6-3 Electrode Potentials

The producing of Electrode Potentials

Standard electrode potential

Standard hydrogen electrode (SHE)

Application of standard electrode
potential
6-3.1 Generating Electrode Potentials
A strip of mental, M, is called an electrode. An
electrode, immersed in a solution containing
the metal ions, M n+ . Two kinds of
interactions are possible between metal atoms
on the electrode and metal ions in solution.
The Producing
of Electrode
Potentials
Figure:
A metal strip,
M,
partially
immersed
in
an
aqueous
solution of its
ions, M n+.
1. A mental atom M on the electrode may lose n
electrons and enter the solution as the ion M n+.
The metal atom is oxidized.
2. A mental ion M n+ may collide with the
electrode, gain n electrons, and be converted to a
metal atom M. The ion is reduced.
An equilibrium is quickly established between the
metal and the solution, which we can represent as
oxidation
M(s)
M n + (a q) + n ereduction
oxidation
M(s)
M n + (aq) + n ereduction
1) v
forward>v reverse
+
+
+
2) v
+
+
+
forward<v reverse
+
+
-- +
-- +
-
+
+
+
+
---
If the oxidation tendency is strong, we expect a
very slight negative charge density to
accumulate on the electrode. The greater the
tendency for the metal to become oxidized, the
greater this negative charge density.
In turn, the solution develops a very slight buildup in
concentration of M n+ ions and a slight positive
charge density. If the reduction tendency is strong,
we expect the reverse situation-a very slight positive
charge density on the electrode and negative
charge density in the solution.
电极电位的产生

金属电极板浸入其盐溶液中，存在相反的过程，速
率相等时，建立动态平衡：

M (s)

Èܽâ
Mn+ (aq) + ne-
Îö ³ö
金属极板表面上带有过剩负电荷；溶液中等量正电
荷的金属离子受负电荷吸引，较多地集中在金属极
板附近，形成所谓双电层结构，其间电位差称为电
极电位。
6-3.2 Determination of Standard
Electrode Potentials (Φ°)
(-)
SHE
Standard Electrode (+)
E =φ+ - φ

Φ°SHE= 0
=E
● Standard Hydrogen Electrode
(SHE)
[H+] = 1mol/L
T = 298K
PH2 = 1atm
Figure: A standard
hydrogen electrode.
φ°SHE = 0 v
H+(l.0M)|H2(1atm)|Pt
● Standard Electrode Potentials
The standard electrode potential, it is
designated by a superscript degree sign:φ°
A standard electrode refers to an electrode
in which the concentration of ions and the
pressure of gases are equal to 1 at 25℃.
Examples:
Determination of standard
zinc electrode potential.
(-) Zn Zn2+(1)
E =φ+ - φΦSHE=0
SHE (+)
φ°= - E0
Zn2+ /Zn versus H+/H2
Volts
Zn
-
+
Salt Bridge
H2
Zn2+
Zn
Zn2+ + 2eOXIDATION
ANODE
H+
2 H+ + 2eH2
REDUCTION
CATHODE
Zn/Zn2+ half-cell combined with a SHE.
Eo for the cell is +0.76 V
Examples:
Determination of standard
copper electrode potential.
(-) SHE
E =φ+ - φΦ°SHE=0
Cu 2+(1) Cu (+)
Φ + ° = E0
Cu2+ / Cu and H+ /H2 Cell
E o = +0.34 V
Volts
Cu
+
Salt Bridge
H2
Cu2+
Cu2+ + 2eCu
REDUCTION
CATHODE
H+
H2
2 H+ + 2eOXIDATION
ANODE

