Animated Solution for Chemistry - Electrochemistry: For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295\text{ V}$ at $25^{\circ }\text{C}$. The equilibrium constant of the reaction at $25^{\circ }\text{C}$ will be

The relationship between the standard cell potential ($E_{cell}^{\circ }$) and the equilibrium constant ($K_{eq}$) is one of the most beautiful intersections of thermodynamics and electrochemistry. It tells us exactly how the electrical driving force of a reaction translates into the final balance of products and reactants.

Imagine an electrochemical cell where a spontaneous redox reaction is taking place. As the reaction proceeds, electrons flow through the external circuit, and the cell does electrical work. In our specific problem, we are told that exactly two electrons are transferred during the balanced reaction, meaning $n=2$.

The standard electromotive force (EMF) of this cell is given as $0.295\text{ V}$ at $25^{\circ }\text{C}$. This positive voltage indicates that under standard conditions, the forward reaction is spontaneous. But how far will this reaction go before it stops? That is exactly what the equilibrium constant will tell us.

To find the equilibrium constant, we turn to the Nernst equation, which relates the cell potential at any moment to the standard potential and the reaction quotient ($Q$):

As the reaction progresses, the reactants are consumed, and the products are formed. The cell potential ($E_{cell}$) gradually drops. Eventually, the system reaches equilibrium. At this magical point, the battery is "dead"—the cell potential becomes exactly zero ($E_{cell}=0$), and the reaction quotient becomes the equilibrium constant ($Q=K_{eq}$).

Substituting these equilibrium conditions into the Nernst equation gives us our master formula:

At a standard temperature of $25^{\circ }\text{C}$ (or $298\text{ K}$), the collection of constants $\frac{2.303RT}{F}$ evaluates to approximately $0.0591\text{ V}$. This is a massive time-saver for calculations. Let's plug in our known values:

Now, we just need to isolate $\log K_{eq}$. Multiplying both sides by 2 gives:

Dividing $0.590$ by $0.0591$ yields a value incredibly close to $10$. For the sake of standard physical chemistry problems, we take this as exactly $10$:

To find $K_{eq}$, we simply take the antilog (base 10) of both sides:

What does this number mean physically? An equilibrium constant of $10^{10}$ is absolutely massive. It means that at equilibrium, the concentration of the products is ten billion times greater than the concentration of the reactants. The reaction doesn't just go forward; it practically goes to completion! This perfectly aligns with our initial observation of a positive standard cell potential.