Animated Solution for Chemistry - Electrochemistry: Emf of the following cell at $298\text{ K}$ in $\text{V}$ is $x\times 10^{-2}$, $\text{Zn}|{\text{Zn}}^{2+}(0.1\text{ M})||{\text{Ag}}^{+}(0.01\text{ M})|\text{Ag}$ The value of $x$ is ......... . (Rounded off to the nearest integer). [Given, $E_{{\text{Zn}}^{2+}/\text{Zn}}^{\circ }=-0.76\text{ V}$, $E_{{\text{Ag}}^{+}/\text{Ag}}^{\circ }=+0.80\text{ V}$, $\frac{2.303RT}{F}=0.059$]

Let's decode the cell representation given in the problem: $\text{Zn}|{\text{Zn}}^{2+}(0.1\text{ M})||{\text{Ag}}^{+}(0.01\text{ M})|\text{Ag}$. This standard notation is a map of the physical galvanic cell. On the left, we have the Zinc anode undergoing oxidation, losing electrons to become ${\text{Zn}}^{2+}$ ions. On the right, the Silver cathode is undergoing reduction, where ${\text{Ag}}^{+}$ ions gain electrons to form solid silver. The double vertical line in the middle represents the salt bridge connecting the two half-cells, maintaining electrical neutrality.

To find the Electromotive Force (EMF), we first need the balanced overall cell reaction. Zinc loses two electrons per atom:

However, each silver ion only gains one electron:

To balance the electron transfer, we must multiply the silver half-reaction by two. This gives us the overall balanced equation:

Crucially, this tells us that the total number of electrons transferred in the balanced reaction, $n$, is $2$.

A Tale of Concentrations

Next, we calculate the reaction quotient, $Q$. It is the ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients. Because solid metals have an activity of 1, they are omitted from the expression.

Notice how the silver ion concentration is squared because of the coefficient $2$ in our balanced equation. Substituting the given values:

The Baseline

Before we can find the actual potential, we need the standard cell potential, $E_{\text{cell}}^{\circ }$. This is the potential 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:

Bridging the Gap

With all our pieces ready, we bring in the master tool: the Nernst Equation. This equation connects the standard potential to the actual potential at non-standard concentrations.

The problem kindly provides the value for the constant term $\frac{2.303RT}{F}=0.059$. Let's substitute our calculated values into the Nernst equation:

Now for the final arithmetic. The logarithm of $10^3$ is simply $3$.

The question asks for the EMF in the specific format of $x\times 10^{-2}\text{ V}$. Converting our result:

Rounding off to the nearest integer, we find that $x=147$.