For a cell reaction involving a two-electron change, the standard EMF of the cell is found to be 0.2 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Using the relation between standard EMF and equilibrium constant: $E^\circ = \frac{0.0591}{n} \log K$. Given $n = 2$ and $E^\circ = 0.295$ V, we have $0.295 = \frac{0.0591}{2} \log K$, so $0.295 = 0.02955 \log K$. Solving for $\log K$ yields 10, which means $K = 10^{10} = 1 \times 10^{10}$.