The reaction of sodium with water is quite infamous.

$\ce{2Na + 2H_2O \rightarrow 2NaOH + H_2}$

But exactly why does $\ce{Na}$ displace $\ce H$ in $\ce{H_2O}$?


2 Answers 2


Alkali metals in general are extremely reactive metals. They are highly reducing in nature. Hence, they react with water forming their corresponding hydroxides evolving dihydrogen gas.

In case of sodium - as cited in your question - its reduction potential is $\pu{-2.7109 V}$ hence it easily reduces hydrogen.

Also, as the density of sodium is quite low it floats on the water. It immediately reacts with water to form a white trail of $\ce{NaOH}$ which further gets dissolved to give a colourless solution.

This reaction is highly exothermic evolving a lot of heat which is sufficient enough to melt sodium due to its low melting point.

Sodium moves in water as the dihydrogen gas formed below the waterline pushes sodium. If sodium is trapped in the container the increasing temperature could induce dihydrogen to catch fire.


Sodium is more "electropositive" than hydrogen as it is down on the periodic table. This makes it more reactive to say elements like fluorine or oxygen. In this same way it is reactive to ions like hydroxide. Since water can turn into hydroxide and hydrogen ion the sodium ion is able to displace the hydrogen ion. This causes sodium to react to hydroxide to form sodium hydroxide and hydrogen. The two hydrogen atoms merge to form a hydrogen molecule.


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