# Energy level of the product in a cell reaction

I'm very new to chemistry, so please try to be as plain as possible with your answers. I'm studying the chemistry of alkaline cells:

The cell I'm interested in has the following chemical reactions going on:

\begin{align} &\text{Anode (oxidation):} &\ce{Zn (s) + 2 OH- (aq) &-> ZnO (s) + H2O (l) + 2 e-} \\ &\text{Cathode (reduction):} &\ce{2 MnO2 (s) + H2O (l) + 2 e- &-> Mn2O3 (s) + \color{red}{2 OH- (aq)}} \\ \end{align}

The part that I do not understand is marked in red. As far as I know, chemical reactions occur only if it results in a lower energy level than the reactants.

However, the second reaction creates a hydroxide that has a negative charge, wouldn't that violate that principle? How or why is that a lower energy state?

• Redox reactions and electrochemistry are but two unnecessary levels of complexity; let's remove them altogether. Throw some NaCl in the water; do you know what will happen next? – Ivan Neretin Nov 23 '17 at 18:03
• The two hydroxides you produce are consumed by the anodic oxidation process. Overall, no hydroxides are required. – Jan Nov 24 '17 at 3:49
• @IvanNeretin yeah they will dissociate into Na and Cl ions and float freely in the water because water is polar. – Joaquin Brandan Nov 24 '17 at 19:48
• @Jan I see that they are inmediately consumed, but why do they form in the first place?. – Joaquin Brandan Nov 24 '17 at 19:48
• Your question seems to implicitly assume that free-floating ions can't be a low energy state which comes about spontaneously. The example of NaCl dissolution in water shows that in fact it can. – Ivan Neretin Nov 26 '17 at 16:57

To quote from the question:

As far as I know, chemical reactions occur only if it results in a lower energy level than the reactants.

This is correct in principle and it is correct if your energy meter is in fact Gibbs free energy $G$. Reactions will occur spontaneously (are exergonic) if $\Delta G < 0$ and will not occur spontaneously if $\Delta G>0$.

However, you are applying that principle wrongly. You are looking at only one single product of one half-reaction to which you arbitrarily slap the label ‘high in energy’ — the hydroxide ion. First, it should be stated that ‘high in energy’ is noninformation unless you compare its energy to something else. For example, the enthalpy ($\Delta_\mathrm f H$, the measure typically associated with the energy content of a compound) of a hydroxide ion will be higher than that of water but lower than that of an oxide ion. Yet oxide ions are well known and form freely; sometimes even in aquaeous solution.[1]

The enthalpy of individual compounds that are formed or consumed in a reaction is, however, mostly irrelevant. What counts is the overall equation. Often, very reactive compounds with a high enthalpy content can be generated easily just be the liberation of something else that happens to be particularly stable.[2] The principal corresponding driving force in your reaction is the formation of zinc oxide from zinc metal.

Furthermore, you claim that hydroxide ions are ‘produced’ at all. That claim may be founded on a microscopic level, but as I mentioned these cancel out macroscopically. While it is common practice to separate oxidation and reduction half-reactions in education, this separation does not happen in real life. Rather, to determine enthalpy or Gibbs free energy, the full reaction $(1)$ must be examined. If you do that, you will notice that the hydroxides cancel out.

\begin{align}\ce{Zn + 2 OH- + 2 MnO2 + H2O + 2e- &-> ZnO + H2O + 2 e- + Mn2O3 + 2 OH-}\\ \ce{Zn + 2 MnO2 &-> ZnO + Mn2O3}\tag{1}\end{align}

When calculating enthalpy and entropy, the total hydroxide ion concentration remains the same and the formed hydroxides thus do not contribute to the final values at all.

Notes:

[1]: For example, if you dissolve an aluminium salt in hydrochloric acid an then slowly increase the pH for example with sodium hydroxide until you have reached a pH of just above seven, an aluminium cluster complex $\ce{[Al13(μ4{-}O)4(μ{-}OH)24(H2O)12]^7+}$ will be formed. (Reference in German)

[2]: For example, consider the peroxide dimer of benoic acid $\ce{PhCOO-OOCPh}$. The high-energy peroxide bond can be cleaved to give two benzoyl radicals. These will liberate carbon dioxide and form the very highly reactive, very unstable phenyl radical. This reaction $(2)$ is possible because of the enthalpy and entropy gain associated with the formation of $\ce{CO2}$

$$\ce{PhCOO-OOCPh -> 2 PhCOO^. -> 2 Ph^. + 2 CO2}\tag{2}$$