Animated Solution for Chemistry - Electrochemistry: Consider the cell at $25^{\circ }\text{C}$ $Zn|Z{n}^{2+}(aq),(1M)||F{e}^{3+}(aq),F{e}^{2+}(aq)|Pt(s)$ The fraction of total iron present as $F{e}^{3+}$ ion at the cell potential of $1.500\text{ V}$ is $x\times 10^{-2}$. The value of $x$ is ……… . (Nearest integer) (Given, $E_{F{e}^{3+}/F{e}^{2+}}^{\circ }=0.77\text{ V}$, $E_{Z{n}^{2+}/Zn}^{\circ }=-0.76\text{ V}$)

Imagine you are looking at a microscopic battery. On the left side, we have a solid zinc electrode submerged in a solution of zinc ions. On the right side, there is a platinum electrode, which is inert, sitting in a mixture of iron(III) and iron(II) ions. The two sides are connected, and a voltmeter reads exactly $1.500\text{ V}$. Our mission is to figure out what fraction of the iron ions are in the $+3$ oxidation state.

To understand what's happening, we first need to write down the half-cell reactions. At the anode, oxidation occurs. Zinc metal loses two electrons to become zinc ions:

At the cathode, reduction takes place. The iron(III) ions grab those electrons to become iron(II) ions. Since zinc gives up two electrons, we need two iron(III) ions to accept them:

Combining these gives us the overall cell reaction:

Notice that exactly $2$ electrons are transferred in this process, so $n=2$.

Before we can use the actual cell potential, we need the standard cell potential, $E_{cell}^{\circ }$. This is the potential the cell would have if all concentrations were exactly $1\text{ M}$. It is calculated as the standard reduction potential of the cathode minus that of the anode:

Plugging in the given values:

Now, we bridge the gap between the standard potential and the actual potential using the Nernst equation. This equation is the heart of electrochemistry when dealing with non-standard concentrations:

The reaction quotient, $Q$, is the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. Remember, pure solids like zinc are excluded from $Q$:

Let's substitute all our known values into the Nernst equation. We know $E_{cell}=1.50\text{ V}$, $E_{cell}^{\circ }=1.53\text{ V}$, $n=2$, and $[Z{n}^{2+}]=1\text{ M}$:

Rearranging the terms to isolate the logarithmic part:

Notice how the square inside the logarithm comes out as a multiplier of $2$, perfectly canceling the $2$ in the denominator! Now, we solve for the log ratio:

Taking the antilog of both sides, we find the ratio of the iron concentrations:

We are almost there! The question asks for the fraction of total iron that is present as $F{e}^{3+}$. This fraction is defined as:

To make use of our ratio, we can divide the numerator and the denominator by $[F{e}^{3+}]$:

Since $\sqrt{10}\approx 3.16$:

Expressing this in the requested format of $x\times 10^{-2}$:

Thus, the value of $x$ is 24.