Question

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

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

Hint:

To solve electrochemical feasibility questions under ten seconds, arrange the given systems vertically by their reduction potentials from highest to lowest. A species on the left side (oxidized form) of a higher system can only react spontaneously with a species on the right side (reduced form) of a system located below it on your list! Because Ag (\(0.80\text{V}\)) sits higher up than \(\text{Fe}^{3+}\) (\(0.77\text{V}\)), they cannot react.

Answer 1

The Correct Option is D

Concept: For any chemical redox reaction to be thermodynamically spontaneous or feasible, the overall standard cell potential (\(\text{E}^\circ_{\text{cell}}\)) calculated from its individual half-reactions must be strictly positive (\(\text{E}^\circ_{\text{cell}} > 0\)). This directly relates to a negative change in Gibbs free energy (\(\Delta G^\circ = -nF\text{E}^\circ_{\text{cell}}\)). The standard formula to evaluate cell potential uses the standard reduction potentials (SRP) of the participating electrodes: \[ \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{cathode (Reduction)}} - \text{E}^\circ_{\text{anode (Oxidation)}} \] An alternative conceptual rule to remember is: A species with a higher reduction potential will easily oxidize a species with a lower reduction potential.

Step 1: Analyzing the chemical behavior of Option D ($\text{Ag}$ and $\text{Fe}^{3+}$).
Let's construct the hypothetical redox system for this pair:
• \(\text{Fe}^{3+}\) is in its highest oxidation state, so it can only undergo reduction: \[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad \left(\text{E}^\circ_{\text{red}} = +0.77\text{ V}\right) \]
• \(\text{Ag}\) is in its reduced metallic form, so it can only undergo oxidation: \[ \text{Ag} \rightarrow \text{Ag}^+ + e^- \quad \left(\text{E}^\circ_{\text{red}} = +0.80\text{ V}\right) \] Here, the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) system acts as the cathode (reduction) and the \(\text{Ag}^+/\text{Ag}\) system acts as the anode (oxidation).

Step 2: Calculating the cell potential ($\text{E}^\circ_{\text{cell}}$) for Option D.
Using the standard formula: \[ \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} - \text{E}^\circ_{\text{Ag}^+/\text{Ag}} \] \[ \text{E}^\circ_{\text{cell}} = (+0.77\text{ V}) - (+0.80\text{ V}) = -0.03\text{ V} \] Since \(\text{E}^\circ_{\text{cell}}\) evaluates to a negative value (\(-0.03\text{ V}\)), the forward reaction is entirely non-spontaneous. Therefore, this specific interaction pair is not feasible.

Step 3: Verifying why other options are feasible (Optional confirmation).
Let's quickly check the remaining pairs using our core rule:
• Option A (\(\text{Fe}^{3+}\) and \(\text{I}^-\)): \(\text{E}^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} (0.77\text{ V}) > \text{E}^\circ_{\text{I}_2/\text{I}^-} (0.54\text{ V})\). Fe\({}^{3+}\) successfully reduces while oxidizing \(\text{I}^-\). (\(\text{E}^\circ_{\text{cell}} = +0.23\text{ V} \rightarrow\) Feasible)
• Option B (\(\text{Ag}^+\) and \(\text{Cu}\)): \(\text{E}^\circ_{\text{Ag}^+/\text{Ag}} (0.80\text{ V}) > \text{E}^\circ_{\text{Cu}^{2+}/\text{Cu}} (0.34\text{ V})\). Ag\({}^+\) successfully reduces while oxidizing \(\text{Cu}\). (\(\text{E}^\circ_{\text{cell}} = +0.46\text{ V} \rightarrow\) Feasible)
• Option C (\(\text{Fe}^{3+}\) and \(\text{Cu}\)): \(\text{E}^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} (0.77\text{ V}) > \text{E}^\circ_{\text{Cu}^{2+}/\text{Cu}} (0.34\text{ V})\). Fe\({}^{3+}\) successfully reduces while oxidizing \(\text{Cu}\). (\(\text{E}^\circ_{\text{cell}} = +0.43\text{ V} \rightarrow\) Feasible)