Metals reacting with water

Three bands, one ladder. Every metal that reacts with water at all does the same basic job — it takes the place of hydrogen in $H_2O$ — but how easily it does so tracks its position in the activity series, the same reactivity order used for reactions with acids and displacement reactions.

• Cold water: sodium, potassium, and calcium are reactive enough to displace hydrogen from water at room temperature, forming the metal hydroxide and releasing hydrogen gas.
• Hot water: magnesium sits just below that band — too weak for cold water, but hot water supplies enough extra energy for the same displacement.
• Steam: aluminium, zinc, and red-hot iron are weaker still; only steam (water vapour at high temperature) is energetic enough to force the reaction, and the high temperature means the product is the metal oxide, not the hydroxide.
• No reaction: metals below iron in the activity series (copper, silver, gold) do not react with water in any state.

The general pattern for the cold- and hot-water cases is:

where: $M$ is a monovalent reactive metal (sodium or potassium); calcium, being divalent, follows the closely related form $Ca + 2H_2O \rightarrow Ca(OH)_2 + H_2\uparrow$. In every case one water molecule contributes one hydroxide group and, together, two water molecules release one molecule of hydrogen gas.

The activity series ranks metals by how easily their atoms lose electrons to form positive ions — the more easily a metal loses electrons, the more strongly it wants to displace a less-willing element (here, the hydrogen in water) from its compound. Sodium and potassium sit at the very top: their outer electron is loosely held, so they tear hydrogen away from cold water almost instantly, releasing enough heat to melt themselves and sometimes ignite the hydrogen produced. Calcium is a little less eager, so its reaction with cold water is steady effervescence rather than a flare-up. Magnesium's electrons are held more tightly, so cold water simply does not supply enough energy for the displacement to start; heating the water gives the reacting particles the extra energy they need to break water's bonds and complete the swap. Aluminium, zinc, and iron are weaker again — heating the water is not enough, so the metal must be exposed to steam, whose particles are both hotter and more energetic than boiling liquid water. At those temperatures the metal hydroxide that would normally form is itself unstable and decomposes, which is why the steam reactions always give the metal oxide, never the hydroxide. Iron's reaction with steam is additionally reversible, because at that same high temperature hydrogen gas can just as easily reduce iron oxide back to iron and steam as the forward reaction can produce it — whichever direction wins depends on whether steam or hydrogen is kept in excess. Below iron, the pull on electrons is too weak for the metal to displace hydrogen from water under any ordinary condition, so copper, silver, and gold show no reaction at all — cold, hot, or as steam.

Question: 39 g of potassium is dropped into excess cold water. (a) Write the balanced equation. (b) Calculate the volume of hydrogen gas produced, measured at STP, where 1 mole of gas occupies 22.4 L. (Atomic mass of K = 39.)

Step 1. Write the balanced equation. Potassium is very high in the activity series, so it reacts vigorously with cold water: $2K + 2H_2O \rightarrow 2KOH + H_2\uparrow$.

Step 2. Convert the given mass of potassium to moles. Moles of $K = \dfrac{\text{mass}}{\text{atomic mass}} = \dfrac{39\ \text{g}}{39\ \text{g/mol}} = 1\ \text{mol}$.

Step 3. Use the mole ratio from the balanced equation. The equation shows 2 mol of $K$ produce 1 mol of $H_2$, a ratio of $2:1$. So 1 mol of $K$ produces $\dfrac{1}{2} = 0.5\ \text{mol}$ of $H_2$.

Step 4. Convert moles of hydrogen to volume at STP. Volume $= 0.5\ \text{mol} \times 22.4\ \text{L/mol} = 11.2\ \text{L}$.

39 g of potassium reacting with excess cold water releases 11.2 L of hydrogen gas at STP.

Competitive stretch

If the same mass of sodium (23 g = 1 mol, from $2Na + 2H_2O \rightarrow 2NaOH + H_2\uparrow$) reacted instead, the mole ratio is identical ($2:1$), so 1 mol of Na would also release 0.5 mol, or 11.2 L, of hydrogen — the volume of hydrogen depends only on the moles of metal reacting and the equation's ratio, not on which metal (Na, K, or Ca) is used, since all three follow the same $2M:1H_2$ pattern.

• Assuming every reactive metal reacts with cold water — magnesium needs hot water, and aluminium, zinc, and iron need steam.
• Writing the product of a steam reaction as a hydroxide — at steam temperatures the hydroxide is unstable, so the product is the metal oxide (for example $Al_2O_3$, $Fe_3O_4$).
• Dropping the reversible arrow for iron with steam — it is the one water reaction in this chapter that goes both ways, and exams specifically test it.
• Forgetting to convert mass to moles before applying the mole ratio from the equation, or using the wrong ratio (it is moles of metal to moles of $H_2$, not metal to $H_2O$).

Ask the tutor: Want to try predicting the products and states needed for a few more metals, or practise a stoichiometry question on hydrogen volume?