Transcript ELECTROCHEMISTRY Chapter 21

ELECTROCHEMISTRY
Chapter 21
• redox reactions
• electrochemical cells
• electrode processes
• construction
• notation
Electric automobile
• cell potential and Go
• standard reduction potentials (Eo)
• non-equilibrium conditions (Q)
• batteries
• corrosion
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Electrochemistry (Ch. 21)
1
TRANSFER REACTIONS
Atom transfer
HCl (g) + H2O (l)  Cl- (aq) + H3O+ (aq)
Electron transfer
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
- 2 e2 x +1 e-
Loss of Electrons =
OXIDATION (LEO)
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Gain of Electrons =
REDUCTION (GER)
Electrochemistry (Ch. 21)
2
Electron Transfer Reactions
• Electron transfer reactions are oxidationreduction or redox reactions.
• Redox reactions can result in :
– generation of an electric current, or
– be caused by imposing an electric current.
• When external electric current is involved,
this field of chemistry is called
ELECTROCHEMISTRY.
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Electrochemistry (Ch. 21)
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Terminology for Redox Reactions
• OXIDATION—loss of electron(s) by a
species; increase in oxidation number.
• REDUCTION—gain of electron(s);
decrease in oxidation number.
• OXIDIZING AGENT—electron acceptor;
species is reduced.
• REDUCING AGENT—electron donor;
species is oxidized.
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Electrochemistry (Ch. 21)
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Direct Redox Reactions
Oxidizing and reducing agents in direct contact.
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
2Al (s) + 3Cu2+ 
2 Al3+ + 3 Cu (s)
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Electrochemistry (Ch. 21)
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Indirect Redox Reactions
A battery functions by transferring electrons
through an external wire from the reducing
agent to the oxidizing agent.
Electron transfer
Reduction
Oxidation
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Ions
Electrochemistry (Ch. 21)
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Balancing Equations
Cu (s) + Ag+ (aq) 
Cu2+ (aq) + Ag (s)
How to balance for both charge and mass ?
Step 1: Identify the oxidation and reduction HALF-REACTIONS:
RED: Ag+ + e-  Ag
OX: Cu  Cu2+ + 2eStep 2: Balance each HALF-REACTION for charge and mass (done)
Step 3:Multiply each half-reaction by a factor that makes the
reducing agent supply as many electrons as the oxidizing agent
requires - ELECTRON TRANSFER NUMBER (2 here)
Oxidation (Reducing agent)
Reduction (Oxidizing agent)
Cu  Cu2+ + 2e2 Ag+ + 2 e-  2 Ag
Step 4:Add half-reactions to give the overall equation.
Cu (s) + 2 Ag+ (aq)
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 Cu2+ (aq) + 2Ag (s)
Electrochemistry (Ch. 21)
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Balancing Equations (2)
Balance the following in acid solution—
VO2+ + Zn  VO2+ + Zn2+
Step 1:
Write the half-reactions
Ox
Zn  Zn2+
Red
VO2+  VO2+
Step 2:
Balance each half-reaction for mass.
Ox
Zn  Zn2+
Red
2 H+ + VO2+  VO2+ + H2O
Add H2O on O-deficient side and add H+ on
other side for H-balance.
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Electrochemistry (Ch. 21)
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Balancing Equations (3)
Step 3:
Balance half-reactions for charge.
Ox
Zn  Zn2+ + 2eRed
e- + 2 H+ + VO2+  VO2+ + H2O
Step 4:
Multiply by an appropriate factor to
balance the electron transfer in OX. and RED.
Ox
Zn  Zn2+ + 2eRed 2e- + 4 H+ + 2 VO2+  2 VO2+ + 2 H2O
Step 5:
Add half-reactions
Zn + 4 H+ + 2 VO2+  Zn2+ + 2 VO2+ + 2 H2O
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Electrochemistry (Ch. 21)
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Tips on Balancing Equations
• Never add O2, O atoms, or
O2- to balance oxygen.
• Never add H2 or H atoms to
balance hydrogen.
• Be sure to write the correct
charges on all the ions.
• Check your work at the end
to make sure mass and
charge are balanced.
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Electrochemistry (Ch. 21)
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Why Study Electrochemistry?
• Batteries
• Corrosion
• Industrial production
of chemicals such as
Cl2, NaOH, F2 and l
• Biological redox
reactions
The heme group
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Electrochemistry (Ch. 21)
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Electrochemical Cells
• An apparatus in which a redox
reaction occurs by
transferring electrons through
an external connector.
• VOLTAIC CELL
• Product favored reaction
chemical reaction  electric current
• ELECTROLYTIC CELL
• Reactant favored reaction
electric current  chemical reaction
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Electrochemistry (Ch. 21)
Batteries are
voltaic cells
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CHEMICAL CHANGE  ELECTRIC CURRENT
Zn metal
Cu2+ ions
With time, Cu plates out
onto Zn metal strip, and
Zn strip “disappears.”
