 1 M
a) Q = 1/1^2 = 1
B) log Q = log1 = 0
c) equal to E° = 0.463
d) if [Cu2+] = 0.1 M [Ag+] = 0.1 M
Q = 0.1/(0.1^2) = 10
E) log Q = log 10 = 1
f) E = 0.463 - (0.0591/2)log 10
= 0.4335 V
less than E°
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