Problem 162 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(E_{\mathrm{cu}^{2+}}^{0}=0.34 \mathrm{~V}\) and \(E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=0.8 \mathrm{~V}\), predict whether the reaction given below is feasible or not ? \(\mathrm{Cu}^{2}+(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) \longrightarrow \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq})\) (a) Not feasible (b) Feasible (c) Feasible on increasing the conc. of \(\mathrm{Ag}^{+}(\mathrm{aq})\). (d) easible at high temp.

Step-by-Step Solution

Answer

(a) Not feasible.

Step 1: Identify the Reaction

The reaction given is \( \mathrm{Cu}^{2+} + 2\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Cu}(\mathrm{s}) + 2\mathrm{Ag}^{+}(\mathrm{aq}) \). Our goal is to determine if this reaction is feasible under standard conditions by calculating the cell potential.

Step 2: Calculate Standard Electrode Potentials

We use the standard reduction potentials for the half-reactions: \( E_{\mathrm{Ag}^{+}/\mathrm{Ag}}^0 = 0.8 \mathrm{~V} \) and \( E_{\mathrm{Cu}^{2+}/\mathrm{Cu}}^0 = 0.34 \mathrm{~V} \). Identify the actual changes in the reaction to calculate the total cell potential.

Step 3: Determine Oxidation and Reduction

In the reaction, \( \mathrm{Ag} \) is oxidized to \( \mathrm{Ag}^+ \) with an oxidation potential of \(-0.8 \mathrm{~V} \) (negative because oxidation is the reverse of reduction), and \( \mathrm{Cu}^{2+} \) is reduced to \( \mathrm{Cu} \) with a potential of \( +0.34 \mathrm{~V} \).

Step 4: Calculate the Cell Potential

The standard cell potential \( E^0_{\text{cell}} \) is found by adding the reduction potential of \( \mathrm{Cu}^{2+}/\mathrm{Cu} \) and the oxidation potential of \( \mathrm{Ag}^{+}/\mathrm{Ag} \). Thus, \( E^0_{\text{cell}} = 0.34 \mathrm{~V} + (-0.8 \mathrm{~V}) = -0.46 \mathrm{~V} \).

Step 5: Evaluate Feasibility

Since the cell potential \( E^0_{\text{cell}} = -0.46 \mathrm{~V} \) is negative, the reaction is not feasible under standard conditions because it is not spontaneous.

Key Concepts

Standard Electrode Potentials

In electrochemistry, understanding standard electrode potentials is fundamental. Electrode potentials help in predicting the direction of redox reactions and in evaluating the feasibility of these reactions.
Standard electrode potential, denoted as \(E^0\), is the measure of the individual potential of a reversible electrode at standard state, which includes solutes at a 1 M concentration, gases at 1 atm pressure, and the system at a specified temperature, usually 25°C (298 K).
To understand standard electrode potentials, it's important to look at half-reactions involving redox processes:

• Reduction reactions, where a species gains electrons.
• Oxidation reactions, where a species loses electrons.

The electrode potential can indicate whether a substance is likely to gain or lose electrons. Higher positive values indicate a strong tendency to gain electrons, making a good oxidizing agent. For example, \(E_{ ext{Ag}^+/ ext{Ag}}^0 = 0.8 \text{ V}\) indicates that \( ext{Ag}^+\) has a strong tendency to gain electrons to form \( ext{Ag}\).
Comparing these potentials allows scientists to predict how readily certain redox reactions will occur under standard conditions.

Cell Potential Calculation

Calculating cell potential is an important step to assess whether a particular redox reaction can occur spontaneously. The cell potential, \(E^0_{\text{cell}}\), refers to the potential difference between the cathode and the anode.
For calculating the overall cell potential, follow these steps:

• Identify the two half-reactions that make up the electrochemical cell.
• From the standard electrode potentials, determine which species is reduced and which is oxidized.
• Note the standard reduction potentials of each: the species with the higher positive potential tends to undergo reduction.
• To get the cell potential, combine the standard reduction potential of the cathodic half-reaction and the oxidation potential of the anodic half-reaction (the latter is the negative of its reduction potential).

In the given exercise, the potential calculation was \(E^0_{\text{cell}} = 0.34 \text{ V} + (-0.8 \text{ V}) = -0.46 \text{ V}\). A negative \(E^0_{\text{cell}}\) value shows that the reaction is non-spontaneous under standard conditions.

Feasibility of Redox Reactions

The feasibility of redox reactions heavily depends on their cell potential. A redox reaction is only spontaneous and considered feasible when the cell potential \(E^0_{\text{cell}}\) is positive. Positive values imply that energy can be gained from the reaction, driving it forward without external input.
In contrast, a negative cell potential suggests that the reaction requires energy input to proceed, suggesting it isn’t feasible under standard conditions.
It's important to remember:

• Spontaneity can sometimes be achieved by altering conditions, such as concentration changes (e.g., through the Nernst equation) or temperature shifts.
• Even if a reaction isn’t spontaneous, adjusting external factors or coupling with other reactions can make the desired process possible.

In the discussed case, since \(E^0_{\text{cell}} = -0.46 \text{ V}\), the reaction of \(\text{Cu}^{2+} + 2\text{Ag}(\text{s}) \to \text{Cu}(\text{s}) + 2\text{Ag}^{+}(\text{aq})\) is not feasible as is. Hence, it reaffirms that careful analysis of the cell potential provides crucial insight into redox reaction feasibility.