Question

Using the standard electrode potential, find out the pair between which redox reaction is not feasible. \( E^{\ominus} \) values: \( \text{Fe}^{3+}/\text{Fe}^{2+} = +0.77\text{V} \); \( \text{I}_2/\text{I}^- = +0.54\text{V} \); \( \text{Cu}^{2+}/\text{Cu} = +0.34\text{V} \); \( \text{Ag}^+/\text{Ag} = +0.80\text{V} \).

• A. \( \text{Fe}^{3+} \) and \( \text{I}^- \)
• B. \( \text{Ag}^+ \) and \( \text{Cu} \)
• C. \( \text{Fe}^{3+} \) and \( \text{Cu} \)
• D. \( \text{Ag} \) and \( \text{Fe}^{3+} \)

Hint:

A simple rule of thumb: A metal with a higher (more positive) reduction potential cannot be oxidized by a species with a lower (less positive) reduction potential. Always check if the oxidizing agent has a higher reduction potential than the reducing agent.

Answer 1

The Correct Option is D

Step 1 : Understanding the Question:
This problem concerns electrochemistry and the spontaneity of redox reactions. We need to analyze which combination of the given species cannot undergo a feasible redox reaction based on their standard reduction potentials (\( E^{\ominus} \)).
Step 2 : Key Formulas and Approach:
For any redox reaction to be spontaneous and feasible, the standard cell potential \( E^{\ominus}_{\text{cell}} \) must be positive. This relates directly to the change in standard Gibbs free energy:
\[ \Delta G^{\ominus} = -n F E^{\ominus}_{\text{cell}} \]
For a positive \( E^{\ominus}_{\text{cell}} \) (and a spontaneous reaction), the species with the higher standard reduction potential must undergo reduction at the cathode, while the species with the lower reduction potential must undergo oxidation at the anode:
\[ E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{reduction}} - E^{\ominus}_{\text{oxidation}} = E^{\ominus}_{\text{cathode}} - E^{\ominus}_{\text{anode}}>0 \]
Step 3 : Detailed Explanation:

Let us analyze the feasibility of Option (A), involving the reaction between \( \text{Fe}^{3+} \) and \( \text{I}^- \). Here, \( \text{Fe}^{3+} \) is reduced to \( \text{Fe}^{2+} \) (\( E^{\ominus} = 0.77\text{V} \)) and \( \text{I}^- \) is oxidized to \( \text{I}_2 \) (\( E^{\ominus} = 0.54\text{V} \)). The cell potential is \( E^{\ominus}_{\text{cell}} = 0.77 - 0.54 = +0.23\text{V} \). Since this value is positive, the reaction is feasible.

Let us analyze the feasibility of Option (B), involving the reaction between \( \text{Ag}^+ \) and \( \text{Cu} \). Here, \( \text{Ag}^+ \) is reduced to \( \text{Ag} \) (\( E^{\ominus} = 0.80\text{V} \)) and \( \text{Cu} \) is oxidized to \( \text{Cu}^{2+} \) (\( E^{\ominus} = 0.34\text{V} \)). The cell potential is \( E^{\ominus}_{\text{cell}} = 0.80 - 0.34 = +0.46\text{V} \). Since this value is positive, the reaction is feasible.

Let us analyze the feasibility of Option (C), involving the reaction between \( \text{Fe}^{3+} \) and \( \text{Cu} \). Here, \( \text{Fe}^{3+} \) is reduced to \( \text{Fe}^{2+} \) (\( E^{\ominus} = 0.77\text{V} \)) and \( \text{Cu} \) is oxidized to \( \text{Cu}^{2+} \) (\( E^{\ominus} = 0.34\text{V} \)). The cell potential is \( E^{\ominus}_{\text{cell}} = 0.77 - 0.34 = +0.43\text{V} \). Since this value is positive, the reaction is feasible.

Let us analyze the feasibility of Option (D), involving the reaction between \( \text{Ag} \) and \( \text{Fe}^{3+} \). In this scenario, \( \text{Fe}^{3+} \) is intended to act as the cathode (reduction, \( E^{\ominus} = 0.77\text{V} \)) and \( \text{Ag} \) is intended to act as the anode (oxidation, \( E^{\ominus} = 0.80\text{V} \)). The standard cell potential is \( E^{\ominus}_{\text{cell}} = 0.77 - 0.80 = -0.03\text{V} \). Since the cell potential is negative, this reaction is not feasible.

Step 4 : Final Answer:
The redox reaction between \( \text{Ag} \) and \( \text{Fe}^{3+} \) is not feasible, which corresponds to option (D).