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Electrochemistry
Mr. Matthew Totaro Legacy High School
AP Chemistry © 2012 Pearson Education, Inc.
Copyright © 2011 Pearson Education, Inc.
Electricity from Chemistry • Many chemical reactions involve the
transfer of electrons between atoms or ions – electron transfer reactions
• all single displacement, and combustion reactions
• some synthesis and decomposition reactions • The flow of electrons is associated
with electricity • Basic research into the nature of this
relationship may, in the near future, lead to cheaper, more efficient ways of generating electricity – fuel cell 2
Electrochemistry
Electrochemical Reactions In electrochemical reactions, electrons are
transferred from one species to another.
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Electrochemistry
Oxidation Numbers
In order to keep track of what loses
electrons and what gains them, we assign
oxidation numbers.
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Electrochemistry
Oxidation and Reduction
• A species is oxidized when it loses electrons. – Here, zinc loses two electrons to go from neutral zinc
metal to the Zn2+ ion.
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Electrochemistry
Oxidation and Reduction
• A species is reduced when it gains electrons. – Here, each of the H+ gains an electron, and they
combine to form H2.
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Electrochemistry
Oxidation and Reduction
• What is reduced is the oxidizing agent. – H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent. – Zn reduces H+ by giving it electrons.
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Electrochemistry
Oxidation–Reduction • Reactions where electrons are transferred
from one atom to another are called oxidation–reduction reactions – redox reactions for short
• Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation
Cl2 + 2 e– → 2 Cl– reduction
Oil Rig
8
Electrochemistry
Rules for Assigning Oxidation States • Rules are in order of priority 1. free elements have an oxidation state = 0
– Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) 2. monatomic ions have an oxidation state equal
to their charge – Na = +1 and Cl = −1 in NaCl
3. (a) the sum of the oxidation states of all the atoms in a compound is 0
– Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0
9
Electrochemistry
Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms
in a polyatomic ion equals the charge on the ion – N = +5 and O = −2 in NO3
–, (+5) + 3(−2) = −1
4. (a) Group I metals have an oxidation state of +1 in all their compounds
– Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in
all their compounds – Mg = +2 in MgCl2
10
Electrochemistry
Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation
states according to the table below – nonmetals higher on the table take priority
Nonmetal Oxidation State Example F −1 CF4 H +1 CH4 O −2 CO2
Group 7A −1 CCl4 Group 6A −2 CS2 Group 5A −3 NH3
11
Electrochemistry
Example: Determine the oxidation states of all the atoms in a propanoate ion, C3H5O2
– • There are no free elements or free ions in
propanoate, so the first rule that applies is Rule 3b (C3) + (H5) + (O2) = −1
• Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order – H = +1 – O = −2
(C3) + 5(+1) + 2(−2) = −1 (C3) = −2 C = −⅔
Note: unlike charges, oxidation states can be fractions!
12
Electrochemistry
Practice – Assign an oxidation state to each element in the following
• Br2
• K+
• LiF
• CO2
• SO42−
• Na2O2
Br = 0 (Rule 1)
K = +1 (Rule 2)
Li = +1 (Rule 4a) & F = −1 (Rule 5)
O = −2 (Rule 5) & C = +4 (Rule 3a)
O = −2 (Rule 5) & S = +6 (Rule 3b)
Na = +1 (Rule 4a) & O = −1 (Rule 3a) 13
Electrochemistry
Example: Assign oxidation states, determine the element oxidized and reduced, and determine the
oxidizing agent and reducing agent in the following reaction
Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O
0 −2 +7 +1 +3 −2 +4 +1 −2
Oxidation Reduction
Oxidizing Agent
Reducing Agent
14
Electrochemistry
Practice – Assign oxidation states, determine the element oxidized and reduced, and determine
the oxidizing agent and reducing agent in the following reactions
Sn4+ + Ca → Sn2+ + Ca2+
F2 + S → SF4
+4 0 +2 +2 Ca is oxidized, Sn4+ is reduced Ca is the reducing agent, Sn4+ is the oxidizing agent
0 0 +4−1 S is oxidized, F is reduced S is the reducing agent, F2 is the oxidizing agent
15
Electrochemistry
Balancing Oxidation-Reduction Equations
Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.
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Electrochemistry
Half-Reactions • We generally split the redox reaction into two separate
half-reactions – a reaction just involving oxidation or reduction – the oxidation half-reaction has electrons as products – the reduction half-reaction has electrons as reactants
3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+
0 −1 +1 −2 −1 +5 −2 +1
oxidation: I− → IO3− + 6 e−
reduction: Cl2 + 2 e− → 2 Cl− 17
Electrochemistry
The Half-Reaction Method
1. Assign oxidation numbers to determine what is oxidized and what is reduced.
2. Write the oxidation and reduction half-reactions.
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Electrochemistry
The Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O. c. Balance H by adding H+. d. Balance charge by adding
electrons.
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Electrochemistry
The Half-Reaction Method 4. Multiply the half-reactions by integers
so that the electrons gained and lost are the same.
5. Add the half-reactions, subtracting things that appear on both sides.
6. Make sure the equation is balanced according to mass.
7. Make sure the equation is balanced according to charge.
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Electrochemistry
The Half-Reaction Method
Consider the reaction between MnO4− and C2O4
2−: MnO4
−(aq) + C2O42−(aq) → Mn2+(aq) + CO2(aq)
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Electrochemistry
The Half-Reaction Method
First, we assign oxidation numbers:
MnO4− + C2O4
2− → Mn2+ + CO2
+7 +3 +4 +2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
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Electrochemistry
Oxidation Half-Reaction
C2O42− → CO2
To balance the carbon, we add a
coefficient of 2:
C2O42− → 2CO2
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Electrochemistry
Oxidation Half-Reaction
C2O42− → 2CO2
The oxygen is now balanced as well.
To balance the charge, we must add 2 electrons to the right side:
C2O4
2− → 2CO2 + 2e−
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Electrochemistry
Reduction Half-Reaction
MnO4− → Mn2+
The manganese is balanced; to balance
the oxygen, we must add 4 waters to the right side:
MnO4
− → Mn2+ + 4H2O
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Electrochemistry
Reduction Half-Reaction
MnO4− → Mn2+ + 4H2O
To balance the hydrogen, we add
8H+ to the left side:
8H+ + MnO4− → Mn2+ + 4H2O
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Electrochemistry
Reduction Half-Reaction
8H+ + MnO4− → Mn2+ + 4H2O
To balance the charge, we add 5e− to
the left side:
5e− + 8H+ + MnO4− → Mn2+ + 4H2O
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Electrochemistry
Combining the Half-Reactions Now we evaluate the two half-reactions
together:
C2O42− → 2CO2 + 2e−
5e− + 8H+ + MnO4− → Mn2+ + 4H2O
To attain the same number of electrons
on each side, we will multiply the first reaction by 5 and the second by 2:
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Electrochemistry
Combining the Half-Reactions
5C2O42− → 10CO2 + 10e−
10e− + 16H+ + 2MnO4− → 2Mn2+ + 8H2O
When we add these together, we get: 10e− + 16H+ + 2MnO4
− + 5C2O42− →
2Mn2+ + 8H2O + 10CO2 +10e−
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Electrochemistry
Combining the Half-Reactions 10e− + 16H+ + 2MnO4
− + 5C2O42− →
2Mn2+ + 8H2O + 10CO2 +10e−
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with: 16H+ + 2MnO4
− + 5C2O42− →
2Mn2+ + 8H2O + 10CO2
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Electrochemistry
1. assign oxidation states and determine element oxidized and element reduced
2. separate into oxidation & reduction half-reactions
Fe2+ + MnO4– → Fe3+ + Mn2+
+2 +7 −2 +3 +2 oxidation
ox: Fe2+ → Fe3+
red: MnO4– → Mn2+
Example: Balancing redox reactions in acidic solution
reduction
31
Electrochemistry
3. balance half-reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
Fe2+ → Fe3+
MnO4– → Mn2+
MnO4– → Mn2+ + 4H2O
MnO4– + 8H+ → Mn2+ + 4H2O
Example: Balancing redox reactions in acidic solution
32
Electrochemistry
4. balance each half-reaction with respect to charge by adjusting the numbers of electrons
a) electrons on product side for oxidation
b) electrons on reactant side for reduction
Fe2+ → Fe3+ + 1 e−
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
MnO4– + 8H+ → Mn2+ + 4H2O
+7 +2
Example: Balancing redox reactions in acidic solution
33
Electrochemistry
5. balance electrons between half-reactions
6. add half-reactions, canceling electrons and common species
7. Check that numbers of atoms and total charge are equal
Fe2+ → Fe3+ + 1 e−
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
} x 5
5 Fe2+ → 5 Fe3+ + 5 e− MnO4
– + 8H+ + 5 e− → Mn2+ + 4H2O
5 Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5 Fe3+
