Transcript 20.2 Voltaic Cell Generating Voltage (Potential)

Electrochemistry
An electrochemical cell produces electricity using
a chemical reaction.
It consists of two half-cells connected via an
external wire with a salt bridge connecting the
solutions
An electrolytic cell uses an external electricity
source to produce a chemical reaction. This is
usually called electrolysis.
1
Historically
Historically oxidation involved reaction with O2.
i.e., Rusting
4 Fe(s) + 3O2 (g) Fe2O3
(s)
Another example
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
In this reaction:
Zn(s)  Zn2+(aq) Oxidation
Cu2+(aq)  Cu(s) Reduction
In a redox reaction, one process can’t occur without the
other. Oxidation and Reduction reactions must
simultaneously occur.
2
Redox Between Zn and Cu
If Zn(s) and Cu2+(aq) is in the
same solution, then the
electron is a transferred
directly between the Zn and
Cu.
No useful work is obtained. However if the reactants are separated
and the electrons shuttle through an external path...
3
Electrochemical Cells
Voltaic / Galvanic Cell Apparatus which produce electricity
Electrolytic Cell Apparatus which consumes electricity
Consider:
Zn
Cu
Initially there is a flow
of eAfter a very short time
the process stops
Electron transport
stops because of
charge build up on
both sides
Build up of
positive charge
The charge separation will lead to
process where it cost too much
energy to transfer electron.
4
Build up of
negative charge
Completing the Circuit
Electron transfer can occur if the circuit is closed
Parts:
Two conductors
Electrolyte solution
Salt Bridge / Porous membrane
Process that must happen if e- is to flow.
A. e- transport through external circuit
B. In the cell, ions must migrate
C. Circuit must be closed using a salt bridge (no
charge build up)
Anode (-)
Green
Negative
electrode
generates
electron
Cathode (+)
A
Red
Positive
electrode
accepts
electron
C
B
Oxidation
Occurs
5
Reduction
Occurs
Anode/Anion (-)
Cathode/Cation(+)
Voltaic Cell
Electron transfer can occur if the circuit is closed
Parts:
Two conductors
Electrolyte solution
Salt Bridge / Porous membrane
6
3 process must happen if e- is to flow.
A. e- transport through external circuit
B. In the cell, ions a must migrate
C. Circuit must be closed (no charge build up)
Anode (-)
Cathode (+)
Black
Red
Negative
electrode
generates
electron
Positive
electrode
accepts
electron
Oxidation
Occur
Reduction
Occur
Anode/Anion (-)
Cathode/Cation(+)
Completing the Circuit: Salt Bridge
In order for electrons
to move through an
external wire, charge
must not build up at
any cell. This is
done by the salt
bridge in which ions
migrate to different
compartments
neutralize any
charge build up.
7
Sign Convention of Voltaic Cell
@ Anode: Negative Terminal (anions).
Source of electron then repels electrons. Oxidation occurs.
Zn(s)  Zn+2(aq) + 2e- : Electron source
@ Cathode: Positive Terminal (cation)
Attracts electron and then consumes electron. Reduction occurs.
Electron target: 2e- + Cu+2(aq) Cu(s)
Overall:
Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s)
E° = 1.10 V
Note when the reaction is reverse: Cu(s) + Zn+2(aq)  Cu+2(aq) + Zn(s)
Sign of E ° is also reversed E° = -1.10 V
Oxidation:
Zn(s)  Zn+2(aq)
Reduction: Cu+2(aq) Cu(s)
E° = 0.76 V
E° = 0.34 V
1.10 V = E°CELL
or E°CELL = E°red (Red-cathode) - E°red (Oxid-anode)
8
Another Voltaic Cell
Zn(s) + 2H+ (aq)  Zn+2(aq) + H2 (g)
E° = 0.76 V
@ Anode: Negative Terminal (anions): Zn(s)  Zn+2(aq) + 2e- :
Source of electron then repels electrons. Oxidation occurs.
