Animated Solution for Chemistry - Electrochemistry: For the galvanic cell, $Zn(s)+C{u}^{2+}(0.02\text{ M})⟶Z{n}^{2+}(0.04\text{ M})+Cu(s)$, $E_{Cell}=\cdots \times 10^{-2}\text{ V}$. (Nearest integer) [Use $E_{Cu/C{u}^{2+}}^{\circ }=-0.34\text{ V}$, $E_{Zn/Z{n}^{2+}}^{\circ }=+0.76\text{ V}$, $\frac{2.303RT}{F}=0.059\text{ V}$]

A Dance of Electrons

Imagine you are standing right inside a beaker of a galvanic cell. On one side, you have a solid Zinc electrode slowly dissolving into the solution, releasing $Z{n}^{2+}$ ions and leaving behind precious electrons. On the other side, a Copper electrode is eagerly waiting to grab those electrons to turn $C{u}^{2+}$ ions from the solution into solid Copper. This beautiful, spontaneous flow of electrons is what generates the cell potential, or voltage, that we measure.

But how do we calculate this voltage when the conditions aren't perfect? That's where the legendary Nernst Equation comes into play.

Before we even touch the Nernst equation, we need to find the standard cell potential, $E_{cell}^{\circ }$. The formula is simple:

However, there is a massive trap waiting for you in the problem statement. The question provides $E_{Cu/C{u}^{2+}}^{\circ }=-0.34\text{ V}$ and $E_{Zn/Z{n}^{2+}}^{\circ }=+0.76\text{ V}$. Notice the order of the species? They are written as Metal going to Metal Ion. These are oxidation potentials!

By IUPAC convention, we must always use standard reduction potentials. To convert an oxidation potential to a reduction potential, we simply flip the sign.

Now, we can safely calculate the standard cell potential:

Nernst

Since our ion concentrations are $0.04\text{ M}$ and $0.02\text{ M}$ instead of the standard $1\text{ M}$, the actual cell potential will deviate from $1.10\text{ V}$. We invoke the Nernst equation:

Here, $n$ is the number of electrons transferred in the balanced redox reaction. Since Zinc loses 2 electrons and Copper gains 2 electrons, $n=2$. The reaction quotient $Q$ is the ratio of the concentration of products to reactants. Solid metals are ignored, so $Q=\frac{[Z{n}^{2+}]}{[C{u}^{2+}]}$.

Let's substitute our values into the Nernst equation:

Simplifying the fraction inside the logarithm gives us $\log 2$, which is approximately $0.30$.

The question cleverly asks for the answer in the format of an integer multiplied by $10^{-2}$.

Thus, our final, triumphant integer is 109. Always stay vigilant with signs and conventions, and the math will naturally flow!