Example: Balancing redox reactions in acidic solution
reactant side Element product side 5 Fe 5 1 Mn 1 4 O 4 8 H 8
+17 charge +17 34
Electrochemistry
Practice – Balance the following equation in acidic solution
I– + Cr2O72− → Cr3+ + I2
35
Electrochemistry
Practice – Balancing redox reactions 1. assign oxidation
states and determine element oxidized and element reduced
2. separate into oxidation & reduction half-reactions
I− + Cr2O72– → I2 + Cr3+
−1 +6 −2 0 +3
oxidation
reduction
ox: I− → I2
red: Cr2O72– → Cr3+
36
Electrochemistry
Practice – Balancing redox reactions 3. balance half-
reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
ox: I− → I2 ox: 2 I− → I2
red: Cr2O72– → Cr3+
red: Cr2O72– → 2 Cr3+
red: Cr2O72– → 2Cr3+ +7H2O
Cr2O72– +14H+ → 2Cr3+ +7H2O
37
Electrochemistry
Practice – Balancing redox reactions 4. balance each
half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on
product side for oxidation
b) electrons on reactant side for reduction
2 I− → I2 2 I− → I2 + 2e−
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O Cr2O72– +14H+ → 2Cr3+ +7H2O
38
Electrochemistry
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O
Practice – Balancing redox reactions 5. balance
electrons between half-reactions
6. add half-reactions, canceling electrons and common species
7. check
} x 3
Cr2O72– +14H+ + 6 I−→ 2Cr3+ +7H2O + 3 I2
2 I− → I2 + 2e− 6 I− → 3 I2 + 6e−
reactant side Element product side 2 Cr 2 6 I 6 7 O 7
14 H 14 +6 charge +6
39
Electrochemistry
Balancing in Basic Solution
• If a reaction occurs in a basic solution, one can balance it as if it occurred in acid.
• Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.
• If this produces water on both sides, you might have to subtract water from each side.
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Electrochemistry
assign oxidation states
separate into half-reactions
ox: I−(aq) → I2(aq)
red: MnO4−(aq) → MnO2(s)
Example: Balance the equation: I−
(aq) + MnO4−
(aq) → I2(aq) + MnO2(s) in basic solution
assign oxidation states
I−(aq) + MnO4
−(aq) → I2(aq) + MnO2(s)
separate into half-reactions
ox: red:
−1 +7 −2 0 +4 −2
Oxidation Reduction
41
Electrochemistry
balance half-reactions by mass
ox: I−(aq) → I2(aq)
red: MnO4−(aq) → MnO2(s)
balance half-reactions by mass first, elements other than H and O
ox: 2 I−(aq) → I2(aq)
red: MnO4−(aq) → MnO2(s)
balance half-reactions by mass then O by adding H2O
ox: 2 I−(aq) → I2(aq)
red: MnO4−(aq) → MnO2(s) + 2 H2O(l)
balance half-reactions by mass then H by adding H+
ox: 2 I−(aq) → I2(aq)
red: 4 H+(aq) + MnO4
−(aq) → MnO2(s) + 2 H2O(l)
balance half-reactions by mass in base, neutralize the H+ with OH−
ox: 2 I−(aq) → I2(aq)
red: 4 H+(aq) + MnO4
−(aq) → MnO2(s) + 2 H2O(l)
4 H+
(aq) + 4 OH−(aq) + MnO4
−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−
(aq)
balance half-reactions by mass in base, neutralize the H+ with OH−
ox: 2 I−(aq) → I2(aq)
red: 4 H+(aq) + MnO4
−(aq) → MnO2(s) + 2 H2O(l)
4 H+
(aq) + 4 OH−(aq) + MnO4
−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−
(aq)
4 H2O(aq) + MnO4−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−
(aq)
balance half-reactions by mass in base, neutralize the H+ with OH−
ox: 2 I−(aq) → I2(aq)
red: 4 H+(aq) + MnO4
−(aq) → MnO2(s) + 2 H2O(l)
4 H+
(aq) + 4 OH−(aq) + MnO4
−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−
(aq)
4 H2O(aq) + MnO4−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−
(aq)
2 H2O(l) + MnO4−(aq) → MnO2(s) + 4 OH−
(aq)
Example: Balance the equation: I−
(aq) + MnO4−
(aq) → I2(aq) + MnO2(s) in basic solution
42
Electrochemistry
balance Half-reactions by charge
ox: 2 I−(aq) → I2(aq) + 2 e−
red: MnO4−(aq) + 2 H2O(l) + 3 e−→ MnO2(s) + 4 OH−
(aq)
Balance electrons between half-reactions
ox: 2 I−(aq) → I2(aq) + 2 e− } x3
red: MnO4−(aq) + 2 H2O(l) + 3 e− → MnO2(s) + 4 OH−
(aq) }x2 ox: 6 I−
(aq) → 3 I2(aq) + 6 e−
red: 2 MnO4−(aq) + 4 H2O(l) + 6 e−→ 2 MnO2(s) + 8 OH−
(aq)
balance Half-reactions by charge
ox: 2 I−(aq) → I2(aq) + 2 e−