@ Cathode: Positive Terminal (cation): 2e- + 2H+(aq) H2 (g)
Attracts electron and then consumes electron. Reduction occurs.
Overall: Zn(s) + 2H+ (aq)  Zn2+ (aq) + H2 (g)
9
Another Voltaic Cell
Zn(s) + 2H+ (aq)  Zn+2(aq) + H2 (g)
E° = 0.76 V
@ Anode: Negative Terminal (anions): Zn(s)  Zn+2(aq) + 2e- :
Source of electron then repels electrons. Oxidation occurs.
@ Cathode: Positive Terminal (cation): 2e- + 2H+(aq) H2 (g)
Attracts electron and then consumes electron. Reduction occurs.
Overall: Zn(s) + 2H+ (aq)  Zn2+ (aq) + H2 (g)
10
Line Notation Convention
basically:
metal | ionic solution || ionic solution | metal
This is a convenient way of representing cells
1.
Anode

Cathode
[oxidation (-) ]
[reduction (+)]
2.
| represents a phase boundary
3.
|| represents the salt bridge
4.
Concentration of component
1
4
Zn(s) ZnSO4 (aq,1.0M) CuSO4 (aq,1.0M) Cu(s)
3
2
11
Line Notation Examples
Consider :
Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s
Anode:
Zn  Zn+2 + 2e-
Cathode:
Cu+2 + 2e-  Cu
Shorthand “Line” notation
Zn (s) Zn+2 (aq)(1.0M) Cu+2(aq) (1.0M) Cu(s)
2nd Example : Zn(s) + 2H+ (aq)  Zn+2(aq) + H2(g)
Anode:
Zn  Zn+2 + 2e-
Cathode:
2H+ + 2e-  H2 (g)
Shorthand “Line” notation
Zn (s) Zn+2 (aq)(1.0M) H+(aq) (1.0M), H2(g, 1atm) Pt(s)
12
EMF - ElectroMotive Force
Potential energy of electron is higher at the anode.
This is the driving force for the reaction (e- transfer)
Anode
(-)
e
e- flow
toward
cathode
Larger the gap,
the greater the
potential (Voltage)
13
(+)
Cathode
D P.E. = V = J
eC
ElectroMotive Force
For electrochemical cells
Emf – electromotive force
Potential energy difference between the two electrodes
Units of emf: V - Volts :
1V - 1 Joule / Coulomb
1 Joule of work per coulomb of charge transferred.
For electrolysis
When a current of one amp flow for 1 second this equals one
coulomb of charge passed.
Charge (C) = Current (A) x Time (s)
96,400 coulombs = 1 Faraday of charge
1 Faraday is equal to 1 mole of electrons
14
Standard Electrode Potentials
Written as reduction
Cell Potential is written as a reduction equation.
M+ + e-  M
E° = std red. potential
15
Zoom of Std. Electrode Potentials
Cell Potential is written as a reduction equation.
Written as reduction
M+ + e-
 M
E° =
F2 (g) + 2e-  2 F - (aq)
2.87 V
Ce4+ + e-  Ce3+ (aq)
1.61 V
2H+ + 2e-  H2 (g)
0.00 V
Li+(aq) + e-  Li(s)
-3.045 V
All reaction written as reduction reaction. But in
electrochemistry, there can’t be just a reduction reaction. It must
be coupled with an oxidation reaction.
16
E°Cell Evaluation
E°Cell Function of the reaction
 Oxidation Process (Anode reaction)
 Reduction Process (Cathode reaction)
E°Cell = E°red (cathode) - E°red (anode)
Simple to remember:
Eº(cell) = Eº(red) – Eº(ox)
E = E redox
The potential difference of the cell is equal to
the electrode potential of the reduced
component minus the electrode potential of the
oxidised component.
