Question

41. The standard reduction potential of Pb and Zn
electrodes are -0.126 and -0.763 volts
respectively. The e.m.f. of the cell
ZnZn2+ (0.1 M) || Pb2+ (1 M) Pb is
(1) 0.637 V
(2) <0.637 v
(3) > 0.637 V
(4) 0.889​

Answer 1

Answer:

The emf of the cell will be more than 0.637 V.

Explanation:

The cell as shown is:

$Zn|Zn^{2+} (0.1 M) || Pb^{2+} (1 M)| Pb$

Here zinc acts as anode and lead will act as cathode

The formula for emf of cell is known as Nernst's equation:

$E_{cell}=E^{0}_{cell}-\frac{0.0592}{n}log(\frac{[Zn^{+2}}{Pb^{+2}} )$

$E^{0}_{cell}=E^{0}_{cathode}-E^{0}_{anode}=-0.126-(-0.763)=0.637V$

Now let us put the values in the equation

$E_{cell}=0.637-\frac{0.0592}{2}log(\frac{0.1}{1})=0.667V$

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The emf of the cell will be:

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Answer 2

The e.m.f. of the cell is $Zn/Zn^{2+}(0.1M)//Pb^{2+}(1M)/Pb$ > 0.637 V

Explanation:

$Zn/Zn^{2+}(0.1M)//Pb^{2+}(1M)/Pb$

Here Zinc acts as anode and lead acts as cathode.

$Zn+Pb^{2+}\rightarrow Zn^{2+}+Pb$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $25^oC=273+25=298K$

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell =$E^0_{cathode}-E^0_{anode}=-0.126-(-0.763)=0.637V$

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.637-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{0.1M}{1M}$

$E_{cell}=0.637-(-0.0295)=0.666V$

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