electrochemistry

AIEEE Answers › Physical Chemistry questions › electrochemistry

find the solubility product of a saturated solution of Ag2CrO4 in water at 298K if the emf of the cell is Ag|Ag+(satd Ag2CrO4 soln.)||Ag+(0.1M)|Ag is 0.164V at 298K.

#Chemistry #Physical Chemistry

UP 0 DOWN 0 0 4
Post on FacebookPost anonymously?

4 Answers

1

Debosmit Majumder ·2011-10-10 21:39:36

@avisikta:is K_sp = 1.02 x 10⁵....pls verify????

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

AVISIKTA UPADHYAY ·2011-10-11 08:47:50

@debosmit, na re, noi eta answer.........cud u please say how u did it????

UP 0 DOWN 0
Post on FacebookPost anonymously?

21

Shubhodip ·2011-10-11 10:04:32

The cell is Ag|Ag⁺||Ag⁺|Ag, so when the concentrations are unity the potential of cathode and anode will be equal and opposite, so E⁰_cell = 0

By Nernst equation

E_cell = E⁰_cell - RTnFln([Ag⁺]in satd. Ag₂CrO₄ solution[Ag⁺])

We know that E_cell = 0.164 V, T = 298K, n= 1, [Ag⁺] (that in the denominator) = 0.1 mol and E⁰_cell = 0

So 0.164 = - 0.059 log ₁₀(10([Ag⁺]in satd. Ag₂CrO₄ solution))

Gives me [Ag⁺]in satd. Ag₂CrO₄ solution = 1.66 * 10^-4

Ag₂CrO₄(aq) <---> 2 Ag⁺ + CrO₄^2- (*)

So , solubility product is = [Ag⁺]²[ CrO₄^2-]

from (*), [Ag⁺]= 1.66 * 10^-4, [CrO₄^2-] = 8.3 *10^-5

so solubility product = 2.287 * 10^-12

Is this the answer ? what i have done seems logical :D

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

AVISIKTA UPADHYAY ·2011-10-11 11:34:56

yes dat is it...d concentrations were where i got confused....:P thanks...:)

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω