The EMF of the cell: at 298 K is 0.2905 V, then the value of equilibrium constant for the cell react — Electrochemistry Chemistry Question

💡 Solution & Explanation

Using Nernst eq: $E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Fe^{2+}]}$. $0.2905 = E^\circ_{cell} - 0.02955 \log(\frac{0.01}{0.001}) = E^\circ_{cell} - 0.02955 \log(10)$. So $E^\circ_{cell} = 0.2905 + 0.02955 = 0.32005$ V. At equilibrium, $E^\circ_{cell} = \frac{0.0591}{2} \log K_c = 0.02955 \log K_c$. Thus, $\log K_c = \frac{0.32}{0.0295}$. Therefore, $K_c = 10^{\frac{0.32}{0.0295}}$. Therefore, correct answer is B.