Chemistry
Session
Electrochemistry - 3
Session Objective
•
Nernst equation
•
Equilibrium constant and Nernst equation
•
Primary cell(Batteries)
•
Secondary cell(Batteries)
•
Fuel Cell
•
Corrosion and its prevention.
Nernst Equation
For a general reduction reaction,
M
n
 ne

 M(s)
The Nernst equation can be written as
RT
1
0
E n+
=E
- 2 .3 0 3
lo g
n+
n+
M
/M
nF
M
/M
[M
]
0 .0 5 9
1
o
E n+
= E
lo
g
M
/M
Mn+ /M
n
M n+ 

(At 298K)

Where n = Number of electrons involved
[Mn+] = molar concentrations at 298K
Illustrative Example
Calculate the electrode potential at a copper
electrode dipped in a 0.1M solution of copper
sulphate at 250C . The standard potential of
Cu2+/Cu system is 0.34 volt at 298 K.
Solution: Cu2+ + 2e-  Cu
W e k n o w th a t E
Cu
2
/ Cu
P u ttin g th e v a lu e s o f E
E re d  0 .3 4 
0 .0 5 9 1
2
 E
0
re d
Cu
2
/ Cu

0 .0 5 9 1
n
lo g1 0 [C u
 0 .3 4 V , n  2 a n d [C u
lo g1 0 [0 .1]
 0.34  0.02955  (  1)
 0 .3 1 0 4 5 v o lt
0
2
2
]
]  0 .1 M
Equilibrium constant from Nernst equation
Consider the following cell reaction in
equilibrium then
Z n (s) + C u
2+
(aq)
Zn
2+
+ C u (s)
0
E = 0 = E (cell) 0
E (cell) =
2.303RT
2.303RT
2F
2F
log
2+
Therefore,
Generally,
]
2+
[Cu
]
0
[Zn
E (cell) =
]
2+
[Cu
]
]
2+
[Cu
]
= Kc
E (cell) =
0
log
2+
[Zn
At equilibrium
2+
[Zn
2.303RT
2F
2.303RT
2F
logK c
logK
Electrode potential
Then Nernst equation is:
Reduction half reaction
Equilibrium point
Oxidation half reaction
Progress of reaction
Illustrative Example
Calculate the equilibrium constant of the reaction:
Cu(s)+2Ag+(aq.)Cu2+(aq.)+Cu(s) E0=0.46 V
Solution :
0
E (ce ll) =
lo g K c =
0 .0 5 9
2
lo g K c
0 .4 6 × 2
0 .0 5 9
Kc = 4 × 10
15
= 1 5 .6
Electrochemical cell and Gibbs
Energy of the reaction
Δ r G = -n F E
e .g . fo r ce ll re a ctio n
Z n (s) + C u
2+
(a q .)  Z n
2+
(a q .) + C u (s)
Δ r G = -2 F E
If th e re a cta n ts a re in sta n d a rd sta te th e n ,
0
Δ rG = Δ rG , E = E
Δ rG
0
= -n F E
0
= -n F ×
Δ rG
0
0
RT
nF
= -R T ln K
ln K
Illustrative Example
Calculate DG° for Zn-Cu cell at standard state conditions
[Given Eo
2+/Zn
Zn
= –0.76 V, E0
2+
Cu
/Cu
Solution

E cell  E

Cu
2
/ Cu
E

Zn
2
/ Zn
= + 0.34 V – (–0.76 V) = 1.10 V
DG° = –nFE°
= 2×96500×1.10
= –212.3 kJ mol–1
= +0.34 V ]
Commercial Cells
Primary Cell
Dry Cell
Commercial Cells
The oxidation taking place at the
negative zinc electrode.
Anode: Z n (s)  Z n2+ (aq)+ 2 e The reduction takes place at positive electrode
Cathode: 2M n O + H O + 2e  M n O + 2O H
-
2
2
2
-
3
The net cell reaction is
Z n(s)  2M n O 2 (s)  H 2O
Zn
2
 M n 2O 3  2O H
The emf of the cell is about 1.45 V.

Secondary Cells
Lead storage battery
Net cell reaction is reversible. Hence, it
can be recharged.
Lead storage battery
At anode:
2-
Pb s  + SO4
 a q .
   

PbSO4
s 
+
2e
-
At cathode:
2-
P b O2 + S O 4
 a q .
+ 4H
+
 a q .
+ 2e
-
 
 P b S O 4  s  + 2 H2 O
Overall reaction:
P b  s  + P b O2  s  + 4 H
+
 a q .  + 2 S O 24  a q .
  2 P b S O 4  s  + 2 H 2 O
Secondary Cell
In the above equation H2SO4 is used up during the
discharge.During recharging the reactions are the
reverse of those that occurs during discharge.
At cathode:
P bS O 4  s   2e

2
  P b  s   S O 4
 a q. 
At anode:
P b S O 4  s   2H 2 O
2
 
 PbO2  S O 4
 a q. 
 4H

 a q. 
 2e

Overall reaction:
2P b S O 4  s   2H 2 O
 

P b  s   P b O 2  s   4H

 a q.   2S O 24   a q. 
Fuel Cells
Galvanic cells which converts energy of combustion of
fuel like hydrogen, methane and methanol etc.
directly into electrical energy are called fuel cells.
Example
One of the most successful fuel cell uses
hydrogen and oxygen reaction to form water.
At cathode: O2(g)+2H2O(l)+4e-  4OH-(aq.)
At anode: 4H2O(l)+4e-  2H2 + 4OH-(aq.)
Overall cell reaction is:
2H2(g)+O2(g)  2H2O(l)
Efficiency of fuel cell is 70% much more as
compared thermal plants(40%).
Corrosion
Process of slowly eating away of the metal
due to attack of atmospheric gases on the
surface of the metal.
Examples of corrosion
•
Rusting of iron
•
Tarnishing of silver
•
Development of green coating on copper and bronze, etc.
Corrosion
Methods of preventing corrosion
•
Barrier protection
•
Using anti rust solutions
•
Sacrificial protection
For example iron surface is covered with a metal
which has higher tendency to get oxidized (larger
negative value of standard reduction potential) than
iron.
Zinc is used for covering iron and the process is called
galvanization.
Illustrative Example
The standard reduction potentials of Sn+2/Sn and
Zn+2/Zn are respectively –0.14V, -0.76V. Predict
whether the corrosion of tin can be prevented by
coating with zinc or not.
Solution :
Zinc lies above tin in the electrochemical series, therefore
it has a lower reduction potential than tin.This property is
employed to prevent corrosion of tin by coating it with zinc
as zinc acts as a sacrificial electrode.
Thank you