For a reaction ; has been found to be . The EMF of the cell is — Electrochemistry Chemistry Question

💡 Solution & Explanation

The relationship between standard EMF and equilibrium constant is $E^\circ = \frac{0.0591}{n} \log K_c$. Here, 2 electrons are transferred ($n=2$). $E^\circ = \frac{0.0591}{2} \log 10^{12} = 0.02955 \times 12 = 0.3546$ V, which approximates to 0.354 V. Therefore, correct answer is A.