Transcript Chem 1202 - LSU Chemistry Home Page

Electrochemistry
Chapter 20
Practical uses of redox reactions
Watkins
Chem 1422, Chapter 20
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REDOX Reactions
Review Chapter 4, section 4
→
Watkins
Chem 1422, Chapter 20
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Redox Reactions
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
The oxidation number of Zn
increases from 0 to +2 (Zn loses electrons)
so Zn(s) has been oxidized
so Zn(s) is the reducing agent
The oxidation number of H
decreases from +1 to 0 (H gains electrons)
so H+(aq) has been reduced
so H+(aq) is the oxidizing agent
Watkins
Chem 1422, Chapter 20
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Balancing Redox Equations
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
Conservation of mass: the number of atoms of an
element on the left and right sides of the equation
must be equal; this is mass balance.
Conservation of charge: the total charge (all species) on
the left must equal the total charge (all species) on the
right; this is charge balance.
Some reactions can be balanced by inspection, but
half-reactions are a convenient way of balancing
complex redox reactions.
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Chem 1422, Chapter 20
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Redox Half-Reactions
•
•
•
•
•
The Rules
Both half-reactions display electrons e–
The reduction half-reaction contains the oxidizing
agent only, with e- on the left.
2e- + 2H+(aq) → H2(g)
The oxidation half-reaction contains the reducing
agent only, with e- on the right.
Ca(s) → Ca2+(aq) + 2eEach half-reaction must balance by mass and by
charge.
The sum of the two half-reactions must equal the
total REDOX reaction (without displaying electrons)
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Chem 1422, Chapter 20
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Redox Half-Reactions
Example: Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq)
Reducing agent: Sn2+(aq) → Sn4+(aq)
Oxidizing agent: 2Fe3+(aq) → 2Fe2+(aq)
These reactions are each balanced by mass.
To balance by charge, add electrons as needed:
Sn2+(aq) → Sn4+(aq) + 2e2e- + 2Fe3+(aq) → 2Fe2+(aq)
Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq)
Note that when these two half-reactions are added
together, the electrons cancel.
Watkins
Chem 1422, Chapter 20
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Balancing Redox Equations
Example: an HCl solution of Na2C2O4 (sodium oxalate,
colorless) is titrated with a solution of KMnO4
(potassium permanganate, deep purple); Mn2+ ions
and carbon dioxide gas are produced.
Start by writing down partial half-reactions, each with
only one reactant and one product:
KMnO4(aq) → Mn2+(aq)
K(s) is not a product, so K+ must be a spectator ion (its
ON does not change).
Na2C2O4(aq) → CO2(g)
Na(s) is not a product, so Na+ must be a spectator ion.
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Chem 1422, Chapter 20
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Balancing Redox Equations
Example: an HCl solution of Na2C2O4 (sodium oxalate,
colorless) is titrated with a solution of KMnO4
(potassium permanganate, deep purple); Mn2+ ions
and carbon dioxide gas are produced.
KMnO4(aq) → Mn2+(aq)
Na2C2O4(aq) → CO2(g)
No products containing chlorine are mentioned, so Clmust be a spectator. Likewise, no H2 gas is produced,
so H+ is not an oxidizing agent (it cannot be a
reducing agent).
Eliminate all spectator ions (for now):
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Chem 1422, Chapter 20
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Balancing Redox Equations
Example: an HCl solution of Na2C2O4 (sodium oxalate,
colorless) is titrated with a solution of KMnO4
(potassium permanganate, deep purple); Mn2+ ions
and carbon dioxide gas are produced.
The two partial half reactions are
MnO4-(aq) → Mn2+(aq)
C2O42-(aq) → CO2(g)
and the unbalanced net ionic reaction is
C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)
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Chem 1422, Chapter 20
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Balancing Redox Equations
C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)
Permanganate (buret) is added to oxalate (beaker); the
solution fizzes (CO2(g) is produced).
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Chem 1422, Chapter 20
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Balancing Redox Equations
C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)
The equivalence point is detected by a slight excess of
MnO4- (light purple).
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Chem 1422, Chapter 20
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Balancing Redox Equations
C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)
If more KMnO4 is added, the solution turns deep
purple due to the large excess of KMnO4.
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Chem 1422, Chapter 20
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Balancing Redox Equations
Step 1: Write down the net ionic reaction
(ignore spectator ions initially)
C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)
Step 2: Determine oxidation numbers
C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)
+3 -2
+7 -2
+2
+4 -2
RA
OA
Step 3: Write the unbalanced half-reactions
Oxidation
C2O42-(aq) → CO2(g)
Reduction
MnO4-(aq) → Mn2+(aq)
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Chem 1422, Chapter 20
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Balancing Redox Equations
Step 4: Balance each half-reaction by mass
(in acidic aqueous solution, add H2O(l) and H+(aq)
wherever needed)
C2O42-(aq) → 2 CO2(g)
8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l)
Step 5: Balance each half-reaction by charge (add ewherever needed)
C2O42-(aq) → 2CO2(g) + 2e5e- + 8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l)
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Chem 1422, Chapter 20
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Balancing Redox Equations
Step 6: Equalize electrons (multiply each half-reaction)
5X {C2O42-(aq) → 2CO2(g) + 2e-}
2X {5e- + 8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l)}
5C2O42-(aq) → 10CO2(g) + 10e-
10e- + 16H+(aq) + 2MnO4-(aq) → 2Mn2+(aq) + 8H2O(l)
5C2O42-(aq) + 16H+(aq) + 2MnO4-(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l)
Step 7: Add the balanced half-reactions
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Chem 1422, Chapter 20
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Balancing Redox Equations
Step 8: Check mass and charge balance
THE MOST IMPORTANT STEP!
5C2O42-(aq) + 16H+(aq) + 2MnO4-(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l)
10
C
10
28
O
28
16
H
16
2
Mn
2
4+
+/-
4+
Net ionic reaction:
5C2O42-(aq) + 16H+(aq) + 2MnO4-(aq) →
10CO2(g) + 2Mn2+(aq) + 8H2O(l)
Watkins
Chem 1422, Chapter 20
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Balancing Redox Equations
Step 9: If required, add spectator ions (to both sides!)
