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

Concept: A redox reaction is spontaneous and feasible only if the standard Gibbs free energy change \( \Delta G^{\ominus} \) is negative. This corresponds to a positive standard cell potential \( E^{\ominus}_{\text{cell}} \). \[ E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{cathode}} - E^{\ominus}_{\text{anode}} \] For a reaction to occur, the species with the higher standard reduction potential must act as the cathode (undergoing reduction), while the species with the lower reduction potential must act as the anode (undergoing oxidation).

Step 1: Analyze the fundamental condition for non-feasibility. A reaction is not feasible if the calculated \( E^{\ominus}_{\text{cell}} \) is negative. This occurs when an attempt is made to force the species with a higher reduction potential to undergo oxidation. If \( E^{\ominus}_{\text{cathode}} < E^{\ominus}_{\text{anode}} \), then \( E^{\ominus}_{\text{cell}} < 0 \), rendering the reaction non-spontaneous.

Step 2: Evaluate each option based on the provided \( E^{\ominus} \) values.
• Option (1): \( 2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2 \)
\( E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E^{\ominus}_{\text{I}_2/\text{I}^-} = 0.77 - 0.54 = +0.23\text{V} \). Since it is positive, the reaction is feasible.
• Option (2): \( 2\text{Ag}^+ + \text{Cu} \rightarrow 2\text{Ag} + \text{Cu}^{2+} \)
\( E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{Ag}^+/\text{Ag}} - E^{\ominus}_{\text{Cu}^{2+}/\text{Cu}} = 0.80 - 0.34 = +0.46\text{V} \). Since it is positive, the reaction is feasible.
• Option (3): \( 2\text{Fe}^{3+} + \text{Cu} \rightarrow 2\text{Fe}^{2+} + \text{Cu}^{2+} \)
\( E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E^{\ominus}_{\text{Cu}^{2+}/\text{Cu}} = 0.77 - 0.34 = +0.43\text{V} \). Since it is positive, the reaction is feasible.
• Option (4): \( \text{Ag} + \text{Fe}^{3+} \rightarrow \text{Ag}^+ + \text{Fe}^{2+} \)
Here, Ag is intended to act as the anode (oxidation, \( E^{\ominus} = 0.80\text{V} \)) and \( \text{Fe}^{3+} \) as the cathode (reduction, \( E^{\ominus} = 0.77\text{V} \)).
\( E^{\ominus}_{\text{cell}} = 0.77 - 0.80 = -0.03\text{V} \). Because the potential is negative, the reaction is not feasible.