The standard reduction potentials in acidic conditions are 0.77 V and 0.53 V, respectively, for and — Electrochemistry Chemistry Question

💡 Solution & Explanation

The standard cell potential $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.77 - 0.53 = 0.24$ V. The number of electrons transferred is $n = 2$. Using the relation $\log K_c = \frac{nE^\circ}{0.06}$, we get $\log K_c = \frac{2 \times 0.24}{0.06} = 8$. Therefore, $K_c = 10^8$. Therefore, correct answer is B.