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

Question

The electrochemical series is the arrangement of various electrode systems in the increasing order of their standard reduction potentials. It has several important features. On moving from the top to the bottom in the series, tendency to gain electrons, i.e., to get reduced increases. The electrode systems having negative values of standard reduction potentials act as anode when connected to a standard hydrogen electrode, while those having positive values act as cathode. If \(E_{\mathrm{cu}}^{0}{\underline{\phantom{xx}}}_{\mathrm{c}}^{2+}=0.34 \mathrm{~V}\) 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 \(\mathrm{Ag}^{+}\)(aq). (d) easible at high temp.

Step-by-Step Solution

Answer

(a) Not feasible

Step 1: Understand the Reaction

The given reaction is \( \mathrm{Cu}^{2+}(\mathrm{aq}) + 2 \mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Cu}(\mathrm{s}) + 2 \mathrm{Ag}^{+}(\mathrm{aq}) \). This reaction can be split into two half-reactions: one for silver and one for copper.

Step 2: Identify Half-Reactions

The oxidation half-reaction can be written as \( 2\mathrm{Ag}(\mathrm{s}) \rightarrow 2\mathrm{Ag}^{+}(\mathrm{aq}) + 2e^- \). The reduction half-reaction is \( \mathrm{Cu}^{2+}(\mathrm{aq}) + 2e^- \rightarrow \mathrm{Cu}(\mathrm{s}) \).

Step 3: Determine Standard Reduction Potentials

Refer to the electrochemical series to determine the standard reduction potentials: \( E^0_{\mathrm{Ag}^+ / \mathrm{Ag}} = 0.8 \; \text{V} \) for silver and \( E^0_{\mathrm{Cu}^{2+} / \mathrm{Cu}} = 0.34 \; \text{V} \) for copper.

Step 4: Calculate Cell Potential (Ecell)

The cell potential is calculated using the formula: \[ E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}.\] Here, the cathode half-reaction (reduction) is copper, and the anode half-reaction (oxidation) is silver. Thus, \[ E^0_{\text{cell}} = 0.34 \; \text{V} - 0.8 \; \text{V} = -0.46 \; \text{V}.\]

Step 5: Analyze Feasibility Based on Ecell

A reaction is deemed feasible if \( E^0_{\text{cell}} > 0 \). The calculated cell potential is \( -0.46 \; \text{V} \); hence, the reaction is not feasible under standard conditions.

Key Concepts

Standard Reduction Potential

The concept of Standard Reduction Potential is pivotal in predicting the direction of redox reactions in electrochemistry. It measures the tendency of a chemical species to gain electrons and, thus, get reduced. The standard reduction potential is denoted by the symbol \( E^0 \) and is measured under standard conditions which include a concentration of 1 M, a pressure of 1 atm, and a temperature of 25°C (298 K). The electrochemical series lists elements based on their standard reduction potentials. Elements at the top, such as fluorine, have a high positive reduction potential meaning they are strong oxidizing agents, ready to accept electrons. Elements lower down, like lithium, possess low or even negative reduction potentials, indicating their preference to donate electrons.In our context, copper and silver were analyzed. Copper’s standard reduction potential is \( 0.34 \; \text{V} \), while silver’s is \( 0.8 \; \text{V} \). This means that silver has a higher tendency to remain reduced compared to copper, a crucial factor when assessing the feasibility of reactions.

Half-Reaction Method

When predicting the feasibility of redox reactions, the Half-Reaction Method is extremely useful. It breaks down a complex reaction into simpler, manageable parts, specifically oxidation and reduction reactions.To implement the Half-Reaction Method:

• Identify the two halves of the redox process: oxidation (loss of electrons) and reduction (gain of electrons).
• In the given reaction, silver undergoes oxidation \( 2 \text{Ag(s)} \rightarrow 2 \text{Ag}^+ \text{(aq)} + 2e^- \) and copper is reduced \( \text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)} \).
• Align the half-reactions to deduce net energy changes and determine which direction the reaction naturally progresses under standard conditions.

This approach helps in calculating the net Ecell, which further tells us if the overall reaction is spontaneous.

Cell Potential Calculation

Calculating the cell potential, \( E^0_{\text{cell}} \), is the final step in assessing a cell's capability to drive a reaction. The formula used is:\[E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}\]This tells us the maximum voltage we can obtain from the half-reactions.In the given example, copper acts as the cathode with the potential \( 0.34 \; \text{V} \) and silver the anode with \( 0.8 \; \text{V} \). On substituting into our equation, we find:\[E^0_{\text{cell}} = 0.34 \; \text{V} - 0.8 \; \text{V} = -0.46 \; \text{V}\]Since \( E^0_{\text{cell}} \) is negative, the process is non-spontaneous under standard conditions and will not occur without external intervention. A positive cell potential indicates a spontaneous reaction, driving the electrons naturally in the direction proposed by the reaction equation.