red: MnO4−(aq) + 2 H2O(l) + 3 e−→ MnO2(s) + 4 OH−
(aq)
Balance electrons between half-reactions
balance Half-reactions by charge
ox: 2 I−(aq) → I2(aq)
red: MnO4−(aq) + 2 H2O(l) → MnO2(s) + 4 OH−
(aq)
Balance electrons between half-reactions
Example: Balance the equation: I−
(aq) + MnO4−
(aq) → I2(aq) + MnO2(s) in basic solution
43
Electrochemistry
add the half-rxns
ox: 6 I−(aq) → 3 I2(aq) + 6 e−
red: 2 MnO4−(aq) + 4 H2O(l) + 6 e− → 2 MnO2(s) + 8 OH−
(aq) tot: 6 I−
(aq)+ 2 MnO4−(aq) + 4 H2O(l) → 3 I2(aq)+ 2 MnO2(s) + 8 OH−
(aq)
check
add the half-rxns
ox: 6 I−(aq) → 3 I2(aq) + 6 e−
red: 2 MnO4−(aq) + 4 H2O(l) + 6 e− → 2 MnO2(s) + 8 OH−
(aq)
check
Example: Balance the equation: I−
(aq) + MnO4−
(aq) → I2(aq) + MnO2(s) in basic solution
Reactant Count Element
Product Count
6 I 6 2 Mn 2
12 O 12 8 H 8 8− charge 8−
44
Electrochemistry
Practice – Balance the following equation in basic solution
I– + CrO42− → Cr3+ + I2
45
Electrochemistry
Practice – Balancing redox reactions 1. assign oxidation
states and determine element oxidized and element reduced
2. separate into oxidation & reduction half-reactions
I− + CrO42– → I2 + Cr3+
−1 +6 −2 0 +3
oxidation
reduction
ox: I− → I2
red: CrO42– → Cr3+
46
Electrochemistry
Practice – Balancing redox reactions 3. balance half-
reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
d) if in basic solution, neutralize the H+ with OH−
ox: I− → I2 ox: 2 I− → I2
red: CrO42– → Cr3+
red: CrO42– → Cr3+ +4H2O
CrO42– +8H+ → Cr3+ +4H2O
CrO42– +8H+ +8OH− → Cr3+ +4H2O +8OH− CrO42– +8H2O → Cr3+ +4H2O +8OH−
CrO42– +4H2O → Cr3+ +8OH−
47
Electrochemistry
Practice – Balancing redox reactions 4. balance each
half-reaction with respect to charge by adjusting the numbers of electrons
a) electrons on product side for oxidation
b) electrons on reactant side for reduction
2 I− → I2 2 I− → I2 + 2e−
CrO42– +4H2O → Cr3+ +8OH− CrO42– +4H2O+ 3e− → Cr3+ +8OH−
48
Electrochemistry
Practice – Balancing redox reactions
2CrO42– +8H2O+ 6e− → 2Cr3+ +16OH− CrO4
2– +4H2O+ 3e− → Cr3+ +8OH−
5. balance electrons between half-reactions
6. add half-reactions, canceling electrons and common species
7. check
} x 3
2CrO42– +8H2O + 6 I− → 2Cr3+ +16OH− + 3 I2
2 I− → I2 + 2e− 6 I− → 3 I2 + 6e−
reactant side Element product side 2 Cr 2 6 I 6
16 O 16 16 H 16
−10 charge −10
} x 2
49
Electrochemistry
Voltaic Cells
In spontaneous oxidation-reduction
(redox) reactions, electrons are
transferred and energy is released.
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Electrochemistry
Voltaic Cells
• We can use that energy to do work if we make the electrons flow through an external device.
• We call such a setup a voltaic cell or a galvanic cell.
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Electrochemistry
Voltaic Cells • A typical cell looks
like this. • The oxidation occurs
at the anode. • The reduction occurs
at the cathode. • Think Red Cat
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Electrochemistry
Voltaic Cell the salt bridge is
required to complete the
circuit and maintain charge
balance
53
Electrochemistry
Electrochemical Cells
54
• Oxidation and reduction half-reactions are kept separate in half-cells
• Electron flow through a wire along with ion flow through a solution constitutes an electric circuit
• Requires a conductive solid electrode to allow the transfer of electrons through external circuit metal or graphite
• requires ion exchange between the two half-cells of the system electrolyte
Electrochemistry
Voltaic Cells Once even one electron flows from
the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons
would stop.
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Electrochemistry
Voltaic Cells • Therefore, we use a
salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. – Cations move toward
the cathode. – Anions move toward
the anode.
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Electrochemistry
Voltaic Cells • In the cell, then,
electrons leave the anode and flow through the wire to the cathode.
