Emf of the cell(ELECTRO CHEMISTRY)
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The document defines electromotive force (emf) as the potential difference between two electrodes of a galvanic cell, driving the flow of electrons. It explains how to calculate the emf using various formulas and illustrates the concept through multiple problems involving different electrode potentials. The calculations demonstrate the method for determining the emf in specific galvanic cells, showcasing examples with magnesium, cadmium, zinc, and nickel.
Emf of the cell(ELECTRO CHEMISTRY)
- 1. Dr. K. RAJENDER REDDY LECTURER IN CHEMISTRY EMAIL:[email protected] 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
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