standard electrode potentials when the concentrations of cu 2+ (aq) and zn 2+ (aq) are both kept at...
Post on 16-Dec-2015
226 Views
Preview:
TRANSCRIPT
1
Standard Electrode Potentials When the concentrations of Cu2+
(aq) and Zn2+(aq) are
both kept at unit activity, the emf of the galvanic cell is 1.10 V at 25 oC. (V is the unit for voltage).
2
Standard Electrode Potentials When the concentrations of Cu2+
(aq) and Zn2+(aq) are
both kept at unit activity, the emf of the galvanic cell is 1.10 V at 25 oC. (V is the unit for voltage).
Recall: The activity of a species X can be written as
where is called the activity coefficient.
[X]a XX γ
Xγ
3
Standard Electrode Potentials When the concentrations of Cu2+
(aq) and Zn2+(aq) are
both kept at unit activity, the emf of the galvanic cell is 1.10 V at 25 oC. (V is the unit for voltage).
Recall: The activity of a species X can be written as
where is called the activity coefficient. For fairly dilute solutions, , so that .
[X]a XX γ
1X γXγ
[X]aX
4
Standard Electrode Potentials When the concentrations of Cu2+
(aq) and Zn2+(aq) are
both kept at unit activity, the emf of the galvanic cell is 1.10 V at 25 oC. (V is the unit for voltage).
Recall: The activity of a species X can be written as
where is called the activity coefficient. For fairly dilute solutions, , so that . We will make the gross assumption (as does the text) that
at a concentration of 1 M, and replace unit activity for Cu2+
(aq) and Zn2+(aq) by concentrations of 1
M.
[X]a XX γ
1X γXγ
[X]aX
1X γ
5
The value of the emf is independent of the amount of solution or the size of the electrodes.
6
The value of the emf is independent of the amount of solution or the size of the electrodes.
The measured emf can be treated as the sum of the two electric potentials arising from the Zn and Cu electrodes.
7
The value of the emf is independent of the amount of solution or the size of the electrodes.
The measured emf can be treated as the sum of the two electric potentials arising from the Zn and Cu electrodes.
It is impossible to measure the potential of a single electrode: any complete circuit must by necessity, contain two electrodes.
8
The value of the emf is independent of the amount of solution or the size of the electrodes.
The measured emf can be treated as the sum of the two electric potentials arising from the Zn and Cu electrodes.
It is impossible to measure the potential of a single electrode: any complete circuit must by necessity, contain two electrodes.
A simple way out of this dilemma is to chose a certain electrode and arbitrarily set its potential value to zero volts.
9
This electrode can then be used to determine the potentials of other electrodes by measuring the emf of various cells.
10
This electrode can then be used to determine the potentials of other electrodes by measuring the emf of various cells.
The standard hydrogen electrode is chosen as the reference (abbreviated as SHE).
11
This electrode can then be used to determine the potentials of other electrodes by measuring the emf of various cells.
The standard hydrogen electrode is chosen as the reference (abbreviated as SHE). The reaction is
2 H+(aq) + 2e- H2(g) E0 = 0 V
12
This electrode can then be used to determine the potentials of other electrodes by measuring the emf of various cells.
The standard hydrogen electrode is chosen as the reference (abbreviated as SHE). The reaction is
2 H+(aq) + 2e- H2(g) E0 = 0 V
(1 M) (1 bar)
13
This electrode can then be used to determine the potentials of other electrodes by measuring the emf of various cells.
The standard hydrogen electrode is chosen as the reference (abbreviated as SHE). The reaction is
2 H+(aq) + 2e- H2(g) E0 = 0 V
(1 M) (1 bar) The symbol for the emf is Ecell (some use just E).
14
This electrode can then be used to determine the potentials of other electrodes by measuring the emf of various cells.
The standard hydrogen electrode is chosen as the reference (abbreviated as SHE). The reaction is
2 H+(aq) + 2e- H2(g) E0 = 0 V
(1 M) (1 bar) The symbol for the emf is Ecell (some use just E). The
superscript 0 denotes standard state conditions, which for the present case refers to H+
(aq) at 1 M, H2(g) at 1 bar, and a reference temperature of exactly 25 oC is assumed.
15
For a half-cell reaction at standard conditions, the notation E0 is employed. Other notation that is employed is or sometimes , this latter one signifying that it is a standard reduction potential.
Standard emf: The potential difference between two electrodes which can be measured for a given cell when all solutes are at a concentration of 1 M and all gases are at 1 bar.
0cell-halfE 0
redE
16
Suppose we want to determine the for the reaction
Cu2+(aq) + 2 e- Cu(s)
then set up the cell with a SHE, so that:
anode: H2(g) 2 H+(aq) + 2e- = 0 V
cathode: Cu2+(aq) + 2 e- Cu(s) = ?
overall reaction: H2(g) + Cu2+(aq) 2 H+
(aq) + Cu(s)
= 0.34 V Since the two values must add to 0.34 V,
therefore = 0.34 V for the Cu2+ half-reaction.