• Zn is oxidized and is the reducing agent
Zn(s)  Zn2+(aq) + 2e• Cu2+ is reduced and is the oxidizing agent
Cu2+(aq) + 2e-  Cu(s)
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Electrochemistry (Ch. 21)
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CHEMICAL CHANGE  ELECTRIC CURRENT (2)
Oxidation: Zn(s)  Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e-  Cu(s)
-------------------------------------------------------Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s)
Zn metal
Cu2+ ions
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Electrons are
transferred from Zn to
Cu2+, but there is no
useful electric current.
Electrochemistry (Ch. 21)
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CHEMICAL CHANGE  ELECTRIC CURRENT (2)
• To obtain a useful
current, we separate the
oxidizing and reducing
agents so that electron
transfer occurs thru an
external wire.
wire
elect rons
Zn
salt
bridge
Zn 2+ ions
Cu
Cu2+ ions
• This is accomplished in a VOLTAIC cell.
(also called GALVANIC cell)
• A group of such cells is called a battery.
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Electrochemistry (Ch. 21)
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wire
ANODE
CATHODE
elect rons
OXIDATION
Zn
Zn 2+ ions
REDUCTION
Cu
salt
bridge
Cu2+ ions
• Electrons travel thru external wire.
• Salt bridge allows anions and cations to
move between electrode compartments.
• This maintains electrical neutrality.
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Electrochemistry (Ch. 21)
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Electrons move
from anode to
cathode in the wire.
Anions & cations
move through the
salt bridge.
Electrochemical Cell
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Electrochemistry (Ch. 21)
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Standard Notation for Electrochemical Cells
Phase boundary
ANODE
Salt bridge
Phase boundary
Zn / Zn2+ // Cu2+ / Cu
Anode electrode
Cathode electrode
Active electrolyte in
oxidation half-reaction
OXIDATION
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CATHODE
Active electrolyte in
reduction half-reaction
REDUCTION
Electrochemistry (Ch. 21)
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CELL POTENTIAL, E
Zn  Zn2+ + 2eANODE
2e- + Cu2+ Cu
CATHODE
• Electrons are “driven” from anode to cathode by an
electromotive force or emf.
• For Zn/Cu cell, this is indicated by a voltage of 1.10
V at 25C and when [Zn2+] and [Cu2+] = 1.0 M.
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Electrochemistry (Ch. 21)
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CELL POTENTIAL, Eo
For Zn/Cu, voltage is 1.10 V at 25C
and when [Zn2+] and [Cu2+] = 1.0 M.
• This is the
STANDARD CELL POTENTIAL, Eo
• Eo is a quantitative measure of the tendency
of reactants to proceed to products when all
are in their standard states at 25 C.
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Electrochemistry (Ch. 21)
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o
E
and
o
G
Eo is related to Go, the free
energy change for the reaction.
Go = - n F Eo
•
F = Faraday constant
= 9.6485 x 104 J/V•mol
•n
= the number of moles of
electrons transferred.
Zn / Zn2+ // Cu2+ / Cu
n for Zn/Cu cell ?
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n=2
Electrochemistry (Ch. 21)
Michael Faraday
1791-1867
Discoverer of
• electrolysis
• magnetic props. of matter
• electromagnetic induction
• benzene and other
organic chemicals
21
Eo and Go (2)
Go = - n F Eo
• For a product-favored reaction
– battery or voltaic cell: Chemistry  electric current
Reactants  Products
Go < 0 and so Eo > 0 (Eo is positive)
• For a reactant-favored reaction
- electrolysis cell: Electric current  chemistry
Reactants  Products
Go > 0 and so Eo < 0 (Eo is negative)
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Electrochemistry (Ch. 21)
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Calculating Cell Voltage
• Balanced half-reactions can be added together
to get the overall, balanced equation.
 I2 + 2e-
Anode:
2 I-
Cathode:
2 H2O + 2e-  2 OH- + H2
Net rxn:
2 I- + 2 H2O  I2 + 2 OH- + H2
• If we know Eo for each half-reaction, we
can calculate Eo for the net reaction.
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Electrochemistry (Ch. 21)
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STANDARD CELL POTENTIALS, Eo
• Can’t measure half- reaction Eo directly.
Therefore, measure it relative to a standard
HALF CELL:
the Standard Hydrogen
Electrode (SHE).
2 H+(aq, 1 M) + 2eEo = 0.0 V
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Electrochemistry (Ch. 21)
H2(g, 1 atm)
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Zn/Zn2+ versus H+/H2
Volts
Zn
-
+
Salt Bridge
H2
H+
Zn2+
Zn
Zn 2+ + 2eOXIDATION
ANODE
2 H+ + 2eH2
REDUCTION
CATHODE
Zn/Zn2+ half-cell combined with a SHE.
Eo for the cell is +0.76 V
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Electrochemistry (Ch. 21)
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Eo for Zn/Zn2+ half-cell
Volts
Zn
Overall reaction is
reduction of H+ by Zn
metal.
