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

0.91 V (option b)

Step 1: Identify the Half Reactions

Identify the oxidation and reduction half-reactions in the given overall reaction: \(\text{Sn} \rightarrow \text{Sn}^{2+} + 2e^-\) (oxidation), \(\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}\) (reduction).

Step 2: Use Standard Reduction Potentials

Look up the standard reduction potentials given: \(E^{\circ}(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \text{ V}\) and \(E^{\circ}(\text{Sn}^{2+}/\text{Sn}) = -0.14 \text{ V}\). Note that these values are for reduction reactions.

Step 3: Determine the Oxidation Potential

Since oxidation is occurring with \(\text{Sn}\), we use the reverse reaction and change the sign of the potential: \(E^{\circ}(\text{Sn}/\text{Sn}^{2+}) = +0.14 \text{ V}\).

Step 4: Calculate the Cell Potential

Calculate the overall cell potential by summing the standard reduction potential of the iron half-reaction and the oxidation potential of tin: \[ E^{\circ}_{\text{cell}} = E^{\circ}(\text{Fe}^{3+}/\text{Fe}^{2+}) + E^{\circ}(\text{Sn}/\text{Sn}^{2+}) = 0.77 \text{ V} + 0.14 \text{ V} = 0.91 \text{ V} \].

Step 5: Match the Calculated Potential to the Choices

Compare the calculated potential \(0.91 \text{ V}\) with the given options to determine the correct answer: (a) \(1.68 \text{ V}\), (b) \(0.91 \text{ V}\), (c) \(0.63 \text{ V}\), (d) \(1.46 \text{ V}\). The correct choice is (b) \(0.91 \text{ V}\).

Key Concepts

Standard Reduction Potential

Standard reduction potentials ( E^{ ext{o}} ) are a measure of the tendency of a chemical species to acquire electrons and be reduced. These values are determined under standard conditions, which typically involve solutes at a concentration of 1 M, gases at a pressure of 1 atm, and a temperature of 25°C (298 K). In electrochemistry, the standard reduction potential is vital because it helps predict the direction of electron flow in an electrochemical cell.
For any half-reaction, the standard reduction potential is noted as the potential difference between a species being reduced and a standard hydrogen electrode (SHE), which is assigned a potential of 0.00 V.

• Positive E^{ ext{o}} values (e.g., E( ext{Fe}^{3+}/ ext{Fe}^{2+}) = +0.77 ext{ V} ) indicate a greater tendency to gain electrons and undergo reduction.
• Negative E^{ ext{o}} values (e.g., E( ext{Sn}^{2+}/ ext{Sn}) = -0.14 ext{ V} ) indicate a lesser tendency to gain electrons, and are less favorable for reduction.

In a given electrochemical cell, identifying the redox pairs using standard reduction potentials allows us to determine which species will act as the oxidizing agent (reduction) and which will act as the reducing agent (oxidation).

Electrochemical Cell

An electrochemical cell is a device that allows the flow of electrons through an external circuit via a redox reaction. It consists of two half-cells, each containing electrodes and electrolyte solutions where separate oxidation and reduction reactions take place. This setup enables the conversion of chemical energy into electrical energy.

Components of an Electrochemical Cell

• Anode: The electrode where oxidation occurs. Electrons are lost by a chemical species.
• Cathode: The electrode where reduction occurs. Electrons are gained by a chemical species.
• Salt Bridge: Maintains electrical neutrality by allowing the flow of ions between the solutions in the two half-cells.
• External Circuit: Pathway allowing electron flow from the anode to the cathode.

In the electrochemical cell discussed, tin is oxidized, which means it acts as the anode, and iron is reduced, which means it acts as the cathode. The cell potential, calculated as the difference between the two half-reactions’ standard potentials, dictates how much electrical energy the cell can supply.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are chemical reactions where electrons are transferred between two substances. In every redox reaction, one species undergoes oxidation (losing electrons) and another undergoes reduction (gaining electrons).

Identifying Oxidation and Reduction

• Oxidation: A process in which a substance loses electrons. For example, in our exercise, tin ( ext{Sn} ) loses electrons to form ext{Sn}^{2+} .
• Reduction: A process involving the gain of electrons. For instance, iron ( ext{Fe}^{3+} ) gains electrons to form ext{Fe}^{2+} .

Every complete redox reaction comprises two half-reactions: an oxidation half-reaction and a reduction half-reaction. The balance between these simultaneously occurring reactions is necessary to maintain the flow of electrons in an electrochemical cell. Moreover, electrodes facilitate this process by providing a surface for these reactions to occur, enabling connectivity within the cell.