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Chapter 20
Chapter 20
Electrochemistry
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Electrochemistry deals with the relationships between electricity and chemical reactions.
Oxidation-reduction (redox) reactions were introduced in Chapter 4
Can be simple displacement reactions:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
20.1 Oxidation-Reduction Reactions
• Redox reactions can also be more complex, with structural and composition changes as well
as an exchange of electrons.
• 5VO2+(aq) + MnO4-(aq) + H2O(l) → 5VO2+(aq) + Mn2+(aq) + 2H+(aq)
• 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Oxidation and Reduction
You can’t have one without the other!
Oxidation
Decrease in number of electrons
(loss of electrons)
Increase in oxidation number
Oxidation number: -3 -2 -1 0 +1 +2 +3
Increase in number of electrons
(gain of electrons)
Decrease in oxidation number
Reduction
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Half-Reactions
Oxidation-reduction reactions can be written as separate oxidation and reduction reactions,
called half-reactions.
Oxidation:
VO2+(aq) + H2O(l) → VO2+(aq) + 2H+(aq) + eVO2+ is called the reducing agent (or reductant), because it causes the reduction of another
substance; the reducing agent is oxidized in the process
Reduction:
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
MnO4- is called the oxidizing agent (or oxidant), because it causes the oxidation of another
substance; the oxidizing agent is reduced in the process
Oxidation-reduction can be considered to be the competition between two substances for
electrons. The one with the greater attraction for additional electrons becomes the oxidizing
agent; the one with the least is the reducing agent.
20-1
Chapter 20
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Spontaneous Oxidation-Reduction Reactions
If we place a metal in a solution of another metal ion, sometimes we get metal deposition,
sometimes not. How do we decide?
Zn + Sn2+ → Zn2+ + Sn
Sn + Zn2+ → no rxn
How do we know which substances will act as oxidizing agents or as reducing agents?
An activity series gives the relative activity of substances as oxidizing or reducing agents.
Review from Chapter 4.
Determine an activity series in several ways:
activity in displacing H2 from water
activity in displacing metals from metal ion solutions (more active metal displaces a less
active metal from solution)
generation of an electrochemical potential or voltage
Spontaneous Redox
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Na displaces H2 from H2O
Zn displaces Pb from a solution of Pb2+
Pb + 2Ag+ → Pb2+ + 2Ag
Ag + Pb2+ → no rxn
The reverse reactions can be made to occur with electric current, but they are not
spontaneous.
General rule, using an activity series:
stronger stronger
weaker
weaker
oxidizing + reducing → reducing + oxidizing
agent
agent
agent
agent
Depending on the relative strengths, the reaction can go to completion, or reach a state of
equilibrium.
Activity Series
20-2
Chapter 20
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Predict which of the following combinations will undergo an oxidation-reduction reaction:
Mg + K+
Mg + Zn2+
H2 + Ni2+
H2 + Cu2+
20.2 Balancing Oxidation-Reduction Equations
• Some redox equations can be balanced by inspection, just like other types of reactions.
Zn + CuCl2 → ZnCl2 + Cu
• Net ionic equation:
Zn + Cu2+ → Zn2+ + Cu
• Need to balance atoms and charge
• What is wrong with the following?
Al + Cu2+ → Al3+ + Cu
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Balancing Equations
5Cr + 3MnO4 + 8H2O → 5CrO42- + 3Mn2+ + 16H+
How do we balance an equation as complex as this?
Two methods:
• half-reaction method
• oxidation number change method
Will focus on the half-reaction method since it is useful in more circumstances.
3+
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Half-Reaction Method
Write an unbalanced half-reaction for either oxidation or reduction.
Balance the half-reaction:
a. Balance all atoms other than H and O.
b. Balance O by adding H2O to the equation.
c. Balance H by adding H+ to the equation.
d. Balance ionic charges by adding electrons to the equation.
e. If in basic solution, add OH- to each side of the equation to neutralize all the H+.
f. Cancel any H2O occurring on both sides.
Write an unbalanced half-reaction for the other process.
Balance by the same procedure.
Equalize the number of electrons lost and gained by multiplying each coefficient in each
half-reaction by the appropriate constant.
Add the two half-reactions and cancel equal amounts of anything occurring on both sides.
Make a final check of atom and charge balances.
