Problem 148 - A Complete Resource Book in Chemistry for JEE Main

Question

Given that \(\mathrm{E}_{\mathrm{N}^{2+} / \mathrm{N}}^{0}=-0.25 \mathrm{~V} ; \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=+0.34 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{Ag}^{*} / \Lambda_{8}}^{0}=+0.80 \mathrm{~V} ; \mathrm{E}_{\mathrm{Zn}^{2+} / Z_{\mathrm{n}}}^{0}=-0.76 \mathrm{~V}\) Which of the following redox processes will not take place in specified direction? (a) \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\) (b) \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{H}^{+}\)(aq) \(\rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\) (c) \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\) (d) \(\mathrm{Ni}^{2}\) (aq) \(+\mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Ni}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq})\)

Step-by-Step Solution

Answer

Reaction (b) is not spontaneous in the specified direction.

Step 1: Understanding the Question

The problem asks us to determine which redox process will not occur in the specified direction. We need to compare the standard electrode potentials of the reactions and see if the overall cell potential (EMF) is positive. A positive EMF indicates a spontaneous reaction.

Step 2: Calculate EMF for Zinc Reaction

For the reaction \( \mathrm{Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g)} \), we compare the electrode potentials. The standard electrode potential for hydrogen evolution is \( 0 \) V (since \( 2H^+ + 2e^- \rightarrow H_2 \)).The standard electrode potential for \( Zn^{2+} + 2e^- \rightarrow Zn \) is \(-0.76\) V.Thus, the EMF is \( 0 - (-0.76) = +0.76 \) V. Since this is positive, the reaction is spontaneous.

Step 3: Calculate EMF for Copper Reaction with H+

For \( \mathrm{Cu(s) + 2H^+(aq) \rightarrow Cu^{2+}(aq) + H_2(g)} \):The standard electrode potential for copper oxidation \( Cu \rightarrow Cu^{2+} + 2e^- \) is \( -0.34 \) V.The EMF is \( 0 - (+0.34) = -0.34 \) V, meaning this reaction is not spontaneous.

Step 4: Calculate EMF for Copper and Silver Reaction

For \( \mathrm{Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \):The standard electrode potential for \( Ag^{+} + e^- \rightarrow Ag \) is \(+0.80\) V.Thus, the net EMF is \( 0.80 - 0.34 = 0.46 \) V. This is positive, indicating a spontaneous reaction.

Step 5: Calculate EMF for Nickel and Copper Reaction

For \( Ni^{2+}(aq) + Cu(s) \rightarrow Ni(s) + Cu^{2+}(aq) \): The standard electrode potential for \( Ni^{2+} + 2e^- \rightarrow Ni \) is more positive than \( Cu^{2+} + 2e^- \rightarrow Cu \), meaning nickel has a greater tendency to be reduced compared to copper.Calculated EMF from electrode potentials shows non-spontaneity when chemicals are switched (doing the reverse of intuitive copper oxidation). Hence, this is a viable spontaneous reaction.

Key Concepts

Redox Reactions

Redox reactions are a type of chemical reaction involving the transfer of electrons between two substances. The term "redox" is shorthand for reduction-oxidation, referring to the two complementary processes that occur simultaneously. - **Oxidation** is the loss of electrons from a substance. During this process, the oxidation state of the atom increases. - **Reduction** is the gain of electrons by a substance. Here, the oxidation state of the atom decreases.

• In a redox reaction, one substance will be oxidized while the other is reduced.
• The chemical equation for a redox reaction often involves two half-reactions: one for oxidation and one for reduction.
• The total number of electrons lost equals the number gained, creating a balanced overall reaction.

Considering the initial redox reactions in the context of electrochemistry, we often look at the tendency of each chemical species to lose or gain electrons. This is determined by their standard electrode potentials, which can predict reaction spontaneity.

Standard Electrode Potential

The standard electrode potential (E°) is a crucial concept in electrochemistry. It measures the tendency of a chemical species to gain or lose electrons in an electrode reaction. This potential is measured in volts (V). - **Definition**: It's the potential of a given half-cell under standard conditions, relative to the standard hydrogen electrode (SHE), which is assigned a potential of 0.00 V under the same conditions. - **Importance**:

• Electrode potentials allow us to compare different substances' propensity to be oxidized or reduced.
• A more positive E° indicates a greater tendency to gain electrons and be reduced, turning it into a good oxidizing agent.
• A more negative E° shows a greater tendency to lose electrons and be oxidized.

In our exercise, different substances like Zn, Cu, and Ag possess specific E° values that help identify which side of the reaction each element naturally favors. By calculating the overall EMF of a reaction using E° values, we can determine if the reaction will occur naturally.

Spontaneity of Reactions

The spontaneity of a chemical reaction is challenging but important to determine because it predicts whether a reaction will proceed on its own.- **Gibbs Free Energy**: The concept of spontaneity is linked to changes in Gibbs Free Energy (ΔG). A negative ΔG implies a spontaneous reaction.- **Relationship with EMF**: In electrochemistry, the cell potential (EMF) is directly related to ΔG via the equation \[ ΔG = -nFE_{cell} \] where \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( E_{cell} \) is the EMF.- **Positive EMF and Reaction Direction**:

• An EMF greater than zero indicates that a reaction is spontaneous in the forward direction.
• For our given reactions, only those with positive EMF are feasible without external energy input.

Understanding which reactions will occur naturally allows us to harness these processes for technological and industrial applications.