The value of equilibrium constant for a feasible cell reaction must be — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Thermodynamic Criterion for Spontaneous (Feasible) Reactions For any chemical or electrochemical reaction to be thermodynamically feasible (spontaneous) under standard conditions, the standard Gibbs free energy change ($\Delta G^\circ$) of the system must be strictly negative: $$\Delta G^\circ < 0$$ Step 2 - Relate Gibbs Free Energy to Standard Cell Potential ($E^\circ_{\text{cell}}$) The standard Gibbs free energy change ($\Delta G^\circ$) is related to the standard electromotive force of the cell ($E^\circ_{\text{cell}}$) by the fundamental relationship: $$\Delta G^\circ = -n F E^\circ_{\text{cell}}$$ Where: * $n$ is the number of moles of electrons transferred in the balanced cell reaction ($n > 0$). * $F$ is Faraday's constant ($F \approx 96500\text{ C mol}^{-1}$), which is always positive. For the cell reaction to be feasible ($\Delta G^\circ < 0$): $$-n F E^\circ_{\text{cell}} < 0 \implies E^\circ_{\text{cell}} > 0$$ Thus, a spontaneous cell reaction must have a positive standard cell potential. Step 3 - Relate Standard Cell Potential to the Equilibrium Constant ($K_c$) The standard cell potential ($E^\circ_{\text{cell}}$) is related to the equilibrium constant ($K_c$) via the Nernst equation when the cell reaction reaches chemical equilibrium: $$E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K_c$$ At $25^\circ\text{C}$ ($298\text{ K}$), converting the natural logarithm to base-10 logarithm gives: $$E^\circ_{\text{cell}} = \frac{0.0591\text{ V}}{n} \log_{10} K_c$$ Step 4 - Determine the Bound for the Equilibrium Constant ($K_c$) Since $E^\circ_{\text{cell}} > 0$ for a feasible cell reaction: $$\frac{0.0591\text{ V}}{n} \log_{10} K_c > 0$$ Because both $n$ and the constant $0.0591$ are positive quantities: $$\log_{10} K_c > 0$$ Taking the antilogarithm of both sides: $$K_c > 10^0 \implies \mathbf{K_c > 1}$$ Therefore, the equilibrium constant ($K_c$) for a thermodynamically feasible cell reaction must be strictly greater than $1$. Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** If $K_c < 1$, then $\log_{10} K_c < 0$, which results in $E^\circ_{\text{cell}} < 0$ and $\Delta G^\circ > 0$. This represents a non-feasible (non-spontaneous) cell reaction. * **Option (B) is incorrect:** An equilibrium constant of exactly zero ($K_c = 0$) means no products are formed at all, representing a reaction that does not proceed. Additionally, $\log_{10}(0)$ is mathematically undefined ($-\infty$). * **Option (C) is incorrect:** If $K_c = 1$, then $\log_{10} K_c = 0$, which yields $E^\circ_{\text{cell}} = 0$ and $\Delta G^\circ = 0$. This corresponds to a system that is already at standard state equilibrium, meaning no net cell reaction can occur to produce electrical work. * **Option (D) is correct:** As mathematically demonstrated, a feasible cell reaction requires a positive standard cell potential ($E^\circ_{\text{cell}} > 0$), which dictates that the equilibrium constant must be $K_c > 1$. $$\text{Correct Option: } \boxed{\text{D}}$$