EMF OF THE CELL, NUMERICAL PROBLEMS ON CELL EMF

1. Dr. K. RAJENDER REDDY
LECTURER IN CHEMISTRY
EMAIL:raj3iict@gmail.com
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2. CONTENTS
Electro Motive Force(EMF)
1. Definition
2. Calculation of EMF of the cell
3. EMF of the cell constructed from Cu and Ag
4. Numerical problems
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3. EMF OF THE CELL
Definition: The potential difference between two
electrodes of a galvanic cell, which causes flow
electrons is called electromotive force(EMF)
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4. Calculation of EMF of Cell(Ecell)
• 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
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5. EMF of the cell constructed from Cu and Ag
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The reading (+0.46 v)of the
voltmeter in picture is the
EMF of the cell constructed
from Cu (anode) and
Ag(cathode)

6. Calculation of EMF
• 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
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ECELL= ERHS+ELHS

7. NUMERICAL PROBLEMS ON EMF OF THE CELL
• PROBLEM 1:
The SRP of Mg and Cd electrodes are -2.37 V and -0.40 V 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
=-0.40 V–(-2.37 V)
= + 1.97V
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8. NUMERICAL PROBLEMS ON EMF OF THE CELL
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• PROBLEM 2:
The Zn/Zn+2 electrode potential is +0.76 V 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 –(- 0.76 V)
= + 1.1 V

9. NUMERICAL PROBLEMS ON EMF OF THE CELL
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• PROBLEM 3:
The Zn/Zn+2 electrode potential is +0.76 V and Ca+2/Ca electrode potential is -
2.87 V . 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
= -0.76 V –(- 2.87 V)
= + 2.10 V

10. NUMERICAL PROBLEMS ON EMF OF THE CELL
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• PROBLEM 4:
The SRP of Ni and Cd electrode are -0.25 V and +0.34 V 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 –(- 0.25 V)
= + 0.59 V

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