Animated Solution for Chemistry - Electrochemistry: The standard emf of a cell, involving one electron change is found to be $0.591\text{ V}$ at $25^{\circ }\text{C}$. The equilibrium constant of the reaction is ($F=96500{\text{ C mol}}^{-1}$)

Imagine a standard galvanic cell, a device that miraculously converts chemical energy into electrical energy. At the heart of this process is a spontaneous redox reaction. The driving force behind this electron flow is measured as the standard electromotive force, or $E_{cell}^{\circ }$. In our specific problem, we are given a cell with an $E_{cell}^{\circ }$ of $0.591\text{ V}$ operating at a standard temperature of $25^{\circ }\text{C}$ (or $298\text{ K}$). Crucially, we are also told that the reaction involves the transfer of exactly one electron, meaning $n=1$.

To find the equilibrium constant ($K_{eq}$) from the cell potential, we need a bridge. That bridge is Thermodynamics, specifically the standard Gibbs free energy change, $\Delta G^{\circ }$.

There are two fundamental equations that define $\Delta G^{\circ }$: 1. From electrochemistry: $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$ 2. From chemical equilibrium: $\Delta G^{\circ }=-2.303RT\log K_{eq}$

By equating these two expressions, we create a direct link between the electrical potential of the cell and the extent of the chemical reaction:

Rearranging this to solve for $E_{cell}^{\circ }$, we get the master formula:

Calculating the term $\frac{2.303RT}{F}$ every time can be tedious. However, at the standard temperature of $25^{\circ }\text{C}$ ($298\text{ K}$), this entire cluster of constants evaluates to a very familiar and convenient number: $0.0591\text{ V}$.

Substituting this into our master formula gives us the simplified Nernst equation at equilibrium:

Now, we simply plug in the values provided in the problem. We know $E_{cell}^{\circ }=0.591\text{ V}$ and $n=1$.

To isolate $\log K_{eq}$, we divide both sides by $0.0591$:

Notice how perfectly these numbers are set up! Shifting the decimal point reveals that the ratio is exactly $10$.

To find the actual equilibrium constant, we must remove the base-10 logarithm by taking the antilogarithm of both sides.

This is a massive equilibrium constant! Physically, a $K_{eq}$ of $10^{10}$ indicates that the forward reaction is highly favored, and at equilibrium, the concentration of products will vastly outnumber the reactants. The reaction essentially goes to completion.

Always remember to check the number of electrons ($n$) transferred in the balanced equation. If this same cell had a two-electron transfer ($n=2$), the equilibrium constant would have been squared, resulting in an astronomical $K_{eq}$ of $10^{20}$!