Chemistry - Electrochemistry Question with Solution | TestHub

Solution:

The given reaction is:

$\mathrm{A}(\mathrm{s})+{\mathrm{B}}^{2+}(\mathrm{aq})\rightleftharpoons \mathrm{B}(\mathrm{s})+{\mathrm{A}}^{2+}(\mathrm{aq})$

The standard cell potential, $E_{cell}^{\circ }$, is given as -0.36 V.

The temperature is $25^{\circ }\mathrm{C}$, which is 298 K.

The initial concentration of ${\mathrm{B}}^{2+}$ is $5\mathrm{M}$.

We need to find the equilibrium concentration of ${\mathrm{A}}^{2+}$.

For the given reaction, the number of electrons transferred, $n$, is 2 (since ${\mathrm{B}}^{2+}$ gains 2 electrons to become B, and A loses 2 electrons to become ${\mathrm{A}}^{2+}$).

At equilibrium, the cell potential $E_{cell}=0$.

The relationship between standard cell potential ($E_{cell}^{\circ }$) and the equilibrium constant ($K_c$) at $25^{\circ }\mathrm{C}$ is given by:

$$E_{cell}^{\circ }=\frac{0.059}{n}\log K_c$$

Using the approximation $0.06$ instead of $0.059$:

$$E_{cell}^{\circ }=\frac{0.06}{n}\log K_c$$

Substitute the given values: $E_{cell}^{\circ }=-0.36\mathrm{V}$ and $n=2$.

$$-0.36=\frac{0.06}{2}\log K_c$$

$$-0.36=0.03\log K_c$$

$$\log K_c=\frac{-0.36}{0.03}$$

$$\log K_c=-12$$

$$K_c=10^{-12}$$

Now, let's write the expression for the equilibrium constant $K_c$ for the reaction:

$$K_c=\frac{[{\mathrm{A}}^{2+}]}{[{\mathrm{B}}^{2+}]}$$

(Note: The concentrations of pure solids A and B are not included in the equilibrium constant expression.)

At equilibrium, let the concentration of ${\mathrm{A}}^{2+}$ be $x$.

Since the initial concentration of ${\mathrm{B}}^{2+}$ is $5\mathrm{M}$ and the reaction proceeds to form ${\mathrm{A}}^{2+}$, the concentration of ${\mathrm{B}}^{2+}$ at equilibrium will be $5-x$.

However, since $K_c$ is very small ($10^{-12}$), the reaction does not proceed significantly to the right. This means that the amount of ${\mathrm{B}}^{2+}$ consumed ($x$) will be negligible compared to its initial concentration ($5\mathrm{M}$).

So, at equilibrium, $[{\mathrm{B}}^{2+}]\approx 5\mathrm{M}$.

Now, substitute the equilibrium concentrations into the $K_c$ expression:

$$K_c=\frac{[{\mathrm{A}}^{2+}]}{[{\mathrm{B}}^{2+}]}$$

$$10^{-12}=\frac{x}{5}$$

$$x=5\times 10^{-12}\mathrm{M}$$

Therefore, the equilibrium concentration of ${\mathrm{A}}^{2+}$ is $5\times 10^{-12}\mathrm{M}$.