5C2O42-(aq) + 16H+(aq) + 2MnO4-(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l)
10Na+
16Cl-
2K+
4Cl10Na+ 10Cl2K+
2Cl-
5Na2C2O4(aq) + 16HCl(aq) + 2KMnO4(aq) →
10CO2(g) + 2MnCl2(aq) + 8H2O(l) + 10NaCl + 2KCl
Watkins
Chem 1422, Chapter 20
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Balancing Redox Equations
Example: Gaseous chlorine dioxide is bubbled into a
sodium hydroxide solution containing hydrogen
peroxide; sodium chlorite and oxygen gas are
produced. Write the balanced redox reaction with the
smallest whole-number coefficients.
ClO2(g) NaOH(aq) H2O2(aq) NaClO2(aq) O2(g)
•
•
•
•
Na+ is a spectator ion
ClO2(aq) is reduced to ClO2-(aq) [ON(Cl): +4 → +3]
H2O2(aq) is oxidized to O2(g) [ON(O): -1 → 0]
In basic solution, OH- and H2O can be added
wherever needed.
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Chem 1422, Chapter 20
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Balancing Redox Equations
Step 1: Write down the net ionic reaction
(ignore OH-, H2O and spectator ions initially)
ClO2(aq) + H2O2(aq) → ClO2-(aq) + O2(g)
Step 2: Determine oxidation numbers
ClO2(aq) + H2O2(aq) → ClO2-(aq) + O2(g)
+4 -2
+1 -1
+3 -2
0
OA
RA
Step 3: Write preliminary half-reactions
Oxidation
H2O2(aq) → O2(g)
Reduction
ClO2(aq) → ClO2-(aq)
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Chem 1422, Chapter 20
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Balancing Redox Equations
Step 4: Balance each half-reaction by mass (in basic
aqueous solution, add H2O(l) and H+(aq) wherever
needed; then titrate with OH-)
ClO2(aq) → ClO2-(aq)
2OH-(aq) + H2O2(aq) → O2(g) + 2H+(aq) + 2OH-(aq)
2OH-(aq) + H2O2(aq) → O2(g) + 2H2O(l)
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Chem 1422, Chapter 20
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Balancing Redox Equations
Step 5: Balance each half-reaction by charge
2OH-(aq) + H2O2(aq) → O2(g) + 2H2O(l) + 2ee- + ClO2(aq) → ClO2-(aq)
Steps 6 & 7: Equalize electrons and add half-reactions
1X {2OH-(aq) + H2O2(aq) → O2(g) + 2H2O(l) + 2e-}
2X {e- + ClO2(aq) → ClO2-(aq)}
2OH-(aq) + H2O2(aq) → O2(g) + 2H2O(l) + 2e2e- + 2ClO2(aq) → 2ClO2-(aq)
2OH-(aq) + H2O2(aq) + 2ClO2(aq) →
O2(g) + 2H2O(l) + 2ClO2-(aq)
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Chem 1422, Chapter 20
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Balancing Redox Equations
Step 8: Check mass and charge balance
2OH-(aq) + H2O2(aq) + 2ClO2(aq) → O2(g) + 2H2O(l) + 2ClO2-(aq)
2
8
4
2
Cl
O
H
+/-
2
8
4
2
Step 9: If required, add spectator ions
2OH-(aq) + H2O2(aq) + 2ClO2(aq) → O2(g) + 2H2O(l) + 2ClO2-(aq)
2Na+
2Na+
2NaOH(aq) + H2O2(aq) + 2ClO2(aq) →
O2(g) + 2H2O(l) + 2NaClO2(aq)
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Chem 1422, Chapter 20
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Balancing Redox Equations
Caution
Half-reactions are written in your text book with forward
reaction arrows (→).
In fact, redox reactions and half-reactions are reversible,
so all equilibrium concepts apply:
C2O42-(aq) ⇌ CO2(g)
MnO4-(aq) ⇌ Mn2+(aq)
5Na2C2O4(aq) + 16HCl(aq) + 2KMnO4(aq) ⇌
10CO2(g) + 2MnCl2(aq) + 8H2O(l) + 10NaCl + 2KCl
Watkins
Chem 1422, Chapter 20
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Summary: Balancing REDOX Reactions
Step 1: Write the net ionic reaction by determining
reactants and products from the problem.
Step 2: Calculate all oxidation numbers and identify
Oxidizing Agent and Reducing Agent.
Step 3: Write half-reactions for OA and RA.
Step 4: Balance each half-reaction by mass (add H+,
OH-, H2O as necessary).
Step 5: Balance each half-reaction by charge (add ewherever necessary).
Step 6: Equalize electrons (multiply half-reactions).
Step 7: Add the half-reactions.
Step 8: Check mass and charge balance.
Step 9: If required, add spectator ions.
Watkins
Chem 1422, Chapter 20
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Electricity
Some things to understand and remember about
electricity.
1. Charge (C, Coulombs): an electron and a Cl– ion
carry 1 unit of negative charge (z– = -1.602×10-19 C); a
proton and a Na+ ion carry 1 unit of positive charge
(z+ = +1.602×10-19 C).
1 C is the charge carried by 6.24×1018 e–
1 mole of e– carries a charge of 96,485 C
2. Current (I, Amps): the net (non-random) flow of
charge carriers (electrons, ions) through a conductor,
from point A to point B, over time t:
1 Amp = 1 Coulomb/second (I = C/t)
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Chem 1422, Chapter 20
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Electricity
3. Voltage (E, Volts): the Electromotive Force (EMF)
driving current through a conductor (f×d = energy):
1 Volt = 1 Joule/Amp.sec (E = J/It)
4. Electrical Energy (J, Joules): J = EIt = Wt
(volt-amp-seconds, watt-second, kilowatt-hour)
5. Electrical Power (W, Watt): W = EI = J/t; energy
(Joules) delivered during time t (seconds);
1 Watt = 1 Joule/second = 1 Volt-Amp
6. Charge Carriers move slowly in a conductor, but
charge flow (current) moves at the speed of light!
