Animated Solution for Chemistry - Electrochemistry: Given, $E_{F{e}^{3+}/Fe}^{\circ }=-0.036\text{ V}$, $E_{F{e}^{2+}/Fe}^{\circ }=-0.439\text{ V}$. The value of standard electrode potential for the charge, $F{e}^{3+}(aq)+e^{-}⟶F{e}^{2+}(aq)$ will be

Why We Can't Just Add Them Up

Imagine you are trying to find the temperature of a mixture of two cups of water. If one cup is at $40^{\circ }\text{C}$ and the other is at $60^{\circ }\text{C}$, you wouldn't just add them to say the mixture is $100^{\circ }\text{C}$, right? Temperature is an intensive property; it doesn't depend on the amount of substance.

Similarly, in electrochemistry, the standard electrode potential ($E^{\circ }$) is an intensive property. A very common trap students fall into is directly adding or subtracting $E^{\circ }$ values when combining half-reactions.

In our problem, we are given: $F{e}^{3+}+3e^{-}⟶Fe{E}_1^{\circ }=-0.036\text{ V}$ $F{e}^{2+}+2e^{-}⟶Fe{E}_2^{\circ }=-0.439\text{ V}$

We need to find the potential for the intermediate step: $F{e}^{3+}+e^{-}⟶F{e}^{2+}{E}_3^{\circ }=?$

Gibbs Free Energy

To navigate around this trap, we must use an extensive property—one that scales with the amount of substance and can be algebraically added or subtracted. Enter standard Gibbs free energy change ($\Delta G^{\circ }$).

The magic bridge connecting the intensive $E^{\circ }$ to the extensive $\Delta G^{\circ }$ is the equation:

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

Let's convert our known potentials into Gibbs free energy.

For the first reaction ($F{e}^{3+}\to Fe$), three electrons are involved ($n=3$): $\Delta G_1^{\circ }=-3\times F\times (-0.036)=+0.108F$

For the second reaction ($F{e}^{2+}\to Fe$), two electrons are involved ($n=2$): $\Delta G_2^{\circ }=-2\times F\times (-0.439)=+0.878F$

Now, how do we construct our target reaction? If we take the first equation and subtract the second equation from it, the solid $Fe$ cancels out perfectly, leaving us with $F{e}^{3+}+e^{-}⟶F{e}^{2+}$.

Since we subtracted the equations, we must subtract their Gibbs free energies: $\Delta G_3^{\circ }=\Delta G_1^{\circ }-\Delta G_2^{\circ }$ $\Delta G_3^{\circ }=0.108F-0.878F=-0.770F$

We have the Gibbs free energy for our target reaction. Now, we just convert it back to an electrode potential. For the target reaction, only one electron is transferred ($n=1$).

$\Delta G_3^{\circ }=-n_{3}F{E}_3^{\circ }$ $-0.770F=-1\times F\times E_3^{\circ }$

The Faraday constant $F$ and the negative signs cancel out beautifully, leaving us with:

This sequence of reactions is a classic example of a Latimer Diagram. Always remember the golden rule: never add potentials directly. Convert to $\Delta G^{\circ }$, do the math, and convert back!