Standard electrode potential data are useful for understanding the suitability of an oxidant in a re — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Principle of Redox Titration of $\ce{Fe^{2+}}$ Ions The quantitative estimation of aqueous iron(II) nitrate, $\ce{Fe(NO3)2}$, is carried out through redox titration, where the analyte $\ce{Fe^2+}$ is oxidized to $\ce{Fe^3+}$. The oxidation half-reaction is represented as: $$\ce{Fe^2+(aq) -> Fe^3+(aq) + e^-} \quad E^\circ_{\text{ox}} = -E^\circ(\ce{Fe^3+/Fe^2+}) = -0.77\text{ V}$$ To successfully titrate $\ce{Fe^2+}$, we require a strong oxidizing agent (oxidant) whose reduction potential is significantly greater than $+0.77\text{ V}$ so that the overall cell potential is positive (spontaneous reaction). Two common powerful oxidants used in volumetric analysis are potassium permanganate ($\ce{KMnO4}$) and potassium dichromate ($\ce{K2Cr2O7}$). Step 2 - Role of the Acidic Medium and Potential Interfering Reactions To undergo complete reduction to their respective lower oxidation states, both permanganate ($\ce{MnO4^-}$) and dichromate ($\ce{Cr2O7^2-}$) require a strong acidic medium ($\ce{H^+}$ ions): $$\ce{MnO4^-(aq) + 8H^+(aq) + 5e^- -> Mn^2+(aq) + 4H2O(l)} \quad E^\circ = 1.51\text{ V}$$ $$\ce{Cr2O7^2-(aq) + 14H^+(aq) + 6e^- -> 2Cr^3+(aq) + 7H2O(l)} \quad E^\circ = 1.38\text{ V}$$ However, the conjugate base of the acid selected to provide the acidic medium must not interfere with the titration. Specifically, the oxidant must not be strong enough to oxidize the anion of the acid, as this side reaction would consume the titrant, leading to a significant positive error in the calculated iron concentration. Step 3 - Analyzing the Feasibility of Titration in Hydrochloric Acid ($\ce{HCl}$) Hydrochloric acid ($\ce{HCl}$) dissociates in water to release chloride ions ($\ce{Cl^-}$), which can potentially undergo oxidation to chlorine gas ($\ce{Cl2}$): $$\ce{2Cl^-(aq) -> Cl2(g) + 2e^-} \quad E^\circ_{\text{ox}} = -1.40\text{ V}$$ Let us analyze the feasibility of using our two oxidants in an $\ce{HCl}$ medium: 1. **Using $\ce{MnO4^-}$:** The standard cell potential ($E^\circ_{\text{cell}}$) for the oxidation of $\ce{Cl^-}$ by $\ce{MnO4^-}$ is: $$E^\circ_{\text{cell}} = E^\circ(\ce{MnO4^-/Mn^2+}) - E^\circ(\ce{Cl2/Cl^-})$$ $$E^\circ_{\text{cell}} = 1.51\text{ V} - 1.40\text{ V} = +0.11\text{ V}$$ Since $E^\circ_{\text{cell}} > 0\text{ V}$, the reaction is thermodynamically spontaneous ($\Delta G^\circ < 0$). Permanganate will spontaneously oxidize the chloride ions of the acid alongside the $\ce{Fe^2+}$ analyte, interfering with the quantitative estimation. Thus, **$\ce{MnO4^-}$ cannot be used in aqueous $\ce{HCl}$**. 2. **Using $\ce{Cr2O7^2-}$:** The standard cell potential ($E^\circ_{\text{cell}}$) for the oxidation of $\ce{Cl^-}$ by $\ce{Cr2O7^2-}$ is: $$E^\circ_{\text{cell}} = E^\circ(\ce{Cr2O7^2-/Cr^3+}) - E^\circ(\ce{Cl2/Cl^-})$$ $$E^\circ_{\text{cell}} = 1.38\text{ V} - 1.40\text{ V} = -0.02\text{ V}$$ Since $E^\circ_{\text{cell}} < 0\text{ V}$, the reaction is non-spontaneous. Under standard conditions, dichromate is not strong enough to oxidize chloride ions. It will selectively oxidize only the $\ce{Fe^2+}$ ions ($E^\circ_{\text{cell}} = 1.38\text{ V} - 0.77\text{ V} = +0.61\text{ V}$). Thus, **$\ce{Cr2O7^2-}$ can be successfully used in aqueous $\ce{HCl}$**. Step 4 - Analyzing the Feasibility of Titration in Sulfuric Acid ($\ce{H2SO4}$) Sulfuric acid ($\ce{H2SO4}$) releases sulfate ions ($\ce{SO4^2-}$). In the sulfate anion, the sulfur atom is in its maximum possible oxidation state of $+6$. Consequently, $\ce{SO4^2-}$ ions are exceptionally stable and cannot be oxidized further by either $\ce{MnO4^-}$ or $\ce{Cr2O7^2-}$. Therefore, no parallel oxidation occurs, and **both $\ce{MnO4^-}$ and $\ce{Cr2O7^2-}$ can be safely used in aqueous $\ce{H2SO4}$**. Step 5 - Evaluating the Options and Selecting the Incorrect Statement * **Option (A) is the correct answer (the only incorrect statement):** It states that $\ce{MnO4^-}$ can be used in aqueous $\ce{HCl}$, which is incorrect because $\ce{MnO4^-}$ spontaneously oxidizes $\ce{Cl^-}$ to $\ce{Cl2}$ gas ($1.51\text{ V} > 1.40\text{ V}$), thereby interfering with the titration. * **Option (B) is a correct statement** because $E^\circ(\ce{Cr2O7^2-/Cr^3+}) = 1.38\text{ V}$ is less than $E^\circ(\ce{Cl2/Cl^-}) = 1.40\text{ V}$, making chloride oxidation non-spontaneous. * **Option (C) is a correct statement** because sulfate ions from $\ce{H2SO4}$ cannot be oxidized. * **Option (D) is a correct statement** because sulfate ions from $\ce{H2SO4}$ cannot be oxidized. $$\text{Incorrect Statement: } \boxed{\text{A}}$$