Question

The $ {E_{cell}} $ for $ Ag(s)|AgI(satd)||A{g^ + }(0.10M)|Ag(s) $ is $ + 0.413\;V $ . What is the value of $ {K_{sp}} $ for $ AgI $ ?
A. $ 1.0 \times {10^{ - 8}} $
B. $ 1.0 \times {10^{ - 7}} $
C. $ 1.0 \times {10^{ - 14}} $
D. $ 1.0 \times {10^{ - 16}} $

Answer 1

Hint: The solubility product is the equilibrium constant for a substance which dissolves in an aqueous solution and breaks into its respective ions. It represents the level at which a solute can be dissolved in the solution. It is denoted by $ {K_{sp}} $ .

Complete answer:
As per question, the given cell is as follows:
 $ Ag(s)|AgI(satd)||A{g^ + }(0.10M)|Ag(s) $
We know that $ AgI $ dissociates to form silver and iodide ion, so the cell can alternatively be represented as follows:
 $ Ag(s)|A{g^ + }(x\;M)||A{g^ + }(0.10M)|Ag(s) $
Here, we have considered the unknown concentration of $ A{g^ + } $ ion as $ x $ .
As the given cell is concentration cell, so the standard reduction potential of the cell will be zero, so according to Nernst equation:
 $ {E_{cell}} = E_{cell}^o - \dfrac{{0.0592}}{n}\log \dfrac{{{c_1}}}{{{c_2}}} $
Where, $ {c_1} $ and $ {c_2} $ are the concentration of unknown and known $ A{g^ + } $ ions respectively and because $ E_{cell}^o = 0 $ for concentration cells therefore, the equation can be represented as follows:
 $ \Rightarrow {E_{cell}} = - \dfrac{{0.0592}}{n}\log \dfrac{{{c_1}}}{{{c_2}}} $
Substituting values of $ {E_{cell}} $ , n, $ {c_1} $ and $ {c_2} $ :
 $ \Rightarrow + 0.413 = - \dfrac{{0.0592}}{1}\log \dfrac{x}{{0.1}} $
 $ \Rightarrow \log \dfrac{x}{{0.1}} = - 6.97 $
Taking antilog on both sides of the expression:
 $ \Rightarrow \dfrac{x}{{0.1}} = {10^{ - 6.97}} $
 $ \Rightarrow x = 1.07 \times {10^{ - 8}}M $
Therefore, the concentration of saturated $ AgI = 1.07 \times {10^{ - 8}}M $ for the given concentration cell. Now, the dissociation of $ AgI $ takes place as follows:
 $ AgI \rightleftharpoons A{g^ + } + {I^ - } $
The ICE table for the above dissociation reaction is as follows:

	$ \left[ {AgI} \right] $	$ \left[ {A{g^ + }} \right] $	$ \left[ {{I^ - }} \right] $
Initial concentration	$ 1.07 \times {10^{ - 8}}M $	$ 0 $	$ 0 $
Change	$ - 1.07 \times {10^{ - 8}}M $	$ + 1.07 \times {10^{ - 8}}M $	$ + 1.07 \times {10^{ - 8}}M $
Equilibrium concentration	$ 0 $	$ 1.07 \times {10^{ - 8}}M $	$ 1.07 \times {10^{ - 8}}M $

The expression for solubility product can be written as follows:
 $ {K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{I^ - }} \right] $
Substituting values as per given in ICE table:
 $ \Rightarrow {K_{sp}} = 1.07 \times {10^{ - 8}} \times 1.07 \times {10^{ - 8}} $
 $ \Rightarrow {K_{sp}} = 1.0 \times {10^{ - 16}} $
Hence, the value of $ {K_{sp}} $ for $ AgI $ is $ 1.0 \times {10^{ - 16}} $ .
Thus, option (D) is the correct answer.

Note:
It is important to note that the value of $ {K_{sp}} $ for silver iodide in the given concentration cell can alternatively be calculated by directly substituting the value of $ {c_1} $ as $ \sqrt {{K_{sp}}} $ in the Nernst equation. Also, remember that a concentration cell is an electrolytic cell that is composed of two half cells with the same electrodes but differ in concentrations.