Question #44548

A redox reaction between $\ce{KMnO4}$ and a weak acid reductant $\ce{HA}$ ($K_a = 2.0 \times 10^{-5}$) proceeds in an acidic medium containing $\ce{HCl}$. The proposed mechanism involves the formation of a transient $\ce{Mn(VI)}$ intermediate:

Step 1: $\ce{MnO4- + HA ->[k_1] MnO4^{2-} + HA^{\bullet+}}$ (slow) Step 2: $\ce{MnO4^{2-} + H+ ->[k_2] Mn^{3+} + \text{Products}}$ (fast)

An electrochemical half-cell is prepared at $298 \text{ K}$ containing $0.05 \text{ M } \ce{KMnO4}$, $0.05 \text{ M } \ce{Mn^{2+}}$, $0.1 \text{ M } \ce{HA}$, and $1.0 \times 10^{-3} \text{ M } \ce{HCl}$.

Given: $k_1 = 4.0 \times 10^{-2} \text{ M}^{-1}\text{s}^{-1}$ $k_2 = 1.0 \times 10^3 \text{ M}^{-1}\text{s}^{-1}$ $E^\circ_{\ce{MnO4- / Mn^{2+}}} = 1.51 \text{ V}$ $\frac{2.303 RT}{F} = 0.06 \text{ V}$, $\log 2 = 0.30$.

Assuming the system is at a steady state for $\ce{MnO4^{2-}}$ and the weak acid equilibrium is established instantaneously, determine the steady-state concentration of $\ce{MnO4^{2-}}$ and the instantaneous reduction potential ($E_{red}$) of the $\ce{MnO4- / Mn^{2+}}$ couple.