Electromotive Force Worksheet





The short form of Electromotive force is EMF. According to the law of Faraday “EMF” is a  voltage generated by a magnetic force.  But in classic physical sense EMF is not a force as it is measured in term of volt, not Newton.

External Work expanded per unit of charge to produce an electrical potential difference accross two open-circuited terminal is known as Electromotive force.

Here, you can find some questions related to “Electromotive force” and their answers:

Questions:

1. The emf of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to -0.54 V, is:

(a)    +1.12 V

(b)   -1.12 V

(c)    +0.44 V

(d)   -0.44 V

 

2. The electrode potentials of half-cell reactions are Zn \to Zn^{2+} +2e^-; E = -0.76 V \text{and} Fe \to Fe^{2+} + 2e^-; E = 0.41 V. The emf of the cell Fe^{2+} + Zn \to Zn^{2+} + Fe is:

(a)    +0.35 V

(b)   -1.17 V

(c)    -0.35 V

(d)   +1.17 V

 

3. The mathematical equation,

E- E_0 = - \dfrac{RT}{nF}ln Q

where Q is reaction quotient is called:

(a)    Hemmet equation

(b)    Finkelstein equation

(c)    Helmholtz equation

(d)   Nernst equation

 

4. Which of the following conditions will increase the voltage of the cell, represented by the reaction,

Cu (s) + 2Ag ^+(aq) \to Cu^{2+} (aq) + 2Ag (s) ?

(a)    increase in the concentration of Ag+ ions

(b)   increase in the dimensions of silver electrode

(c)    increase in the dimensions of copper electrode

(d)   increase in the concentration of ions

 

5. A cell is constituted as follows:

Pt, H_2 (1 atm) | HA_1 | HA_2 | H_2 (1 atm), Pt

 

The pH of two acid solutions HA_1 \text{and} HA_2 is 5 and 3, respectively. The emf of the cell is:

(a)    -O.118V

(b)   0.0295V

(c)    0.118V

(d)   -0.0598V

 

 

6. The emf of a galvanic cell with electrode potentials of silver equal to +0.80 V and that of copper equal to + 0.34 V is:

(a)    +0.4 V

(b)   +2.34 V

(c)    -0.46 V

(d)   -1.14 V

Answers:

1.(b)     2. (b)     3. (d)     4. (a)     5. (c)     6. (a)



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