The equilibrium constant at 298 K for the following reaction is 1.2 x 10⁵.
H₂(g) + Sn⁴⁺(aq) $\rightarrow$ 2 H⁺(aq) + Sn²⁺(aq)
What is the value of E° for the reaction?

Question

The equilibrium constant at 298 K for the following reaction is 1.2 x 10⁵. H₂(g) + Sn⁴⁺(aq) $\rightarrow$ 2 H⁺(aq) + Sn²⁺(aq) What is the value of E° for the reaction?

Quizplus · Accepted Answer

The Nernst equation at standard conditions (E°) can be simplified to calculate the standard electrode potential for a reaction given the equilibrium constant (K). The equation is E° = (0.0592/n) log K, where n is the number of moles of electrons transferred in the reaction. In this case, n = 2 (from Sn^4+ to Sn^2+, a 2-electron process), and K = 1.2 x 10^15. Plugging these values into the equation gives E° = (0.0592/2) log(1.2 x 10^15) ≈ 0.15 V.

The Equilibrium Constant at 298 K for the Following Reaction →\rightarrow→

The Equilibrium Constant at 298 K for the Following Reaction $\rightarrow$