Balancing Equations
Cr3+ + MnO4- → CrO42- + Mn2+
How do we balance this equation in acidic solution?
1. Cr3+ → CrO422a. Cr already balanced
2b. Cr3+ + 4H2O → CrO422c. Cr3+ + 4H2O → CrO42- + 8H+
20-3
Chapter 20
2d. Cr3+ + 4H2O → CrO42- + 8H+ + 3e3. MnO4- → Mn2+
4a. Mn already balanced
4b. MnO4- → Mn2+ + 4H2O
4c. MnO4- + 8H+ → Mn2+ + 4H2O
4d. MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
5. 3 e- lost, but 5 e- gained. Equalize by multiplying half-reactions:
5(Cr3+ + 4H2O → CrO42- + 8H+ + 3e-)
3(MnO4- + 8H+ + 5e- → Mn2+ + 4H2O)
5Cr3+ + 20H2O → 5CrO42- + 40H+ + 15e3MnO4- + 24H+ + 15e- → 3Mn2+ + 12H2O
6. Add the equations:
5Cr3+ + 3MnO4- + 20H2O + 24H+ + 15e- → 5CrO42- + 3Mn2+ + 12H2O + 40H+ + 15eCancel any substances on both sides of the equation:
5Cr3+ + 3MnO4- + 8H2O → 5CrO42- + 3Mn2+ + 16H+
7. Check atom and charge balance:
5 Cr, 3 Mn, 20 O, 16 H, 12 + charges on each side
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Balancing Equations in Basic Solution
Neutralize any H with OH- to form H2O and add the same number of OH- to the other side
of the equation.
Cr3+ + 4H2O → CrO42- + 8H+ + 3eAdd 8OH- to each side of the equation:
Cr3+ + 4H2O + 8OH- → CrO42- + 8H+ + 8OH- + 3eForm water by neutralization:
Cr3+ + 4H2O + 8OH- → CrO42- + 8H2O + 3eCancel any water occurring on both sides:
Cr3+ + 8OH- → CrO42- + 4H2O + 3e+
Balance the following equation in acidic solution:
MnO4- + Cl- → Mn2+ + ClO3-
20.3 Voltaic Cells
• Voltaic (or galvanic) cells: spontaneous redox → electricity (or electrical work)
• Electrolysis cells: electricity → redox
• To produce electricity, we must direct the electron flow through an external circuit. We
cannot have direct redox.
• Daniell cell: Zn - Cu
Voltaic Cells
• To produce electricity, we need:
• Isolated half-reactions, using half-cells
• Conductive solids (electrodes) connected by external circuits
• May consist of a reactant/product or be an inert substance such as platinum or
graphite
• Anode: oxidation half-reaction
20-4
Chapter 20
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• Cathode: reduction half-reaction
Externally, the anode has the negative charge; internally, it has a positive charge)
Anions flow towards the anode; cations move away from it and towards the cathode.
Two half-cells must be connected to pass ions
Salt bridge or porous glass
20.4 Cell EMF
• If the half-reactions are carried out separately (but coupled), we find they generate an
electrical current characterized by a voltage (or electromotive force or electrical potential).
This is the force pushing electrons through the circuit.
The voltage produced by a voltaic cell is called the cell potential, Eocell (also the reaction
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potential, Eorxn, when the half-reactions are not separated)
• The voltage is called the standard electromotive force (emf), Eo, under standard conditions.
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Eocell
o
Eored
Electromotive Force
= E ox +
Values are determined relative to the standard hydrogen electrode, with Eo = 0 V
Eored = emf for the reduction half-rxn
Eoox = emf for the oxidation half-rxn
Eocell = Eored + Eoox
We can only measure Eocell, so we must define a standard reference to get Eoox and Eored.
Eo Values
• Reference for Eo is the standard hydrogen electrode, using the reaction:
• 2H+(aq) + 2e- → H2(g) Eo = 0
1M
1 atm
(std conditions)
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Cell Potential
Other values are then measured from Eocell with the standard hydrogen electrode or with
other known half-cells.
Eo = 0.22 V
• AgCl + e- → Ag + Cl• Hg2Cl2 + 2e- → 2Hg + 2Cl- Eo = 0.2802 V
Half-cells such as these are used as reference electrodes. The Ag/AgCl electrode, along
with a glass electrode, is used in a pH meter.
Eo Values
Then get other half-reaction potentials from measured Eocell values.