Cu2+ (a q) + H2(g)  Cu(s) + 2 H+(a q)
Measured E o = +0.34 V

Therefore, Φ°for Cu2+ + 2e-  Cu is

+0.34 V
6-3.3 Applications of Standard
Electrode Potentials
● Strength of Oxidizing or
Reducing Agents
The more positive φ°is, the stronger
the oxidizing agent.
The more negative φ°is, the stronger
the reducing agent.
Example:
Oxidizing ability of ion
Half-Reaction
φ°
Cu2+ + 2e-
 Cu
+ 0.34
2 H+ + 2e-
 H2
0.00
Zn2+ + 2e-
 Zn
-0.76
BEST Oxidizing agent ?
Cu2+
BEST Reducing agent ?
Zn
Reducing ability
of element
Example :
Which of the following is the most powerful
oxidizing agent?
Cr3+(aq) ,Br2(l) , Cu(s)
Solution:
Look at the values for φ° for the reduction of each
of the above species: the largest value is the most
powerful oxidizing agent
Cr3+(a q) +3 e → C r(s) φ° = -0.74 V
Cu2+ + 2e → Cu(s)
φ° = 0.34 V
Br2(l ) + 2e → 2Br -(a q) φ° = +1.07 V
● E °and Spontaneity
─ Predicting the direction of a
redox reaction from the φ°
E o ＞ 0, indicates the reaction is spontaneous
E o ＜ 0, then the reaction is nonspontaneous
Reaction is spontaneous in the
reverse direction.
E o = 0, at equilibrium
E °= φox° - φ red°
If E o is positive, φox°
＞ φ red°,
a reaction occurs spontaneously
in the forward direction.
If E o is negative, φox°
＜ φ red°,
a reaction occurs spontaneously
in the reverse direction.
If E o = 0, φox°
=φred°,
a reaction is at equilibrium.
Example
Consider the reaction
Zn2+(a q)+2Fe2+(a q)→ Zn(s)+2Fe3+(a q)
Does the reaction go spontaneously
in the direction indicated, under
standard conditions?
Solution:


Zn 2 / Zn


Fe3 / Fe2
 0.76V
 0.77V


3
2
Fe / Fe


2
Zn / Zn
Thus, the electrode potential is greater
for Fe3+/Fe2+, so that Fe3+ is the stronger
oxidizing
agent.
The
nonspontaneous as written.
reaction
is
● Calculation of equilibrium constant
─ K , K sp
From electrochemistry
ΔG°= - n F E°
In addition, ΔG°is related to the
equilibrium constant (K) through
ΔG° = - RT ln K
so we conclude that
RT lnK = nFE°
● Calculation of equilibrium constant
─ K , K sp
nF 
ln k 
E
RT
n

log k 
E
0.0592
n(   )
log k 
0.0592




Example
Calculate the equilibrium constant of the
redox reaction at 25℃.
2MnO4- (a q) +5 Zn(s) + 16H+(a q) →
2Mn2+ (a q) +5 Zn2+(a q) + 8H2O ( l )
Solution:



2
MnO4 / Mn


2
Zn / Zn
 1.51V
 0.76V
n(   )
log k 
0.0592




10  (1.51  0.76)

 383
0.0592
k  10
383
Example
Estimate the extent of the following
redox reaction at 25℃.
Z n + Cu2+
Z n2+ + Cu
Solution:


Zn
2
/ Zn
 0.76V


2
Cu / Cu  0.337V
log k =16.9×2 ×(0.337+0.76)=37.18
K = 1.5 ×10 37 The reaction is very
completely from left to right.
6-3.4 Diagram and Its Applications
of Element Potentials
Fe3+ + e
→ Fe2+
φ°= 0.77 V
Fe2+ + 2 e → Fe(s)
φ° = - 0.41 V
Fe3+ + 3e → Fe
φ°= - 0.0368 V
Electrode potentials diagram (Diagram of
element potentials)
Diagrams display the standard reduction
potentials (φ°) connecting various
oxidation states of an element.
● Diagram of element potentials
The potential diagram of iron
Fe3+
0.771
Fe2+
-0.4402
Fe
- 0.0368
oxidizing
agent
Reducing
agent
● Applications of diagram of
element potentials
─ Calculate unknown φ°
M2+
φ1°
M+
φ2°
M
φ3°
n3φ3°= n1φ1°+ n2φ2°
n   n2
 
n3

3

1 1

2
n  n   n2   ni


1 1

2
n  n1  n2   ni


unknown
n   n2   ni

n1  n2   ni

1 1

2

i

i
Example
According to the diagram, calculate 
Cu2+
0.16
Cu+
0.52

Cu 2 / Cu
?
Cu
?
Solution:


