Among the following, identify the correct statement? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Relationship between Standard Reduction Potential and Oxidizing/Reducing Power The standard reduction potential ($E^\circ$) is a quantitative measure of the tendency of a chemical species to acquire electrons and thereby undergo reduction. * **Oxidizing Agents:** A species with a higher (more positive) standard reduction potential has a stronger tendency to undergo reduction, meaning it acts as a **stronger oxidizing agent** and can spontaneously oxidize other species. * **Reducing Agents:** A species with a lower (less positive or more negative) standard reduction potential has a weaker tendency to undergo reduction. Consequently, its conjugate reduced form has a stronger tendency to undergo oxidation, acting as a **stronger reducing agent**. The given standard reduction potentials ($E^\circ$) in an acidic medium are: 1. $$\ce{Mn^3+(aq) + e^- -> Mn^2+(aq)} \quad E^\circ = +1.50\text{ V}$$ 2. $$\ce{Cl2(g) + 2e^- -> 2Cl^-(aq)} \quad E^\circ = +1.36\text{ V}$$ 3. $$\ce{O2(g) + 4H^+(aq) + 4e^- -> 2H2O(l)} \quad E^\circ = +1.23\text{ V}$$ 4. $$\ce{Fe^3+(aq) + e^- -> Fe^2+(aq)} \quad E^\circ = +0.77\text{ V}$$ 5. $$\ce{I2(s) + 2e^- -> 2I^-(aq)} \quad E^\circ = +0.54\text{ V}$$ Step 2 - Establish the Thermodynamic Criterion for Spontaneity For a redox reaction to occur spontaneously under standard conditions, the overall standard cell potential ($E^\circ_{\text{cell}}$) must be strictly positive: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} > 0\text{ V}$$ $$\implies E^\circ_{\text{cathode (reduction)}} > E^\circ_{\text{anode (oxidation)}}$$ This indicates that an oxidizing agent (at the cathode) can spontaneously oxidize any reducing agent (at the anode) whose corresponding conjugate redox couple has a **lower** standard reduction potential. Step 3 - Evaluate Each Option Separately * **Option (A) - Chloride ion is oxidized by \ce{O2}:** Here, the potential cathode reaction is the reduction of oxygen gas ($\ce{O2}$): $$\ce{O2(g) + 4H^+(aq) + 4e^- -> 2H2O(l)} \quad E^\circ_{\text{cathode}} = +1.23\text{ V}$$ The potential anode reaction is the oxidation of chloride ions ($\ce{Cl^-}$): $$\ce{2Cl^-(aq) -> Cl2(g) + 2e^-} \quad E^\circ_{\text{anode}} = +1.36\text{ V}$$ Calculating the standard cell potential: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = 1.23\text{ V} - 1.36\text{ V} = -0.13\text{ V}$$ Since $E^\circ_{\text{cell}} < 0\text{ V}$, this reaction is non-spontaneous. Thus, chloride ions cannot be oxidized by oxygen under standard conditions. Option (A) is **incorrect**. * **Option (B) - \ce{Fe^2+} is oxidized by iodine (\ce{I2}):** Here, the potential cathode reaction is the reduction of iodine ($\ce{I2}$): $$\ce{I2(s) + 2e^- -> 2I^-(aq)} \quad E^\circ_{\text{cathode}} = +0.54\text{ V}$$ The potential anode reaction is the oxidation of iron(II) ions ($\ce{Fe^2+}$): $$\ce{Fe^2+(aq) -> Fe^3+(aq) + e^-} \quad E^\circ_{\text{anode}} = +0.77\text{ V}$$ Calculating the standard cell potential: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = 0.54\text{ V} - 0.77\text{ V} = -0.23\text{ V}$$ Since $E^\circ_{\text{cell}} < 0\text{ V}$, this reaction is non-spontaneous. Thus, iron(II) ions cannot be oxidized by iodine. Option (B) is **incorrect**. * **Option (C) - Iodide ion (\ce{I^-}) is oxidized by chlorine (\ce{Cl2}):** Here, the potential cathode reaction is the reduction of chlorine gas ($\ce{Cl2}$): $$\ce{Cl2(g) + 2e^- -> 2Cl^-(aq)} \quad E^\circ_{\text{cathode}} = +1.36\text{ V}$$ The potential anode reaction is the oxidation of iodide ions ($\ce{I^-}$): $$\ce{2I^-(aq) -> I2(s) + 2e^-} \quad E^\circ_{\text{anode}} = +0.54\text{ V}$$ Calculating the standard cell potential: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = 1.36\text{ V} - 0.54\text{ V}$$ $$E^\circ_{\text{cell}} = \boxed{+0.82\text{ V}}$$ Since $E^\circ_{\text{cell}} = +0.82\text{ V} > 0\text{ V}$, this reaction is highly spontaneous. Thus, iodide ions are spontaneously oxidized by chlorine gas. Option (C) is **correct**. * **Option (D) - \ce{Mn^2+} is oxidized by chlorine (\ce{Cl2}):** Here, the potential cathode reaction is the reduction of chlorine gas ($\ce{Cl2}$): $$\ce{Cl2(g) + 2e^- -> 2Cl^-(aq)} \quad E^\circ_{\text{cathode}} = +1.36\text{ V}$$ The potential anode reaction is the oxidation of manganese(II) ions ($\ce{Mn^2+}$): $$\ce{Mn^2+(aq) -> Mn^3+(aq) + e^-} \quad E^\circ_{\text{anode}} = +1.50\text{ V}$$ Calculating the standard cell potential: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = 1.36\text{ V} - 1.50\text{ V} = -0.14\text{ V}$$ Since $E^\circ_{\text{cell}} < 0\text{ V}$, this reaction is non-spontaneous. Thus, manganese(II) ions cannot be oxidized by chlorine gas under standard conditions. Option (D) is **incorrect**. $$\text{Correct Option: } \boxed{C}$$