• Zn + 2H+ → Zn2+ + H2
Eocell = 0.76 V
2+
• Zn → Zn + 2e
Eoox = 0.76 V since Eored = 0.00 V
Reference to Hydrogen Electrode
The E of the hydrogen electrode is defined as 0.00 V. What would be the values of Eo for
other half-reactions if this were defined as 1.00 V?
The Eo of the hydrogen electrode is defined as 0.00 V. What would be the values of Eo for
other half-reactions if this were defined as 1.00 V?
o
20-5
Chapter 20
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What would be the Eo values of overall redox reactions if the reference were defined as 1.00
V?
What would be the Eo values of overall redox reactions if the reference were defined as 1.00
V?
Eo Values
If we reverse a half-reaction, we change the sign of Eo.
Zn2+ + 2e- → Zn
Eored = -0.76 V
o
Measure a value of E cell = 0.63 V for the following reaction.
Zn + Pb2+ → Zn2+ + Pb
What is the Eored of
Pb2+ + 2e- → Pb ?
Eoox = 0.76 V
Zn → Zn2+ + 2ePb2+ + 2e → Pb
Eored = ?
2+
2+
Eocell = 0.63 V
Zn + Pb → Zn + Pb
Eocell = Eored + Eoox
0.63 V = Eored + 0.76 V
Eored = 0.63 V - 0.76 V = -0.13 V
Values determined in this way are listed in Table 20.1 and Appendix E.
Reduction Potentials
20.5 Spontaneity of Redox Reactions
• A redox reaction is spontaneous if Eorxn > 0.
• To determine spontaneity, add the two half-reactions and their Eo values to see if Eorxn has a
positive value.
• What will happen if we place a piece of Zn and a piece of Cu in a solution that contains a
mixture of Zn2+ and Cu2+?
20-6
Chapter 20
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Two possibilities:
• Zn + Cu2+ → Zn2+ + Cu
• Cu + Zn2+ → Cu2+ + Zn
Spontaneous Redox
Two possible reduction half-reactions:
Eored = -0.76 V
Zn2+ + 2e- → Zn
2+
Cu + 2e → Cu
Eored = 0.34 V
Two possible oxidation half-reactions:
Zn → Zn2+ + 2eEoox = 0.76 V
Cu → Cu2+ + 2eEoox = -0.34 V
Two ways to combine them:
Zn2+ + 2e- → Zn
Eored = -0.76 V
2+
Cu → Cu + 2e
Eoox = -0.34 V
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2+
2+
Eorxn = -1.10 V
Cu + Zn → Cu + Zn
Two ways to combine them:
Cu2+ + 2e- → Cu
Eored = 0.34 V
2+
Zn → Zn + 2e
Eoox = 0.76 V
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Eorxn = 1.10 V
Zn + Cu2+ → Zn2+ + Cu
The combination with Eorxn > 0 is spontaneous.
Predicting Reactions
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Vanadium Reduction Potentials:
VO2+ + 2H+ + e- → VO2+ + H2O E°red = +1.00 V
VO2+ + 2H+ + e- → V3+ + H2O
E°red = +0.36 V
E°red = -0.26 V
V3+ + e- → V2+
2+
V + 2e → V
E°red = -1.20 V
Chromium Reduction Potentials:
Cr2O72- + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
Cr3+(aq) + e- → Cr2+(aq)
Cr2+(aq) + 2e- → Cr(s)
Manganese Reduction Potentials:
MnO4-(aq) + e- → MnO42-(aq)
MnO42-(aq) + 4H+(aq) + 2e- → MnO2(s) + 2H2O(l)
MnO2(s) + 4H+(aq) + e- → Mn3+(aq) + 2H2O(l)
Mn3+(aq) + e- → Mn2+(aq)
Mn2+(aq) + 2e- → Mn(s)
Predicting Reactions
Add excess Cr to VO2 . What is the product?
Cr3+(aq) + e- → Cr2+(aq)
E°red = -0.41 V
2+
3+
Cr (aq) → Cr (aq) + e
E°ox = 0.41 V
VO2+ + 2H+ + e- → VO2+ + H2O E°red = +1.00 V
VO2+ + 2H+ + e- → V3+ + H2O E°red = +0.36 V
2+
+
20-7
E°red = +1.33 V
E°red = -0.41 V
E°red = -0.91 V
E°red = +0.56 V
E°red = +2.26 V
E°red = +0.95 V
E°red = +1.51 V
E°red = -1.18 V
Chapter 20
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V3+ + e- → V2+
E°red = -0.26 V
2+
V + 2e → V
E°red = -1.20 V
Product is V2+, because successive Eorxn = 1.41, 0.77, 0.15, -0.79 V
Predicting Reactions
Add excess Cr to Cr2O72-. What is the product?