Watkins
Chem 1422, Chapter 20
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Electrochemical Energy
A spontaneous redox reaction releases energy
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
A strip of Zn is placed
The Zn metal
oxidizes,
forming Zn2+
ions in solution;
Zn(s) is the
reducing agent.
in a 1 M solution of Cu(NO3)2
Zn
Cu
Zn2+
Cu2+
NO3-
Cu2+ ions are
reduced and
deposit directly on
the Zn strip as
copper metal;
Cu2+(aq) is the
oxidizing agent.
This direct electron transfer , due to random motion
of ions, releases energy which just heats the solution
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Chem 1422, Chapter 20
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Electrochemical Energy
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
However, this energy could be used to perform useful
electrical work
Electrical work: the work performed by the non-random
motion of charge carriers:
– electrons moving through a wire electric
current
– ions moving through a solution
The device which converts the electrochemical energy of
a spontaneous redox reaction into electrical work is
called a Voltaic Cell
Watkins
Chem 1422, Chapter 20
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Voltaic Cell
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
Zn half-cell
Cu half-cell
Zn(s)/Zn2+(aq)
Cu(s)/Cu2+(aq)
Each metal and its cation is called a redox couple; the
couple is placed in a separate compartment called a
half-cell
Watkins
Chem 1422, Chapter 20
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Voltaic Cell
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
The metals are called electrodes
Zn electrode
Cu electrode
Zn(s)/Zn2+(aq)
Cu(s)/Cu2+(aq)
Each half-cell is filled with a salt solution (electrolyte)
containing the redox couple and a spectator anion.
Watkins
Chem 1422, Chapter 20
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Voltaic Cell
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
The electrodes are connected by a wire
(external circuit) through which electrons flow
Zn electrode
Cu electrode
Zn(s)/Zn2+(aq)
Cu(s)/Cu2+(aq)
The half-cells are connected by a Salt Bridge
(internal circuit) through which ions flow. These ions
replace charges lost from the half-cells during reaction.
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Chem 1422, Chapter 20
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Voltaic Cell
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
Reduction occurs
spontaneously at the
Cathode (CATRED)
Oxidation occurs
spontaneously at
the Anode(ANOX)
Zn anode
Cu cathode
+
–
Zn(s)/Zn2+(aq)
Zn(s)
Cu(s)/Cu2+(aq)
Zn2+(aq) + 2e-
Cu2+(aq) + 2e-
Cu(s)
“Anode” is defined as the electrode at which
oxidation occurs. “Cathode” is defined as the
electrode at which reduction occurs.
Watkins
Chem 1422, Chapter 20
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Voltaic Cell
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
In the external circuit
electrons flow from Anode (–) to Cathode (+)
Zn anode
Cu cathode
+
–
Zn(s)/Zn2+(aq)
Cu(s)/Cu2+(aq)
In the salt bridge:
cations move toward the cathode
anions move toward the anode
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Chem 1422, Chapter 20
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Voltaic Cell
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
Copper half-cell:
Cu/Cu2+ couple
Zinc half-cell:
Zn/Zn2+ couple
Watkins
Chem 1422, Chapter 20
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Voltaic Cell
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
DG < 0
Electrical
Potential Energy
Electrons flow spontaneously from the anode to the
cathode because
– the cathode has a lower electrical potential energy
– The difference in electrical potentials is measured
in volts (sometimes called a voltage drop).
A volt is an Electromotive
eforce (E) pushing charge
Anode
carriers (e–, ions) from
electrode to electrode. The
EMF creates electrical work
Watkins
eCathode
Chem 1422, Chapter 20
EIt = J
35
Voltaic Cell EMF
• The Electromotive Force (EMF) of a voltaic cell
pushes electrons through the external circuit and ions
through the salt bridge.
• The EMF of the cell is symbolized Ecell
(a.k.a. the cell voltage or cell potential)
• Under standard conditions:
– All solids and liquids are pure
– All electrolyte solutions are 1 M
– All gases are 1 atm
– The Standard Cell Potential is denoted Eocell
Watkins
Chem 1422, Chapter 20
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ANOX CATRED
Half-Cells
• A half-cell is composed of a chemical substance in
both its oxidized and reduced forms (a redox couple):
– M(s) ⇌ M+x(aq) + xe- (Anode = Oxidation)
– M+y(aq) + ye- ⇌ M(s) (Cathode = Reduction)
– H2(g) ⇌ 2H+(aq) + 2e- (Anode)
A half-reaction can
– 2H+(aq) + 2e- ⇌ H2(g) (Cathode) always be reversed.
• Each half-reaction can be assigned a Standard HalfCell Potential, Eoox or Eored.
• When an oxidation half-reaction is reversed and
written as a reduction: Eored = –Eoox.
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Chem 1422, Chapter 20
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ANOX CATRED
Half-Cells
• A half-cell is composed of a chemical substance in
both its oxidized and reduced forms (a redox couple):
– M(s) ⇌ M+x(aq) + xe- (Anode = Oxidation)
– M+y(aq) + ye- ⇌ M(s) (Cathode = Reduction)
– H2(g) ⇌ 2H+(aq) + 2e- (Anode)
A half-reaction can
– 2H+(aq) + 2e- ⇌ H2(g) (Cathode) always be reversed.
• The combination of a standard oxidation and a
standard reduction produces a standard voltaic cell:
Eocell = Eoox + Eored
• The voltage of a voltaic cell is easy to measure, and
also easy to calculate
Watkins
Chem 1422, Chapter 20
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ANOX CATRED
Half-Cells
Consider a voltaic cell, under standard conditions,
composed of these two half-cells:
M(s) ⇌ M2+(aq) + 2e- (Anode)
Eoox
2H+(aq) + 2e- ⇌ H2(g) (Cathode)
Eored
M(s) + 2H+(aq) ⇌ M2+(aq) + H2(g) Eocell = Eoox + Eored
If the REDOX reaction, as written, is spontaneous in the
forward direction so that DGo < 0, then by convention
Eocell > 0.
Metals which spontaneously reduce hydrogen ions to
hydrogen gas are called active metals.
