Given:
Zn (s) + Sn²⁺→Sn (s) + Zn²⁺ with E ° = +0.62 V What is the equilibrium constant for this reaction at 298 K?

Question

Given: Zn (s) + Sn²⁺→Sn (s) + Zn²⁺ with E ° = +0.62 V What is the equilibrium constant for this reaction at 298 K?

Quizplus · Accepted Answer

The equilibrium constant (K) for a reaction can be calculated from the standard electrode potential (E°) using the Nernst equation or its derived formula: $K = 10^{\frac{nE°}{0.0592}}$ at 298 K, where n is the number of moles of electrons transferred in the reaction. Given E° = +0.62 V and assuming a 2-electron transfer (common for reactions involving Zn and Sn), $K = 10^{\frac{2 \times 0.62}{0.0592}}$, which results in a value on the order of $10^{21}$, making option B correct.

Given: Zn (S) + Sn2+→Sn (S) + Zn2+ with E °

Given: Zn (S) + Sn²⁺→Sn (S) + Zn²⁺ with E °