Problem 137 - A Complete Resource Book in Chemistry for JEE Main

Question

The half cell reaction for the corrosion \(2 \mathrm{H}^{+}+1 / 2 \mathrm{O}_{2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}, E^{\circ}=1.23 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; E^{\circ}=-0.44 \mathrm{~V}\) Find the \(\Delta \mathrm{G}^{\circ}\) (in \(\mathrm{kJ}\) ) for the overall reaction. (a) \(-76\) (b) \(-322\) (c) \(-161\) (d) \(-152\)

Step-by-Step Solution

Answer

(b) -322

Step 1: Write Overall Reaction

First, combine the given half-reactions to get the overall redox reaction. The reactions are:- Reduction: \(2 \mathrm{H}^{+} + \frac{1}{2} \mathrm{O}_{2} + 2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}\)- Oxidation: \(\mathrm{Fe}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Fe(s)}\)The electrons are already balanced, so the overall reaction is:\(2 \mathrm{H}^{+} + \frac{1}{2} \mathrm{O}_{2} + \mathrm{Fe}^{2+} \rightarrow \mathrm{H}_{2} \mathrm{O} + \mathrm{Fe(s)}\)

Step 2: Combine Standard Electrode Potentials

Next, determine the standard cell potential \(E^{\circ}_{\text{cell}}\) by subtracting the anode reaction potential (oxidation) from the cathode reaction potential (reduction):\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 1.23 \mathrm{V} - (-0.44 \mathrm{V})\]\[E^{\circ}_{\text{cell}} = 1.23 \mathrm{V} + 0.44 \mathrm{V} = 1.67 \mathrm{V}\]

Step 3: Calculate Gibbs Free Energy Change

Use the relation between standard cell potential and Gibbs free energy change \(\Delta G^{\circ}\):\[\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}\]where:- \(n = 2\) (number of moles of electrons exchanged)- \(F = 96485 \text{ C/mol}\) (Faraday's constant)Substitute the values to calculate \(\Delta G^{\circ}\):\[\Delta G^{\circ} = -2 \times 96485 \text{ C/mol} \times 1.67 \text{ V}\]This gives:\[\Delta G^{\circ} = -322766.9 \text{ J/mol}\]Convert to kJ/mol:\[\Delta G^{\circ} = -322.77 \text{ kJ/mol}\]

Step 4: Select the Correct Answer

Compare the calculated \(\Delta G^{\circ}\) with the given options. Our value of \(-322.77 \text{ kJ/mol}\) is closest to \(-322 \text{ kJ/mol}\). Therefore, the correct answer is (b) \(-322\).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G \), is a thermodynamic quantity that helps determine whether a reaction will occur spontaneously under constant pressure and temperature. In simple terms, it tells us if the energy available from a reaction is enough to do work, like forming new chemical bonds.
If \( \Delta G \) is negative, the reaction will likely proceed without outside intervention, which we call 'spontaneous'. Positive \( \Delta G \) means the reaction needs additional energy input to occur.

• \( \Delta G^\circ \) is the change in free energy under standard conditions (1 atm, 298 K).
• It's calculated from the equation \( \Delta G^\circ = -nFE^\circ_{\text{cell}} \), where \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant, and \( E^\circ_{\text{cell}} \) is the standard cell potential.

Using the provided values, we calculated \( \Delta G^\circ \) to be -322 kJ/mol, indicating a spontaneous reaction.

Standard Electrode Potential

Standard Electrode Potential, represented as \( E^\circ \), is a measure of the tendency of a chemical species to be reduced. How good a substance is at accepting electrons or being "reduced" is what the electrode potential tells us.
Each half-reaction has its own \( E^\circ \). When two half-reactions combine in a cell, we can calculate the total potential power of that electrochemical cell.
To find the standard cell potential \( E^\circ_{\text{cell}} \):

• Take the \( E^\circ \) of the cathode reaction (the more positive value, which gets reduced).
• Subtract the \( E^\circ \) of the anode reaction (the less positive value, which gets oxidized).
• This yields \( E^\circ_{\text{cell}} = 1.23 \, \text{V} + 0.44 \, \text{V} = 1.67 \, \text{V} \).

A greater positive value of \( E^\circ_{\text{cell}} \) indicates a stronger drive for the reaction to proceed, as it means lower energy is needed to make the electrons move or the reaction to occur.

Redox Reactions

Redox (reduction-oxidation) reactions are chemical processes involving the transfer of electrons between two species. In these reactions, one reactant gets oxidized (loses electrons), and the other gets reduced (gains electrons).
For the given problem:

• The cathode half-reaction is the reduction of hydrogen ions and oxygen: \(2 \mathrm{H}^{+} + \frac{1}{2} \mathrm{O}_{2} + 2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O} \).
• The anode half-reaction is the oxidation of iron: \(\mathrm{Fe}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Fe(s)} \).

In the overall reaction, electrons are shared equally between oxidation and reduction, allowing for a balanced and energetically favorable outcome.
The challenge for students often lies in accurately balancing and combining these equations, which is critical in calculating the \( \Delta G^\circ \) values.