Cr2O72- + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
E°red = +1.33 V
Cr3+(aq) + e- → Cr2+(aq)
E°red = -0.41 V
2+
Cr (aq) + 2e → Cr(s)
E°red = -0.91 V
Product is Cr2+; the successive values of Eorxn are 2.24 and 0.50 V.
Predicting Reactions
Add excess Cr2+ to Cr2O72-. What is the product?
Cr2O72- + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
E°red = +1.33 V
Cr3+(aq) + e- → Cr2+(aq)
E°red = -0.41 V
2+
Cr (aq) + 2e → Cr(s)
E°red = -0.91 V
Predicting Reactions
Add excess V2+ to MnO4-. What is the product?
V2+ → V3+ + eMnO4-(aq) + e- → MnO42-(aq)
MnO42-(aq) + 4H+(aq) + 2e- → MnO2(s) + 2H2O(l)
MnO2(s) + 4H+(aq) + e- → Mn3+(aq) + 2H2O(l)
Mn3+(aq) + e- → Mn2+(aq)
Mn2+(aq) + 2e- → Mn(s)
E°ox = 0.26 V
E°red = +0.56 V
E°red = +2.26 V
E°red = +0.95 V
E°red = +1.51 V
E°red = -1.18 V
Predicting Reactions
Add Mn to MnO4 . What is the product?
MnO4-(aq) + e- → MnO42-(aq)
MnO42-(aq) + 4H+(aq) + 2e- → MnO2(s) + 2H2O(l)
MnO2(s) + 4H+(aq) + e- → Mn3+(aq) + 2H2O(l)
Mn3+(aq) + e- → Mn2+(aq)
Mn2+(aq) + 2e- → Mn(s)
E°red = +0.56 V
E°red = +2.26 V
E°red = +0.95 V
E°red = +1.51 V
E°red = -1.18 V
2+
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Predict no reaction, but reaction actually occurs to form Mn3+ or MnO2. We will deal with
this discrepancy in the next section. It arises from the fact that there are more half-reactions
that could be considered, which arise from combinations of these half-reactions.
Stability in Aqueous Systems
Disproportionation Reactions
A substance reacts with itself to form new substances with higher and lower oxidation
numbers.
No examples with V or Cr.
MnO42- will disproportionate:
MnO4-(aq) + e- → MnO42-(aq)
E°red = +0.56 V
+
2MnO4 (aq) + 4H (aq) + 2e → MnO2(s) + 2H2O(l)
E°red = +2.26 V
20-8
Chapter 20
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––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
3MnO42- + 4H+ → 2MnO4- + MnO2 + 2H2O
Eorxn = 1.70 V
Stability in Aqueous Systems
3+
Mn will disproportionate:
MnO2(s) + 4H+(aq) + e- → Mn3+(aq) + 2H2O(l)
E°red = +0.95 V
3+
2+
Mn (aq) + e → Mn (aq)
E°red = +1.51 V
–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
2Mn3+ + 2H2O → Mn2+ + MnO2 + 4H+
Eorxn = 0.56 V
Stability in Aqueous Systems
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Reaction with Water
Reduce hydronium ion to release hydrogen gas:
2H+(aq) + 2e- → H2(g)
E°red = 0.000 V
Any substance with Eoox > 0 will reduce H+ to H2
Examples are V, V2+, Cr, Cr2+, Mn
The ions will react, but tend to react only very slowly. There seems to be a kinetic factor
that results in a fast reaction only if Eorxn > 0.4-0.5 V (called an overvoltage).
Reaction with Water
Oxidize water to release oxygen gas:
2H2O(l) → O2(g) + 4H+(aq) + 4e- E°ox = -1.23 V
Any Eored > 1.23 V will result in production of O2. Generally need Eorxn > 0.4-0.5 V for fast
reaction.
