The standard potentials of MnO4^- \ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Balanced Half-Cell Reactions In an acidic solution, we can write the balanced reduction half-reactions and determine the number of electrons transferred ($n$-factor) for each electrode system: 1. **Reduction of permanganate ($\ce{MnO4^-}$) to manganese(II) ($\ce{Mn^{2+}}$):** $$\ce{MnO4^-(aq) + 8H^+(aq) + 5e^- -> Mn^{2+}(aq) + 4H2O(l)} \quad E^\circ_1 = +1.51\text{ V}$$ The number of electrons transferred is: $$n_1 = 5$$ 2. **Reduction of manganese dioxide ($\ce{MnO2}$) to manganese(II) ($\ce{Mn^{2+}}$):** $$\ce{MnO2(s) + 4H^+(aq) + 2e^- -> Mn^{2+}(aq) + 2H2O(l)} \quad E^\circ_2 = +1.23\text{ V}$$ The number of electrons transferred is: $$n_2 = 2$$ 3. **Target Reduction: permanganate ($\ce{MnO4^-}$) to manganese dioxide ($\ce{MnO2}$):** $$\ce{MnO4^-(aq) + 4H^+(aq) + 3e^- -> MnO2(s) + 2H2O(l)} \quad E^\circ_3 = x\text{ V}$$ The number of electrons transferred is: $$n_3 = 3$$ Step 2 - Relate the Free Energy Changes ($\Delta G^\circ$) Using Hess's Law Standard electrode potentials ($E^\circ$) are intensive properties and cannot be added or subtracted directly. However, standard Gibbs free energy change ($\Delta G^\circ$) is an extensive property and can be combined algebraically. The relationship is given by: $$\Delta G^\circ = -nFE^\circ$$ By comparing the three half-reactions, we can obtain the target reaction (Reaction 3) by subtracting Reaction 2 from Reaction 1: $$\text{Reaction 3} = \text{Reaction 1} - \text{Reaction 2}$$ Therefore, according to Hess's Law, the free energy changes are related as: $$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$$ Step 3 - Calculate the Unknown Standard Potential ($E^\circ_3$) Substitute the expression $\Delta G^\circ = -nFE^\circ$ into the thermodynamic relation: $$-n_3 F E^\circ_3 = -n_1 F E^\circ_1 - \left(-n_2 F E^\circ_2\right)$$ Since Faraday's constant ($F$) is common to all terms, it can be cancelled out from both sides: $$-n_3 E^\circ_3 = -n_1 E^\circ_1 + n_2 E^\circ_2$$ $$n_3 E^\circ_3 = n_1 E^\circ_1 - n_2 E^\circ_2$$ Now, substitute the known numerical values with units into the formula first: $$3 \times E^\circ_3 = \left(5 \times 1.51\text{ V}\right) - \left(2 \times 1.23\text{ V}\right)$$ $$3 \times E^\circ_3 = 7.55\text{ V} - 2.46\text{ V}$$ $$3 \times E^\circ_3 = 5.09\text{ V}$$ Solve for the target potential $E^\circ_3$: $$E^\circ_3 = \frac{5.09\text{ V}}{3}$$ $$E^\circ_3 = 1.6967\text{ V} \approx \mathbf{1.70\text{ V}}$$ Thus, the standard electrode potential for the $\ce{MnO4^- / MnO2}$ couple is approximately $+1.70\text{ V}$. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** $+1.697\text{ V}$ represents the unrounded theoretical value. While mathematically highly precise, multiple-choice options conventionally round standard electrode potentials to two decimal places, which makes Option (B) the most appropriate choice. * **Option (B) is correct:** As shown by our Gibbs free energy calculations, the standard electrode potential for the target half-reaction is $+1.697\text{ V}$, which rounds to $+1.70\text{ V}$. * **Option (C) is incorrect:** $+0.28\text{ V}$ is obtained by directly subtracting the electrode potentials ($1.51\text{ V} - 1.23\text{ V} = 0.28\text{ V}$). This is incorrect because electrode potentials are intensive properties and cannot be subtracted directly without multiplying by their respective number of electrons. * **Option (D) is incorrect:** $+2.74\text{ V}$ is obtained by directly adding the electrode potentials ($1.51\text{ V} + 1.23\text{ V} = 2.74\text{ V}$), which is incorrect for the same thermodynamic reasons. $$\text{Correct Option: } \boxed{\text{B}}$$