Question

Given the data at \[25^\circ C\]
  \[
  Ag + {I^ - } \to Agl + {e^ - }E^\circ = 0.152{\text{ }}V \\
  Ag \to A{g^ + } + {e^ - }E^\circ = - 0.800{\text{ }}V \\
 \]
What is the value of log \[\;{K_{sp}}\] for AgI [2.303RT = 0.059 V]?
A)-8.12
B)+8.612
C)-37.83
D)-16.10

Answer 1

Hint: Electrochemistry is a wide subject and Nernst equation and corrosion are a part of it. The Nernst equation is used to calculate the cell potential of an electrochemical cell at any given temperature, pressure, and reactant concentration Nernst equation can be understood as an equation that helps us relate the reduction potential of an electrochemical cell to the standard electrode potential as well as the temperature of the system.

Formula used: \[{E_{cell}} = E_{cell}^0 - \dfrac{{0.0591}}{n}\log {k_{sp}}\].

Complete step by step answer:
Before we move forward with the solution to the given question, let us first understand some important basic concepts.
The Nernst equation gives a relation between the cell potential of an electrochemical cell, the standard cell potential, temperature, and the reaction quotient. The cell potentials of electrochemical cells can be determined even under non-standard conditions with the help of the Nernst equation.
Nernst equation is an equation that relates the capacity of an atom or ion to take up one or more electrons and measure reduction potential at any conditions to that measured at standard conditions i.e. standard reduction potential at 298K, one molar, and one atmospheric pressure.
 \[{E_{cell}} = {E^0} - \left[ {\dfrac{{RT}}{{nf}}} \right]\,\,\ln Q\] is the Nernst equation that we study in electrochemistry.
Here, \[{E_{cell}}\] is the cell of the potential of cell
 \[{E^0}\] is the cell potential under standard conditions
R is the universal gas constant
T is temperature, F is Faraday constant, Q is reaction quotient and n are the number of electrons transferred in the redox reaction.
When AgI is dissociated into its constituent ions. This reaction can be given as:
 \[AgI \rightleftharpoons A{g^ + } + {I^ - }\]
Now, the Since AgI is a binary electrolyte, the solubility product can be given as
 \[{K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{I^ - }} \right] = {S^2}\]
Where \[{K_{sp}}\] is the solubility product and S is the solubility of AgI.
Now from the given reactions
 \[
  AgI + {e^ - } \to Ag + {I^ - }E^\circ = - 0.152{\text{ }}V \\
  Ag \to A{g^ + } + {e^ - }E^\circ = - 0.800{\text{ }}V \\
 \]
the overall reaction is, \[AgI(s) \rightleftharpoons A{g^ + } + {I^ - };\;{E^o} = - 0.952\]
we know that the Nernst equation can be calculated as:
\[{E_{cell}} = E_{cell}^0 - \dfrac{{0.0591}}{n}\log {k_{sp}}\]. Where \[{k_{sp}}\] is the equilibrium constant. Now at equilibrium, the value of the EMF of the cell is zero. Therefore, the equation is,
\[E_{cell}^0 = \;\dfrac{{0.0591}}{n}\log {k_{sp}}\]. Where, \[E_{cell}^0\] is the standard EMF of the cell. And n is the number of total electrons used in the reaction.
So, the value of \[{k_{sp}}\] is,
\[
  E_{cell}^0 = \;\dfrac{{0.0591}}{n}\log {k_{sp}} \\
\Rightarrow - 0.952 = \;\dfrac{{0.0591}}{1}\log {k_{sp}} \\
\Rightarrow \dfrac{{ - 0.952}}{{0.0591}} = \log {k_{sp}} \\
\Rightarrow - 16.10 = \log {k_{sp}} \\
 \]

So the correct answer is D.

Note: Solubility can be understood as the maximum amount of solute that is dissolved in the given solvent at equilibrium. It can be regarded as the maximum degree to which the given compound can dissolve itself in the given solvent. The solubility product constant can be understood as the equilibrium constant for the dissolution of a given solute into a given solvent.