Question

a galvanic cell consists of a metallic zinc plate immersed in 0.1M Zn(NO3)2 solution and a metallic plate of lead in 0.02M Pb(NO3)2 solution. calculate the EMF of the cell. Given E Pb2 +/Pb=0.13v EZn2 +/Zn=-0.76v

Answer 1

it may be pb+zo+sn+o2

Answer 2

Answer : The EMF of the cell is, 0.87 V

Solution :

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

Here Zn undergoes oxidation by loss of electrons, thus act as anode. Lead undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard cell potential.

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0_{[Pb^{2+}/Pb]}=0.13V$

$E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Zn^{2+}/Zn]}$

$E^0=0.13V- (-0.76V)=0.89V$

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Zn^{2+}]}$

where,

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

$E_{cell}=0.89-\frac{0.0592}{2}\log \frac{0.02}{0.1}=0.8693=0.87V$

Therefore, the EMF of the cell is, 0.87 V