-
+
Salt Bridge
H2
H+
Zn2+
Zn
Zn 2+ + 2eOXIDATION
ANODE
Zn(s) + 2 H+ (aq)  Zn2+ + H2(g)
2 H+ + 2eH2
REDUCTION
CATHODE
Eo = +0.76 V
Therefore,
+0.76 V.
Eo for Zn  Zn2+ (aq) + 2e- is ??
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Electrochemistry (Ch. 21)
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Standard REDUCTION potentials
Zn  Zn2+ (aq) + 2eQ.
A.
Eo = +0.76 V
Relative to H2 is Zn a (better/worse) reducing agent ?
Zn is a better reducing agent than H2.
What is Eo for the reverse reaction ?
Zn2+ + 2e-  Zn
The value for the REDUCTION 1/2-cell is the negative of
that for the OXIDATION 1/2-cell:
Zn  Zn2+ (aq) + 2e-
Eo = +0.76 V
THUS
Zn2+ + 2e-  Zn
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Eo = -0.76 V
Electrochemistry (Ch. 21)
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Cu/Cu2+ and H2/H+ Cell
Eo = +0.34 V
Volts
Cu
+
Salt Bridge
H2
H+
Cu2+
Cu2+ + 2eCu
REDUCTION
CATHODE
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H2
2 H+ + 2eOXIDATION
ANODE
Electrochemistry (Ch. 21)
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Cu/Cu2+ half cell Eo
Volts
Cu
+
Salt Bridge
H2
Overall reaction is
reduction of Cu2+ by H2
gas.
Cu2+
Cu2+ + 2eCu
REDUCTION
CATHODE
H+
H2
2 H+ + 2eOXIDATION
ANODE
• Cu2+ (aq) + H2(g)  Cu(s) + 2 H+(aq)
• Measured Eo = +0.34 V
+0.34 V
• Therefore, Eo for Cu2+ + 2e-  Cu is ??
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Electrochemistry (Ch. 21)
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Zn/Cu Electrochemical Cell
What is Eo for the Zn/Cu cell (Daniel’s cell) ??
wire
Anode,
negative,
source of
electrons
Anode:
Cathode:
Net:
elect rons
Zn
Zn 2+ ions
Cathode,
positive,
sink for
electrons
Cu
salt
bridge
Cu2+ ions
Zn(s)  Zn2+(aq) + 2eCu2+(aq) + 2e-  Cu(s)
Eo = +0.76 V
Eo = +0.34 V
Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s)
Eo = Eo(anode) + Eo(cathode) = 0.76 + 0.34 = +1.10 V
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Electrochemistry (Ch. 21)
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Uses of Eo Values
This shows we can
a) decide on relative ability of elements to act
as reducing agents (or oxidizing agents)
b) assign a voltage to a half-reaction that
wire
reflects this ability.
elect rons
Zn
Zn 2+ ions
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Electrochemistry (Ch. 21)
salt
bridge
Cu
Cu2+ ions
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STANDARD REDUCTION POTENTIALS
Oxidizing ability of ion
Half-Reaction
Cu2+ + 2e-
 Cu
+ 0.34
2 H+ + 2e-
 H2
0.00
Zn2+ + 2e-
 Zn
-0.76
BEST Oxidizing agent Cu
? ?2+
BEST Reducing agent ?Zn
?
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Eo (Volts)
Reducing ability
of element
Electrochemistry (Ch. 21)
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Standard Redox Potentials, Eo
• Any substance on the right will
reduce any substance higher
than it on the left.
• Zn can reduce H+ and Cu2+.
• H2 can reduce Cu2+ but not Zn2+
• Cu cannot reduce H+ or Zn2+.
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
reducing ability
of element
Use tabulated reduction potentials to analyse
spontaneity of ANY REDOX REACTION
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Electrochemistry (Ch. 21)
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Determining Eo for a Voltaic Cell
Volts
Cd
Fe2+
Cd2+
Cd  Cd2+ + 2eor
Cd2+ + 2e-  Cd
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Fe
Salt Bridge
Fe  Fe2+ + 2eor
Fe2+ + 2e-  Fe
Electrochemistry (Ch. 21)
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Eo for Fe/Cd Cell
Volts
Cd
LHS species
Fe
Salt Bridge
is better oxidizing agent
Cd 2+ + 2e-  Cd
-0.40
Fe 2+ + 2e-  Fe
-0.44
Cd2+
Fe2+
RHS species
is better reducing agent
• Fe is a better reducing agent than Cd
• Cd 2+ is a better oxidizing agent than Fe 2+
• Overall reaction as written is spontaneous:
Fe + Cd 2+  Cd + Fe 2+
Eo = +0.04 V
• The reverse reaction is not spontaneous:
Cd + Fe 2+  Fe + Cd 2+
Eo = -0.04 V
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Electrochemistry (Ch. 21)
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