Transcript Pb 2+ - ChemGod.com

st
1
you need the BALANCED
EQUATION
Pb2+ + 2 e- Pb(s) E0red = -0.13 V
Cu2+ + 2 e- Cu(s) E0red = 0.34 V
You need an OXIDATION and a
REDUCTION.
E0cell >0 or nothing happens!
st
1
you need the BALANCED
EQUATION
Pb(s)  Pb2+ + 2 e- E0ox = +0.13 V
Cu2+ + 2 e- Cu(s) E0red = 0.34 V
Pb(s) + Cu2+ Pb2+ + Cu(s) E0cell =
0.13+0.34V
That assumes 1 M concentrations.
Since they aren’t 1 M: NERNST EQUATION
st
1
you need the BALANCED
EQUATION
Pb(s) + Cu2+ Pb2+ + Cu(s) E0cell = 0.47 v
Ecell = E0cell – 0.0592/n log Q
Ecell = E0cell – 0.0592 log Pb2+
n
Cu2+
Ecell = 0.47V – 0.0592 log 0.050
2
1.50
Ecell = 0.51 V
What is the cell potential when the
concentration of Cu2+ has fallen to
0.200 M?
Ecell = E0cell – 0.0592 log Pb2+
n
Cu2+
Ecell = 0.47V – 0.0592 log
1.35
2
0.2
Ecell = 0.45 V
Pb(s) + Cu2+ Pb2+ + Cu(s)
I
-
1.5 0.0 5
C
E
-
-x
+x
0.2 1.3 5
What are the concentrations of Pb2+
and Cu2+ when the cell potential
falls to 0.35 V?
Ecell = E0cell – 0.0592 log Pb2+
n
Cu2+
I
0.35 = 0.47V – 0.0592 log
0.05+x
C
2
1.5-x
E
-0.12 = -0.0296 log (0.05+x/1.5x)
4.054 = log (0.05+x/1.5-x)
104.054 = 0.05+x/1.5-x
1.1324x104 = 0.05+x/1.5-x
1.6986x104 – 1.1324x104 x =
0.05 +x
1.6986x104 = 1.1325x104 x
Pb(s) + Cu2+ Pb2+ + Cu(s)
-
1.5
-x
0.05
1.5-x
0.05+
x
+x
-
What are the concentrations of Pb2+ and
Cu2+ when the cell potential falls to
0.35 V?
Ecell = E0cell – 0.0592 log Pb2+
n
Cu2+
I
0.35 = 0.47V – 0.0592 log
0.05+x
C
2
1.5-x
E
-0.12 = -0.0296 log (0.05+x/1.5x)
4.054 = log (0.05+x/1.5-x)
104.054 = 0.05+x/1.5-x
1.1324x104 = 0.05+x/1.5-x
1.6986x104 – 1.1324x104 x =
0.05 +x
1.6986x104 = 1.1325x104 x
Pb(s) + Cu2+ Pb2+ + Cu(s)
-
1.5
0.05
-1.4999
+1.4999
0.0001 1.54
-
A.
B.
C.
D.
E.
Makes sense
Still don’t get it
Is it time to click already?
Love you
Hate you