Transcript 4-5-2011

4-5-2011
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Calculation of the standard emf of
an electrochemical cell
The procedure is simple:
1. Arrange the two half reactions placing
the one with the larger Eo up (the
cathode).
2. The half reaction with the lower Eo is
placed down (the anode).
3. Eocell = Eocathode - Eoanode
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A galvanic cell consists of a Mg electrode in
a 1.0 M Mg(NO3)2 and a Ag electrode in a
1.0 M AgNO3 solution. Find the overall
reaction and the standard cell emf.
From table of standard reduction potentials,
we have:
Ag+ + e g Ag(s)
Eo = 0.80V
Mg2+ + 2e g Mg(s)
Eo = -2.37V
From diagonal rule Ag+ will react with Mg(s),
therefore the reaction will be as follows:
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Multiply the first half reaction by 2 to account for
the number of electrons in the second one and
reverse the second half reaction:
2Ag+ + 2e g 2Ag(s)
Eo = 0.80V
Mg(s) g Mg2+ + 2e
Eo = +2.37V
2Ag+ + Mg(s) g Mg2+ + 2Ag(s)
Eorxn = Eocathode - Eoanode
Eorxn = 0.80 – (- 2.37) = 3.17V
Eo = 3.17V
Remember to use Eo values with sign as it
appears in the table (reduction potential).
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What is the standard emf of an electrochemical cell
made of a Cd electrode in a 1.0 M Cd(NO3)2 solution
and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e-
Cd (s) E0 = -0.40 V
Cr3+ (aq) + 3e-
Cr (s)
E0 = -0.74 V
Anode (oxidation):
Cr (s)
Cr3+ (1 M) + 3e- x 2
Cathode (reduction): 2e- + Cd2+ (1 M)
2Cr (s) + 3Cd2+ (1 M)
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Cd (s) x 3
3Cd (s) + 2Cr3+ (1 M)
0
0
E0 = Ecathode
- Eanode
cell
0 = -0.40 – (-0.74)
Ecell
0 = 0.34 V
Ecell
Remember that multiplying a half reaction
with any coefficient will not change the
value of Eo.
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Spontaneity of Redox Reactions
DG = -nFEcell
DG0
=
0
-nFEcell
n = number of moles of electrons in reaction
J
F = 96,500
= 96,500 C/mol
V • mol
0
DG0 = -RT ln K = -nFEcell
(8.314 J/K•mol)(298 K)
RT
0 =
ln K =
ln K
Ecell
nF
n (96,500 J/V•mol)
0
Ecell
=
0
Ecell
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0.0257 V
ln K
n
0.0592 V
log K
=
n
What is the equilibrium constant for the following
reaction at 250C? Fe2+ (aq) + 2Ag (s) D Fe (s) + 2Ag+ (aq)
0
Ecell
=
0.0257 V
ln K
n
From the reaction as written, Fe half cell is the cathode, and
Ag half cell is the anode:
From diagonal
rule, the rxn is
2Ag+ + 2e g Ag(s)
Eo = 0.80V
non spontaneous
Fe2+ + 2e g Fe(s)
0
0
E0 = EFe
2+/Fe – EAg+ /Ag
E0 = -0.44 – (0.80)
E0 = -1.24 V
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n=2
Eo = -0.44V
K =e
0
Ecell
xn
0.0257 V
K = 1.23 x 10-42
=e
-1.24 V x 2
0.0257 V
Calculate the equilibrium constant for the
following reaction at 25 oC:
Sn(s) + 2Cu2+ D Sn2+ + 2Cu+
From the reaction as written, Cu half cell is the
cathode, and Sn half cell is the anode:
2Cu2+ + 2e g 2Cu+
Eo = 0.15V
Sn2+ + 2e g Sn(s)
Eo = -0.14V
According to the diagonal rule the reaction
occurs spontaneously as written:
Eocell = Eocathode (Cu2+, Cu+) – Eoanode(Sn2+, Sn(s))
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Eo = 0.15 – (-0.14) = 0.29 V
K= e
E0cell x n
0.0257 V
0.29 x 2
K=e
0.0257
K = 6.3*109
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Calculate the standard free energy for the following reaction at 25
oC:
2Au(s) + 3Ca2+ D 2Au3+ + 3Ca(s)
First, we write the two half cells and calculate the cell potential as
they appear in the reaction. From the reaction as written, Ca
half cell is the cathode, and Au half cell is the anode:
2Au3+ + 6e g 2Au(s)
Eo = +1.50V
anode
3Ca2+ + 6e g 3Ca(s)
Eo = -2.87V
cathode
According to the diagonal rule the reaction will not occur spontaneously
as written:
Eocell = Eocathode– Eoanode
Eocell = -2.87 – 1.50 = -4.37 V
DGo = - nFEocell
DGo = - (6)(96500)(-4.37) = 2.53*106 J
DGo = 2.53*103 kJ
The large +ve DGo suggests a non-spontaneous reaction at the standard
states.
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The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q
DG = -nFE
DG0 = -nFE 0
-nFE = -nFE0 + RT ln Q
Nernst equation
E = E0 -
RT
ln Q
nF
At 298 K
E = E0 -
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0.0257 V
ln Q
n
E = E0 -
0.0592 V
log Q
n
Will the following reaction occur spontaneously at
250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M?
Fe2+ (aq) + Cd (s)
Fe (s) + Cd2+ (aq)
From the reaction as written, Fe half cell is the
cathode, and Cd half cell is the anode, use these to
calculate the standard electrode potential:
0
0
E0 = EFe
2+/Fe – ECd2+ /Cd
E0
= -0.44 – (-0.40)
E0 = -0.04 V
We cannot use Eo to
predict whether the rxn
is spontaneous or not
since we do not have
standard conditions
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0.0257 V
ln Q
n
0.010
0.0257 V
ln
E = -0.04 V 2
0.60
E = 0.013
E = E0 -
E>0
Spontaneous
Will the following reaction occur spontaneously at
250C if [Fe2+] = 0.68 M and [Co2+] = 0.15 M?
Fe2+ (aq) + Co (s)
Fe (s) + Co2+ (aq)
From the reaction as written, Fe half cell is the
cathode, and Co half cell is the anode, use these to
calculate the standard electrode potential:
0
0
E0 = EFe
2+/Fe – ECo2+ /Co
E0 = -0.44 – (-0.28)
E0 = -0.16 V
E = E0 14
0.0257 V
ln ([Co2+]/[Fe2+])
n
0.15
0.0257 V
ln
E = -0.16 V 2
0.68
E = - 0.14
E<0 a
Non-spontaneous
Now think at what ratio of [Co2+]/[Fe2+] will
the reaction be spontaneous?
To find such a ratio we must pass by the
point where Eocell = 0
0 = - 0.16 - 0.0257 V ln ([Co2+]/[Fe2+])
n
[Co2+]/[Fe2+] = 4*10-6, the ratio must be
less than 4*10-6
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Concentration Cells
Galvanic cell from two half-cells composed of the same
material but differing in ion concentrations.
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Selected Problems
2, 3, 6, 11-13, 15, 16, 17, 18, 21, 22, 24, 25,
29, 32, 34.
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