# Why does sodium react with water to produce a hydroxide, while zinc produces an oxide?

Sodium belongs to the so-called active metals and produces a hydroxide on contact with water. Zinc produces an oxide. Why is it so?

I was given a hint: to read up about electrochemical series. I've read about it, and naturally zinc is way below sodium in the series. Sodium is more enthusiastic about giving away its electrons and forming positive ions.

Zinc is slightly below this bolded line in the series, but I'm afraid I do not fully understand what that means. "Hydrogen gas bubbled through an alkaline solution, shedding two electrons to produce water and neutralize the solution"? The overall potential of an electrochemical cell is equal to the reduction potential for the substance being reduced plus the oxidation potential for the substance being oxidized. In other words, $ΔE_{rxn} = E°_{red} + E°_{ox}$. Now, all that we need to do is find a list of reduction potentials. From here, I found the following reduction potentials:

$$\ce{Na+(aq) + e- -> Na(s)}: E°_{red}=-2.71 V$$

$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH^{-}(aq)}: E°_{red}=-0.83 V$$

$$\ce{Zn^2+(aq) + 2e- -> Zn(s)}: E°_{red}=-0.76 V$$

A redox reaction occurs between one species being oxidized (losing electrons) and another being reduced (gaining electrons). Since all of these values are reduction potentials, we can flip reactants and products to get oxidation potentials. The oxidation potential for a reaction is negative one times the reduction potential (due to conservation of energy) and since we possess sodium and zinc metal rather than ions, we want for these reagents to be oxidized. This gives us the following equations with oxidation potentials:

$$\ce{Na(s) -> Na+(aq) + e-}: E°_{ox}=2.71 V$$

$$\ce{Zn(s)->Zn^2+(aq) + 2e-}: E°_{ox}=0.76 V$$

Now, you might be wondering what any of this has to do with your original question. It turns out that a reaction is spontaneous (will occur) when the $ΔE_{rxn}>0$. Calculating $ΔE_{rxn}$ for the oxidation of sodium gives us $ΔE_{rxn} = E°_{red} + E°_{ox}=-0.83V + 2.71V=1.88V$, which is greater than zero, so this reaction will be spontaneous. On the other hand, if we calculate the cell potential for zinc being oxidized and water being reduced, we get a negative number, so this reaction is not spontaneous. In other words, sodium will readily reduce water, but zinc will not. Since zinc cannot directly reduce water, it will react with water to produce zinc hydroxide and hydrogen gas. The zinc will then decompose into zinc oxide and water. The reaction mechanism for this is given here:

$$\ce{Zn + 2 H2O → Zn(OH)2 + H2}$$ $$\ce{Zn(OH)2 → ZnO + H2O }$$

You might be wondering why this reaction occurs. It turns out that the standard electrode is:

$$\ce{2H+(aq) + 2e- -> H2(g) + e-}: E°_{red}=0 V$$

which means that while zinc cannot reduce $\ce{H2O}$, it can reduce $\ce{H+}$, reacting with the protons present due to the natural decomposition of water. After losing two electrons, the positively charged zinc will react with the hydroxide found in solution. This creates zinc hydroxide, which is an insoluble compound. Sodium will remain dissociated from the hydroxide, as sodium hydroxide is water soluble. From here, the decomposition of zinc hydroxide is spontaneous, producing water and zinc oxide.

It has something to do with which potential product is more stable under the conditions required for the reaction, notably temperature. Active metals like sodium (also all other alkali metals, and magnesium and heavier alkaline earth metals) react with water at a low enough temperatures where the hydroxide is stable. Less active metals like zinc require heating to high temperatures where the hydroxide of the metal, if it were formed, would decompose to the oxide.