The standard reduction potentials of Pt\ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Correlating Standard Reduction Potential with Oxidizing Power An oxidizing agent is a chemical species that gains electrons and undergoes reduction. The standard reduction potential ($E^\circ$) directly measures this tendency: $$\text{Oxidizing Power} \propto E^\circ_{\text{red}}$$ A higher, more positive standard reduction potential indicates a more powerful oxidizing agent. Step 2 - Comparing the Given $E^\circ$ Values We are given the standard reduction potentials at $25^\circ\text{C}$ in acidic medium: 1. $E^\circ(\ce{Ce^4+/Ce^3+}) = 1.61\text{ V}$ 2. $E^\circ(\ce{MnO4^-/Mn^2+}) = 1.51\text{ V}$ 3. $E^\circ(\ce{Cr2O7^{2-}/Cr^3+}) = 1.33\text{ V}$ Step 3 - Determining the Decreasing Order of Oxidizing Power Comparing: $1.61\text{ V} > 1.51\text{ V} > 1.33\text{ V}$ $$E^\circ(\ce{Ce^4+/Ce^3+}) > E^\circ(\ce{MnO4^-/Mn^2+}) > E^\circ(\ce{Cr2O7^{2-}/Cr^3+})$$ Therefore, the decreasing order of oxidizing strength is: $$\ce{Ce^4+ > MnO4^- > Cr2O7^{2-}}$$ Step 4 - Evaluation of the Options * **Option (A)** $\ce{Cr2O7^{2-} > MnO4^- > Ce^4+}$ — incorrect (reversed order). * **Option (B)** $\ce{Ce^4+ > MnO4^- > Cr2O7^{2-}}$ — correct. * **Option (C)** $\ce{MnO4^- > Ce^4+ > Cr2O7^{2-}}$ — incorrect (permanganate above cerium). * **Option (D)** $\ce{MnO4^- > Cr2O7^{2-} > Ce^4+}$ — incorrect. $$\text{Correct Option: } \boxed{\text{B}}$$