Animated Solution for Chemistry - Electrochemistry: The Gibbs energy change (in J) for the given reaction at $[C{u}^{2+}]=[S{n}^{2+}]=1\text{ M}$ and $298\text{ K}$ is $Cu(s)+S{n}^{2+}(aq)⟶C{u}^{2+}(aq)+Sn(s)$; $(E_{S{n}^{2+}/Sn}^{\circ }=-0.16\text{ V},E_{C{u}^{2+}/Cu}^{\circ }=0.34\text{ V}$, Take, $F=96500{\text{ C mol}}^{-1})$

Let's visualize the electrochemical cell for this reaction. We are given the chemical equation:

By looking closely at the oxidation states, we can see that solid copper ($Cu$) is turning into copper ions ($C{u}^{2+}$). This means it is losing electrons, which is the definition of oxidation. Therefore, copper acts as our anode.

Meanwhile, the tin ions ($S{n}^{2+}$) are gaining electrons to become solid tin ($Sn$). This process is reduction, making tin our cathode.

To find the Gibbs free energy change ($\Delta G$), we first need to determine the standard cell potential ($E_{cell}^{\circ }$). The formula is beautifully simple:

We know tin is the cathode and copper is the anode. Let's substitute their given standard reduction potentials into our equation.

Calculating this, we find the standard cell potential is $-0.50\text{ V}$. Notice the negative sign? It hints that this reaction isn't going to happen spontaneously on its own.

Now, since the concentrations of both ions are exactly $1\text{ M}$, the reaction quotient $Q$ is $1$, and the Nernst equation tells us that the cell potential equals the standard cell potential ($E_{cell}=E_{cell}^{\circ }$).

To find the Gibbs free energy change, we use the master equation:

How many electrons are transferred? Copper goes from $0$ to $+2$, so $n$ is $2$. Let's plug in $2$ for $n$, $96500$ for Faraday's constant ($F$), and $-0.50\text{ V}$ for the cell potential.

Multiply these together. The two negative signs cancel out, and $2\times 0.50$ is exactly $1$. So, we are left with exactly $96500\text{ Joules}$.

And there we have it! The Gibbs energy change for this reaction is $96500\text{ J}$. A positive $\Delta G$ confirms our earlier suspicion: this reaction is non-spontaneous. If we wanted this reaction to actually happen, we would need to supply external electrical energy, turning this into an electrolytic cell.