Examples are Cr2O72- (very slow), MnO42- (disproportionates faster), Mn3+
(disproportionates faster)
Stability in Aqueous Systems
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Oxidation by O2 in Air
O2(g) + 4H+(aq) + 4e- → 2H2O(l) E°red = 1.23 V
Any Eoox > -1.23 V will result in oxidation by air. Many substances fall into this category
(Eorxn > 0.4-0.5 V for fast reaction).
V
Cr
Mn
V2+
Cr2+
not Mn2+
3+
3+
V
not Cr
Mn3+
VO2+ (very slow)
not MnO2
MnO42- (disproportionates faster)
EMF and Free Energy Change
We have seen three criteria for spontaneity:
• Eo > 0
1 J = 1 coul x 1 V
• ∆Go < 0
• K >> 1
These criteria are related:
• ∆Go = -RT lnK
• ∆Go = - nFEo, ∆G = - nFE
where n = number of e- transferred and F = Faraday constant (charge on 1 mole e-)
1 F = 96,500 coul/mol e- = 96,500 J/V mol e20-9
Chapter 20
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Thermodynamics
These relationships work for half-reactions or complete redox reactions.
Zn + Cu2+ → Zn2+ + Cu
Eo = 1.10
n=2
∆Go = -2mol e- x 96500 J/V mol e- x 1.10 V
∆Go = -212,300 J = -212.3 kJ
∆Go depends on the number of moles, but Eo does not
Voltage and Moles
Note that different size alkaline cells all deliver the same voltage, in spite of different
number of moles of reactants.
o
Eored
Thermodynamics
or Eorxn in the same way that we can add half-reactions
Eocell
We can add E ox to
to give
to give an overall reaction.
• Fe → Fe2+ + 2eEoox = +0.44 V
o
∆G ox = - 2 x 96500 x 0.44 = -84900 J
• Cl2 + 2e- → 2ClEored = 1.36 V
∆Gored = - 2 x 96500 x 1.36 = -262500 J
• Fe + Cl2 → Fe2+ + 2ClEorxn = 1.80 V
∆Go = - 2 x 96500 x 1.80 = -347400 J
∆Gorxn = ∆Goox + ∆Gored
= -84.9 + -262.5 = -347.4 kJ
From Chapter 19, we know that ∆Go values are additive when we add reactions.
Eos are additive when we add half-reactions to give a complete reaction because the value of
n is the same for the half-reactions and the complete reaction.
Eos are not additive when adding two half-reactions to give a third half-reaction because the
value of n is not constant.
We can add ∆Go under all circumstances:
∆Go3 = ∆Go1 + ∆Go2
-n3FEo3 = -n1FEo1 - n2FEo2
n3Eo3 = n1Eo1 + n2Eo2
Eo3 = (n1Eo1 + n2Eo2)/n3
V → V2+ + 2eEo1 = 1.20 V
V2+ → V3+ + eEo2 = 0.26 V
3+
V → V + 3e
Eo3 < 1.20 + 0.26
Eo3 = (2 x 1.20 + 1 x 0.26)/3 = 0.887 V
Given the first two half-reactions, what is the value of E°red for the third?
MnO4-(aq) + e- → MnO42-(aq)
E°1,red = +0.56 V
MnO42-(aq) + 4H+(aq) + 2e- → MnO2(s) + 2H2O(l)
E°2.red = +2.26 V
MnO4-(aq) + 4H+(aq) + 3e- → MnO2(s) + 2H2O(l)
E°3,red = ?
E°3,red = 1.69 V
This is why we could not make correct predictions for this system earlier.
20-10
Chapter 20
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o
Thermodynamics
We can calculate Keq from E :
∆Go = - nFEo = - RT ln K
Eo = (RT/nF) ln K = 2.303 (RT/nF) log K
At 25oC, 2.303 RT/F = 0.05916
Eo = (0.05916/n) log K at 25oC
Thus, we can measure Eo for a redox reaction and then calculate the equilibrium constant for
that reaction.
20.6 Effect of Concentraiton on Cell EMF
• So far, we have been using standard state conditions, but we don’t always have 1 M
solutions. We can correct Eo to E by using the Nernst equation.
∆G = ∆Go + RT ln Q
But, ∆G = - nFE and ∆Go = -nFEo, so
- nFE = -nFEo + 2.303 RT log Q
E = Eo - (2.303 RT/nF) log Q
• At 25oC, E = Eo - (0.05916/n) log Q
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Nernst Equation
At 25oC, E = Eo - (0.05916/n) log Q
E = Eo if Q = 1
When the system reaches equilibrium, Q = K, and E = 0, because Eo = (0.05916/n) log K,
and the cell has “run down”.
