Problem 193 - A Complete Resource Book in Chemistry for JEE Main

Question

Given the data at \(25^{\circ} \mathrm{C}\) \(\mathrm{Ag}+\mathrm{I}^{-} \longrightarrow \mathrm{AgI}+\mathrm{e}^{-} ; E^{\circ}=0.152 \mathrm{~V}\) \(\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; E^{\circ}=-0.800 \mathrm{~V}\) What is the value of \(\log \mathrm{K}_{\text {sp }}\) for \(\mathrm{AgI}\) ? \((2.303 \mathrm{RT} / F=0.059 \mathrm{~V})\) (a) \(-8.12\) (b) \(+8.612\) (c) \(-37.83\) (d) \(-16.13\)

Step-by-Step Solution

Answer

The value of \( \log K_{sp} \) for \( \mathrm{AgI} \) is \(-16.13\).

Step 1: Understand the Reactions

We are given two reactions and their standard electrode potentials:1. \( \mathrm{Ag} + \mathrm{I}^- \longrightarrow \mathrm{AgI} + \mathrm{e}^- \); \( E^{\circ} = 0.152 \mathrm{~V} \)2. \( \mathrm{Ag} \longrightarrow \mathrm{Ag}^+ + \mathrm{e}^- \); \( E^{\circ} = -0.800 \mathrm{~V} \)These are half-reaction potentials used to find the overall reaction potential.

Step 2: Combine the Half-Reactions

To find the net reaction, we combine both half-reactions such that electrons cancel out.First reaction, revised: \( \mathrm{Ag} + \mathrm{I}^- \longrightarrow \mathrm{AgI} + \mathrm{e}^- \)Actually, combine them directly to eliminate electrons:\( \mathrm{Ag}^+ + \mathrm{I}^- \longrightarrow \mathrm{AgI} \)Since one electron is transferred, the potentials add up as:\( E^{\circ}_{\text{net}} = E^{\circ}_{1} - E^{\circ}_{2} = 0.152 - (-0.800) = 0.952 \mathrm{~V} \)

Step 3: Apply the Nernst Equation

We use the relationship\[ E^{\circ} = \frac{2.303\ \text{RT}}{F} \log K_{sp} \]Substitute the known values: \( E^{\circ} = 0.952 \mathrm{~V} \), and \( \frac{2.303\ \text{RT}}{F} = 0.059 \mathrm{~V} \).This gives us:\[ 0.952 = 0.059 \log K_{sp} \]

Step 4: Solve for \( \log K_{sp} \)

Rearrange the equation to find \( \log K_{sp} \):\[ \log K_{sp} = \frac{0.952}{0.059} \]Calculate this division to get:\[ \log K_{sp} = 16.127 \]

Step 5: Finalize the Answer

The calculated \( \log K_{sp} \) is \(+16.127\). The negative sign is important (indicating solubility product), thus the option closely matching is \(-16.13\).Thus, the value of \( \log K_{sp} \) for \( \mathrm{AgI} \) is \(-16.13\).

Key Concepts

Standard Electrode Potential

In electrochemistry, the standard electrode potential, denoted as \(E^{\circ}\), represents the inherent tendency of a chemical species to acquire electrons and undergo reduction. It is measured in volts and evaluated under standard conditions: a pressure of 1 atm, a concentration of 1 M for each species involved in the reaction, and a temperature of 25°C (298 K).
In this context, the given half-reactions have their own \(E^{\circ}\) values: +0.152 V for Ag reacting with I- to form AgI and -0.800 V for Ag transforming into Ag+. The positive value indicates a favorable process for the formation of AgI, while the negative value shows the non-spontaneous dissociation of Ag into Ag+.
Understanding these values helps predict the direction of electron flow between the species involved in an electrochemical cell.

• A positive \(E^{\circ}\) suggests a stronger tendency for the reduction (gain of electrons).
• A negative \(E^{\circ}\) indicates a tendency for the oxidation process (loss of electrons).

These potentials are foundational for calculating the overall cell potential by combining half-reactions, which in turn assists in determining reaction spontaneity.

Nernst Equation

The Nernst equation allows us to calculate the cell potential at any state, not just under standard conditions. It integrates the effect of concentration (or pressure) of reactants and products into electromotive force (EMF) calculations.
In simpler terms, this equation connects the equilibrium condition of a redox reaction with its reaction quotient, essentially relating changes in concentration to changes in voltage.
The equation is expressed as:
\[E = E^{\circ} - \frac{2.303\,RT}{nF} \log Q\]
Where:

• \(E\) is the cell potential under non-standard conditions.
• \(E^{\circ}\) is the standard cell potential.
• \(R\) is the universal gas constant, \(8.314 \text{ J mol}^{-1} \text{K}^{-1}\).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is Faraday’s constant, about \(96485 \text{ C mol}^{-1}\).
• \(Q\) is the reaction quotient, the ratio of concentrations of products to reactants.

In the given exercise, we used the Nernst equation to connect the standard electrode potential \((E^{\circ})\) with the solubility product constant \((K_{sp})\) through an adaptation:\[ E^{\circ} = \frac{2.303RT}{F} \log K_{sp} \]This version directly relates the overall cell potential to the equilibrium constant for a sparingly soluble salt like \(\text{AgI}\).

Solubility Product Constant (Ksp)

The solubility product constant, \(K_{sp}\), is a specialized type of equilibrium constant used in sparingly soluble ionic compounds. It quantifies the extent to which a compound will dissolve in water. For AgI, the dissolution can be described by the reaction:
\[\text{AgI} \rightleftharpoons \text{Ag}^+ + \text{I}^-\]
In this expression, \(K_{sp}\) is calculated as:
\[K_{sp} = [\text{Ag}^+][\text{I}^-]\]
Where the brackets denote the molar concentration of the ions in solution at equilibrium.
The exercise aims to find \(\log K_{sp}\), which is effectively \(\log_{10}\) of the solubility product constant. The sign and value of \(\log K_{sp}\) provide insightful understanding:

• Negative values imply low solubility, significant for predicting if a precipitant will form.
• A more negative \(\log K_{sp}\) indicates an extremely low concentration of dissolved ions for very insoluble compounds.

This concept is crucial in contexts such as predicting precipitation reactions and solving equilibrium problems in chemical systems.