Which one of the following does not get oxidized by bromine water? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Principle of Redox Spontaneity using Standard Reduction Potentials For a redox reaction to occur spontaneously under standard conditions, the standard cell potential ($E^\circ_{\text{cell}}$) must be positive ($E^\circ_{\text{cell}} > 0$), which corresponds to a negative standard Gibbs free energy change ($\Delta G^\circ < 0$). In any redox system involving an oxidizing agent and a reducing agent: * The oxidizing agent undergoes reduction at the cathode. * The reducing agent undergoes oxidation at the anode. The standard cell potential is calculated as: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}$$ Therefore, an oxidizing agent can spontaneously oxidize a species only if the standard reduction potential of the oxidizing agent ($E^\circ_{\text{reduction}}$) is greater than the standard reduction potential of the species being oxidized ($E^\circ_{\text{oxidation}}$): $$E^\circ_{\text{oxidizing agent}} > E^\circ_{\text{species to be oxidized}}$$ Step 2 - Analyze Bromine Water as an Oxidizing Agent Bromine water ($\ce{Br2(aq)}$) acts as an oxidizing agent by accepting electrons and undergoing reduction to bromide ions ($\ce{Br^-}$): $$\ce{Br2(aq) + 2e^- -> 2Br^-(aq)} \quad E^\circ_{\ce{Br2|Br^-}} = +1.09\text{ V}$$ For bromine water to successfully oxidize a species, the standard reduction potential of that species' redox couple must be less than $+1.09\text{ V}$: $$E^\circ_{\text{red}} < +1.09\text{ V}$$ Step 3 - Evaluate Each Option * **Option (A) - Oxidation of $\ce{Fe^2+}$ to $\ce{Fe^3+}$:** The reduction potential for the $\ce{Fe^3+/Fe^2+}$ couple is: $$E^\circ_{\ce{Fe^3+|Fe^2+}} = +0.77\text{ V}$$ We set up the redox reaction: $$\ce{2Fe^2+(aq) + Br2(aq) -> 2Fe^3+(aq) + 2Br^-(aq)}$$ $$E^\circ_{\text{cell}} = E^\circ_{\ce{Br2|Br^-}} - E^\circ_{\ce{Fe^3+|Fe^2+}} = +1.09\text{ V} - (+0.77\text{ V}) = +0.32\text{ V}$$ Since $E^\circ_{\text{cell}} > 0$, bromine water can spontaneously oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$. Thus, this option is incorrect. * **Option (B) - Oxidation of $\ce{Cu^+}$ to $\ce{Cu^2+}$:** The reduction potential for the $\ce{Cu^2+/Cu^+}$ couple is: $$E^\circ_{\ce{Cu^2+|Cu^+}} = +0.153\text{ V}$$ We set up the redox reaction: $$\ce{2Cu^+(aq) + Br2(aq) -> 2Cu^2+(aq) + 2Br^-(aq)}$$ $$E^\circ_{\text{cell}} = E^\circ_{\ce{Br2|Br^-}} - E^\circ_{\ce{Cu^2+|Cu^+}} = +1.09\text{ V} - (+0.153\text{ V}) = +0.937\text{ V}$$ Since $E^\circ_{\text{cell}} > 0$, bromine water can spontaneously oxidize $\ce{Cu^+}$ to $\ce{Cu^2+}$. Thus, this option is incorrect. * **Option (C) - Oxidation of $\ce{Mn^2+}$ to $\ce{MnO4^-}$:** The reduction potential for the permanganate to manganese(II) couple in acidic medium is: $$E^\circ_{\ce{MnO4^-|Mn^2+}} = +1.51\text{ V}$$ We set up the hypothetical redox reaction: $$\ce{2Mn^2+(aq) + 5Br2(aq) + 8H2O(l) -> 2MnO4^-(aq) + 10Br^-(aq) + 16H+(aq)}$$ $$E^\circ_{\text{cell}} = E^\circ_{\ce{Br2|Br^-}} - E^\circ_{\ce{MnO4^-|Mn^2+}} = +1.09\text{ V} - (+1.51\text{ V}) = -0.42\text{ V}$$ Since $E^\circ_{\text{cell}} < 0$, the reaction is non-spontaneous ($\Delta G^\circ > 0$). Bromine water is a weaker oxidizing agent than permanganate and cannot oxidize $\ce{Mn^2+}$ to $\ce{MnO4^-}$. Thus, this option is correct. * **Option (D) - Oxidation of $\ce{Sn^2+}$ to $\ce{Sn^4+}$:** The reduction potential for the $\ce{Sn^4+/Sn^2+}$ couple is: $$E^\circ_{\ce{Sn^4+|Sn^2+}} = +0.15\text{ V}$$ We set up the redox reaction: $$\ce{Sn^2+(aq) + Br2(aq) -> Sn^4+(aq) + 2Br^-(aq)}$$ $$E^\circ_{\text{cell}} = E^\circ_{\ce{Br2|Br^-}} - E^\circ_{\ce{Sn^4+|Sn^2+}} = +1.09\text{ V} - (+0.15\text{ V}) = +0.94\text{ V}$$ Since $E^\circ_{\text{cell}} > 0$, bromine water can spontaneously oxidize $\ce{Sn^2+}$ to $\ce{Sn^4+}$. Thus, this option is incorrect. Step 4 - Conclusion Bromine water cannot oxidize species whose standard reduction potentials are higher than $+1.09\text{ V}$. Because the reduction potential of the permanganate-manganese(II) couple is $+1.51\text{ V}$, manganese(II) cannot be oxidized to permanganate by bromine water. $$\text{Correct Option: } \boxed{\text{C}}$$