Step 1 : Understanding the Question:
This problem asks us to determine which of the given pairs of chemical species cannot undergo a spontaneous redox reaction under standard conditions. A reaction is thermodynamically feasible if its standard cell potential is positive. We are given the standard reduction potentials for several half-cells, and we need to evaluate the cell potential for the proposed reactions to find the one with a negative cell potential.
Step 2 : Key Formulas and Approach:
For any redox reaction, the standard cell potential ($E^\circ_{\text{cell}}$) is calculated using the standard reduction potentials (SRP) of the cathode (where reduction occurs) and the anode (where oxidation occurs):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode (reduction)}} - E^\circ_{\text{anode (oxidation)}} \]
A reaction is spontaneous and thermodynamically feasible if:
\[ E^\circ_{\text{cell}}>0 \]
If $E^\circ_{\text{cell}}<0$, the reaction is non-spontaneous and not feasible.
Alternatively, a species with a higher reduction potential will spontaneously oxidize a species with a lower reduction potential. We will test the feasibility of the options using this principle.
Step 3 : Detailed Explanation:
We begin by analyzing the reaction between $\text{Ag}$ and $\text{Fe}^{3+}$ proposed in Option (D). In this potential reaction, $\text{Fe}^{3+}$ is in its oxidized state and must be reduced to $\text{Fe}^{2+}$, while $\text{Ag}$ is in its reduced state and must be oxidized to $\text{Ag}^+$.
The reduction half-reaction is: $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ with $E^\circ_{\text{reduction}} = +0.77\text{ V}$.
The oxidation half-reaction is: $\text{Ag} \rightarrow \text{Ag}^+ + e^-$ with $E^\circ_{\text{reduction}} = +0.80\text{ V}$.
We designate the iron half-cell as the cathode and the silver half-cell as the anode to calculate the standard cell potential: $E^\circ_{\text{cell}} = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E^\circ_{\text{Ag}^+/\text{Ag}} = +0.77\text{ V} - (+0.80\text{ V}) = -0.03\text{ V}$.
Since the calculated cell potential is negative ($E^\circ_{\text{cell}} = -0.03\text{ V}$), this reaction is non-spontaneous under standard conditions and is not feasible.
For comparison, let us verify Option (A) involving $\text{Fe}^{3+}$ and $\text{I}^-$. Here, iron is reduced ($+0.77\text{ V}$) and iodide is oxidized ($+0.54\text{ V}$): $E^\circ_{\text{cell}} = 0.77\text{ V} - 0.54\text{ V} = +0.23\text{ V}$, which is positive and feasible.
For Option (B) involving $\text{Ag}^+$ and $\text{Cu}$, silver is reduced ($+0.80\text{ V}$) and copper is oxidized ($+0.34\text{ V}$): $E^\circ_{\text{cell}} = 0.80\text{ V} - 0.34\text{ V} = +0.46\text{ V}$, which is positive and feasible.
For Option (C) involving $\text{Fe}^{3+}$ and $\text{Cu}$, iron is reduced ($+0.77\text{ V}$) and copper is oxidized ($+0.34\text{ V}$): $E^\circ_{\text{cell}} = 0.77\text{ V} - 0.34\text{ V} = +0.43\text{ V}$, which is positive and feasible.
Step 4 : Final Answer:
The pair between which the redox reaction is not feasible is $\text{Ag}$ and $\text{Fe}^{3+}$, which corresponds to Option (D).