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Question: The \( {E_{cell}} \) for \( Ag(s)|AgI(satd)||A{g^ + }(0.10M)|Ag(s) \) is \( + 0.413\;V \) . What is ...

The Ecell{E_{cell}} for Ag(s)AgI(satd)Ag+(0.10M)Ag(s)Ag(s)|AgI(satd)||A{g^ + }(0.10M)|Ag(s) is +0.413  V+ 0.413\;V . What is the value of Ksp{K_{sp}} for AgIAgI ?
A. 1.0×1081.0 \times {10^{ - 8}}
B. 1.0×1071.0 \times {10^{ - 7}}
C. 1.0×10141.0 \times {10^{ - 14}}
D. 1.0×10161.0 \times {10^{ - 16}}

Explanation

Solution

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 Ksp{K_{sp}} .

Complete answer:
As per question, the given cell is as follows:
Ag(s)AgI(satd)Ag+(0.10M)Ag(s)Ag(s)|AgI(satd)||A{g^ + }(0.10M)|Ag(s)
We know that AgIAgI dissociates to form silver and iodide ion, so the cell can alternatively be represented as follows:
Ag(s)Ag+(x  M)Ag+(0.10M)Ag(s)Ag(s)|A{g^ + }(x\;M)||A{g^ + }(0.10M)|Ag(s)
Here, we have considered the unknown concentration of Ag+A{g^ + } ion as xx .
As the given cell is concentration cell, so the standard reduction potential of the cell will be zero, so according to Nernst equation:
Ecell=Ecello0.0592nlogc1c2{E_{cell}} = E_{cell}^o - \dfrac{{0.0592}}{n}\log \dfrac{{{c_1}}}{{{c_2}}}
Where, c1{c_1} and c2{c_2} are the concentration of unknown and known Ag+A{g^ + } ions respectively and because Ecello=0E_{cell}^o = 0 for concentration cells therefore, the equation can be represented as follows:
Ecell=0.0592nlogc1c2\Rightarrow {E_{cell}} = - \dfrac{{0.0592}}{n}\log \dfrac{{{c_1}}}{{{c_2}}}
Substituting values of Ecell{E_{cell}} , n, c1{c_1} and c2{c_2} :
+0.413=0.05921logx0.1\Rightarrow + 0.413 = - \dfrac{{0.0592}}{1}\log \dfrac{x}{{0.1}}
logx0.1=6.97\Rightarrow \log \dfrac{x}{{0.1}} = - 6.97
Taking antilog on both sides of the expression:
x0.1=106.97\Rightarrow \dfrac{x}{{0.1}} = {10^{ - 6.97}}
x=1.07×108M\Rightarrow x = 1.07 \times {10^{ - 8}}M
Therefore, the concentration of saturated AgI=1.07×108MAgI = 1.07 \times {10^{ - 8}}M for the given concentration cell. Now, the dissociation of AgIAgI takes place as follows:
AgIAg++IAgI \rightleftharpoons A{g^ + } + {I^ - }
The ICE table for the above dissociation reaction is as follows:

| [AgI]\left[ {AgI} \right] | [Ag+]\left[ {A{g^ + }} \right] | [I]\left[ {{I^ - }} \right]
---|---|---|---
Initial concentration| 1.07×108M1.07 \times {10^{ - 8}}M | 00 | 00
Change | 1.07×108M- 1.07 \times {10^{ - 8}}M | +1.07×108M+ 1.07 \times {10^{ - 8}}M | +1.07×108M+ 1.07 \times {10^{ - 8}}M
Equilibrium concentration| 00 | 1.07×108M1.07 \times {10^{ - 8}}M | 1.07×108M1.07 \times {10^{ - 8}}M

The expression for solubility product can be written as follows:
Ksp=[Ag+][I]{K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{I^ - }} \right]
Substituting values as per given in ICE table:
Ksp=1.07×108×1.07×108\Rightarrow {K_{sp}} = 1.07 \times {10^{ - 8}} \times 1.07 \times {10^{ - 8}}
Ksp=1.0×1016\Rightarrow {K_{sp}} = 1.0 \times {10^{ - 16}}
Hence, the value of Ksp{K_{sp}} for AgIAgI is 1.0×10161.0 \times {10^{ - 16}} .
Thus, option (D) is the correct answer.

Note:
It is important to note that the value of Ksp{K_{sp}} for silver iodide in the given concentration cell can alternatively be calculated by directly substituting the value of c1{c_1} as Ksp\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.