Animated Solution for Chemistry - Electrochemistry: The reduction potential ($E^0$, in V) of ${\text{MnO}}_4^{-}(\text{aq})/\text{Mn}(\text{s})$ is _______. [Given : $E_{({\text{MnO}}_4^{-}(\text{aq})/{\text{MnO}}_2(\text{s}))}^0=1.68\text{ V}$ ; $E_{({\text{MnO}}_2(\text{s})/{\text{Mn}}^{2+}(\text{aq}))}^0=1.21\text{ V}$ ; $E_{({\text{Mn}}^{2+}(\text{aq})/\text{Mn}(\text{s}))}^0=-1.03\text{ V}$ ]

Mapping the Journey

Imagine you are tracking the journey of a manganese atom as it slowly gains electrons, stepping down from a highly oxidized state to its pure metallic form. This step-by-step reduction is beautifully captured in a Latimer diagram. We start with the permanganate ion, ${\text{MnO}}_4^{-}$, where manganese sits at a $+7$ oxidation state. It first reduces to ${\text{MnO}}_2$ ($+4$), then to ${\text{Mn}}^{2+}$ ($+2$), and finally to solid $\text{Mn}$ ($0$).

The problem asks us to find the overall standard reduction potential, $E^{\circ }$, for the direct leap from ${\text{MnO}}_4^{-}$ all the way down to $\text{Mn}$.

Why We Can't Just Add Potentials

Here is where many students fall into a classic trap. You might look at the individual potentials—$1.68\text{ V}$, $1.21\text{ V}$, and $-1.03\text{ V}$—and think, "Why not just add them up?"

Don't make this silly mistake! Standard reduction potentials are intensive properties. Just like you cannot add the temperatures of two cups of water to find the final temperature, you cannot directly add voltages of consecutive half-reactions. To combine these steps, we must cross over into the realm of thermodynamics and use a property that is additive: the standard Gibbs free energy change, $\Delta G^{\circ }$.

Gibbs Free Energy

Gibbs free energy is an extensive property, meaning it depends on the amount of substance and the total number of electrons transferred. The relationship is given by the master equation:

Because $\Delta G^{\circ }$ is a state function, the total free energy change for the overall reaction is exactly equal to the sum of the free energy changes of the individual steps:

Let's substitute our master equation into this additivity rule. Remember, $n$ represents the number of electrons transferred. For the overall reaction from $+7$ to $0$, a total of $7$ electrons are transferred. For the individual steps, the electron transfers are $3$, $2$, and $2$ respectively.

Notice how the Faraday constant, $F$, and the negative signs appear in every single term. We can cleanly divide the entire equation by $-F$, leaving us with a beautiful, simple algebraic expression:

Now, it is just a matter of careful arithmetic. Let's multiply the values out:

Finally, we divide by $7$ to isolate our overall standard reduction potential:

Rounding to two decimal places as required by typical numerical formats, we get $0.77\text{ V}$. This problem is a perfect demonstration of why we must always rely on fundamental thermodynamic principles rather than taking intuitive, but incorrect, shortcuts.