Question #44792

The electrochemical reduction of a metal ion \ce{M^{2+}(aq) + 2e^- -> M(s)} is carried out in a batch electrolytic cell at $298 \text{ K}$. The instantaneous rate of consumption of \ce{M^{2+}} (in $\text{mol L}^{-1} \text{s}^{-1}$) is empirically given by: $-\frac{d[\ce{M^{2+}}]}{dt} = k_0 \exp\left( \frac{\alpha z F E_{\text{red}}}{RT} \right)$ where $E_{\text{red}}$ is the instantaneous Nernst reduction potential of the \ce{M^{2+}/M} half-cell, $z=2$ is the number of electrons transferred, and the transfer coefficient $\alpha = 0.50$. Initially, the concentration of \ce{M^{2+}} is $1.00 \text{ M}$ and the initial rate of reaction is $2.0 \times 10^{-3} \text{ M s}^{-1}$. Assuming standard conditions apply and the activity of solid \ce{M} is unity, what is the time required for the concentration of \ce{M^{2+}} to drop to $0.25 \text{ M}$? (Given: $\ln 4 = 1.386$)