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

Question

Given the data at \(25^{\circ} \mathrm{C}\) [2006] \(\mathrm{Ag}+\mathrm{I}^{-} \longrightarrow \mathrm{Ag} \mathrm{I}+\mathrm{e} ; E^{\circ}=0.152 \mathrm{~V}\) \(\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; E^{\circ}=-0.800 \mathrm{~V}\) What is the value of \(\log \mathrm{K}_{\mathrm{sp}}\) for \(\mathrm{Ag}\) I? \((2.303 \mathrm{RT} / F=0.059 \mathrm{~V})\) (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: Write the Nernst Equation for Redox Reaction

For a redox reaction, the Nernst equation relates the cell potential to the concentrations of the reactants and products. For the solid state reaction \( \mathrm{AgI (solid)} \longrightarrow \mathrm{Ag}^{+} (aq) + \mathrm{I}^{-} (aq) \), the standard cell potential \( E^{\circ} \) can be calculated using:\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]

Step 2: Identify Anode and Cathode Reactions

The two half-reactions are:1. \( \mathrm{Ag} + \mathrm{I}^{-} \longrightarrow \mathrm{AgI (solid)} + \mathrm{e}^{-}; \quad E^{\circ} = 0.152 \mathrm{~V} \)2. \( \mathrm{Ag} \longrightarrow \mathrm{Ag}^{+} + \mathrm{e}^{-}; \quad E^{\circ} = -0.800 \mathrm{~V} \)\( \mathrm{Ag} + \mathrm{I}^{-} \longrightarrow \mathrm{AgI (solid)} + \mathrm{e}^{-} \) is the cathode with \( E^{\circ}_{\text{cathode}} = 0.152 \mathrm{~V} \) and \( \mathrm{Ag} \longrightarrow \mathrm{Ag}^{+} + \mathrm{e}^{-} \) is the anode with \( E^{\circ}_{\text{anode}} = -0.800 \mathrm{~V} \).

Step 3: Calculate the Standard Cell Potential

Use the equation from Step 1 to find the standard cell potential:\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.152 - (-0.800) = 0.952 \mathrm{~V}\]

Step 4: Use Nernst Equation to Solve for Ksp

The Nernst equation relates \( E^{\circ}_{\text{cell}} \) to the solubility product \( K_{sp} \) via:\[E^{\circ}_{\text{cell}} = \left( \frac{2.303RT}{F} \right) \log K_{sp}\]Substitute \( E^{\circ}_{\text{cell}} = 0.952 \mathrm{~V} \) and \( \frac{2.303RT}{F} = 0.059 \mathrm{~V} \):\[0.952 = 0.059 \log K_{sp}\]

Step 5: Solve for log Ksp

Rearrange the Nernst equation to find \( \log K_{sp} \):\[\log K_{sp} = \frac{0.952}{0.059} = 16.1271\]Rounding to appropriate significant figures: \( \log K_{sp} = +16.13 \). Thus, the correct option matches \((d)\).

Key Concepts

Redox Reaction

Redox reactions, short for reduction-oxidation reactions, involve a transfer of electrons between two species. In a redox reaction, one species gets oxidized, losing electrons, while another gets reduced by gaining those electrons. Consider the process:

• Oxidation: The loss of electrons.
• Reduction: The gain of electrons.

For example, in the reaction \( \mathrm{Ag} + \mathrm{I}^{-} \longrightarrow \mathrm{AgI (solid)} + \mathrm{e}^{-} \), silver (Ag) cations form silver iodide (AgI) by gaining an electron, effectively reducing iodine from its ionic form. Understanding which species undergoes oxidation or reduction helps in identifying the anode and cathode in electrochemical cells.

Cell Potential

Cell potential, also known as electromotive force (emf), is the measure of the energy available from a redox reaction as it occurs in an electrochemical cell. Measured in volts (V), it is calculated from the difference in potential energy between the electrodes:

• Positive value: Indicates a spontaneous reaction.
• Negative value: Implies a non-spontaneous reaction as written.

To find the standard cell potential \( E^{\circ}_{\text{cell}} \), use the formula: \[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]For the reaction given in the problem, the calculated \(E^{\circ}_{\text{cell}}\) is 0.952 V. This positive value suggests that the reaction is spontaneous under standard conditions.

Solubility Product Constant (Ksp)

The solubility product constant, \(K_{sp}\), is a measure used to express the solubility of sparingly soluble compounds. It provides insight into how much a solid can dissolve in a solution, reaching equilibrium.

• High \(K_{sp}\): Indicates a relatively soluble compound.
• Low \(K_{sp}\): Indicates a sparingly soluble or nearly insoluble compound.

In the context of the Nernst equation, the solubility product \(K_{sp}\) is related to cell potential. Using the equation:\[E^{\circ}_{\text{cell}} = \left( \frac{2.303RT}{F} \right) \log K_{sp}\]allows us to calculate \(\log K_{sp}\) using the potential change observed in the cell reaction. Solving this in our example gives a value of 16.13, reflecting on the solubility of \(\mathrm{AgI}\).

Electrode Potentials

Electrode potentials refer to the intrinsic voltage between an electrode and its surrounding solution. Each electrode in an electrochemical cell has its own potential:

• Cathode: Reduction occurs; typically gains electrons.
• Anode: Oxidation takes place; typically loses electrons.

To determine which electrode is the anode or cathode, one should compare their standard reduction potentials. The reaction with a higher reduction potential standardly serves as the cathode. For instance, in the problem reaction, \( \mathrm{Ag} + \mathrm{I}^{-} \rightleftharpoons \mathrm{AgI} + \mathrm{e}^{-} \) occurs at the cathode with \( E^{\circ}_{\text{cathode}} = 0.152 \mathrm{~V} \), making it the site of reduction. Conversely, \( \mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + \mathrm{e}^{-} \) occurs at the anode, with \( E^{\circ}_{\text{anode}} = -0.800 \mathrm{~V} \). Understanding these potentials is crucial to calculating the overall cell potential.