Problem 46 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The standard reduction potential data at \(25^{\circ} \mathrm{C}\) is given below : [Adv. 2013] \(E^{\circ}\left(\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}\right)=+0.77 \mathrm{~V} ; E^{\circ}\left(\mathrm{Fe}^{2+}, \mathrm{Fe}\right)=-0.44 \mathrm{~V} ; E^{\circ}\left(\mathrm{Cu}^{2+}, \mathrm{Cu}\right)=+\) \(0.34 \mathrm{~V} ; E^{\circ}\left(\mathrm{Cu}^{+}, \mathrm{Cu}\right)=+0.52 \mathrm{~V}\) \(E^{\circ}\left[\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\right]=+1.23 \mathrm{~V} ; E^{\circ}\left[\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \rightarrow\right.\) \(\left.4 \mathrm{OH}^{-}\right]=+0.40 \mathrm{~V}\) \(E^{\circ}\left(\mathrm{Cr}^{3+}, \mathrm{Cr}\right)=-0.74 \mathrm{~V} ; E^{\circ}\left(\mathrm{Cr}^{2+}, \mathrm{Cr}\right)=-0.91 \mathrm{~V}\) Match \(E^{\circ}\) of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists: List I \(\quad\) List II P. \(E^{\circ}\left(\mathrm{Fe}^{3+}, \mathrm{Fe}\right)\) 1\. \(-0.18 \mathrm{~V}\) Q. \(\quad E^{\circ}\left(4 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 4 \mathrm{H}^{+}+4 \mathrm{OH}^{-}\right) \quad\) 2. \(-0.4 \mathrm{~V}\) R. \(\quad E^{\circ}\left(\mathrm{Cu}^{2+}+\mathrm{Cu} \rightarrow 2 \mathrm{Cu}^{+}\right)\) 3\. \(-0.04 \mathrm{~V}\) S. \(E^{\circ}\left(\mathrm{Cr}^{3+}, \mathrm{Cr}^{2+}\right)\) 4\. \(-0.83 \mathrm{~V}\) Codes : \(\begin{array}{lllll} & \mathrm{P} & \mathrm{Q} & \mathrm{R} & \mathrm{S} \\\ \text { (a) } & 4 & 1 & 2 & 3\end{array}\) (b) \(2 \quad 3 \quad 4\) (c) \(1 \quad 2 \quad 3 \quad 4\) (d) 3 4 1 2

Step-by-Step Solution

Answer

(c) 1 2 3 4

Step 1: Calculate E⁰(Fe³⁺, Fe)

Combine the reactions for Fe³⁺/Fe²⁺ and Fe²⁺/Fe to find the standard potential for Fe³⁺/Fe. Use the formula:\[ E^{ ext{cell}} = E^{ ext{cathode}} - E^{ ext{anode}} \]Here, - \( E^{ ext{cathode}} = E^{ ext{cell}}(Fe^{3+}, Fe^{2+}) = +0.77 \, \text{V} \),- \( E^{ ext{anode}} = E^{ ext{cell}}(Fe^{2+}, Fe) = -0.44 \, \text{V} \).Thus,\[ E^{ ext{cell}}(Fe^{3+}, Fe) = 0.77 \, \text{V} - (-0.44 \, \text{V}) = 0.33 \, \text{V} \].

Key Concepts

Standard Reduction Potentials

Standard reduction potentials are a crucial concept in electrochemistry. They provide a way to predict how chemical substances will react in redox reactions. Each potential is measured under standard conditions and refers to the tendency of a species to gain electrons and be reduced. The values are comparable only when measured under standard conditions, which include 25°C, 1 M concentrations for each ion, and a partial pressure of 1 atm for any gases involved. These potentials are tabulated for each half-reaction based on a standard hydrogen electrode, which is set at 0 volts. Let's look at some practical applications:

• They are useful in determining the direction of electron flow in electrochemical cells.
• The higher (more positive) the reduction potential, the greater the species' tendency to gain electrons and be reduced.
• Conversely, a low (more negative) potential indicates a tendency to lose electrons and be oxidized.

In practice, you can combine standard reduction potentials to calculate the cell potential of an electrochemical cell. This calculation helps predict whether a redox reaction will occur spontaneously.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes where the oxidation state of atoms changes. This involves the transfer of electrons between two substances. To break it down: - **Oxidation** is the loss of electrons. For instance, when Fe²⁺ becomes Fe³⁺, it loses an electron. - **Reduction** is the gain of electrons. As in Fe³⁺ gaining an electron to become Fe²⁺. These reactions are essential in many biological and industrial processes. They drive the production of energy in batteries, metabolism within cells, and even the rusting of metals. In any redox reaction, the substance that loses electrons becomes oxidized, while the one that gains electrons becomes reduced. To determine the feasibility of these reactions, you can consult the standard reduction potentials. Remember:

• If the net cell potential, calculated using standard reduction potentials, is positive, the reaction will occur spontaneously.
• If negative, the reaction will need the input of energy to proceed.

Understanding redox reactions is vital to mastering electrochemistry as they form the basis of countless chemical and electrochemical processes.

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa. They consist of two electrodes immersed in electrolytes, where redox reactions occur. There are two main types of electrochemical cells:

• **Galvanic (or Voltaic) cells**: These cells store energy and convert chemical energy into electrical energy used to do work. For instance, in a battery, spontaneous redox reactions occur to generate electricity.
• **Electrolytic cells**: Use electrical energy to drive non-spontaneous chemical reactions. An example is the process of electrolysis used to decompose chemical compounds.

In these cells, the electrode at which oxidation occurs is called the anode, and where reduction happens is called the cathode. A salt bridge or porous barrier maintains the flow of ions and completes the circuit in these cells. Understanding the flow of electrons in these cells is crucial: - Electrons flow from the anode to the cathode in a galvanic cell. - In an electrolytic cell, electrons are pushed from the cathode to the anode. This fundamental understanding is instrumental in building and designing batteries and other electrochemical processes, showing the fascinating interplay between chemical reactions and electrical energy.