Consider the Zn/Cu2+ reaction if more Cu2+ is added to the cell. The voltage becomes
greater than 1.10 V.
What is E of the Zn/Cu2+ reaction if [Cu2+] = 0.010 M and [Zn2+] =1.99 M? Note that this
corresponds to starting with standard conditions and changing to 99% completion of
reaction. Eo = 1.10 V (with [Cu2+] = [Zn2+] = 1.00 M)
Zn + Cu2+ → Zn2+ + Cu
For this reaction, n = 2.
Q = [Zn2+]/[Cu2+]
E = Eo - (0.05916/n) log Q
E = 1.10 V - (0.05916/2) log (1.99/0.010)
E = 1.10 V - (0.05916/2) log 199
E = 1.10 - 0.068 = 1.03 V
For 99.9% reaction (1.999 M Zn2+, 0.001 M Cu2+), E = 1.10 - 0.098 = 1.00 V
For 99.99% reaction, 1.9999 M Zn2+, 0.0001 M Cu2+), E = 1.10 - 0.127 = 0.97 V
Concentration Cells
We can generate a voltage with a cell that contains the same materials in the cathode and
anode compartments, but at different concentrations.
20.7 Batteries (Voltaic Cells)
Voltaic cells are used as portable electricity sources.
one unit = cell
20-11
Chapter 20
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several cells = battery
Commercial cells:
• Anode: usually a metal that can be oxidized
• Cathode: usually an inert conductor surrounded by a substance with a high oxidation
number that can be reduced
• Electrolyte: aqueous solution or moist paste for ion movement
• Electrolytes: acidic, alkaline, organic liquid, molten salt, solid state
Primary cells: one use, then discard
Secondary cells: rechargeable, reusable
Consider some commercial cells and batteries
• What are the anode and cathode?
• What is used as an electrolyte?
• What half-reactions are used?
• What voltage is generated?
LeClanche Dry Cell
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Used in flashlight batteries.
Electrolyte is NH4Cl
Zn → Zn2+ + 2eEo = 0.763 V
+
Eo = 0.5 V
2MnO2 + 2NH4 + 2H2O + 2e → 2NH4OH + 2MnO(OH)
Zn + 2MnO2 + 2NH4+ + 2H2O → Zn2+ + 2NH4OH + 2MnO(OH)
Eo = 1.26 V
The measured voltage is actually 1.5 V, because concentrations are higher than standard
state concentrations.
Alkaline Dry Cell
Electrolyte is KOH
Zn + 2OH- → ZnO + H2O + 2e2MnO2 + H2O + 2e- → 2OH- + Mn2O3
Zn + 2MnO2 → ZnO + Mn2O3 E = 1.5 V
Reactions are similar to the acidic dry cell
Can be recharged a few times
Lead Storage Battery
Used in automobiles
Eo = 1.69 V
PbO2 + 4H+ + SO42- + 2e- → PbSO4 + 2H2O
2Pb + SO4 → PbSO4 + 2e
Eo = 0.35 v
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2Pb + PbO2 + 2SO4 + 4H → 2PbSO4 + 2H2O Eo = 2.04 V
With 6 M H2SO4, the cell produces 2.0 V; the voltage decreases as the cell reactions occur.
Six cells connected in series makes up the common 12 V car battery.
Can be recharged:
2PbSO4 + 2H2O → Pb + PbO2 + 2SO42- + 4H+
Must pass 2 V per cell back through the battery to recharge.
Rapid recharging can electrolyze the acid, producing hydrogen gas, which might explode, or
which can dislodge the lead oxide or sulfate, which shortens the battery life.
20-12
Chapter 20
NiCad Cell
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Used in rechargeable power tools.
Rechargeable because the reaction products stick to the electrodes.
Cd + 2OH- → Cd(OH)2 + 2eEo = 0.81 V
NiO2 (on Ni) + 2H2O + 2e → Ni(OH)2 + 2OH
Eo = 0.49 V
Cd + NiO2 + 2H2O → Cd(OH)2 + Ni(OH)2
Eo = 1.30 V
Silver Cell
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Used in cameras, watches, hearing aids
Electrolyte is 20-40% KOH
Zn + 2OH- → ZnO + H2O + 2e2AgO + H2O + 2e- → Ag2O + 2OHZn + 2AgO + H2O → Zn(OH)2 + Ag2O
Rechargeable if don’t pass this point.
