Animated Solution for Chemistry - Electrochemistry: For an electrochemical cell $Sn(s)|S{n}^{2+}(aq,1M)||P{b}^{2+}(aq,1M)|Pb(s)$ the ratio $\frac{[S{n}^{2+}]}{[P{b}^{2+}]}$ when this cell attains equilibrium is ......... . Given : $E_{S{n}^{2+}|Sn}^{\circ }=-0.14\text{ V}$, $E_{P{b}^{2+}|Pb}^{\circ }=-0.13\text{ V}$, $\frac{2.303RT}{F}=0.06$

The magic of electrochemistry lies in its ability to harness the energy of chemical reactions to do electrical work. But what happens when the reaction runs its course and the battery "dies"? In this problem, we explore the fascinating state of equilibrium in a galvanic cell and uncover the hidden mathematical relationship between cell potential and ion concentrations.

Imagine you are looking at a classic galvanic cell. On the left, we have a Tin ($Sn$) anode submerged in a solution of $S{n}^{2+}$ ions. On the right, a Lead ($Pb$) cathode sits in a solution of $P{b}^{2+}$ ions. A salt bridge connects the two beakers, and a wire allows electrons to flow from the anode to the cathode.

The cell notation is given as:

From this, we can immediately write the overall cell reaction. Tin is undergoing oxidation (losing electrons), and Lead is undergoing reduction (gaining electrons):

The equilibrium constant $K$ for this reaction is the ratio of the concentration of the products to the reactants. Since solids do not appear in the equilibrium expression, we get:

This is exactly the ratio we are asked to find!

To find $K$, we need to determine the standard cell potential, $E_{cell}^{\circ }$. This is the driving force of the cell under standard conditions. We calculate it by subtracting the standard reduction potential of the anode from that of the cathode:

Substituting the given values:

It's a tiny positive voltage, meaning the reaction is barely spontaneous under standard conditions.

Now, we bring in the heavy artillery: the Nernst Equation. This equation connects the cell potential at any moment to the standard potential and the reaction quotient.

Here is the crucial conceptual leap: when the cell attains equilibrium, it stops producing a voltage. The battery is dead. Therefore, $E_{cell}=0$, and the reaction quotient $Q$ becomes the equilibrium constant $K$.

Let's substitute our known values into the Nernst equation at equilibrium:

Notice that $n=2$ because two electrons are transferred in the balanced redox reaction. The term $\frac{2.303RT}{F}$ is conveniently given as $0.06$.

Rearranging the equation to solve for $\log K$:

To find $K$, we take the antilogarithm:

Calculating the cube root of 10 gives us approximately $2.154$.

Rounding to two decimal places, the ratio of $[S{n}^{2+}]$ to $[P{b}^{2+}]$ at equilibrium is $2.15$.

This result tells us a beautiful physical truth: at equilibrium, the concentration of Tin ions will be slightly more than double the concentration of Lead ions, perfectly balancing the slight difference in their standard reduction potentials!