• As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.
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Electrochemistry
Voltaic Cells • As the electrons reach
the cathode, cations in the cathode are attracted to the now negative cathode.
• The electrons are taken by the cation, and the neutral metal is deposited on the cathode.
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Electrochemistry
Electromotive Force (emf) • Water only
spontaneously flows one way in a waterfall.
• Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.
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Electrochemistry
Electromotive Force (emf)
• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).
• It is also called the cell potential and is designated Ecell.
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Electrochemistry
Cell Potential
Cell potential is measured in volts (V).
1 V = 1 J C
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Electrochemistry
Standard Reduction Potentials Reduction
potentials for many electrodes have
been measured and tabulated.
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Electrochemistry
Standard Hydrogen Electrode
• Their values are referenced to a standard hydrogen electrode (SHE).
• By definition, the reduction potential for hydrogen is 0 V:
2 H+(aq, 1M) + 2e− → H2(g, 1 atm)
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Electrochemistry
Standard Cell Potentials
The cell potential at standard conditions can be found through this equation:
Ecell ° = Ered (cathode) − Ered (anode) ° °
Because cell potential is based on the potential energy per unit of charge, it is an intensive property.
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Electrochemistry
Cell Potentials • For the oxidation in this cell,
• For the reduction,
Ered = −0.76 V °
Ered = +0.34 V °
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Electrochemistry
Cell Potentials
Ecell ° = Ered ° (cathode) − Ered ° (anode) = +0.34 V − (−0.76 V) = +1.10 V
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Electrochemistry
Which Way Will Electrons flow?
Zn → Zn2+ + 2 e− E°= +0.76 Cu → Cu2+ + 2 e− E°= −0.34
Under standard conditions, zinc has a stronger tendency to oxidize than copper
67
Therefore the electrons flow from zinc; making zinc the anode
Zn → Zn2+ + 2 e−
Cu2+ + 2 e− → Cu
Pote
ntia
l Ene
rgy
Electron Flow
Electrochemistry
Oxidizing and Reducing Agents • The strongest oxidizers
have the most positive reduction potentials.
• The strongest reducers have the most negative reduction potentials.
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Electrochemistry
Oxidizing and Reducing Agents
The greater the difference between the
two, the greater the voltage of the cell.
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Electrochemistry 70
Electrochemistry 71
Electrochemistry
Example: Calculate E°cell for the reaction at 25 °C Al(s) + NO3
−(aq) + 4 H+
(aq) → Al3+(aq) + NO(g) + 2 H2O(l)
separate the reaction into the oxidation and reduction half-reactions
ox (anode): Al(s) → Al3+(aq) + 3 e−
red: NO3
−(aq) + 4 H+
(aq) + 3 e− → NO(g) + 2 H2O(l)
find the E° for each half-reaction and sum to get E°cell
E°ox of Al = −E°red of Al3+ = +1.66 V
E°red of NO3− = +0.96 V
E°cell = (+1.66 V) + (+0.96 V) = +2.62 V
72
Electrochemistry
Practice – Calculate E°cell for the reaction at 25 °C IO3
–(aq) + 6 H+(aq) + 5 I−(aq) → 3 I2(s) + 3 H2O(l)
Reduction Half-Reaction E°red, V F2(g) + 2e−→ 2 F−(aq) +2.87 IO3
−(aq) + 6 H++ 5e−→ ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e−→ Ag(s) +0.80 I2(s) + 2e−→ 2 I−(aq) +0.54 Cu2+(aq) + 2e−→ Cu(s) +0.34 Cr3+(aq) + 1e−→ Cr2+(aq) −0.50 Mg2+(aq) + 2e−→ Mg(s) −2.37
73
Electrochemistry
Practice – Calculate E°cell for the reaction at 25 °C 2 IO3
– + 12 H+ + 10 I−→ 6 I2 + 6 H2O
separate the reaction into the oxidation and reduction half-reactions
ox (anode): 2 I−(s) → I2(aq) + 2 e− red: IO3
−(aq) + 6 H+
(aq) + 5 e− → ½ I2(s) + 3 H2O(l)
find the E° for each half-reaction and sum to get E°cell
E°ox of I− = −E°red of I = −0.54 V
E°red of IO3− = +1.20 V
E°cell = (−0.54 V) + (+1.20 V) = +0.66 V
2
74
Electrochemistry
Practice – Sketch and label a voltaic cell in which one half-cell has Ag(s) immersed in 1 M AgNO3, and the
other half-cell has a Pt electrode immersed in 1 M Cr(NO3)2 and 1 M Cr(NO3)3 Write the half-reactions and overall reaction, and
determine the cell potential under standard conditions.