After the AgO is depleted:
Zn + Ag2O + H2O → Zn(OH)2 + 2Ag
Eo = 1.25 V
Eo = 0.61 V
Eo = 1.86 V
Eo = 1.58 V
Fuel Cell
Used for space travel, and in new hydrogen fuel automobiles
Reactants are stored external to the cell and introduced to the electrodes as they are needed.
The reactants are usually gaseous, such as O2 and H2, or O2 and CH4, or O2 and NH3.
O2 + 2H2O + 4e- → 4OHEo = 0.40 V
Eo = 0.83 V
2H2 + 4OH → 4H2O + 4e
2H2 + O2 → 2H2O
Eo = 1.23 V
Recent development uses butane
20.8 Corrosion
• Corrosion of Iron
• Since E°red(Fe2+) < E°red(O2) iron can be oxidized by oxygen.
• Cathode: O2(g) + 4H+(aq) + 4e- → 2H2O(l).
• Anode: Fe(s) → Fe2+(aq) + 2e-.
• Dissolved oxygen in water usually causes the oxidation of iron.
• Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s).
• Oxidation occurs at the site with the greatest concentration of O2.
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Preventing the Corrosion of Iron
Corrosion can be prevented by coating the iron with paint or another metal.
Galvanized iron is coated with a thin layer of zinc.
Zinc protects the iron since Zn is the anode and Fe the cathode:
Zn2+(aq) +2e- → Zn(s), E°red = -0.76 V
Fe2+(aq) + 2e- → Fe(s), E°red = -0.44 V
With the above standard reduction potentials, Zn is easier to oxidize than Fe.
To protect underground pipelines, a sacrificial anode is added.
The water pipe is turned into the cathode and an active metal is used as the anode.
Often, Mg is used as the sacrificial anode:
20-13
Chapter 20
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Mg2+(aq) +2e- → Mg(s), E°red = -2.37 V
Fe2+(aq) + 2e- → Fe(s), E°red = -0.44 V
Also used in water heaters, on gasoline storage tanks, and on ship hulls
20.9 Electrolysis
• Since chemical oxidation-reduction involves the transfer of electrons from one substance to
another, it should be possible to harness the flow of electrons to produce electricity. We do
this with voltaic cells.
• Electricity can also be used to cause non-spontaneous chemical reactions. This process is
called electrolysis.
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Electrolytic Processes with Metals
A variety of metals can be prepared by electrolysis, if a cheap source of electricity is
available. In addition, some metals* are purified by electrolysis.
aluminum
cadmium
calcium
copper*
gold*
lead*
magnesium
sodium
zinc
Purification of Copper
Recovered from its ores by chemical reduction.
Purified by electrolysis.
Recover impurities:
Mo (25%)
Se (93%)
Te (96%)
Au (32%)
Ag (28%)
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Production of Aluminum
Recovered as Al2O3 nH2O from clays and bauxite.
The oxide is difficult to reduce.
The metal forms a protective coating of the oxide.
Al is very active but can be used in air because of the oxide coating.
The oxide coating can be thickened and colored by anodizing.
Electrolysis of Al2O3 (melts at 2045oC) dissolved in cryolite, Na3AlF6 (melts at 1000oC).
Uses 5% of U.S. electricity production
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Electrolysis
Electrolysis is used for isolating active elements, purifying metals, and electroplating.
Pure compounds: H2O, molten salts
Use inert electrodes in the liquid and pass electricity through the system
The negative electrode (cathode) attracts cations; reduction occurs.
The positive electrode (anode) attracts anions; oxidation occurs.
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20-14
Chapter 20
Electrolysis of NaCl
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Cathode:
Na+(l) + e- → Na(l)
Eo = -2.71 V
Anode:
2Cl-(l) → Cl2(g)+ 2eEo = -1.36 V
+
2Na (l) + 2Cl (l) → 2Na(l) + Cl2(g) Eo = -4.08 V
Must supply at least 4.08 V to electrolyze molten sodium chloride.
NaCl melts at 804oC, where Na vaporizes and burns.
