By how much would the oxidizing power of MnO4–/Mn2+ couple change if the H+ ions concentration is de — Electrochemistry Chemistry Question

💡 Solution & Explanation

Nernst equation for the couple: $E = E^\circ - (0.0591/5) \log([Mn^{2+}] / ([MnO_4^-][H^+]^8))$. Decreasing $[H^+]$ by 100 times ($10^{-2}$) changes the log term by $\log(1/(10^{-2})^8) = \log(10^{16}) = 16$. The change in E is $-(0.0591/5) \times 16 = -0.189$ V = -189 mV. Thus, it decreases by 189 mV.