Question

Standard cell voltage for the cell Pb | Pb²⁺ || Sn²⁺ | Sn is
– 0.01 V. If the cell is to exhibit E꜀ₑₗₗ = 0, the value of
[Sn²⁺] / [Pb²⁺] should be antilog of –
(a) + 0.3 (b) 0.5
(c) 1.5 (d) – 0.5

Answer 1

Explanation:

dissociation constant of this acid is:

(a) 1.25 × 10⁻⁶ is answer to your question ..

Answer 2

(a) +0.3

Solved by Nernst Equation

E_{cell}=E^{0}_{cell} -2.303RT/nF log[Red]/[Oxi]

Explanation:

E_{cell}=E^{0}_{cell} -2.303RT/nF log[Sn^{+2}]/[Pb^{+2}]

Given   E_{cell}=0, E^{0}_{cell}= -0.01 V

n=2

F=96500

R=8.314 J/ mol K

In question temperature is not mention, so we take room temperature T=25^{0}C= 298 K

E^{0}_{cell}= 2.303RT/nF log[Sn^{+2}]/[Pb^{+2}]

Put the values in above equation

-0.01 = 2.303RT/nF log[Sn^{+2}]/[Pb^{+2}]

-0.01 = 2.303\times8.314\times 298/2\times 96500 log[Sn^{+2}]/[Pb^{+2}]

log[Sn^{+2}]/[Pb^{+2}]= -0.33

antilog of -(+0.33)