0cell-halfE
0cell-halfE
0cell-halfE
0cellE
0cell-halfE
0cell-halfE
17
18
The standard electrode potential for the reaction
Zn2+(aq) + 2 e- Zn(s)
0cell-halfE
19
The standard electrode potential for the reaction
Zn2+(aq) + 2 e- Zn(s)
can be measured with a SHE, so that:
0cell-halfE
20
The standard electrode potential for the reaction
Zn2+(aq) + 2 e- Zn(s)
can be measured with a SHE, so that:
anode: Zn(s) Zn2+(aq) + 2 e- = ?
0cell-halfE
0cell-halfE
21
The standard electrode potential for the reaction
Zn2+(aq) + 2 e- Zn(s)
can be measured with a SHE, so that:
anode: Zn(s) Zn2+(aq) + 2 e- = ?
cathode: 2 H+(aq) + 2e- H2(g) = 0 V
0cell-halfE
0cell-halfE
0cell-halfE
22
The standard electrode potential for the reaction
Zn2+(aq) + 2 e- Zn(s)
can be measured with a SHE, so that:
anode: Zn(s) Zn2+(aq) + 2 e- = ?
cathode: 2 H+(aq) + 2e- H2(g) = 0 V
overall reaction: 2 H+
(aq) + Zn(s) H2(g) + Zn2+(aq)
= 0.76 V
0cell-halfE
0cell-halfE
0cellE
0cell-halfE
23
The standard electrode potential for the reaction
Zn2+(aq) + 2 e- Zn(s)
can be measured with a SHE, so that:
anode: Zn(s) Zn2+(aq) + 2 e- = ?
cathode: 2 H+(aq) + 2e- H2(g) = 0 V
overall reaction: 2 H+
(aq) + Zn(s) H2(g) + Zn2+(aq)
= 0.76 V Since the two values must add to 0.76 V,
therefore = 0.76 V for the Zn half-reaction.
0cell-halfE
0cell-halfE
0cellE
0cell-halfE
0cell-halfE
0cell-halfE
24
25
26
27
Standard reduction potential: The voltage associated with a reduction at an electrode when all solutes are 1 M and all gases are at 1 bar.
28
Standard reduction potential: The voltage associated with a reduction at an electrode when all solutes are 1 M and all gases are at 1 bar.
It is most common to table information as reductions potentials.
0cell-halfE
29
30
Standard reduction potential: The voltage associated with a reduction at an electrode when all solutes are 1 M and all gases are at 1 bar.
It is most common to table information as reductions potentials.
Standard oxidation potential: The voltage associated with an oxidation at an electrode when all solutes are 1 M and all gases are at 1 bar.
0cell-halfE
31
The standard oxidation potential for the Zn electrode reaction:
Zn(s) Zn2+(aq) + 2 e-
32
The standard oxidation potential for the Zn electrode reaction:
Zn(s) Zn2+(aq) + 2 e-
is = 0.76 V
0cell-halfE
33
The standard oxidation potential for the Zn electrode reaction:
Zn(s) Zn2+(aq) + 2 e-
is = 0.76 V When we reverse the half-cell reaction, we must
change the sign of .
0cell-halfE
0cell-halfE
34
The standard oxidation potential for the Zn electrode reaction:
Zn(s) Zn2+(aq) + 2 e-
is = 0.76 V When we reverse the half-cell reaction, we must
change the sign of . Thus the standard reduction potential for the reaction:
Zn2+(aq) + 2 e- Zn(s)
0cell-halfE
0cell-halfE
35
The standard oxidation potential for the Zn electrode reaction:
Zn(s) Zn2+(aq) + 2 e-
is = 0.76 V When we reverse the half-cell reaction, we must
change the sign of . Thus the standard reduction potential for the reaction:
Zn2+(aq) + 2 e- Zn(s)
is = -0.76 V
0cell-halfE
0cell-halfE
0cell-halfE
36
Calculation of 0cellE
37
Calculation of Example: Calculate for the reaction Cu2+
(aq) + Zn(s) Zn2+(aq) + Cu(s)
assuming a table of is available.
0cellE
0cellE
0cell-halfE
38
Calculation of Example: Calculate for the reaction Cu2+
(aq) + Zn(s) Zn2+(aq) + Cu(s)
assuming a table of is available. From the table of values the following is
available
0cellE
0cellE
0cell-halfE
0cell-halfE
39
Calculation of Example: Calculate for the reaction Cu2+
(aq) + Zn(s) Zn2+(aq) + Cu(s)
assuming a table of is available. From the table of values the following is
available Zn2+
(aq) + 2 e- Zn(s) = -0.76 V
0cellE
0cellE
0cell-halfE
0cell-halfE
0cell-halfE
40
Calculation of Example: Calculate for the reaction Cu2+
(aq) + Zn(s) Zn2+(aq) + Cu(s)
assuming a table of is available. From the table of values the following is
available Zn2+
(aq) + 2 e- Zn(s) = -0.76 V
Cu2+(aq) + 2 e- Cu(s) = 0.34 V
0cellE
0cellE
0cell-halfE
0cell-halfE
0cell-halfE
0cell-halfE
41
The overall reaction is stripped down to the two half-equations:
42
The overall reaction is stripped down to the two half-equations:
Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
43
The overall reaction is stripped down to the two half-equations:
Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
Cu2+(aq) + 2 e- Cu(s) = 0.34 V
0cell-halfE
44
The overall reaction is stripped down to the two half-equations:
Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
Cu2+(aq) + 2 e- Cu(s) = 0.34 V
Zn(s) Zn2+(aq) + 2 e- = 0.76 V
0cell-halfE
0cell-halfE
45
The overall reaction is stripped down to the two half-equations:
Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
Cu2+(aq) + 2 e- Cu(s) = 0.34 V
Zn(s) Zn2+(aq) + 2 e- = 0.76 V
Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s) = 1.10 V
0cellE
0cell-halfE
0cell-halfE
46
The overall reaction is stripped down to the two half-equations:
Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
Cu2+(aq) + 2 e- Cu(s) = 0.34 V
Zn(s) Zn2+(aq) + 2 e- = 0.76 V
Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s) = 1.10 V
This is the simplest approach to calculate values, and the approach I recommend using.