Cu 2 / Cu
n   n2
 
n3

3

1 1

2
1 0.16  1 0.52

 0.34(V )
11
─ Disproportionation
Disproportionation
is a process in which a single chemical
species is both oxidized and reduced.
Cu2+(a q) + e → Cu+(a q) φ°=0.l6 V
Cu+(a q) + e → Cu(s)
φ°=0.52 V
2 Cu+(a q) → Cu2+(a q) + Cu(s)
E °=φ°+ - φ°- =0.52–0.16=0.36V
Example
Use data in Appendix φ°to decide whether
the iron(Ⅱ) ion is unstable with respect to
disproportionation under standard conditions.
Solution: Fe3+ + e → Fe2+ φ° = 0.770 V
Fe2+ + e → Fe (s) φ°=- 0.409 V
The Fe 2+appears on the left in one and on
the right in the other half-reaction.
The Fe 2+ ion is stable against
disproportionation under standard conditions.




Decide whether the following
substance can
right
left
Example
occur disproportionation under standard
conditions: H2O2, Hg22+
0.68
H2O2
1.77
H2O
(1)
O
(2)
Hg2+ 0.905 Hg22+ 0.796 Hg
2
Solution:
H2O2 can occur disproportionation , but
Hg22+ can not occur disproportionation.
6-4 Electrode Potentials for
N o n s t a n d a r d
C o n d i t i o n s
The Nernst Equation
The
Application
Equation
of
Nernst
• The Nernst Equation
The Nernst equation can be used to calculate E
values at other conditions.
For a reduction half-reaction
Oxidation state + n e
reduction state
RT
[ox]
  
ln
nF
[red ]

This is famous Nernst equation, where:
φ is the electrode potential at nonstandard condition;
R is the gas constant, R=8.314J/(mol·K) ;
T is the absolute temperature;
n is the number of electrons transferred between
the species;
F is the Faraday constant, equal to 96500 C/mol;
[ox], [red] represent concentration respectively.
The Nernst equation for the φ of a
half-cell at 25℃ is
0.0592
[ox]
  
log
n
[red ]

Notice that in the expression for Nernst
equation ─
• reducing species (right side of a half
reaction) appear in the denominator.
• oxidizing species (left side of a half
reaction) appear in the numerator.
• each concentration or partial pressure of a
substance is raised to a power equal to its
coefficient in the half reaction.
• Pure liquids and solids do not appear in
the Nernst equation.
• H+ or OH- enter Nernst equation.
Br2 (l) + 2e
 

Br2 / Br 
2Br -
0.0592
1

log
 2
2
[ Br ]
Cr2O72- +14H++6e
2Cr3+ +7H2O
2
7
 14
[Cr2O ][ H ]
0.0592
  
log
3 2
6
[Cr ]

• The Application of Nernst Equation
1. Calculate the electrode
potential
Example 6-1
Calculate the electrode potential of following
electrode in pH =5 solution ?
Pt|MnO4- (1.0), Mn2+(1.0), H+ (10-5)
Solution:
Electrode half-reaction is
MnO4- +8 H+ +5 e → Mn2+ + 4 H2O
  MnO
O
 MnO

