Animated Solution for Chemistry - Electrochemistry: Calculate the standard cell potential (in V) of the cell in which following reaction takes place $F{e}^{2+}(aq)+A{g}^{+}(aq)⟶F{e}^{3+}(aq)+Ag(s)$ Given that, $E_{A{g}^{+}/Ag}^{\circ }=x\text{ V}$, $E_{F{e}^{2+}/Fe}^{\circ }=y\text{ V}$ $E_{F{e}^{3+}/Fe}^{\circ }=z\text{ V}$

Welcome to a classic electrochemistry problem that tests your fundamental understanding of thermodynamics! We are tasked with finding the standard cell potential ($E_{cell}^{\circ }$) for a specific reaction:

To find the cell potential, we need to identify the cathode and the anode. By looking at the oxidation states, we can see that silver ($A{g}^{+}$) is being reduced to solid silver ($Ag$), making it the cathode. Conversely, iron(II) ($F{e}^{2+}$) is being oxidized to iron(III) ($F{e}^{3+}$), making it the anode.

The standard formula for cell potential is:

Substituting our specific half-reactions, we get:

We are given that $E_{A{g}^{+}/Ag}^{\circ }=x$. However, we hit a roadblock: the problem does not directly provide the standard reduction potential for $F{e}^{3+}\to F{e}^{2+}$. Instead, we are given the potentials for $F{e}^{2+}\to Fe$ (which is $y$) and $F{e}^{3+}\to Fe$ (which is $z$). We must calculate the missing potential!

Gibbs Free Energy

This is where many students make a critical error. You might be tempted to simply add or subtract the given potentials ($y$ and $z$) to find the missing one. Do not do this! Standard electrode potentials ($E^{\circ }$) are intensive properties, meaning they do not depend on the amount of substance and cannot be algebraically added.

Instead, we must use an extensive property that can be added: the standard Gibbs free energy ($\Delta G^{\circ }$). The relationship between the two is given by the master equation:

Here, $n$ is the number of electrons transferred, and $F$ is the Faraday constant.

Let's visualize the reduction of iron as a thermodynamic cycle. Iron(III) can be reduced to solid iron in a single step, or it can be reduced in two steps: first to iron(II), and then to solid iron. According to Hess's Law, the total energy change must be the same regardless of the path taken.

Therefore, we can write:

Let's define the terms for each step:

Step 1: $F{e}^{2+}+2e^{-}⟶Fe$ Here, $n=2$ and $E^{\circ }=y$. So, $\Delta G_1^{\circ }=-2Fy$.

Step 2: $F{e}^{3+}+3e^{-}⟶Fe$ Here, $n=3$ and $E^{\circ }=z$. So, $\Delta G_2^{\circ }=-3Fz$.

Step 3 (Our Target): $F{e}^{3+}+e^{-}⟶F{e}^{2+}$ Here, $n=1$ and $E^{\circ }=E_3^{\circ }$ (unknown). So, $\Delta G_3^{\circ }=-1\cdot F\cdot E_3^{\circ }$.

Now, we substitute these expressions back into our Hess's Law equation:

Notice how the Faraday constant ($F$) appears in every term? We can divide the entire equation by $-F$ to simplify it:

Rearranging to solve for our unknown potential, $E_3^{\circ }$:

We have successfully found the standard reduction potential for $F{e}^{3+}/F{e}^{2+}$! Now, we simply plug this back into our original cell potential equation:

Distributing the negative sign, we arrive at our final answer:

This perfectly matches option (a). The key takeaway from this problem is to always rely on Gibbs free energy when combining half-reactions with different numbers of electrons. Never add potentials directly!