Download 20.2 Voltaic Cell Generating Voltage (Potential)

Electrochemistry
Generating Voltage (Potential)
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Historically
Historically oxidation involved reaction with O2.
i.e., Rusting
4 Fe(s) + 3O2 (g) Fe2O3
(s)
Other 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-Reduction reaction must
simultaneously occurs.
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Redox Between
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...
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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 some time the
process stops
Electron transport
stops because of
charge build up
Build up of
positive charge
The charge separation will lead to
process where it cost too much
energy to transfer electron.
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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
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 (-)
Black
Negative
electrode
generates
electron
Cathode (+)
A
Red
Positive
electrode
accepts
electron
C
B
Oxidation
Occur
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Reduction
Occur
Anode/Anion (-)
Cathode/Cation(+)
Voltaic Cell
Electron transfer can occur if the circuit is closed
Parts:
Two conductors
Electrolyte solution
Salt Bridge / Porous membrane
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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.
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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)
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Other 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.
Net: Zn(s) + 2H+ (aq)  Zn2+ (aq) + H2 (g)
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Other 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.
Net: Zn(s) + 2H+ (aq)  Zn2+ (aq) + H2 (g)
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Line Notation Convention
Line notation: Convenient convention for electrochemical cell
Schematic Representation
1.
Anode

Cathode
[oxidation (-) ]
[reduction (+)]
2.
“ “ phase boundary
(where potential may develop)
3.
“ “ Liquid junction
4.
Concentration of component
4
1
Zn(s) ZnSO4 (aq,1.0M) CuSO4 (aq,1.0M) Cu(s)
3
2
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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)
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Other Voltaic Cell & Their Line Notation
Oxidation half-reaction
2I- (aq)  I2 (s) + 2e-
Oxidation half-reaction
Zn(s)  Zn+2(aq) + 2e-
Oxidation half-reaction
Cr(s)  Cr+3(aq) + 3e-
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5e Mn 2+(aq) +4H2O(l)
Zn(s) | Zn+2 (aq)||H+(aq) , H2 (g,1atm)|Pt
Reduction half-reaction
Ag+(aq) + e-  Ag (s)
Cr(s) | Cr+3 (aq)||Ag+(aq) | Ag(s)
C(s)| I-(aq) , I2 (g,1atm) || MnO4-(aq) , Mn+2 (aq)| C(s)
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Line Notation Examples
Example 1: B&L 20.13
Zn(s) + Ni2+(aq)  Zn+2(aq) + Ni (aq)
Example 2: B&L 20.19
Tl+3(aq) + 2Cr2+(aq)  Tl+(aq) + Cr+3(aq)
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Voltage of Galvanic / Voltaic Cell
Transport of any object requires a net force.
Consider water flowing through pipes. This occurs
because of pressure gradient.
Pressure
(h)
Flow
(Fluid Transport)
Pressure
(i)
Or
Object falling
or transport
down due to Dh
Similarly, electron are
transported through wires
because of the electromotive
force EMF or Ecell.
(-) e 15
(+)
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)
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(+)
Cathode
D P.E. = V = J
eC
ElectroMotive Force
EMF - Electro Motive Force
Potential energy difference between the two electrodes
The larger the DP.E. the larger EMF value.
The magnitude of P.E. for the reaction (half reaction) is an
intensive property)
i.e., Size independent: r, Tbpt, Cs.
Therefore EMF is also an intensive property.
Analogy:
Size of rock not important, only the height from ground.
(Electron all have the same mass)
Unit: EMF: V - Volts :
1V - 1 Joule / Coulomb
1 Joule of work per coulomb of charge transferred.
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Stoichiometry Relationship to E°
EMF - Intensive Property
E°cell Standard state conditions 25°C, 1atm, 1.0 M
E°cell Intensive property, Size Independent
Consider:
Li+ + e-  Li (s)
x2
2 Li+ + 2 e-  2 Li (s)
E°Cell = -3.045 V
E°Cell = (-3.045 V) x 2 = ??
But E° = Voltage per electron
E° ‘ = E° x 2 = ?  - 3.045 V • 2 = -3.045 V
2 e-
\ Stoichiometry does not change E°, but reversing the
reaction does change the sign of E°.
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Standard Reduction Potential
Written as reduction
Cell Potential is written as a reduction equation.
M+ + e-  M
E° = std red. potential
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Zoom View of Std. Reduction Potential
Cell Potential is written as a reduction equation.
Written as reduction
M+ + e-
 M
E° =
F2 (g) + 2e-  2 F - (aq)
Ce4+ + e-  Ce3+ (aq)
2.87 V
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.
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E°Cell Evaluation
E°Cell Function of the reaction
 Oxidation Process (Anode reaction)
 Reduction Process (Cathode reaction)
or
E°Cell = E°Cathode & E°Anode
Cathode (+)
Anode (-) Most Negative Reduction reaction
Therefore,
E°Cell = E°red (Cathode) - E°red (anode)
Very Positive
\ Very Spontaneous
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Neg Minus (Large negative)
(Very Positive Value)
Standard Reduction 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
How can a scale of reduction potential be determine ?
Use a half reaction as reference and assign it a potential of zero.
Electrochemical reaction more spontaneous than this reference will
have positive E°, and those less spontaneous will have negative E°.
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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.
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Reference Potential
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 coupled to this SHE. The potential measured
is then assigned to the half-reaction under investigation.
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 rxn
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Determining Other Half-Cell Potential
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
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Example: Half-Cell Potential
Example BBL20.19:
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+
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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 +
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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.