4
/ Mn
 MnO

4

4
2
/ Mn
2
 1.51V
   MnO / Mn2
4
0.0592

log
n
n=5
+ 8
[MnO
][H
[MO4 ][4 H ]
2
[Mn ]
5 8
/ Mn2
0.0592
1 (10 )
 1.51 
log
5
1
=1.036 (v)
Example 6-2
Calculate φ for the half-reaction
Sn4+ + 2 e-
Sn2+
At 25 ℃ if the concentration of Sn2+ is 0.40mol/L
and the concentration of Sn4+ is 0.10mol/L.
Solution:

 Sn
 Sn
4
/ Sn 2
4
/ Sn
2
 0.15V
0.0592
0. 1
 
log
 0.13V
2
0.4
2. Calculate the cell emf (E)
Example 6-3
The copper-zinc cell shown below. If the reaction
is done in a cell in 5.00 mol/L Zn2+ and 0.30 mol/L
Cu2+ at 25℃,
Zn(s) + Cu2+(a q) → Cu(s) + Zn2+(a q)
(a) write the cell formulation (notation)
(b) what is the cell E ?
Solution:
Anode:
Zn2+ + 2e → Zn(s) φ° = - 0.76V
Cathode: Cu2+ + 2e → Cu(s) φ° = 0.34V
According to the Nernst equation
 Zn
0
.
0592
2
2
  Zn / Zn 
log[ Zn ]
n

2
/ Zn
0.0592
 0.76 
 log 5  0.7393V
2
 Cu
2
/ Cu


Cu 2 / Cu
0.0592
2

log[ Cu ]
n
0.0592
 0.34 
log 0.3  0.3245V
2
 Cu
2
/ Cu
  Zn 2 / Zn
Cu electrode is cathode; Zn electrode is anode.
(a) The cell formulation (notation) is:
(-) Zn(s)│Zn2+(aq)
Cu2+(aq)│Cu (s) (+)
(b) E= φ+-φ-=0.3245-(-0.7393)=1.0638V
Example
Applying the Nernst equation for determining E .
(a) What is the value of the cell E for the
diagrammed bellow ?
(b) write out the cell reaction and electrode reaction.
Pt│Fe3+(0.05),Fe2+(0.5q) Ce4+(1)│Ce3+(0.01) │Pt
Solution:

 1.61V

4
3
(a) On the right
Ce / Ce
half-reaction: Ce4+ + e
Ce3+
Ce
4
[Ce ]
   0.0592 log
3
[Ce ]

4
/ Ce 3
1
 1.61  0.0592 log
 1.73V
0.01
On the left
half-reaction:
 Fe
3
/ Fe2


Fe3 / Fe2
Fe3+ + e
 0.771V
Fe2+
0.05
 0.771  0.0592 log
 0.712V
0.5
 Ce 4 / Ce 3   Fe3 / Fe2

Ce4+/Ce3+ is positive electrode
Fe3+/Fe2+ is negative electrode
E = φ+ - φ- =1.73-0.712=1.018(V)

(b)
Cathode: Ce4+ + e →Ce3+
Anode: Fe2+ - e →Fe3+
Cell reaction: Ce4+ + Fe2+ →Ce3+ + Fe3+
3. Use of the Nernst Equation to
Calculate Ion Concentrations
Example 6-4
If the voltage of the cell bellow is 0.389 V in a
0.500mol/L solution of Fe2+ , 1 atm pressure of
H2 at 25 ℃ , what is the hydrogen concentration
in the solution ?
Fe│Fe2+ H+│H2 │Pt
Solution:
Fe2+ + 2e → Fe φ° = - 0.41V
2H+ + 2e → H2 φ° = 0.00V
0
.
0592
2
2
 Fe2 / Fe   Fe / Fe 
log[ Fe ]
n
0.0592
 0.4 
log 0.5  0.1489V
2

H

/ H2

 2

H  / H2
0.0592
[H ]

log
n
1
0.0592
 2

 0.00 
log[ H ]  0.0592 log[ H ]
2
E= φ+-φ-= 0.0592log[H+]-(-0.4189)
= 0.389V
[H+]=0.3125mol/L
4. Predicting Spontaneous Reactions
for Nonstandard Conditions.

