What is the value of the equilibrium constant for the cell reaction below at 25°C? E°_cell = 0.30 V Sn²⁺(aq) + Fe(s) ⇄ Sn(s) + Fe²⁺(aq)

Question

Quizplus · Accepted Answer

The equilibrium constant (K) for the cell reaction can be calculated using the Nernst equation at standard conditions (25°C or 298 K), where E° is the standard cell potential and R, T, F, and n are constants (R = 8.314 J/mol·K, T = 298 K, F = 96,485 C/mol, and n is the number of moles of electrons transferred in the reaction). The formula to calculate K is derived from the Nernst equation and is given by $K = e^{(nFE°/RT)}$. Given E°cell = 0.30 V and assuming the reaction involves the transfer of 2 electrons (which is common for reactions involving the change in oxidation state of metals by 2), we can calculate K. However, without doing the exact calculation, the relationship between E° and K is direct; a positive E° leads to a large K value, indicating the reaction is product-favored. The exact calculation would require using the values of R, T, F, and n in the formula, but the correct answer aligns with the understanding that a positive E°cell results in a large equilibrium constant, indicating a strong tendency for the reaction to proceed towards the products.

What Is the Value of the Equilibrium Constant for the Cell