Problem 16 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The rusting of iron takes place as follows \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-}+1 / 2 \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}(1) ; E^{\circ}=+1.23 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; E^{\circ}=-0.44 \mathrm{~V}\) Calculate \(\Delta G^{\circ}\) for the net process (a) \(-322 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-161 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-152 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-76 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The answer is (c) \(-152 \text{ kJ/mol}\).

Step 1: Write Down the Reaction

In the rusting process, combine the two half-reactions provided:\[\begin{align*}2 \mathrm{H}^{+} + 2 \mathrm{e}^{-} + \frac{1}{2} \mathrm{O}_2 & \longrightarrow \mathrm{H}_2 \mathrm{O}(l)\\mathrm{Fe}^{2+} + 2 \mathrm{e}^{-} & \longrightarrow \mathrm{Fe}(s)\end{align*}\]

Step 2: Calculate the Net Cell Potential

Add the standard electrode potentials for the two half-reactions:\[ E^{\circ}_{\text{net}} = E^{\circ}_{\text{reduction}} + E^{\circ}_{\text{oxidation}} = (+1.23 \text{ V}) + (-0.44 \text{ V}) = 0.79 \text{ V}\]

Step 3: Use Gibbs Free Energy Equation

To find \( \Delta G^{\circ} \), use the equation:\[ \Delta G^{\circ} = -nFE^{\circ}\]where \( n = 2 \) (the number of electrons transferred), and \( F = 96500 \text{ C/mol} \).

Step 4: Plug in the Values

Substitute the values: \[ \Delta G^{\circ} = -(2)(96500 \text{ C/mol})(0.79 \text{ V}) \]

Step 5: Perform the Calculation

Calculate the value of \( \Delta G^{\circ} \):\[ \Delta G^{\circ} = -152770 \text{ J/mol} = -152.77 \text{ kJ/mol} \]

Step 6: Compare with Provided Options

Compare the calculated \( \Delta G^{\circ} = -152.77 \text{ kJ/mol} \) with the given options. It is closest to (c) \(-152 \text{ kJ/mol}\).

Key Concepts

Rusting of Iron

Rusting is a well-known example of a redox reaction that occurs when iron is exposed to moisture and oxygen. The process leads to the formation of iron oxides, commonly seen as rust. Here’s the breakdown of the rusting reaction:

• The first half-reaction involves oxygen and hydrogen ions combining to form water:

\[ 2 \mathrm{H}^{+} + \frac{1}{2} \mathrm{O}_2 + 2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O}(l) \]

• The second half-reaction involves the reduction of iron ions to solid iron:

\[ \mathrm{Fe}^{2+} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) \]These reactions occur simultaneously, leading iron to gradually transform into rust when exposed to humid air. The electrochemical nature of rusting is key to understanding this phenomenon, as it involves both oxidation (loss of electrons) and reduction (gain of electrons). Essentially, iron atoms lose electrons and become aqueous ions. As these electrons are transferred, oxygen from the air gains these electrons, forming water.

Electrochemistry

Electrochemistry is the study of chemical processes that cause electrons to move. This electron movement leads to chemical changes through oxidation and reduction reactions, broadly known as redox reactions.
In the rusting of iron, electron flow is vital, bridging oxygen reduction and iron oxidation. The essence of electrochemistry resides in these half-reactions. Here, one species gets oxidized by losing electrons, enhancing its oxidation state, while another type gets reduced by gaining electrons, reducing its oxidation state.
Essential elements of electrochemistry include:

• Electrode: Conducts electrons into or out of a solution.
• Electrolyte: Usually a solution that can conduct electricity by moving charged ions.
• Redox Reactivity: Electron transfer is essential for both corrosion, like rusting, and other reactions driven by electrical energy.

Understanding electrochemistry also provides insight into methods to prevent rusting, such as using coatings or alloying with other metals.

Standard Electrode Potential

Standard electrode potential (\( E^{\circ} \)) is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is expressed in volts (V).
In the rusting of iron:

• The reduction potential for the reaction forming water from oxygen and hydrogen ions is high (\(+1.23 \, \text{V}\)), indicating a strong tendency to gain electrons.
• The oxidation potential for the reaction from iron ions to iron metal is low (\(-0.44 \, \text{V}\)), suggesting a lower tendency to lose electrons.

The combination of these potentials gives us the net cell potential:
\[ E^{\circ}_{\text{net}} = (+1.23 \, \text{V}) + (-0.44 \, \text{V}) = 0.79 \, \text{V} \]
This net potential helps us understand the overall force driving the reaction. The value, when positive, points towards a spontaneous reaction under standard conditions. This is essential for calculating Gibbs Free Energy (\( \Delta G^{\circ} \)), giving insight into the energy available from or required for a reaction. Such calculations are critical in predicting the spontaneity and extent of chemical processes.