Decide the following reaction
weather is a spontaneous process:
Pb2+ (c=0.01) + Sn = Pb +
2+
Sn
Solution: (c=1.0)



Pb2 / Pb

Sn2 / Sn
 0.126V
 0.136V
 Pb
2
/ Pb


Pb 2 / Pb
0.0592
2

log[ Pb ]
n
0.0592
 0.126 
log( 0.01)  0.185V
2
 Sn
2
/ Sn


Sn2 / Sn


Sn2 / Sn
0.0592
2

log[ Sn ]
n
 0.136V
 Pb
2
2+
Sn
/ Pb
  Sn2 / Sn
+ Pb =
2+
Pb
+ Sn
Will the redox reaction proceed spontaneously as
written for the following reaction ?
Fe2+(0.015) + Ag+(0.25) →Fe3+(0.35)+Ag(s)
Answer:
The reaction is nonspontaneous as written.
Practice:
Calculate E for the following
voltaic cell.
Ag(s)│Ag+(1.0)
Hg2+(0.001) , Hg(l) │Pt
(a) What is the value of the cell E for the
diagrammed bellow ?
(b) write out the cell reaction and electrode reaction.
6-5 Determination of pH
Reference electrode
cell
e.g., SCE electrode, SHE
Ag-AgCl electrode
Indicator electrode
e.g., hydrogen electrode
pH glass electrode
• Reference electrode
Calomel electrode
Notation of calomel electrode
Pt│ Hg │ Hg2Cl2(s) │C l-(c)
Hg2Cl2(s)+2e
 Hg Cl
2
2 / Hg


Hg 2Cl 2 / Hg
2Hg(l)+2Cl-
2.303RT
1

log
 2
2F
[Cl ]
 Hg Cl
2
2 / Hg
T=298K,
 Hg Cl
2



Hg 2Cl 2 / Hg
2.303RT


log[ Cl ]
F

Hg2Cl 2 / Hg
 0.268V

2 / Hg
 0.268V  0.0592 log[ Cl ]
Saturated calomel electrode
 SCE  0.2412V
Saturated
KCl
solution
KCl crystals
• Indicator electrode
hydrogen electrode
2H ++2e
H2
T=298K, PH2=101.3kPa
H

/ H2

 2

H  / H2
0.0592 [ H ]

lg
2
pH 2
 0.0592 lg[ H  ]  0.0592 pH
pH glass electrode
Figure: A glass electrode
for measuring hydrogenion concentrations.
Ag - AgCl electrode
HCl
Thin glass
membrane
Notation of glass electrode
Ag ,AgCl(s) |H +(1M) | glass membrane | measured solution
glass electrode | measured solution (pHx )
ΦG=φ°G + 0.0592log[H+]
=φ°G - 0.0592pH
• Measurement pH value of a solution
by hydrogen electrode
Cell notation
(-) Pt|H2(1atm) | H+ (pH x) | | SCE (+)
25℃
Φ+= φ SCE = + 0.2412V
   H

/ H2
 0.0592 pH
E= Φ+-Φ-= 0.2412+0.0592 pH
E  0.2412
pH 
0.0592
• Measurement pH value of a solution
by glass electrode
Construction of
glass electrode
First step
(-) glass electrode | pHs standard buffer| | SCE (+)
Es         SCE
2.303RT
 ( 
pH s )
F
pH s  ( Es   SCE

G
F
 )
2.303RT

G
(1)
Second step
(-) glass electrode | pH x solution| | SCE (+)
E x        SCE
2.303RT
 ( 
pH x )
F
pH x  ( E x   SCE
F
 )
2.303RT
F

 G )
2.303RT


pH s  ( Es   SCE

G

G
(2)
(1)
(2) -（1）：
F
pH x  ( E x  Es )
 pH s
2.303RT
T=298K
E x  Es
pH x  pH s 
0.0592