17
Standard Electrode Potential
How is E°red (Cathode) and E°red (Anode) determine.
E° (EMF) - State Function; there is no absolute scale
Absolute E° value can’t be measured experimentally
The method of establishing a scale is to measure the difference
in potential between two half-cells.
Consider:
Zn  Zn+2 + 2e- E°=?
Can’t determine because the
reaction must be coupled
Question? - How can a scale of reduction potential be determined ?
Answer - Use a half reaction as reference and assign it a potential of
zero.
Measure all other half cells in comparison to this reaction.
18
Side-Bar: Relative Scale
Consider a baby whose weight is to be determine but will not
remain still on top of a scale. How can the parents determine the
babies weight?
Carry the child
in arms and
weight both
child and parent
then subtract
the weight of
the parent from
the total to yield
the baby
weight.
19
Standard Reference Half Cell
Selected half reaction is:
H+ / H2 (g) couple half reaction: 2H+ (aq, 1.0M) + 2e-  H2 (g,1atm)
by definition c E° = 0.0 V, the reverse is also 0.0 V
H+/H2 couple - Standard Hydrogen Electrode (SHE)
To determine E° for a another half reaction, the reaction of
interest needs to be connected to this SHE. The potential
measured is then assigned to the half-reaction under
investigation.
Example:
E°Cell = 0.76 V = E°red (Cat) - E°red (Anode)
0.0 V - (?)
E°red (Anode) = - 0.76 V
\Zn+2/Zn E° = -0.76 V reduction reaction
20
Determining Other Half-Cell Potentials
Now consider the reaction:
Zn(s)|Zn+2 (1.0 M)||Cu+2(1.0 M)|Cu(s)
E°Cell = 1.10 V
E°Cell = E°red (Cat) - E°red (Anode)
recall, E° Zn+2/Zn = - 0.76 V
Therefore,
E°Cell = E°Cu+2/Cu - E° Zn+2/Zn
1.10 V = (?) - (- 0.76 V)
E°Cu+2/Cu = + 0.34 V
21
Example: Half-Cell Potential
Example:
For the reaction: Tl+3 + 2Cr2+  Tl+ + 2Cr3+
E°Cell = 1.19 V
i) Write both half reaction and balance
ii) Calculate the E°Cell Tl+3  Tl+
iii) Sketch the voltaic cell and line notation
i) Tl+3 + 2e-  Tl+
(Cr2+  2Cr3+ + 2e- ) x 2
E° = 0.41 V
1.19 V
ii) E°Cell = 1.19 V = E°red (Cat) - E°red (Anode)
1.19 V= E°red (Cat) - 0.041 V
Pt
for Tl+3 + 2e-  Tl+ :
1.19 V - 0.41 = E°red (Cat) = 0.78 V
Pt
Cr2+  Cr3+
22
Tl3+  Tl+
Voltaic Vs. Electrolytic Cells
Voltaic Cell
Energy is released from
spontaneous redox reaction
System does work on
load (surroundings)
Electrolytic Cell
Energy is absorbed to drive
nonspontaneous redox reaction
Surrounding (power supply) do
work on system (cell)
Anode
(Oxidation)
Oxidation Reaction
X  X+ + e-
Oxidation Reaction
A-  A + e-
Reduction Reaction
e- + Y+  Y
X +
23
Overall (Cell) Reaction
 X+ + Y, DG = 0
Y+
Reduction Reaction
e- + B+  B
A-
Overall (Cell) Reaction
+ B+ A + B, DG> 0
General characteristics of
voltaic and electrolytic
cells. A voltaic cell
generates energy from a
spontaneous reaction
(DG<0), whereas an
electrolytic cell requires
energy to drive a
nonspontaneous reaction
(DG>0). In both types of
cell, two external circuits
provides the means or
electrons to flow.
Oxidation takes place all
the anode, and reduction
takes place at the cathode,
but the relative electrode
changes are opposite in the
two cells.