Watkins
Chem 1422, Chapter 20
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ANOX CATRED
Half-Cells
Consider a voltaic cell, under standard conditions,
composed of these two half-cells:
M(s) ⇌ M2+(aq) + 2e- (Anode)
Eoox
2H+(aq) + 2e- ⇌ H2(g) (Cathode)
Eored
M(s) + 2H+(aq) ⇌ M2+(aq) + H2(g) Eocell = Eoox + Eored
If the REDOX reaction, as written, is not spontaneous in
the forward direction (DGo > 0), then by convention
Eocell < 0.
Metals which do not spontaneously reduce hydrogen ions
to hydrogen gas are called passive metals.
Watkins
Chem 1422, Chapter 20
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ANOX CATRED
Half-Cells
Consider a voltaic cell, under standard conditions,
composed of these two half-cells:
M(s) ⇌ M2+(aq) + 2e- (Anode)
Eoox
2H+(aq) + 2e- ⇌ H2(g) (Cathode)
Eored
M(s) + 2H+(aq) ⇌ M2+(aq) + H2(g) Eocell = Eoox + Eored
These REDOX reactions can be arranged in order of
decreasing Eocell. Such an arrangement is called an
activity series (see Chapter 4).
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Chem 1422, Chapter 20
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M(s) ⇌ M2+(aq) + 2e2H+(aq) + 2e- ⇌ H2(g)
Eoox
Eore
d
M(s) + 2H+(aq) ⇌ M2+(aq) + H2(g)
Eocel
l
Watkins
Chem 1422, Chapter 20
Li+
+
e-
K
⇌
K+
+
e-
Ba
⇌
Ba2+
+
2e-
Ca
⇌
Ca2+
+
2e-
Na
⇌
Na+
+
e-
Mg
⇌
Mg2+
+
2e-
Al
⇌
Al3+
+
3e-
Mn
⇌
Mn2+
+
2e-
Zn
⇌
Zn2+
+
2e-
Cr
⇌
Cr3+
+
3e-
Fe
⇌
Fe2+
+
2e-
Co
⇌
Co2+
+
2e-
Ni
⇌
Ni2+
+
2e-
Sn
⇌
Sn2+
+
2e-
Pb
⇌
Pb2+
+
2e-
H2
⇌
2H+
+
2e-
Cu
⇌
Cu2+
+
2e-
Ag
⇌
Ag+
+
e-
Hg
⇌
Hg2+
+
2e-
Pt
⇌
Pt2+
+
2e-
Au
⇌
Au3+
+
3e-
Active Metals
Eocell > 0
increasing →
⇌
Passive Metals
Eocell < 0
← decreasing
Activity Series for the REDOX
reaction of a pure metal in an
acid solution in equilibrium with
its cations and hydrogen gas:
[H+] = [M+n] = 1 M
p{H2(g)} = 1 atm
For example:
Li
42
ANOX CATRED
Half-Cells
Consider a voltaic cell, under standard conditions,
composed of these two half-cells:
M(s) ⇌ M2+(aq) + 2e- (Anode)
Eoox
2H+(aq) + 2e- ⇌ H2(g) (Cathode)
Eored
M(s) + 2H+(aq) ⇌ M2+(aq) + H2(g) Eocell = Eoox + Eored
The half-reaction with redox couple H+/H2 at 25 oC
under standard conditions ( [H+] = 1 M, p(H2) = 1 atm)
is called the Standard Hydrogen Half-Reaction. The
physical electrode which uses this half-reaction is called
the Standard Hydrogen Electrode (SHE). By definition
Eored{H+(aq)/H2(g)} = Eoox{H+(aq)/H2(g)} = 0
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Chem 1422, Chapter 20
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ANOX CATRED
Half-Cell SOP
Zn(s) + 2H+(aq)
Zn2+(aq) + H2(g)
Eocell = Eoox + Eored = Eoox + 0 = +0.76 V
Active
Metal
e–
e–
Zn
anode
Anode
compartment
Eoox
Watkins
Cathode
compartment
(SHE)
Zn(s) ⇌ Zn2+(aq) + 2e–
2H+(aq) + 2e– ⇌ H2(g)
Chem 1422, Chapter 20
Eored
44
ANOX CATRED
Half-Cell SRP
2Ag+(aq) + H2(g)
2Ag(s) + 2H+(aq)
Eocell = Eoox + Eored = 0 + Eored = +0.80 V
Passive
Metal
e–
e–
Ag
cathode
Cathode
compartment
Eored
Watkins
Anode
compartment
(SHE)
Ag+(aq) + e– ⇌ Ag(s)
H2(g) ⇌ 2H+(aq) + 2e–
Chem 1422, Chapter 20
Eoox
45
Half-Cell Potentials
ANOX CATRED
Reduction
2H+(aq) + 2e– ⇌ H2(g)
Ion good oxidizing agent Ag+(aq) + e– ⇌ Ag(s)
Ion poor oxidizing agent Zn2+(aq) + 2e– ⇌ Zn(s)
Eored
0.00
+0.80
–0.76
Oxidation
Eoox
H2(g) ⇌ 2H+(aq) + 2e– 0.00
Metal poor reducing agent Ag(s) ⇌ Ag+(aq) + e– –0.80
Metal good reducing agent Zn(s) ⇌ Zn2+(aq) + 2e– +0.76
DG < 0: reaction is spontaneous in the forward direction.
E > 0: reaction is spontaneous in the forward direction.
The Standard Reduction Potentials of many redox
couples are tabulated in Appendix E.
Watkins
Chem 1422, Chapter 20
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ANOX CATRED
See Appendix E
Half-Cell Potentials
Watkins
Chem 1422, Chapter 20
47
ANOX CATRED
Interpretation of SRP's or SOP's in Appendix E
Half-Cell Potentials
F2(g) + 2e- → 2F-(aq)
Eored = +2.87 V
F2(g) is a very strong oxidizing agent
2F-(aq) → F2(g) + 2e-
Eoox = -2.87 V
F-(aq) is a very weak reducing agent
Li+(aq) + e- → Li(s)
Eored = -3.05 V
Li+(aq) is a very, very weak oxidizing agent
Li(s) → Li+(aq) + e-
Eoox = +3.05 V
Li(s) is a very, very strong reducing agent
Watkins
Chem 1422, Chapter 20
48
ANOX CATRED
Interpretation of SRP's or SOP's in Appendix E
Half-Cell Potentials
Both chlorine and ozone are used in chemical processes as oxidizing
agents. Compare their oxidizing power.
