Animated Solution for Chemistry - Electrochemistry: For the following cell : $\text{Zn(s)}\mid {\text{ZnSO}}_4\text{ (aq.)}\parallel {\text{CuSO}}_4\text{ (aq.)}\mid \text{Cu(s)}$ when the concentration of ${\text{Zn}}^{2+}$ is $10$ times the concentration of ${\text{Cu}}^{2+}$, the expression for $\Delta G$ (in ${\text{J mol}}^{-1}$) is [$\text{F}$ is Faraday constant, $\text{R}$ is gas constant, $\text{T}$ is temperature, $E^{\circ }(\text{cell})=1.1\text{ V}$]

Electrochemistry is a fascinating bridge between chemical reactions and electrical energy. When we look at a galvanic cell, like the classic Daniell cell presented in this problem, we are witnessing a spontaneous chemical reaction being harnessed to do electrical work. But how do we quantify this work? How do we know if a cell will actually produce a voltage under non-standard conditions? This is where the elegant principles of thermodynamics step in, specifically through the concept of Gibbs Free Energy ($\Delta G$).

The Daniell Cell

The problem provides us with the cell notation:

This notation tells a complete story. On the left, we have the anode where oxidation occurs. Solid zinc ($\text{Zn}$) loses electrons to become aqueous zinc ions (${\text{Zn}}^{2+}$). On the right, we have the cathode where reduction takes place. Aqueous copper ions (${\text{Cu}}^{2+}$) gain those electrons to form solid copper ($\text{Cu}$).

Let's write down the half-reactions to see exactly what's happening with the electrons:

Anode (Oxidation): $\text{Zn}\to {\text{Zn}}^{2+}+2e^{-}$ Cathode (Reduction): ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$

By adding these together, we get the net cell reaction:

Notice that exactly $2$ electrons are transferred in this process. This is a crucial piece of information: $n=2$.

We are given a specific condition: the concentration of ${\text{Zn}}^{2+}$ is $10$ times the concentration of ${\text{Cu}}^{2+}$. In thermodynamics, we use the reaction quotient ($Q$) to describe the ratio of product concentrations to reactant concentrations at any given moment.

For our net reaction, the solids ($\text{Zn}$ and $\text{Cu}$) have an activity of $1$, so they don't appear in the expression. Thus:

Since $[{\text{Zn}}^{2+}]=10\times [{\text{Cu}}^{2+}]$, we can easily see that $Q=10$.

Nernst and Gibbs

To find the expression for the change in Gibbs free energy ($\Delta G$), we rely on the fundamental thermodynamic equation that links it to the standard Gibbs free energy ($\Delta G^{\circ }$) and the reaction quotient ($Q$):

Because working with base-10 logarithms is often more intuitive (especially when $Q$ is a power of $10$), we convert the natural logarithm ($\ln$) to a base-10 logarithm (${\log }_{10}$) by multiplying by $2.303$:

We also know the relationship between standard Gibbs free energy and the standard cell potential ($E_{\text{cell}}^{\circ }$):

Substituting this into our main equation gives us the master formula we need:

Putting It All Together

Now, it's just a matter of carefully substituting our known values into the master formula. We have: - $n=2$ - $E_{\text{cell}}^{\circ }=1.1\text{ V}$ - $Q=10$

Let's plug them in:

Since ${\log }_{10}(10)=1$, the equation simplifies beautifully:

Rearranging the terms to match the options provided in the question, we get:

This perfectly matches option (B). By systematically breaking down the cell notation, identifying the electron transfer, and applying the core thermodynamic equations, we've arrived at the correct expression with absolute certainty.