Reduction Half-Reaction E°red, V F2(g) + 2e−→ 2 F−(aq) +2.87 IO3
−(aq) + 6 H++ 5e−→ ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e−→ Ag(s) +0.80 I2(s) + 2e−→ 2 I−(aq) +0.54 Cu2+(aq) + 2e−→ Cu(s) +0.34 Cr3+(aq) + 1e−→ Cr2+(aq) −0.50 Mg2+(aq) + 2e−→ Mg(s) −2.37
75
Electrochemistry
ox: Cr2+(aq) → Cr3+(aq) + 1 e− E = +0.50 V red: Ag+(aq) + 1 e− → Ag(s) E = +0.80 V
tot: Cr2+(aq) + Ag+(aq) → Cr3+(aq) + Ag(s) E = +1.30 V
anode = Pt
Cr2+
Cr3+
e− →
e− →
e− →
e− →
cathode = Ag
Ag+
salt bridge
76
Electrochemistry
Cell Notation • Shorthand description of a voltaic cell • Electrode | electrolyte || electrolyte |
electrode • Oxidation half-cell on the left, reduction half-
cell on the right • Single | = phase barrier
– if multiple electrolytes in same phase, a comma is used rather than |
– often use an inert electrode • Double line || = salt bridge
77
Electrochemistry
Voltaic Cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode = Zn(s) the anode is oxidized to Zn2+
Cathode = Cu(s) Cu2+ ions are reduced at the cathode
78
Electrochemistry Zn(s) | Zn2+(1 M) || H+(1 M) | H2(g) | Pt(s) 79
Electrochemistry
Predicting Spontaneity of Redox Reactions
• A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table
• If paired the other way, the reverse reaction is spontaneous
Cu2+(aq) + 2 e− → Cu(s) E°red = +0.34 V Zn2+(aq) + 2 e− → Zn(s) E°red = −0.76 V
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) spontaneous Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s) nonspontaneous
80
Electrochemistry
Example: Predict if the following reaction is spontaneous under standard conditions
Fe(s) + Mg2+(aq) → Fe2+
(aq) + Mg(s) separate the reaction into the oxidation and reduction half-reactions
ox: Fe(s) → Fe2+(aq) + 2 e−
red: Mg2+
(aq) + 2 e− → Mg(s)
look up the relative positions of the reduction half-reactions
red: Fe2+(aq) + 2 e− → Fe(s)
red: Mg2+(aq) + 2 e− → Mg(s)
because Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written
81
Electrochemistry
the reaction is spontaneous in the reverse direction
Mg(s) + Fe2+(aq) → Mg2+
(aq) + Fe(s) ox: Mg(s) → Mg2+
(aq) + 2 e− red: Fe2+
(aq) + 2 e− → Fe(s)
sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode
82
Electrochemistry
Practice – Decide whether each of the following will be spontaneous as written or in the reverse direction
Reduction Half-Reaction
F2(g) + 2e−→ 2 F−(aq)
IO3−(aq) + 6 H++ 5e−→ I2(s) + 3H2O(l)
Ag+(aq) + 1e−→ Ag(s)
I2(s) + 2e−→ 2 I−(aq)
Cu2+(aq) + 2e−→ Cu(s)
Cr3+(aq) + 1e−→ Cr2+(aq)
Mg2+(aq) + 2e−→ Mg(s)
F2(g) + 2 I−(aq) → I2(s) + 2 F−(aq)
Mg(s) + 2 Ag+(aq) → Mg2+
(aq) + 2 Ag(s)
Cu2+(aq) + 2 I−(aq) → I2(s) + Cu(s)
Cu2+(aq) + 2 Cr2+
(aq) → Cu(s) + 2 Cr3+(aq)
spontaneous as written
spontaneous as written
spontaneous as written
spontaneous in reverse
83
Electrochemistry
Free Energy
∆G for a redox reaction can be found by using the equation
∆G = −nFE
where n is the number of moles of
electrons transferred, and F is a constant, the Faraday:
