In this you will learnt about, emf of the cell and calculation of emf of the cell(numerical problems)

1. Dr. K. RAJENDER REDDY
LECTURER IN CHEMISTRY
EMAIL:raj3iict@gmail.com
4/20/2020 1

2. Defination: The potential difference between
two electrodes of a galvanic cell, which causes
flow electrons is called electromotive force(EMF)
4/20/2020 2

3.  The emf of the cell can be calculated from the
elctrode potential values
 Ecell= Ecathode-Eanode
Or
Ecell = RP of cathode-RP of anode
or
4/20/2020 3

4. The reading (+0.46 v)of the
voltmeter in picture is the
EMF of the cell constructed
from Cu (anode) and
Ag(cathode)
4/20/2020 4

5.  Ecell= ERHS-ELHS
Where RHS = Right Hand Side
LHS = Left Hand Side
Ecell=RP of cathode + OP of anode
If EMF of the cell is calculated from standard
electrode potentials, then it is indicated as Eo
cell
ECELL= ERHS+ELHS
4/20/2020 5

6.  PROBLEM 1:
The SRP of Mg and Cd electrodes are -2.37V and -0.40V respectively.
Calculate the EMF of the cell Mg/Mg+2//Cd+2/Cd.
Solution:
In this cell Mg is anode and Cd is Cathode.
Eo
cell = Eo
cathode- Eo
anode
=-0.40V–(-2.37V)
= + 1.97V
4/20/2020 6

7. • PROBLEM 2:
The Zn/Zn+2 electrode potential is +0.76V and Cu+2/Cu electrode potential is
+0.34V respectively. Calculate the EMF of the cell Zn/Zn+2//Cu+2/Cu.
Solution:
In this cell Zn is anode and Cu is Cathode.
Eo
cell = Eo
cathode- Eo
anode
= +0.34V –(- 0.76V)
= + 1.1V
4/20/2020 7

8. • PROBLEM 3:
The Zn/Zn+2 electrode potential is +0.76V and Ca+2/Ca electrode potential is -
2.87V . Calculate the EMF of the cell Ca/Ca+2//Zn+2/Zn.
Solution:
In this cell Ca is anode and Zn is Cathode.
Eo
cell = Eo
cathode- Eo
anode
= -0.76V –(- 2.87V)
= + 2.10V
4/20/2020 8

9. • PROBLEM 4:
The SRP of Ni and Cd electrode are -0.25V and +0.34V respectively.What
is the EMF of the cell constructed from these electrodes.
Solution:
In this cell Ni is anode and Cd is Cathode.
Eo
cell = Eo
cathode- Eo
anode
= +0.34V –(- 0.25V)
= + 0.59V
4/20/2020 9

10. 4/20/2020 10