By how much would the oxidizing power of MnO4^-/Mn^2+ couple change if the H^+ ions concentration is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Balanced Reduction Half-Reaction The reduction half-reaction of the permanganate/manganese(II) couple in an acidic medium is written as: $$\ce{MnO4^-(aq) + 8H^+(aq) + 5e^- -> Mn^2+(aq) + 4H2O(l)}$$ From this balanced equation, the number of moles of electrons transferred is: $$n = 5$$ Step 2 - Formulate the Nernst Equation According to the Nernst equation at $298\text{ K}$ ($25^\circ\text{C}$), the reduction potential ($E$) of this half-cell is expressed as: $$E = E^\circ - \frac{2.303 RT}{nF} \log Q$$ Where: * $E^\circ$ is the standard reduction potential of the $\ce{MnO4^-/Mn^2+}$ couple. * $n = 5$ is the number of electrons transferred. * $\frac{2.303 RT}{F} \approx 0.059\text{ V}$ at $298\text{ K}$. * $Q$ is the reaction quotient: $$Q = \frac{[\ce{Mn^2+}]}{[\ce{MnO4^-}][\ce{H+}]^8}$$ Substituting these parameters into the equation yields: $$E = E^\circ - \frac{0.059\text{ V}}{5} \log \left( \frac{[\ce{Mn^2+}]}{[\ce{MnO4^-}][\ce{H+}]^8} \right)$$ Using logarithmic properties, we can isolate the hydrogen ion concentration term: $$E = E^\circ - \frac{0.059\text{ V}}{5} \log \left( \frac{[\ce{Mn^2+}]}{[\ce{MnO4^-}]} \right) - \frac{0.059\text{ V}}{5} \log \left( \frac{1}{[\ce{H+}]^8} \right)$$ $$E = E^\circ - \frac{0.059\text{ V}}{5} \log \left( \frac{[\ce{Mn^2+}]}{[\ce{MnO4^-}]} \right) + \frac{0.059\text{ V} \times 8}{5} \log [\ce{H+}]$$ Step 3 - Analyze the Effect of Decreasing $[\ce{H+}]$ Let the initial concentration of hydrogen ions be $[\ce{H+}]_1$. The initial reduction potential ($E_1$) is: $$E_1 = E^\circ - \frac{0.059\text{ V}}{5} \log \left( \frac{[\ce{Mn^2+}]}{[\ce{MnO4^-}]} \right) + \frac{0.059\text{ V} \times 8}{5} \log [\ce{H+}]_1$$ When the concentration of $\ce{H^+}$ ions is decreased $100$ times, the new concentration ($[\ce{H+}]_2$) becomes: $$[\ce{H+}]_2 = \frac{[\ce{H+}]_1}{100}$$ The new reduction potential ($E_2$) is: $$E_2 = E^\circ - \frac{0.059\text{ V}}{5} \log \left( \frac{[\ce{Mn^2+}]}{[\ce{MnO4^-}]} \right) + \frac{0.059\text{ V} \times 8}{5} \log \left( \frac{[\ce{H+}]_1}{100} \right)$$ Step 4 - Calculate the Change in Potential ($\Delta E$) The change in reduction potential ($\Delta E = E_2 - E_1$) directly represents the change in oxidizing power: $$\Delta E = \frac{0.059\text{ V} \times 8}{5} \left[ \log \left( \frac{[\ce{H+}]_1}{100} \right) - \log [\ce{H+}]_1 \right]$$ Applying the logarithmic property $\log(a) - \log(b) = \log\left(\frac{a}{b}\right)$: $$\Delta E = \frac{0.059\text{ V} \times 8}{5} \log \left( \frac{[\ce{H+}]_1 / 100}{[\ce{H+}]_1} \right)$$ $$\Delta E = \frac{0.059\text{ V} \times 8}{5} \log \left( \frac{1}{100} \right)$$ $$\Delta E = \frac{0.059\text{ V} \times 8}{5} \log(10^{-2})$$ $$\Delta E = \frac{0.059\text{ V} \times 8}{5} \times (-2)$$ $$\Delta E = -\frac{0.059\text{ V} \times 16}{5} = -\frac{0.944\text{ V}}{5}$$ $$\Delta E = -0.1888\text{ V} \approx \boxed{-0.189\text{ V}}$$ Converting volts to millivolts ($1\text{ V} = 1000\text{ mV}$): $$\Delta E \approx -189\text{ mV}$$ The negative sign indicates that the reduction potential decreases, which means the oxidizing power decreases. Step 5 - Explanation of Options * **Option (A)** is incorrect because decreasing the reactant concentration ($\ce{H+}$) shifts the equilibrium to the left according to Le Chatelier's principle, which thermodynamically reduces the tendency to undergo reduction (i.e., decreases the oxidizing power, rather than increasing it). * **Option (B)** is correct because our Nernst calculation shows that a $100$-fold decrease in $[\ce{H+}]$ causes the reduction potential to decrease by exactly $189\text{ mV}$. * **Option (C)** is incorrect because the magnitude of the change is $189\text{ mV}$, not $19\text{ mV}$, and the power is a decrease. * **Option (D)** is incorrect because although it correctly identifies a decrease, the magnitude of $19\text{ mV}$ is incorrect (it misses the multiplication factor of $8$ from the reaction stoichiometry in the logarithm). $$\text{Correct Option: } \boxed{\text{B}}$$