1 F = 96,485 C/mol = 96,485 J/V-mol © 2012 Pearson Education, Inc.
Electrochemistry
red: I2(l) + 2 e− → 2 I−(aq) E° = +0.54 V
ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 V
tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 V
Example: Calculate ∆G° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
85
because ∆G° is +, the reaction is not spontaneous in the forward direction under standard conditions
Answer:
Solve:
Conceptual Plan:
Relationships:
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
DG°, (J) Given:
Find:
E°ox, E°red E°cell ∆G°
Electrochemistry
Practice – Calculate ∆G° for the reaction at 25°C 2IO3
–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
Reduction Half-Reaction E°red, V F2(g) + 2e−→ 2 F−(aq) +2.87 IO3
−(aq) + 6 H++ 5e−→ ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e−→ Ag(s) +0.80 I2(s) + 2e−→ 2 I−(aq) +0.54 Cu2+(aq) + 2e−→ Cu(s) +0.34 Cr3+(aq) + 1e−→ Cr2+(aq) −0.50 Mg2+(aq) + 2e−→ Mg(s) −2.37
86
Electrochemistry
ox : 2 I−(s) → I2(aq) + 2 e− Eº = −0.54 V red: IO3
−(aq) + 6 H+(aq) + 5 e− → ½ I2(s) + 3 H2O(l) Eº = 1.20 V tot: 2IO3
−(aq) + 12H+
(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V
Practice – Calculate ∆G° for the reaction at 25 °C 2IO3
–(aq) + 12H+(aq) + 10 I−(aq) →6 I2(s) + 6 H2O(l)
because ∆G° is −, the reaction is spontaneous in the forward direction under standard conditions
Answer:
Solve:
Conceptual Plan:
Relationships:
2IO3−
(aq)+12H+(aq)+10I−(aq) → 6 I2(s) + 6 H2O(l)
∆G°, (J) Given:
Find:
E°ox, E°red E°cell ∆G°
87
Tro: Chemistry: A Molecular Approach, 2/e
Electrochemistry
• For a spontaneous reaction – one that proceeds in the forward direction with
the chemicals in their standard states -∆G° < 1 (negative) – E° > 1 (positive) – K > 1
• ∆G° = −RTlnK = −nFE°cell – n is the number of electrons – F = Faraday’s Constant = 96,485 C/mol e−
E°cell, ∆G° and K
88
Electrochemistry
tot: Cu(s) + 2H+(aq) → Cu2+
(aq) + H2(g) E° = −0.34 V
Example: Calculate K at 25 °C for the reaction Cu(s) + 2 H+
(aq) → H2(g) + Cu2+(aq)
89
because K < 1, the position of equilibrium lies far to the left under standard conditions
Answer:
Solve:
Conceptual Plan:
Relationships:
Cu(s) + 2 H+(aq) → H2(g) + Cu2+
(aq) K
Given: Find:
E°ox, E°red E°cell K
ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 V
red: 2 H+(aq) + 2 e− → H2(aq) E° = +0.00 V
Tro: Chemistry: A Molecular Approach, 2/e
Electrochemistry
Practice – Calculate K for the reaction at 25°C 2IO3
–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
Reduction Half-Reaction E°red, V F2(g) + 2e−→ 2 F−(aq) +2.87 IO3
−(aq) + 6 H++ 5e−→ ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e−→ Ag(s) +0.80 I2(s) + 2e−→ 2 I−(aq) +0.54 Cu2+(aq) + 2e−→ Cu(s) +0.34 Cr3+(aq) + 1e−→ Cr2+(aq) −0.50 Mg2+(aq) + 2e−→ Mg(s) −2.37
90
Electrochemistry
ox : 2 I−(s) → I2(aq) + 2e− Eº = −0.54 V red: IO3
−(aq) + 6 H+(aq) + 5e− → ½ I2(s) + 3H2O(l) Eº = 1.20 V tot: 2IO3
−(aq) + 12H+
(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V
Practice – Calculate K for the reaction at 25 °C 2IO3
–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
because K >> 1, the position of equilibrium lies far to the right under standard conditions
Answer:
Solve:
Conceptual Plan:
Relationships:
2 IO3−
(aq)+12 H+(aq)+10 I−(aq) → 6 I2(s) + 6 H2O(l)
K Given:
Find:
E°ox, E°red E°cell K
91
Electrochemistry
Cell Potential when Ion Concentrations Are Not 1 M
• We know there is a relationship between the reaction quotient, Q, the equilibrium constant, K, and the free energy change, ∆Gº
• Changing the concentrations of the reactants and products so they are not 1 M will affect the standard free energy change, ∆Gº
• Because ∆Gº determines the cell potential, Ecell, the voltage for the cell will be different when the ion concentrations are not 1 M 92
Electrochemistry
Ecell When Ion Concentrations are Not 1 M
93
Electrochemistry
Ecell When Ion Concentrations are Not 1 M
• The standard cell potential, Eo, corresponds to the standard conditions of Q = 1. As the system approaches equilibrium, the magnitude (i.e., absolute value) of the cell potential decreases, reaching zero at equilibrium (when Q K). Deviations from standard conditions that take the cell further from equilibrium than Q = 1 will increase the magnitude of the cell potential relative to E. Deviations from standard conditions that take the cell closer to equilibrium than Q = 1 will decrease the magnitude of the cell potential relative to E. In concentration cells, the direction of spontaneous electron flow can be determined by considering the direction needed to reach equilibrium.
