Problem 184 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following \(E^{\circ}\) values \(E\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=+0.77 \mathrm{~V}\) \(E\left(\mathrm{Sn}^{2+} / \mathrm{Sn}\right)=-0.14 \mathrm{~V}\) Under standard conditions, the potential for the reaction \(\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\) is (a) \(1.68 \mathrm{~V}\) (b) \(0.91 \mathrm{~V}\) (c) \(0.63 \mathrm{~V}\) (d) \(1.46 \mathrm{~V}\)

Step-by-Step Solution

Answer

The potential for the reaction is 0.91 V, which corresponds to option (b).

Step 1: Understand the given reactions

Identify the half-reactions involved in the full reaction. The half-reactions for this process are:1. Fe³⁺ + e⁻ → Fe²⁺ with a standard potential, \( E^{ ext{°}} = +0.77 \text{ V} \).2. Sn²⁺ + 2e⁻ → Sn with a standard potential, \( E^{ ext{°}} = -0.14 \text{ V} \).The overall reaction involves the oxidation of Sn to Sn²⁺ and the reduction of Fe³⁺ to Fe²⁺. For the overall reaction, these must be reversed and balanced correctly.

Step 2: Reverse the oxidation reaction

Since Sn(s) is oxidized to Sn²⁺, reverse the given reaction involving Sn. This means Sn(s) → Sn²⁺ + 2e⁻. When a half-reaction is reversed, the sign of the standard potential is changed.Thus, the potential for Sn(s) being oxidized is \( E^{ ext{°}} = +0.14 \text{ V} \).

Step 3: Calculate the overall cell potential

The overall cell potential \( E_{ ext{cell}}^{ ext{°}} \) is calculated by adding the standard potentials of the reduction and oxidation processes:\[ E_{ ext{cell}}^{ ext{°}} = E_{ ext{red}}^{ ext{°}} + E_{ ext{ox}}^{ ext{°}} \]Substituting the values:\[ E_{ ext{cell}}^{ ext{°}} = 0.77 \text{ V} + 0.14 \text{ V} = 0.91 \text{ V} \]

Step 4: Choose the correct answer

The calculated overall cell potential is 0.91 V. Among the available options, this corresponds to option (b).

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are integral processes in chemistry where electrons are transferred between two substances. In every redox reaction, one substance is oxidized (loses electrons) and the other is reduced (gains electrons). This electron transfer allows energy to be produced or consumed as the atoms or molecules involved change their oxidation states.

Consider the example given: tin (Sn) and iron ( Fe^{3+} ). Here, Sn is oxidized to Sn^{2+} , meaning it loses electrons. Specifically, Sn changes its oxidation state by losing two electrons and becomes Sn^{2+} . Meanwhile, Fe^{3+} is reduced to Fe^{2+} by gaining electrons. Understanding these two processes is fundamental to solving problems related to redox reactions in electrochemistry.

• **Oxidation:** The process where a substance loses electrons.
• **Reduction:** The process where a substance gains electrons.
• **Electron Transfer:** The core of all redox reactions, balancing number of electrons lost and gained.

Standard Reduction Potentials

Standard reduction potentials ( E^ {°}) quantify the tendency of a chemical species to gain electrons and be reduced. Each half-reaction in a redox process has an associated standard reduction potential defined under standard conditions (25°C, 1 atm pressure, and 1 M concentration). Higher E^ {°} values mean a greater tendency to be reduced.

In the exercise, the standard reduction potential for Fe^{3+} (gaining electrons to become Fe^{2+} ) is given as +0.77 V, showing a relatively strong tendency to undergo reduction. Conversely, the reduction potential for Sn^{2+} gaining electrons to become Sn is given as -0.14 V, indicating a less favorable reduction process.

• Assigning a positive E^ {°} value when a half-reaction is written as a reduction.
• When reversing a reaction from reduction to oxidation, the sign of E^ {°} changes.

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa. They are built around redox reactions occurring between different substances. Each electrochemical cell consists of two half-cells, each responsible for either oxidation or reduction.

In the problem, two half-cells are described: one where Sn is oxidized, and another where Fe^{3+} is reduced. By calculating the standard cell potential ( E_ {cell}^ {°} ), you can determine the voltage generated by the cell under standard conditions. This involves summing the standard reduction potential and the standard oxidation potential (reverse of reduction potential), giving insight into the effectiveness and feasibility of the electrochemical reaction.

• The notation for calculating cell potential: E_ {cell}^ {°} = E_ {red}^ {°} + E_ {ox}^ {°}
• A positive cell potential indicates a spontaneous reaction, where the redox process can proceed without external energy.
• In practice, these principles are applied in batteries, electrolytic processes, and corrosion prevention.