A gas 'X' at 1 atm is bubbled through a solution containing a mixture of 1 M - Y^- and 1 M - Z^- at — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Conceptual Understanding of Reduction Potential and Oxidizing/Reducing Power Standard reduction potential ($E^\circ$) is a thermodynamic measure of the tendency of a chemical species to acquire electrons and thereby undergo reduction: $$\text{Oxidized Form} + n e^{-} \rightarrow \text{Reduced Form}$$ * A higher, more positive standard reduction potential ($E^\circ_{\text{red}}$) indicates a stronger thermodynamic tendency to undergo reduction. Consequently, the oxidized form of that species acts as a stronger oxidizing agent. * A lower standard reduction potential indicates a weaker tendency to undergo reduction, which means its corresponding reduced form acts as a stronger reducing agent. Step 2 - Analyze the Relative Oxidizing Power We are given the standard reduction potentials order of the species at $25^\circ\text{C}$: $$E^\circ(Z \mid Z^-) > E^\circ(Y \mid Y^-) > E^\circ(X \mid X^-)$$ Since the oxidizing power of the oxidized form is directly proportional to its standard reduction potential: $$\text{Oxidizing Power: } \ce{Z2} > \ce{Y2} > \ce{X2}$$ Similarly, the reducing power of the reduced forms is in reverse order: $$\text{Reducing Power: } \ce{X^-} > \ce{Y^-} > \ce{Z^-}$$ Step 3 - Evaluate the Spontaneity of Oxidation of $\ce{X}$ by $\ce{Y2}$ $$\text{Cathode: } \ce{Y2 + 2e^- -> 2Y^-} \quad E^\circ_{\text{cathode}} = E^\circ(Y \mid Y^-)$$ $$\text{Anode: } \ce{X -> X2 + 2e^-} \quad E^\circ_{\text{anode}} = E^\circ(X \mid X^-)$$ $$E^\circ_{\text{cell}} = E^\circ(Y \mid Y^-) - E^\circ(X \mid X^-) > 0 \quad \text{(spontaneous)}$$ Since $E^\circ_{\text{cell}} > 0$, $\Delta G^\circ < 0$, so **$\ce{Y}$ will spontaneously oxidize $\ce{X}$**. Step 4 - Evaluate the Spontaneity of Oxidation of $\ce{Z^-}$ by $\ce{Y2}$ $$\text{Cathode: } \ce{Y2 + 2e^- -> 2Y^-} \quad \text{Anode: } \ce{2Z^- -> Z2 + 2e^-}$$ $$E^\circ_{\text{cell}} = E^\circ(Y \mid Y^-) - E^\circ(Z \mid Z^-) < 0 \quad \text{(non-spontaneous)}$$ Since $E^\circ_{\text{cell}} < 0$, $\Delta G^\circ > 0$, so **$\ce{Y}$ cannot oxidize $\ce{Z}$**. Step 5 - Analyze the Options * **Option (A)** states "Y will oxidize X and not Z" — correct, consistent with Steps 3 and 4. * **Option (B)** states "Y will oxidize Z and not X" — incorrect, exactly opposite to thermodynamic spontaneity. * **Option (C)** states "Y will oxidize both X and Z" — incorrect, Y cannot oxidize $\ce{Z^-}$. * **Option (D)** states "Y will reduce both X and Z" — incorrect, $\ce{Y2}$ is an oxidizing agent, not a reducing agent. Thus, the correct choice is $\boxed{\text{A}}$.