Animated Solution for Chemistry - Electrochemistry: For the redox reaction $Zn(s)+C{u}^{2+}(0.1\text{ M})⟶Z{n}^{2+}(1\text{ M})+Cu(s)$ taking place in a cell, $E_{cell}^{\circ }$ is $1.10\text{ V}$. $E_{cell}$ for the cell will be $\left(2.303\frac{RT}{F}=0.0591\right)$

Imagine you are looking at a classic galvanic cell, specifically the Daniell cell. On one side, we have a Zinc electrode submerged in a solution of Zinc ions ($Z{n}^{2+}$). On the other side, a Copper electrode sits in a solution of Copper ions ($C{u}^{2+}$).

In a standard state, both solutions would have a concentration of exactly $1\text{ M}$. Under those perfect conditions, the cell generates a standard potential, denoted as $E_{cell}^{\circ }$, which is given to us as $1.10\text{ V}$. However, the real world is rarely standard. In our problem, the Zinc ion concentration is indeed $1\text{ M}$, but the Copper ion concentration is only $0.1\text{ M}$. Because the concentrations deviate from the standard $1\text{ M}$, the actual potential of the cell, $E_{cell}$, will be different from $1.10\text{ V}$.

To bridge the gap between standard conditions and our specific scenario, we rely on one of the most powerful tools in electrochemistry: the Nernst Equation. It allows us to calculate the exact cell potential at any given concentration.

The Nernst equation is written as:

At room temperature ($298\text{ K}$), the constant term $\frac{2.303RT}{F}$ simplifies beautifully to $0.0591$. This transforms our equation into a much more manageable form:

To use this equation, we need to uncover two critical parameters: $n$ and $Q$.

First, let's find $n$, which represents the number of moles of electrons transferred in the balanced redox reaction. Looking at the reaction $Zn(s)+C{u}^{2+}(aq)⟶Z{n}^{2+}(aq)+Cu(s)$, we can see that Zinc oxidizes from an oxidation state of $0$ to $+2$, losing two electrons. Simultaneously, Copper reduces from $+2$ to $0$, gaining those exact two electrons. Therefore, $n=2$.

Next, we need the reaction quotient, $Q$. This is the ratio of the concentrations of the product ions to the reactant ions. Remember, we strictly ignore pure solids like $Zn(s)$ and $Cu(s)$ because their active concentrations remain constant.

Now, let's substitute the given concentrations into our expression for $Q$:

With $n=2$ and $Q=10$, we are fully equipped to solve the Nernst equation. Let's plug everything in:

We know that $\log (10)$ is simply $1$. The math becomes a straightforward subtraction:

Rounding to two decimal places, we get $1.07\text{ V}$.

Notice how the potential dropped from $1.10\text{ V}$ to $1.07\text{ V}$. This makes perfect physical sense! Because we started with less reactant ($C{u}^{2+}$) than product ($Z{n}^{2+}$), the driving force of the forward reaction is slightly weaker than it would be under standard conditions. As the cell continues to operate, $[Z{n}^{2+}]$ will increase and $[C{u}^{2+}]$ will decrease, causing $Q$ to grow larger and $E_{cell}$ to drop further until it eventually hits zero, at which point the battery is dead.