Cl2(g) + 2e- → 2Cl-(aq)
Eored = +1.36 V
O3(g) + 2H+(aq) + 2e- → O2(g) + H2O(l) Eored = +2.07 V
Ozone is a more powerful oxidizing agent than chlorine.
Both magnesium and sodium are use as reducing agents in organic
chemistry. Compare their reducing power.
Mg(s) → Mg2+(aq) + 2eEoox = +2.37 V
Na(s) → Na+(aq) + eEoox = +2.71 V
Mg(s) is a very good reducing agent, but Na(s) is a more
powerful reducing agent.
Watkins
Chem 1422, Chapter 20
49
Half-Cell Potentials
ANOX CATRED
Points to Know and Remember
• A REDOX reaction is always the sum of an oxidation
half-reaction and a reduction half-reaction.
• A voltaic cell is the physical realization of a
spontaneous REDOX reaction; it is the combination
of an oxidation half-cell (anode) and a reduction halfcell (cathode).
• For any half-cell reaction, Eoox = -Eored
• The anode voltage (Eoox) and the cathode voltage
(Eored) combine to produce the cell voltage (Eocell)
Eocell = Eoox + Eored
Watkins
Chem 1422, Chapter 20
50
Half-Cell Potentials
ANOX CATRED
Points to Know and Remember
• Changing stoichiometric coefficients does not affect Eo
Zn2+(aq) + 2e- ⇌ Zn(s)
Eored = -0.76 V
2Zn2+(aq) + 4e- ⇌ 2Zn(s)
Eored = -0.76 V
(voltage is measured by a volt meter, and is
independent of how you write the reaction)
• Eocell > 0 means the reaction is spontaneous in
the forward direction
• Eocell < 0 means the reaction is not spontaneous in the
forward direction (but the reverse reaction
is spontaneous)
Watkins
Chem 1422, Chapter 20
51
Spontaneity of Redox Reactions
Question: under standard conditions, will nickel metal reduce
silver ions in water?
? Ni2+(aq) + 2Ag(s)
Ni(s) + 2Ag+(aq) →
Eocell > 0?
Solution: The half reactions (Appendix E SRP’s) are
Ni(s) → Ni2+(aq) + 2eEoox = -Eored = -(-0.28) = +0.28
2Ag+(aq) + 2e- → 2Ag(s)
Eored= +0.80
Eocell = Eoox + Eored = (+0.28) + (+0.80) = +1.08
Since Eocell is positive, the reaction is spontaneous; nickel metal
does reduce silver ions in solution under standard conditions.
Watkins
Chem 1422, Chapter 20
52
Spontaneity of Redox Reactions
Question: For the voltaic cell which uses the Al/Al3+and Co/Co2+
couples, write the spontaneous REDOX reaction, calculate Eocell,
identify anode and cathode, and determine the flow of electrons
in the external circuit and the flow of ions in the salt bridge.
Solution: From Appendix E:
Al3+(aq) + 3e- → Al(s)
Eored = -1.66 V
Co2+(aq) + 2e- → Co(s)
Eored = -0.28 V
Watkins
Chem 1422, Chapter 20
One of these reductions
must be reversed such
that the sum is positive
53
Spontaneity of Redox Reactions
Question: For the voltaic cell which uses the Al/Al3+and Co/Co2+
couples, write the spontaneous REDOX reaction, calculate Eocell,
identify anode and cathode, and determine the flow of electrons
in the external circuit and the flow of ions in the salt bridge.
Solution: From Appendix E:
Al3+(aq) + 3e- → Al(s)
Eored = -1.66 V
Co2+(aq) + 2e- → Co(s)
Eored = -0.28 V
2Al(s) → 2Al3+(aq) + 6e- Eoox = +1.66 V (Anode)
3Co2+(aq) + 6e- → 3Co(s) Eored = -0.28 V (Cathode)
2Al(s) + 3Co2+(aq) → 2Al3+(aq) + 3Co(s); Eocell = +1.38 V.
Al is the anode, Co is the cathode. Electrons flow in the wire from
the Al electrode to the Co electrode. In the salt bridge, cations
flow toward Co, anions flow toward Al.
Watkins
Chem 1422, Chapter 20
54
Spontaneity of Redox Reactions
Under standard conditions ...
a redox reaction is spontaneous if Eo > 0
any reaction is spontaneous if DGo < 0
Is there a connection between DGo and Eo?
DGo = -nFEo
n is the number of moles of e- transferred in the
balanced redox reaction
F is the charge carried by one mole of electrons:
96,485 Coulombs/mol eEo is measured in volts (V = J/C)
Note:
Watkins
– (n mol e-)(F C/mol e-)(E J/C) = DGo(J)
Chem 1422, Chapter 20
55
Spontaneity of Redox Reactions
DGo = -nFEo
Example: Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s)
The half reactions are
Ni(s) → Ni2+(aq) + 2e-
Eoox = +0.28
2Ag+(aq) + 2e- → 2Ag(s)
Eored = +0.80
}
n=2
Eocell = Eoox + Eored = (+0.28) + (+0.80) = +1.08 V
DGo = -(2)(96485)(+1.08) = –208408 J
DGo = -208.41 kJ
Watkins
Chem 1422, Chapter 20
56
Half-Cell Reactions
A complete REDOX reaction is derived from an oxidation
half-reaction and a reduction half-reaction, and the cell
potential is the direct sum of the half-cell potentials.
New half-reactions can also be derived
from known half-reactions.
Consider the two reduction reactions from Appendix E:
(1) Fe2+(aq) + 2e- ⇌ Fe(s)
Eored1 = –0.440 V
(2) Fe3+(aq) + e- ⇌ Fe2+(aq)
Eored2 = +0.771 V
(3) Fe3+(aq) + 3e- ⇌ Fe(s)
Eored3 = ?
Clearly, reduction (3) = reduction (1) + reduction (2);
this is the half-cell reaction missing from Appendix E
However, Eored3  +0.331!!!!
Watkins
Chem 1422, Chapter 20
57
Half-Cell Reactions
Two or more reduction (or oxidation) potentials cannot be
directly summed.