Lower the temperature by adding CaCl2. (Why does this work?)
Na reacts with Cl2, even at room temperature.
Commercial operations use a Downs Cell.
How does the Downs Cell solve the problem of reaction between Na and Cl2?
Electrolysis of H2O
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Cathode:
2H2O + 2e- → H2(g) + 2OHEo = -0.82 V at pH 14, -0.41 V at pH 7, 0.00 V at pH 0
Anode:
2H2O → O2(g) + 4H+ + 4eEo = -0.41 V at pH 14, -0.82 V at pH 7, -1.23 V at pH 0
2H2O → 2H2(g) + O2(g)
Eo = -1.23 V at any pH
Must supply at least 1.23 V to electrolyze water. Add an electrolyte to increase electrical
conductivity
Electrolysis of Aqueous Solutions
Products depend on whether it is easier to oxidize or reduce the dissolved ions or water.
Consider a solution of VCl3 under standard conditions at pH 7.
Cathode:
V3+ + e- → V2+
Eo = -0.26 V
2H2O + 2e → H2 + 2OH
Eo = -0.41 V at pH 7
Because of lower voltage, will reduce V3+, not H2O.
Anode:
2Cl- → Cl2 + 2eEo = -1.36 V
2H2O → O2 + 4H+ + 4eEo = -0.82 V at pH 7
Because of lower voltage, will oxidize H2O, not Cl-.
Products are V2+ and O2.
What are the products of electrolysis of VBr3 at pH 7?
Cathode:
V3+ + e- → V2+
Eo = -0.26 V
2H2O + 2e- → H2 + 2OHEo = -0.41 V at pH 7
20-15
Chapter 20
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Anode:
2Br- → Br2 + 2e2H2O → O2 + 4H+ + 4e-
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What are the products of electrolysis of VI3 at pH 7?
Cathode:
V3+ + e- → V2+
Eo = -0.26 V
2H2O + 2e- → H2 + 2OHEo = -0.41 V at pH 7
Anode:
2I- → I2 + 2eEo = -0.54 V
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2H2O → O2 + 4H + 4e
Eo = -0.82 V at pH 7
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Eo = -1.07 V
Eo = -0.82 V at pH 7
What are the products of electrolysis of a mixture of CuCl2 and VCl3 at pH 7?
Cathode:
Eo = -0.26 V
V3+ + e- → V2+
2+
Cu + 2e → Cu
Eo = 0.34 V
Eo = -0.41 V at pH 7
2H2O + 2e → H2 + 2OH
Anode:
2Cl- → Cl2 + 2eEo = -1.36 V
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2H2O → O2 + 4H + 4e
Eo = -0.82 V at pH 7
Faraday’s Law
Faraday’s Law: the mass of product produced by a given amount of current is proportional
to the equivalent weight
Equivalent weight: molar mass/no. e- transferred
Equivalent weight is the mass of substance oxidized or reduced by 1 mole of electrons
We can pass electrons through a series of cells and compare the amount of substance
deposited.
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What is the equivalent weight of … ?
AgNO3
CuSO4
AuCl3
HCl
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Moles deposited:
• Ag+, 1 mole e- → 1 mol Ag
• Cu2+, 1 mole e- → 1/2 mol Cu
• Au3+, 1 mole e- → 1/3 mol Au
• H+, 1 mole e- → 1/2 mol H2
F = charge on 1 mol e- = 96500 coul/mol
charge = current x time
1 coul = 1 A s
moles e- = charge (coul) x 1 mol/96500 coul
moles e- = current (A) x time (s) x 1 coul/1 A s x 1 mol/96500 coul
If we electrolyze molten NaCl with a current of 5000 A for 30 min (or 1800 s), what mass of
Na is produced?
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20-16
Chapter 20
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Na+ + e- → Na
moles e- = 5000 A x 1800 s x 1 mol/96500 coul = 93.26 mol
moles Na = 93.26 mol e- x 1 mol Na/1 mol e- = 93.26 mol
mass Na = 93.26 mol x 22.99 g/mol = 2144 g
We can also calculate how much electrical energy it will take for an electrolysis. We will
not pursue these calculations.
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moles e- = current (A) x time (s) x 1 coul/1 A s x 1 mol/96500 coul
How long would we have to electrolyze molten NaCl with a current of 3000 A to produce
150 g of Na?
20-17