Electrochemistry
Deriving the Nernst Equation
95
Electrochemistry
Electrochemical Cells • In all electrochemical cells, oxidation occurs at the
anode, reduction occurs at the cathode • In voltaic cells
– anode is the source of electrons and has a (−) charge – cathode draws electrons and has a (+) charge
• In electrolytic cells – electrons are drawn off the anode, so it must have a place
to release the electrons, the + terminal of the battery – electrons are forced toward the anode, so it must have a
source of electrons, the − terminal of the battery
96
Electrochemistry 97
Electrochemistry 98
Electrolysis • Electrolysis is the process of
using electrical energy to break a compound apart
• Electrolysis is done in an electrolytic cell
• Electrolytic cells can be used to separate elements from their compounds
Electrochemistry
Electrolysis • In electrolysis we use electrical energy to overcome
the energy barrier of a non-spontaneous reaction, allowing it to occur
• The reaction that takes place is the opposite of the spontaneous process
2 H2(g) + O2(g) → 2 H2O(l) spontaneous 2 H2O(l) → 2 H2(g) + O2(g) electrolysis • Some applications are (1) metal extraction from
minerals and purification, (2) production of H2 for fuel cells, (3) metal plating
99
Electrochemistry
Electrolytic Cells • The electrical energy is supplied by a Direct
Current power supply – AC alternates the flow of electrons so the
reaction won’t be able to proceed • Some electrolysis reactions require more
voltage than Ecell predicts. This is called the overvoltage
100
Electrochemistry
Electrolytic Cells • The source of energy is a battery or DC power supply • The + terminal of the souce is attached to the anode • The − terminal of the source is attached to the cathode • Electrolyte can be either an aqueous salt solution or a
molten ionic salt • Cations in the electrolyte are attracted to the cathode
and anions are attracted to the anode • Cations pick up electrons from the cathode and are
reduced, anions release electrons to the anode and are oxidized
101
Electrochemistry
Electrolysis of Pure Compounds • The compound must be in molten (liquid)
state • Electrodes normally graphite • Cations are reduced at the cathode to metal
element • Anions oxidized at anode to nonmetal
element
102
Electrochemistry
Electrolysis of Water
103
Electrochemistry
Electroplating In electroplating, the work piece is the cathode
Cations are reduced at cathode and plate to the surface of the work piece The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution.
104
Electrochemistry
Current • Current is the number of electrons that
flow through the system per second – unit = Ampere
• 1 A of current = 1 Coulomb of charge flowing by each second – 1 A = 6.242 x 1018 electrons per second
• Electrode surface area dictates the number of electrons that can flow – larger batteries produce larger currents
105
Electrochemistry
Stoichiometry of Electrolysis • In an electrolytic cell, the amount of product
made is related to the number of electrons transferred – essentially, the electrons are a reactant
• The number of moles of electrons that flow through the electrolytic cell depends on the current and length of time – 1 Amp = 1 Coulomb of charge/second – 1 mole of e− = 96,485 Coulombs of charge
• Faraday’s constant 106
Electrochemistry
Faraday’s Laws – The first law of electrolysis
• We can link this product to the mass of metal produced at the cathode as being directly proportional to the quantity of electricity passed through the cell.
• This relationship is known as Faraday’s first law of electrolysis.
• It may be written as: m
Electrochemistry
Example: Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction
Au3+(aq) + 3 e− → Au(s)
units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−
Check:
Solve:
Conceptual Plan:
Relationships:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g
Given:
Find:
t(s), amp charge (C) mol e− mol Au g Au
108
Electrochemistry
Practice – Calculate the amperage required to plate 2.5 g of gold in 1 hour (3600 sec) Au3+(aq) + 3 e− → Au(s)
109
Electrochemistry
Practice – Calculate the amperage required to plate 2.5 g of Au in 1 hour (3600 sec)
Au3+(aq) + 3 e− → Au(s)
units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−
Check:
Solve:
Conceptual Plan:
Relationships:
3 mol e− : 1 mol Au, mass = 2.5 g, time = 3600 s current, A
Given:
Find:
g Au mol Au mol e− charge amps
110
Electrochemistry
Applications of Oxidation-Reduction
Reactions
© 2012 Pearson Education, Inc.
Electrochemistry
Batteries
© 2012 Pearson Education, Inc.
Electrochemistry
Alkaline Batteries
© 2012 Pearson Education, Inc.
Electrochemistry
Hydrogen Fuel Cells
© 2012 Pearson Education, Inc.
Electrochemistry
Corrosion and…
© 2012 Pearson Education, Inc.
Electrochemistry
…Corrosion Prevention
© 2012 Pearson Education, Inc.
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