However, the free energies of two or more reactions can
be summed directly, and for REDOX reactions:
DG = –nFE
(1) Fe2+(aq) + 2e- ⇌ Fe(s)
DGo1 = –2FEored1
(2) Fe3+(aq) + e- ⇌ Fe2+(aq)
DGo2 = –1FEored2
(3) Fe3+(aq) + 3e- ⇌ Fe(s)
DGo3 = –3FEored3
DGo3 = DGo1 + DGo2
3Eored3 = 2Eored1 + 1Eored2
Eored3 = 2/3(-0.440) + 1/3Eored2(+0.771)
Eored3 = –0.036 V
Watkins
Chem 1422, Chapter 20
58
Non-Standard Cell EMF
Eo is the voltage of a cell under standard conditions
([ions] = 1 M, p(gases) = 1 atm, Q = 1).
What is the voltage of a cell when Q  1?
From thermodynamics, we know that
DG = DGo + RT ln Q
DG = –nFE & DGo = –nFEo
–nFE = –nFEo + RT ln Q
E = Eo – RT ln Q
nF
Nernst Equation
1 J = 1 volt.coulomb, so R = 8.314 volt.coulomb/mol.K
Watkins
Chem 1422, Chapter 20
59
Non-Standard Cell EMF
E = Eo - RT ln Q
nF
E298 = Eo298 - 0.02569 ln Q
n
At room temperature, RT/F = 0.02569
Text: ln Q = (ln 10) log Q
= 2.3026 log Q
E298 = Eo298 - 0.0592 log Q
n
As equilibrium is approached (at any temperature):
Q  K DG  0
E  0 (the battery “dies”!)
At equilibrium, Ecell = 0 and
-DGo
nFEo
Eo = RT ln K
RT = e RT
K
=
e
nF
nF Eo –DGo
=
ln K =
R = 8.314 R = 0.008314
RT
RT
Watkins
Chem 1422, Chapter 20
60
Non-Standard Cell EMF
E = Eo - RT ln Q
nF
What is the voltage of a Ni/Ag cell at 25 oC when
[Ni2+] = 0.1M and [Ag+] = 0.05 M?
Ni(s) ⇌ Ni2+(aq) + 2eEoox = +0.28 V
2e- + 2Ag+(aq) ⇌ 2Ag(s)
Eored = +0.80 V
Ni(s) + 2Ag+(aq) ⇌ Ni2+(aq) + 2Ag(s) Eocell = +1.08 V
[Ni2+] (0.1)
=
Q=
= 40
+
2
2
[Ag ] (0.05)
Ecell = +1.08 – (8.314)(298) ln(40) = +1.033 V
(2)(96485)
Watkins
Chem 1422, Chapter 20
61
Non-Standard Cell EMF
E = Eo - RT ln Q
nF
What is the voltage of a Ni/Ag cell at 25 oC when
[Ni2+] = 0.1M and [Ag+] = 0.05 M?
Ni(s) ⇌ Ni2+(aq) + 2eEoox = +0.28 V
2e- + 2Ag+(aq) ⇌ 2Ag(s)
Eored = +0.80 V
Ni(s) + 2Ag+(aq) ⇌ Ni2+(aq) + 2Ag(s) Eocell = +1.08 V
What is K?
nF Eo (2)(96485)(+1.08)
= 84.11
=
ln K =
(8.314)(298)
RT
K = 3.39×1036
Watkins
Chem 1422, Chapter 20
62
Non-Standard Half-Cell EMF
E = Eo - RT ln Q
nF
The Nernst equation can also be applied to half-reaction
potentials. For example, consider the reduction of
oxygen at 25 oC in acid:
O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l)
Eored = +1.229 V
Let P be the pressure (atm) of O2(g) and pH = -log[H+].
Then it is easy to show that the half-cell potential for
this reduction reaction, as a function of pH and P, is
Ered = +1.229 – 0.02569[pH ln(10) – ¼ ln(P)]
Watkins
Chem 1422, Chapter 20
63
Non-Standard Half-Cell EMF
O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l)
Eored = +1.229 V
Ered = +1.229 – 0.02569[pH ln(10) – ¼ ln(P)]
Watkins
Chem 1422, Chapter 20
64
Practical Voltaic Cells
A “battery” contains two or more voltaic cells
connected in series.
Lead-Acid Battery
The 12 volt lead-acid battery
consists of six voltaic cells
each producing about
2 volts.
Invented in 1859 by Gaston
Planté, it is the oldest
“storage battery”. It was
used to power the telegraph
during the American Civil
War.
Watkins
Chem 1422, Chapter 20
65
Practical Voltaic Cells
A “battery” contains two or more voltaic cells
connected in series.
Lead-Acid Battery
Cathode: lead(IV) is reduced to lead(II) - PbO2 on a
metal grid in sulfuric acid;
Eocathode = +1.685 V
PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- ⇌ PbSO4(s) + 2H2O(l)
Anode: lead(0) is oxidized to lead(II) - Pb plate in
sulfuric acid;
Eoanode = +0.356 V
Pb(s) + SO42-(aq) ⇌ PbSO4(s) + 2e-
Overall REDOX reaction:
PbO2(s) + Pb(s) + 2H2SO4(aq) ⇌ 2PbSO4(s) + 2H2O(l)
Eocell = Eored+ Eoox = (+1.685) + (+0.356) = +2.041 V
Watkins
Chem 1422, Chapter 20
66
Practical Voltaic Cells
A “battery” contains two or more voltaic cells
connected in series.
Lead-Acid Battery
• Wood or fiberglass spacers
are used to prevent the
electrodes from touching.
• This redox reaction can be
reversed (run backwards) by
pumping electrons to store
electrical energy.
PbO2(s) + Pb(s) + H2SO4(aq)
2PbSO4(s) + 2H2O(l)
Watkins
Chem 1422, Chapter 20
67
Practical Voltaic Cells
Georges Leclanché, 1866
Dry Cell
Anode: Zn cap: Zn(s) ⇌ Zn2+(aq) + 2eCathode: Graphite rod in a MnO2/NH4Cl paste:
2NH4+(aq) + 2MnO2(s) + 2e- → Mn2O3(s) + 2NH3(aq) + 2H2O(l)
Some NH3(g) escapes, so this reaction is irreversible.
