Question

Calculate the standard potential of the cell: Pb(s)/PbSO4(aq)//AgSO4(aq)/Ag(s)​

Answer 1

Answer:

Pb(s)+Hg

2

SO

4

⇌PbSO

4

(s)+2Hg(l) design the cell if both electrolytes are present in their saturated solution state. Given E

Pb/Pb

2+

o

and E

Hg/Hg

2

2+

o

are 0.126 and -0.789 V respectively and K

sp

of PbSO

4

and Hg

2

SO

4

are 2.43×10

−8

and 1.46×10

−6

respectively

The e.m.f. of given cell reaction to nearest integer

I hope that helps you

Answer 2

The standard potential of the cell: Pb(s)/PbSO4(aq)//AgSO4(aq)/Ag(s)​ is +0.93V

E° =E°oxi +E° red

Explanation:

Anode : Pb²⁺+2e⁻→Pb where  E° =+0.13

Cathode : Ag⁺ +e⁻→Ag where  E° =+0.80V

Cell reaction is Pb(s)/PbSO4(aq)//AgSO4(aq)/Ag(s)

Pb/Pb²⁺and Ag²⁺/Ag are +0.13 and -0.80 V respectively

Reduction potential of Ag⁺ is large than Pb²⁺ so Ag will be at cathode.

Cathode :-2Ag⁺ +2e⁻→2Ag   E°oxi =+0.13 V

Anode :-Pb→Pb²⁺+2e⁻   E° red=+0.80V

E° =E°oxi +E° red

Standard potential  E°= +0.80V ++0.13 V

Standard potential  E° =+0.93V

The standard potential of the cell: Pb(s)/PbSO4(aq)//AgSO4(aq)/Ag(s)​ is +0.93V