Question

Ksp of AgCl = 10^–10 M^2. The EMF of the
galvanic cell
Ag, AgCl(S)/ KCl(1M)// AgNO3 (1M) | Ag is:
1) 1.52V
2) –0.465V
3) 0.592V
4) 1.39V

Answer 1

Thus the emf of galvanic cell is − 0.037 V

Explanation:

For the galvanic cell, Ag|AgCl(s) | KCl(0.2 M) || KBr(0.001 M) | AgBr(s) | Ag

Calculate the emf generated and assign polarity to each electrode for a spontaneous process after taking into account the cell at 25 Centigrade.

Given Ksp(AgCl) = 2.8 * 10-10 and Ksp​(AgBr) = 3.3  x 10-13

Solution:

Ag / AgCl(s), KCl(0.2 M) ∥ KBr(0.001 M), Ag Br(s)/Ag

Ag(1) → Ag+ (1) + e−      ( at anode)

Ag+ (2) + e− → Ag(2)     (at cathode)

Ag(1) + Ag+(2) → Ag(2) + Ag+(1)    (cell reaction)

According to Nernst equation:

E(cell) = E°cell + 0.0591 n log [reactants] / [products]

E(cell) = 0.0 + 0.05911 log [Ag(1)] [Ag+(2)] / [Ag(2)] [Ag+(1)]

Ksp (Ag Cl) = [Ag+(1)] [Cl−] = 2.8 × 10^−10

[Ag+(1)] = 2.8 × 10^−10 / 0.2 = 1.4 × 10^−9 M

Ksp (Ag Br) = [Ag+(2)] [Br−] = 3.3 × 10^−13

[Ag+(2)] = 3.3 × 10^−13 / 0.001 = 3.3 × 10^−10 M

E(cell) = 0.0591 log [3.3 × 10^−10] / [1.4 × 10^−9] = − 0.037 V

Thus the emf of galvanic cell is − 0.037 V

Answer 2

Answer :-

Thus the emf of galvanic cell is − 0.037 V

Explanation:

For the galvanic cell, Ag|AgCl(s) | KCl(0.2 M) || KBr(0.001 M) | AgBr(s) | Ag

Calculate the emf generated and assign polarity to each electrode for a spontaneous process after taking into account the cell at 25 Centigrade.

Given Ksp(AgCl) = 2.8 * 10-10 and Ksp(AgBr) = 3.3  x 10-13

Solution:

Ag / AgCl(s), KCl(0.2 M) ∥ KBr(0.001 M), Ag Br(s)/Ag

Ag(1) → Ag+ (1) + e−      ( at anode)

Ag+ (2) + e− → Ag(2)     (at cathode)

Ag(1) + Ag+(2) → Ag(2) + Ag+(1)    (cell reaction)

According to Nernst equation:

E(cell) = E°cell + 0.0591 n log [reactants] / [products]

E(cell) = 0.0 + 0.05911 log [Ag(1)] [Ag+(2)] / [Ag(2)] [Ag+(1)]

Ksp (Ag Cl) = [Ag+(1)] [Cl−] = 2.8 × 10^−10

[Ag+(1)] = 2.8 × 10^−10 / 0.2 = 1.4 × 10^−9 M

Ksp (Ag Br) = [Ag+(2)] [Br−] = 3.3 × 10^−13

[Ag+(2)] = 3.3 × 10^−13 / 0.001 = 3.3 × 10^−10 M

E(cell) = 0.0591 log [3.3 × 10^−10] / [1.4 × 10^−9] = − 0.037 V

Thus the emf of galvanic cell is − 0.037 V