For the cell reaction at 25 °C Cu²⁺(aq)+ Zn(s)→ Cu(s)+ Zn²⁺(aq),E°_cell = +1.100 V.What is the equilibrium constant for this reaction?

Question

Quizplus · Accepted Answer

The equilibrium constant (K) for a reaction can be calculated from the standard cell potential (E°cell) using the Nernst equation at 25 °C (298 K) in the form: ΔG° = -nFE°cell, where ΔG° is the standard Gibbs free energy change, n is the number of moles of electrons transferred in the reaction, F is the Faraday constant (96485 C/mol), and E°cell is the standard cell potential. The relationship between ΔG° and K is given by ΔG° = -RT ln(K), where R is the gas constant (8.314 J/(mol·K)). Combining these equations gives: ln(K) = nFE°cell / RT. For the given E°cell of +1.100 V, assuming n = 2 (as is common for reactions involving the transfer of two electrons, such as the reduction of Cu^2+ and oxidation of Zn), and using the values for F, R, and T, we can solve for K. The calculation yields a very large value for K, indicating a strong tendency for the reaction to proceed as written, which corresponds to choice B, 2 × 10^137.

For the Cell Reaction at 25 °C Cu2+(aq)+ Zn(s)→ Cu(s)

For the Cell Reaction at 25 °C Cu²⁺(aq)+ Zn(s)→ Cu(s)