CELL POTENTIAL, E
1
1.10 V
Zn and Zn2+,
anode
Cu and Cu2+,
cathode
1.0 M
1.0 M
• 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.
2
CELL POTENTIAL, E
• For Zn/Cu cell, potential is +1.10 V at 25 ˚C
and when [Zn2+] and [Cu2+] = 1.0 M.
• This is the STANDARD CELL
POTENTIAL, Eo
• —a quantitative measure of the tendency of
reactants to proceed to products when all
are in their standard states at 25 ˚C.
3
Calculating Cell Voltage
• Balanced half-reactions can be added
together to get overall, balanced
equation.
Zn(s) ---> Zn2+(aq) + 2eCu2+(aq) + 2e- ---> Cu(s)
-------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
If we know Eo for each half-reaction, we
could get Eo for net reaction.
4
CELL POTENTIALS, Eo
Can’t measure 1/2 reaction Eo directly.
Therefore, measure it relative to a
STANDARD HYDROGEN CELL, SHE.
2 H+(aq, 1 M) + 2e- <----> H2(g, 1 atm)
Eo = 0.0 V
5
Zn/Zn2+ half-cell hooked to a SHE.
Eo for the cell = +0.76 V
Negative
electrode
Supplier
of
electrons
Zn --> Zn2+ + 2eOxidation
Anode
Positive
electrode
Acceptor
of
electrons
2 H+ + 2e- --> H2
Reduction
Cathode
Reduction of H+ by Zn
6
Figure 20.10
7
Overall reaction is reduction of H+ by Zn metal.
Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g)
Eo = +0.76 V
Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is +0.76 V
Zn is a (better) (poorer) reducing agent than H2.
Cu/Cu2+ and H2/H+ Cell
Eo = +0.34 V
Positive
Acceptor
of
electrons
Cu2+ + 2e- --> Cu
Reduction
Cathode
Negative
Supplier
of
electrons
H2 --> 2 H+ + 2eOxidation
Anode
8
Cu/Cu2+
and
H2/H+
Cell
Overall reaction is reduction of Cu2+ by H2 gas.
Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H+(aq)
Measured Eo = +0.34 V
Therefore, Eo for Cu2+ + 2e- ---> Cu is
+0.34 V
9
Zn/Cu Electrochemical Cell
10
+
Anode,
negative,
source of
electrons
Cathode,
positive,
sink for
electrons
Zn(s) ---> Zn2+(aq) + 2eEo = +0.76 V
Cu2+(aq) + 2e- ---> Cu(s)
Eo = +0.34 V
--------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Eo (calc’d) = +1.10 V
Uses of Eo Values
• Organize halfreactions by
relative ability to
act as oxidizing
agents
• Table 20.1
• Use this to predict
cell potentials and
direction of redox
reactions.
11
TABLE OF STANDARD
REDUCTION POTENTIALS
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
12
13
Potential Ladder for Reduction Half-Reactions
Figure 20.11
14
Using Standard Potentials, Eo
Table 20.1
• Which is the best oxidizing agent:
O2, H2O2, or Cl2? _________________
• Which is the best reducing agent:
Hg, Al, or Sn? ____________________
15
Standard Redox Potentials,
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
Eo
Any substance on the right will reduce any
substance higher than it on the left.
Northwest-southeast rule: product-favored
reactions occur between reducing agent at
southeast corner (anode) and oxidizing agent
at northwest corner (cathode).
16
Standard Redox Potentials, Eo
17
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+.
Using Standard Potentials, Eo
Table 20.1
• In which direction do the following reactions
go?
• Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
• 2 Fe2+(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s)
• What is Eonet for the overall reaction?
18
19
Standard Redox Potentials, Eo
E˚net = “distance” from “top” half-reaction
(cathode) to “bottom” half-reaction (anode)
E˚net = E˚cathode - E˚anode
Eonet for Cu/Ag+ reaction = +0.46 V
Eo
for a Voltaic Cell
Cd --> Cd2+ + 2eor
Cd2+ + 2e- --> Cd
Fe --> Fe2+ + 2eor
Fe2+ + 2e- --> Fe
All ingredients are present. Which way does
reaction proceed?
20
Eo for a Voltaic Cell
From the table, you see
• Fe is a better reducing
agent than Cd
• Cd2+ is a better
oxidizing agent than
Fe2+
Overall reaction
Fe + Cd2+ ---> Cd + Fe2+
Eo = E˚cathode - E˚anode
= (-0.40 V) - (-0.44 V)
= +0.04 V
21
More About
Calculating Cell Voltage
22
Assume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OHCathode
2 I- ---> I2 + 2eAnode
------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
Assuming reaction occurs as written,
E˚net = E˚cathode - E˚anode
= (-0.828 V) - (+0.535 V) = -1.363 V
Minus E˚ means rxn. occurs in opposite direction
23
Is E˚ related to ∆G?
YES!
Michael Faraday
1791-1867
Originated the terms anode,
cathode, anion, cation,
electrode.
Discoverer of
• electrolysis
• magnetic props. of matter
• electromagnetic induction
• benzene and other organic
chemicals
Was a popular lecturer.
24
25
o
E
and
o
∆G
Eo is related to ∆Go, the free
energy change for the reaction.
∆Go = - n F Eo
where F = Faraday constant
= 9.6485 x 104 J/V•mol
and n is the number of moles of
electrons transferred
Michael Faraday
1791-1867
Eo and ∆Go
∆Go = - n F Eo
For a product-favored reaction
Reactants ----> Products
∆Go < 0 and so Eo > 0
Eo is positive
For a reactant-favored reaction
Reactants <---- Products
∆Go > 0 and so Eo < 0
Eo is negative
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E at Nonstandard Conditions
0.0257 V
[Products]
E  EÞ ln
n
[Reactants]
• The NERNST EQUATION
• E = potential under nonstandard conditions
• n = no. of electrons exchanged
• ln = “natural log”
• If [P] and [R] = 1 mol/L, then E = E˚
• If [R] > [P], then E is ______________ than E˚
• If [R] < [P], then E is ______________ than E˚
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