The graphite rod is an inert electrode and has no part in
the redox reaction; it just transfers electrons.
Overall redox reaction:
2NH4+(aq) + 2MnO2(s) + Zn(s) →
Mn2O3(s) + 2NH3(aq) + 2H2O(l) + Zn2+(aq)
Watkins
Chem 1422, Chapter 20
68
Practical Voltaic Cells
Alkaline Cell
Anode: Zn powder mixed with KOH:
Zn(s) + 2OH-(aq) ⇌ Zn(OH)2(s) + 2e-
Cathode: MnO2(s) and graphite (inert)
2MnO2(s) + 2H2O(l) + 2e- ⇌ 2MnO(OH)(s) + 2OH-(aq)
Anode
cap
Lewis Urry
Eveready Battery Co.
1950’s
E = 1.55 V
Cell
can
Watkins
Gasket
Anode
(Zn + KOH)
Chem 1422, Chapter 20
Separator
Cathode
(MnO2 + graphite)
69
Practical Voltaic Cells
Lithium Ion Cell
Invented and developed during the 1970’s, voltaic cells
in this large family all depend on Li+ ions to carry the
internal current, but cells differ from one another in the
materials used for the anode, cathode, electrolyte and
salt bridge.
The most common anode is C(s) (graphite) with
intercalated LiC(s). The cathode is a mixed metal oxide
such as LiCoO2(s). The non-aqueous solvent and salt
bridge are often proprietary. Typical reactions:
LiC(s) ⇌ Li+(sol) + C(s) + e–
CoO2(s) + Li+(sol) + e– ⇌ LiCoO2(s)
Watkins
Chem 1422, Chapter 20
70
Practical Voltaic Cells
All voltaic cells eventually run down because reactants
reach their equilibrium concentrations. Even “storage”
cells wear out and cannot be recharged.
Why not just replenish the reactants?
Fuel Cell
The H2/O2 fuel cell is a primary source of
electricity of space flights
Cathode: reduction of oxygen gas (pH 14)
2H2O(l) + O2(g) + 4e- ⇌ 4OH-(aq)
Eo = +0.40
Anode: oxidation of hydrogen gas (pH 14)
2H2(g) + 4OH-(aq) ⇌ 4H2O(l) + 4e- Eo = +0.83
Watkins
Chem 1422, Chapter 20
71
Practical Voltaic Cells
H2/O2 Fuel Cell
Overall redox reaction
2H2(g) + O2(g) ⇌ 2H2O(l) Eo = 1.23 V
(combustion of hydrogen)
Any fuel could be used. For example,
a hydrocarbon/O2 fuel cell “burns”
the fuel but produces energy much
more efficiently (70%) than an
internal combustion engine (30%).
CH4(g) + 2O2(g) ⇌ CO2(g) + 2H2O(l)
Watkins
Chem 1422, Chapter 20
Eo = 1.06 V
DGo = -818 kJ
DHo = -891 kJ
72
Voltaic Cells
A voltaic cell uses a spontaneous REDOX reaction:
Molten sodium metal and chlorine
gas react spontaneously with a
great amount of heat:
2Na(l) + Cl2(g) → 2NaCl(l)
Voltaic Cell
Eo = +4.07 V
DGo = -785 kJ
DHo = -822 kJ
A voltaic cell using this REDOX reaction would “roll
down the energy hill” while electrons flow in the
external wire from anode (Na/Na+) to cathode (Cl2/Cl-).
Watkins
Chem 1422, Chapter 20
73
Non-Spontaneous Redox Reactions
A non-spontaneous REDOX reaction must be pushed
up the energy hill:
Electrolysis
Molten salt (m.p. = 801 oC) does
not spontaneously decompose into
Eo = -4.07 V
its elements:
DGo = +785 kJ
o = +822 kJ
DH
2NaCl(l) →
2Na(l)
+
Cl
(g)
X
2
This non-spontaneous reaction can be forced to occur
by pumping electrons into molten salt.
This decomposition reaction is called electrolysis, and
the container in which this reaction takes place is
called an electrolytic cell.
Watkins
Chem 1422, Chapter 20
74
Non-Spontaneous Redox Reactions
Na+(l) + Cl-(l) → Na(l) + ½Cl2(g)
Eoox = -1.36 V Eored = -2.71 V
Electrolysis is used
to produce active metals like
Na & Al;
to recharge reversible
(“storage”) cells;
to deposit a layer of metal on
a cathode (electroplating).
Watkins
Chem 1422, Chapter 20
75
Non-Spontaneous Redox Reactions
Electrolysis Problems always have 2 parts:
Anode (oxidation) Reaction
always write the oxidation half-reaction so that
1 mol of product is on the right:
2Cl-(l) → Cl2(g) + 2e-
Cathode (reduction) Reaction
always write the reduction half-reaction so that
1 mol of product is on the right:
Na+(l) + e- → Na(l)
Some problems involve only one of the half-reactions,
so the other can be ignored.
Watkins
Chem 1422, Chapter 20
76
Non-Spontaneous Redox Reactions
How many grams of copper metal will be deposited
if we run an elecroplating cell containing CuSO4
for 20 min at 30 A?
Copper(II) ions are reduced to copper metal, so
the cathodic (reduction) reaction is
Cu2+(aq) + 2e- → Cu(s)
There is no oxidation reaction mentioned, so we
can ignore the anodic half-reaction.
Faraday’s Law of Electrolysis
MM
It
m=
neF
Watkins
Chem 1422, Chapter 20
77
Faraday's Law of Electrolysis
MM
It
m=
neF
m is the mass of the product (gp)
MM is the molar mass of the product (gp/molp)
I is the current (Amps = C/s)
t is the number of seconds the current flows
ne is the number of moles of electrons required to
produce 1 mole of product (mole/molp)
F is the Faraday Constant, 96,485 (C/mole)
Every electrolysis problem has these 6 variables –
one of them is the unknown
Watkins
Chem 1422, Chapter 20
78
Faraday's Law of Electrolysis m = MM It
neF
How many grams of copper metal will be deposited
if we run an elecroplating cell containing CuSO4
for 20 min at 30 A?