0cellE
0cell-halfE
0cell-halfE
0cellE
47
There is an alternative approach, that is based on the formula:
where and are the values for the cathode and anode reactions, pulled directly from a standard table of reduction potentials.
0redanode,
0redcathode,
0cell EEE
0redcathode,E 0
redanode,E 0cell-halfE
48
There is an alternative approach, that is based on the formula:
where and are the values for the cathode and anode reactions, pulled directly from a standard table of reduction potentials.
The reaction Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
has the two half-cell reactions:
0redanode,
0redcathode,
0cell EEE
0redcathode,E 0
redanode,E 0cell-halfE
49
There is an alternative approach, that is based on the formula:
where and are the values for the cathode and anode reactions, pulled directly from a standard table of reduction potentials.
The reaction Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
has the two half-cell reactions: anode: Zn(s) Zn2+
(aq) + 2 e-
0redanode,
0redcathode,
0cell EEE
0redcathode,E 0
redanode,E 0cell-halfE
50
There is an alternative approach, that is based on the formula:
where and are the values for the cathode and anode reactions, pulled directly from a standard table of reduction potentials.
The reaction Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
has the two half-cell reactions: anode: Zn(s) Zn2+
(aq) + 2 e-
cathode: Cu2+(aq) + 2 e- Cu(s)
0redanode,
0redcathode,
0cell EEE
0redcathode,E 0
redanode,E 0cell-halfE
51
There is an alternative approach, that is based on the formula:
where and are the values for the cathode and anode reactions, pulled directly from a standard table of reduction potentials.
The reaction Cu2+(aq) + Zn(s) Zn2+
(aq) + Cu(s)
has the two half-cell reactions: anode: Zn(s) Zn2+
(aq) + 2 e-
cathode: Cu2+(aq) + 2 e- Cu(s)
Therefore 0.34 V – (–0.76 V) = 1.10 V
0redanode,
0redcathode,
0cell EEE
0redcathode,E 0
redanode,E 0cell-halfE
0cellE
52
A large number of mistakes are made when using this approach. The most common one is that the reaction involving Zn is an oxidation, so students reverse the sign of the value in the table for the Zn half-reaction, but retain the minus sign in the formula, thereby getting the wrong answer of
-0.42 V.
53
Spontaneity of Redox Reactions
54
Spontaneity of Redox Reactions Under standard state conditions, a redox reaction is
spontaneous in the forward direction if the standard emf of the cell is positive.
55
Spontaneity of Redox Reactions Under standard state conditions, a redox reaction is
spontaneous in the forward direction if the standard emf of the cell is positive.
The more positive the value, the greater the tendency for the substance to be reduced. For example, F2(g) + 2 e- 2 F-
(aq) = 2.87 V,
is one of the largest values,
0cell-halfE
0cell-halfE
0cell-halfE
56
Spontaneity of Redox Reactions Under standard state conditions, a redox reaction is
spontaneous in the forward direction if the standard emf of the cell is positive.
The more positive the value, the greater the tendency for the substance to be reduced. For example, F2(g) + 2 e- 2 F-
(aq) = 2.87 V,
is one of the largest values, which makes F2 one of the strongest oxidizing agents available.
0cell-halfE
0cell-halfE
0cell-halfE
57
Li+(aq) + e- Li = - 3.05 V
0cell-halfE
58
Li+(aq) + e- Li = - 3.05 V
This reaction has the one of the most negative values, making Li+ one of the weakest oxidizing
agents.
0cell-halfE
0cell-halfE
59
Li+(aq) + e- Li = - 3.05 V
This reaction has the one of the most negative values, making Li+ one of the weakest oxidizing
agents. If we reverse the reaction: Li Li+
(aq) + e- = 3.05 V
0cell-halfE
0cell-halfE
0cell-halfE
60
Li+(aq) + e- Li = - 3.05 V
This reaction has the one of the most negative values, making Li+ one of the weakest oxidizing
agents. If we reverse the reaction: Li Li+
(aq) + e- = 3.05 V
Li is one of the strongest reducing agents available.
0cell-halfE
0cell-halfE
0cell-halfE
top related