Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(E^{\circ}\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)=-0.441 \mathrm{~V}\) and \(E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.771 \mathrm{~V}\), the standard emf of the reaction \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \longrightarrow 3 \mathrm{Fe}^{2+}\) will be (a) \(1.212 \mathrm{~V}\) (b) \(0.111 \mathrm{~V}\) (c) \(0.330 \mathrm{~V}\) (d) \(1.653 \mathrm{~V}\)

Step-by-Step Solution

Answer

The standard emf of the reaction is \(1.212\, \mathrm{V}\), option (a).

Step 1: Write the Half-Reactions

The given overall reaction is \( \mathrm{Fe} + 2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+} \). This can be divided into two half-reactions: \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} \) and \( \mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}^{2+} \).

Step 2: Identify Half-Reaction Potentials

From the problem, we know the standard reduction potentials: \( E^{\circ}\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right) = -0.441 \, \mathrm{V} \) and \( E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right) = 0.771 \, \mathrm{V} \).

Step 3: Reverse the Appropriate Half-Reaction

The first half-reaction \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} \) must be reversed to \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe} \) to align with the oxidation process in the overall reaction, changing the sign of its potential: \( E^{\circ} = 0.441 \, \mathrm{V} \).

Step 4: Combine Potentials for Overall EMF

To find the standard emf for the entire reaction, add the corrected potentials: \( E^{\circ}_{\text{cell}} = 0.441 \, \mathrm{V} + 0.771 \, \mathrm{V} = 1.212 \, \mathrm{V} \).

Step 5: Choose the Correct Answer Choice

Compare the calculated standard emf \( 1.212 \, \mathrm{V} \) with the given options. Option (a) \( 1.212 \, \mathrm{V} \) is the correct answer.

Key Concepts

Standard Reduction Potential

Standard reduction potential is an essential concept in electrochemistry. It measures the tendency of a chemical species to be reduced, i.e., gain electrons, under standard conditions. The standard condition is usually at 25°C (298 K), 1 atmosphere of pressure, and with 1 M concentration of all aqueous species. These potentials are measured in volts (V).
In our exercise, known standard reduction potentials are provided:

• \(E^{\circ}(\mathrm{Fe}^{2+} / \mathrm{Fe}) = -0.441 \, \mathrm{V}\)
• \(E^{\circ}(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}) = 0.771 \, \mathrm{V}\)

With reduction potentials, the more positive the value, the greater the tendency for the species to accept electrons and be reduced.
This tendency is used to determine which half-reaction will proceed in an electrochemical cell and helps predict the direction of electron flow.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often called redox reactions, involve the transfer of electrons between chemical species. In these reactions, one species donates electrons while another accepts them.
In the reaction from our exercise, \(\mathrm{Fe}+2\mathrm{Fe}^{3+} \rightarrow 3\mathrm{Fe}^{2+}\), iron (Fe) loses electrons (is oxidized), and iron(III) ions (Fe³⁺) gain those electrons (are reduced).
This electron transfer is what drives the chemical change, and it occurs across the interface of electrodes in an electrochemical cell. Understanding which species is oxidized and which is reduced is crucial for balancing redox equations and predicting reaction directions.

Half-Reactions

Half-reactions are a way of spliting a redox reaction into two separate equations that represent the oxidation and reduction processes.

• Each half captures how electrons are transferred – one for electron loss (oxidation) and one for electron gain (reduction).
• In the exercise, the full reaction \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+}\) is split into:
  • Oxidation: \(\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2\mathrm{e}^-\)
  • Reduction: \(\mathrm{Fe}^{3+} + \mathrm{e}^- \rightarrow \mathrm{Fe}^{2+}\)

These half-reactions help in calculating the potential and identifying the direction of electron flow in the cell. Accurately understanding half-reactions is required to perform electrochemical cell calculations and predict cell behavior.

Electromotive Force (emf)

Electromotive force (emf), sometimes referred to as cell potential, is a measure of the potential energy difference between two electrodes in an electrochemical cell. It represents the energy that drives electrons from the anode (where oxidation occurs) to the cathode (where reduction takes place).
The emf of a cell under standard conditions (standard emf) can be calculated by adding up the standard reduction potentials of the half-reactions.
In our example, after reversing the oxidation half-reaction potential and combining it with the reduction half-reaction:\[E^{\circ}_{\text{cell}} = E^{\circ}(\mathrm{Fe}^{2+} / \mathrm{Fe}) + E^{\circ}(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}) = 0.441 \mathrm{~V} + 0.771 \mathrm{~V} = 1.212\mathrm{~V}\]

• This value, 1.212 V, represents the standard emf of the cell reaction.

Calculating emf is central in determining the work that can be done by the redox reaction in the cell.

Electrochemical Cell Calculations

Electrochemical cell calculations involve determining the standard emf of a redox reaction based on given half-reactions and their potentials.

• Firstly, identify the half-reactions involved in the overall reaction.
• Next, assign the correct standard reduction potentials to each half-reaction.
• Reverse any half-reaction that represents oxidation, changing the sign of its potential.
• Finally, sum the potentials to find the standard emf.

In our given exercise, after reversing the oxidation half-reaction, we added the potentials to find the emf of 1.212 V.
These calculations ensure that an accurate determination of the cell's voltage can predict how much electrical work the reaction can perform, which is key in designing batteries and other electrochemical applications.