Cu2+(aq) + 2e- → Cu(s)
m=?
MM = 63.55
I = 30
t = 20×60
ne = 2
(63.55 g/molCu)(30 C/s)(20×60 s)
(2 mole/molCu)(96485 C/mole)
11.86 g Cu(s)
F = 96485
Watkins
Chem 1422, Chapter 20
79
Faraday's Law of Electrolysis m = MM It
neF
How much chlorine gas and how much sodium
metal will be produced if a molten salt electrolysis
cell is run for 8 hours at 200 A?
Anode: 2Cl-(l) → Cl2(g) + 2e(2×35.45)(200)(8×60×60)
m{Cl2(g)} =
= 2116 g
(2)(96485)
Cathode: Na+(l) + e- → Na(l)
(22.99)(200)(8×60×60)
= 1372 g
m{Na(l)} =
(1)(96485)
Watkins
Chem 1422, Chapter 20
80
Faraday's Law of Electrolysis m = MM It
neF
The nitrate salt of metal M, M(NO3)3, is used to
electroplate a layer of metal M onto another object.
If 16.03 grams of metal M are deposited in 3 hours
and fifteen minutes by a current of 7.22 Amp, what
is metal M?
mneF
3+
Cathode: M (aq) + 3e → M(s) MM =
It
(16.03 g)(3 mole/molM)(96485 C/mol)
MM =
(7.22 C/s)(11700 s)
= 54.93 M = Mn
Watkins
Chem 1422, Chapter 20
81
Corrosion
Iron is strong, cheap, and used extensively by all
modern societies. Every navy has iron war ships; there
are millions of iron bodied cars and trucks; all major
household appliances have sheet-iron coverings. Iron
bridges span countless waterways worldwide. Every
major city in the world has countless buildings with
iron frames.
Unfortunately, iron is a reactive metal: iron plus
oxygen and water in the air produces “rust” which
destroys the metal.
Corrosion is costly; trillions of dollars every year are
spent fighting it. Corrosion can be deadly (I-35W
bridge in Minneapolis, 2007, 13 dead).
Watkins
Chem 1422, Chapter 20
82
Corrosion
Oxygen is an oxidizing agent. Its reduction is pH
dependent.
acidic solution, standard conditions
[H+] = 1 M, p(O2) = 1 atm
O2(g) + 4H+(aq) + 4e– ⇌ 2H2O(l)
Eored = +1.23 V
basic solution, standard conditions
[OH–] = 1 M, p(O2) = 1 atm
O2(g) + 2H2O(l) + 4e– ⇌ 4OH–(aq) Eored = +0.40 V
normal conditions
[OH–] = [H+] = 10-7 M, p(O2) = 0.21 atm
O2(g) + 4H+(aq) + 4e– ⇌ 2H2O(l)
Ered = +0.80 V
O2(g) + 2H2O(l) + 4e– ⇌ 4OH–(aq) Ered = +0.80 V
Watkins
Chem 1422, Chapter 20
83
Corrosion
Iron metal reacts with oxygen and water.
Cathode:
O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l)
Eored = +1.23 V
Anode:
2Fe(s) ⇌ 2Fe2+(aq) + 4e-
Eoox = +0.44 V
The Fe2+ initially formed is further oxidized to Fe3+
which reacts with OH-(aq) to form rust, Fe2O3.xH2O(s).
Watkins
Chem 1422, Chapter 20
84
Corrosion
Iron metal reacts with oxygen and water.
Watkins
Chem 1422, Chapter 20
85
Corrosion
Corrosion can be prevented (or at least delayed)
in two ways:
1. Form a non-porous physical barrier between
the environment and the surface of the metal:
• oil, grease or paint
• electroplate a passive metal such as gold
2. Use a sacrificial electrochemical reaction such
as galvanizing.
Watkins
Chem 1422, Chapter 20
86
Corrosion
Galvanized iron: Fe coated with a thin layer of Zn
because Zn is more easily oxidized than Fe:
Zn(s) ⇌ Zn2+(aq) +2eEoox = +0.76 V
Fe(s) ⇌ Fe2+(aq) + 2eEoox = +0.44 V
The reaction with the greater potential takes place
first, so the cheap Zn metal is “sacrificed”.
Watkins
Chem 1422, Chapter 20
87
Corrosion
Galvanized
Iron
Watkins
Fe acts as an inert cathode
Zn is the sacrificial anode
Chem 1422, Chapter 20
88
Corrosion
Mg(s) ⇌ Mg2+(aq) +2eFe(s) ⇌ Fe2+(aq) + 2e-
Eoox = +2.37 V
Eoox = +0.44 V
To protect buried iron
pipes, any metal which
is more reactive than
Fe (e.g., Mg) can be
used. The reactive
metal acts as a
sacrificial anode, and
the pipe is the cathode.
Watkins
Chem 1422, Chapter 20
89
Corrosion
Aluminum oxidizes even more readily than iron:
4Al(s) ⇌ 4Al3+(aq) + 12e3O2(g) + 12H+(aq) + 12e- ⇌ 6H2O(l)
Eoox = +1.66 V
Eored = +1.23 V
The Al3+(aq) reacts with OH-(aq) to form a surface
layer of corrundum, Al2O3(s) (aluminum “rust”). This
material is non-porous and very hard; it forms a
natural water/oxygen barrier which prevents further
corrosion.
Watkins
Chem 1422, Chapter 20
90
Electrochemistry Summary
REDOX terms & definitions
Oxidation Numbers
Empirical rules
Lewis structures
Balancing REDOX reactions: half-reaction method
Electricity - definitions
Voltaic cells
Spontaneous REDOX reactions
Elements & definitions
Half-cells & half-reactions
Electromotive force - cell voltage
Activity series
SHE
Watkins
Chem 1422, Chapter 20
91
Electrochemistry Summary
Interpretation of SOP & SRP
Calculation of cell potentials
electrodes, direction of flow of charge carriers
E and DG
Deriving unknown half-cell potentials
Nernst Equation
non-standard potentials, K
Batteries & Fuel cells
Electrolysis
Faraday’s law of electrolysis
electroplating
Corrosion
- finis Watkins
Chem 1422, Chapter 20
92