Animated Solution for Chemistry - Electrochemistry: For the given cell; $Cu(s)|C{u}^{2+}(C_{1}M)||C{u}^{2+}(C_{2}M)|Cu(s)$ change in Gibbs energy ($\Delta G$) is negative, if

Imagine a battery where both the positive and negative terminals are made of the exact same metal. Sounds counterintuitive, right? How can a battery work if there's no difference in the materials? Welcome to the fascinating world of Concentration Cells.

In this problem, we are given a cell represented as: $Cu(s)|C{u}^{2+}(C_{1}M)||C{u}^{2+}(C_{2}M)|Cu(s)$

Here, both electrodes are made of Copper ($Cu$). The only difference lies in the concentration of the $C{u}^{2+}$ ions in the two half-cells. The left side (anode) has a concentration of $C_1$, and the right side (cathode) has a concentration of $C_2$.

Gibbs Free Energy

The question asks for the condition under which the change in Gibbs free energy ($\Delta G$) is negative. In thermodynamics, a negative $\Delta G$ is the ultimate green light—it means the reaction is spontaneous and will happen on its own, generating electrical energy in the process.

We know the fundamental relationship between Gibbs free energy and cell potential ($E_{cell}$): $\Delta G=-nF{E}_{cell}$

For $\Delta G$ to be negative, the cell potential $E_{cell}$ must be strictly positive ($E_{cell}>0$). This is our primary condition.

Let's break down what's happening at each electrode. At the Anode (Oxidation), solid copper dissolves into the solution: $Cu(s)\to C{u}^{2+}(C_1)+2e^{-}$

At the Cathode (Reduction), copper ions from the solution deposit onto the electrode: $C{u}^{2+}(C_2)+2e^{-}\to Cu(s)$

If we add these two half-reactions, the solid copper cancels out, and we get the net cell reaction: $C{u}^{2+}(C_2)\to C{u}^{2+}(C_1)$

Notice something interesting? The standard cell potential ($E_{cell}^{\circ }$) is exactly zero because both electrodes are identical. The only thing driving this reaction is the desire of the universe to equalize the concentrations on both sides.

To find the actual cell potential, we use the Nernst equation: $E_{cell}=E_{cell}^{\circ }-\frac{2.303RT}{nF}\log Q$

Since $E_{cell}^{\circ }=0$ and the reaction quotient $Q$ is the ratio of product concentration to reactant concentration ($Q=\frac{C_1}{C_2}$), the equation simplifies to: $E_{cell}=-\frac{2.303RT}{2F}\log \left(\frac{C_1}{C_2}\right)$

We established earlier that for a spontaneous reaction, $E_{cell}$ must be positive. Let's set up the inequality: $-\frac{2.303RT}{2F}\log \left(\frac{C_1}{C_2}\right)>0$

Since the constants $\frac{2.303RT}{2F}$ are positive, the negative sign means the logarithmic term itself must be negative: $\log \left(\frac{C_1}{C_2}\right)<0$

A logarithm is negative only when its argument is less than 1. Therefore: $\frac{C_1}{C_2}<1$

Which beautifully simplifies to: $C_1<C_2$

This makes perfect physical sense! The cell will spontaneously operate to transfer copper from the more concentrated side ($C_2$) to the less concentrated side ($C_1$) until they are equal.

Looking at our options, we need to find the one where $C_2$ is greater than $C_1$. Option (d) states $C_2=\sqrt{2}{C}_1$. Since $\sqrt{2}\approx 1.414$, this means $C_2$ is indeed greater than $C_1$, satisfying our condition perfectly.