While Fe^3+ is stable, Mn^3+ is not stable in acid solution because — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Stability of Aquated Cations in Acidic Solutions For a transition metal cation like $\ce{Fe^3+}$ or $\ce{Mn^3+}$ to be thermodynamically stable in an acidic aqueous solution, it must not spontaneously react with the solvent (water, $\ce{H2O}$) or its ions ($\ce{H^+}$, $\ce{OH^-}$). The most common decomposition pathway for strongly oxidizing metal cations in water is the oxidation of water to oxygen gas ($\ce{O2}$): $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \quad E^\circ_{\text{ox}} = -1.23\text{ V}$$ The corresponding standard reduction half-reaction for the oxygen-water couple is: $$\ce{O2(g) + 4H^+(aq) + 4e^- -> 2H2O(l)} \quad E^\circ_{\ce{O2/H2O}} = +1.23\text{ V}$$ If a metal cation $\ce{M^{z+}}$ has a standard reduction potential ($E^\circ_{\ce{M^{z+}/M^{(z-1)+}}}$) strictly greater than $+1.23\text{ V}$, it is thermodynamically strong enough to oxidize water to $\ce{O2}$ under standard acidic conditions, rendering the $\ce{M^{z+}}$ ion unstable in aqueous solutions. Step 2 - Analyze the Thermodynamic Stability of \ce{Mn^3+} We are given the standard reduction potential for the $\ce{Mn^3+/Mn^2+}$ couple: $$\ce{Mn^3+(aq) + e^- -> Mn^2+(aq)} \quad E^\circ_{\ce{Mn^3+/Mn^2+}} = +1.50\text{ V}$$ To determine if $\ce{Mn^3+}$ can spontaneously oxidize water, we set up a redox cell where $\ce{Mn^3+}$ undergoes reduction at the cathode and water undergoes oxidation at the anode: * **Cathode (Reduction):** $$\ce{4Mn^3+(aq) + 4e^- -> 4Mn^2+(aq)} \quad E^\circ_{\text{cathode}} = +1.50\text{ V}$$ * **Anode (Oxidation):** $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \quad E^\circ_{\text{anode}} = +1.23\text{ V}$$ * **Overall Reaction:** $$\ce{4Mn^3+(aq) + 2H2O(l) -> 4Mn^2+(aq) + O2(g) + 4H^+(aq)}$$ We calculate the standard cell potential ($E^\circ_{\text{cell}}$): $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = 1.50\text{ V} - 1.23\text{ V}$$ $$E^\circ_{\text{cell}} = \boxed{+0.27\text{ V}}$$ Since $E^\circ_{\text{cell}} = +0.27\text{ V} > 0\text{ V}$, the reaction is thermodynamically spontaneous ($\Delta G^\circ < 0$). This means that $\ce{Mn^3+}$ spontaneously oxidizes water to release oxygen gas while reducing itself to stable $\ce{Mn^2+}$. Thus, $\ce{Mn^3+}$ is unstable in an acidic aqueous solution. Step 3 - Analyze the Thermodynamic Stability of \ce{Fe^3+} We are given the standard reduction potential for the $\ce{Fe^3+/Fe^2+}$ couple: $$\ce{Fe^3+(aq) + e^- -> Fe^2+(aq)} \quad E^\circ_{\ce{Fe^3+/Fe^2+}} = +0.77\text{ V}$$ We set up a similar redox cell where $\ce{Fe^3+}$ would oxidize water: * **Cathode (Reduction):** $$\ce{4Fe^3+(aq) + 4e^- -> 4Fe^2+(aq)} \quad E^\circ_{\text{cathode}} = +0.77\text{ V}$$ * **Anode (Oxidation):** $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \quad E^\circ_{\text{anode}} = +1.23\text{ V}$$ * **Overall Reaction:** $$\ce{4Fe^3+(aq) + 2H2O(l) -> 4Fe^2+(aq) + O2(g) + 4H^+(aq)}$$ We calculate the standard cell potential ($E^\circ_{\text{cell}}$): $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = 0.77\text{ V} - 1.23\text{ V}$$ $$E^\circ_{\text{cell}} = \boxed{-0.46\text{ V}}$$ Since $E^\circ_{\text{cell}} = -0.46\text{ V} < 0\text{ V}$, this reaction is thermodynamically non-spontaneous ($\Delta G^\circ > 0$). Therefore, $\ce{Fe^3+}$ cannot oxidize water, which explains why $\ce{Fe^3+}$ remains stable in acidic aqueous solutions. Step 4 - Evaluate the Options and Explain the Distractors * **Option (A) is incorrect:** Oxygen gas oxidizes $\ce{Mn^2+}$ to $\ce{Mn^3+}$. For this reaction to occur spontaneously, the standard cell potential would have to be positive. However, since $E^\circ_{\ce{O2/H2O}} = +1.23\text{ V}$ is less than $E^\circ_{\ce{Mn^3+/Mn^2+}} = +1.50\text{ V}$, the oxidation of $\ce{Mn^2+}$ by oxygen is non-spontaneous ($E^\circ_{\text{cell}} = 1.23\text{ V} - 1.50\text{ V} = -0.27\text{ V}$). * **Option (B) is incorrect:** While oxygen can spontaneously oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ ($E^\circ_{\text{cell}} = 1.23\text{ V} - 0.77\text{ V} = +0.46\text{ V} > 0$), it cannot oxidize $\ce{Mn^2+}$ to $\ce{Mn^3+}$ due to the high potential of $\ce{Mn^3+}$. * **Option (C) is incorrect:** As calculated in Step 3, the oxidation of water by $\ce{Fe^3+}$ is thermodynamically non-spontaneous ($E^\circ_{\text{cell}} = -0.46\text{ V}$). * **Option (D) is correct:** As calculated in Step 2, the reduction potential of the $\ce{Mn^3+/Mn^2+}$ couple ($1.50\text{ V}$) is greater than that of the oxygen-water couple ($1.23\text{ V}$), which means $\ce{Mn^3+}$ spontaneously oxidizes water to oxygen. $$\text{Correct Option: } \boxed{D}$$