Types of Redox Reactions: Displacement

If combination reactions build something from pieces and decomposition reactions break something apart, displacement reactions are more like a swap: one element shoulders its way into a compound and kicks another element out. The incoming element is a better electron donor (stronger reducing agent), and that difference in reducing strength is what drives the entire reaction forward.

What Is a Displacement Reaction?

In a displacement reaction, an atom or ion inside a compound gets replaced by a different atom or ion from outside. The general form looks like this:

$X + YZ \rightarrow XZ + Y$

Here, $X$ pushes $Y$ out of the compound $YZ$ and takes its place. For this to happen, $X$ must be a stronger reducing agent than $Y$, meaning $X$ has a greater tendency to lose electrons.

Displacement reactions fall into two broad groups: metal displacement (one metal replaces another) and non-metal displacement (hydrogen or, rarely, oxygen gets replaced).

Metal Displacement: Extracting Pure Metals

When a free metal is dropped into a solution (or heated with a compound) of a less reactive metal, it drives that metal out. This principle is not just a classroom demonstration; it sits at the heart of many industrial metallurgical (metal-extraction) processes where pure metals are pulled from their ores.

Let us track the oxidation numbers through several key examples:

Zinc displacing copper from copper sulphate:

$\overset{+2 \quad +6 \quad -2}{CuSO_4}(aq) + \overset{0}{Zn}(s) \rightarrow \overset{0}{Cu}(s) + \overset{+2 \quad +6 \quad -2}{ZnSO_4}(aq) \qquad \text{(7.29)}$

• Zinc goes from $0$ to $+2$: it is oxidised (it loses electrons).
• Copper drops from $+2$ to $0$: it is reduced (it gains the electrons zinc lost).
• Zinc is the reducing agent; copper(II) is the oxidising agent.

Calcium reducing vanadium(V) oxide to vanadium metal:

$\overset{+5 \quad -2}{V_2O_5}(s) + \overset{0}{5Ca}(s) \xrightarrow{\Delta} \overset{0}{2V}(s) + \overset{+2 \quad -2}{5CaO}(s) \qquad \text{(7.30)}$

• Calcium rises from $0$ to $+2$ (oxidised).
• Vanadium falls from $+5$ to $0$ (reduced), emerging as pure metal.

Magnesium reducing titanium(IV) chloride (the Kroll process):

$\overset{+4 \quad -1}{TiCl_4}(l) + \overset{0}{2Mg}(s) \xrightarrow{\Delta} \overset{0}{Ti}(s) + \overset{+2 \quad -1}{2MgCl_2}(s) \qquad \text{(7.31)}$

• Magnesium goes from $0$ to $+2$ (oxidised).
• Titanium drops from $+4$ to $0$ (reduced), giving pure titanium metal used in aerospace and medical implants.

Aluminium reducing chromium(III) oxide (thermite-type reaction):

$\overset{+3 \quad -2}{Cr_2O_3}(s) + \overset{0}{2Al}(s) \xrightarrow{\Delta} \overset{+3 \quad -2}{Al_2O_3}(s) + \overset{0}{2Cr}(s) \qquad \text{(7.32)}$

• Aluminium climbs from $0$ to $+3$ (oxidised).
• Chromium falls from $+3$ to $0$ (reduced).

The Pattern Behind Every Metal Displacement

Notice the common thread in all four reactions above: the metal doing the displacing is always a better reducing agent than the metal being freed. It has a stronger tendency to give up electrons. Zinc beats copper, calcium beats vanadium, magnesium beats titanium, and aluminium beats chromium. This ranking of metals by their reducing ability is what chemists call the activity series (or reactivity series).

Non-Metal Displacement: Swapping Out Hydrogen or Oxygen

Non-metal displacement reactions mainly involve kicking hydrogen out of water or acids. A rarer variety involves displacing oxygen from water.

Hydrogen Displacement from Cold Water

The most reactive metals, all alkali metals (Group 1: $Li$, $Na$, $K$, etc.) and a few alkaline earth metals ($Ca$, $Sr$, and $Ba$), are powerful enough reducing agents to rip hydrogen right out of cold water at room temperature:

$\overset{0}{2Na}(s) + \overset{+1 \quad -2}{2H_2O}(l) \rightarrow \overset{+1 \quad -2 \quad +1}{2NaOH}(aq) + \overset{0}{H_2}(g) \qquad \text{(7.33)}$

$\overset{0}{Ca}(s) + \overset{+1 \quad -2}{2H_2O}(l) \rightarrow \overset{+2 \quad -2 \quad +1}{Ca(OH)_2}(aq) + \overset{0}{H_2}(g) \qquad \text{(7.34)}$

In both cases, hydrogen drops from $+1$ in water to $0$ in $H_2$ (it is reduced), while the metal is oxidised from $0$ to a positive state.

Hydrogen Displacement from Steam

Metals that are moderately active, like magnesium and iron, are not reactive enough for cold water. They need the extra energy provided by steam (water at high temperature):

$\overset{0}{Mg}(s) + \overset{+1 \quad -2}{2H_2O}(l) \xrightarrow{\Delta} \overset{+2 \quad -2 \quad +1}{Mg(OH)_2}(s) + \overset{0}{H_2}(g) \qquad \text{(7.35)}$

$\overset{0}{2Fe}(s) + \overset{+1 \quad -2}{3H_2O}(l) \xrightarrow{\Delta} \overset{+3 \quad -2}{Fe_2O_3}(s) + \overset{0}{3H_2}(g) \qquad \text{(7.36)}$

The principle is the same: the metal is oxidised, hydrogen is reduced.

Hydrogen Displacement from Acids

Even metals that will not react with steam can sometimes displace hydrogen from dilute acids. Cadmium ($Cd$) and tin ($Sn$) are examples of metals that sit too low on the activity series for water or steam but still manage to pull hydrogen out of acid solutions. Some common laboratory reactions for preparing $H_2$ gas:

$\overset{0}{Zn}(s) + \overset{+1 \quad -1}{2HCl}(aq) \rightarrow \overset{+2 \quad -1}{ZnCl_2}(aq) + \overset{0}{H_2}(g) \qquad \text{(7.37)}$

$\overset{0}{Mg}(s) + \overset{+1 \quad -1}{2HCl}(aq) \rightarrow \overset{+2 \quad -1}{MgCl_2}(aq) + \overset{0}{H_2}(g) \qquad \text{(7.38)}$

$\overset{0}{Fe}(s) + \overset{+1 \quad -1}{2HCl}(aq) \rightarrow \overset{+2 \quad -1}{FeCl_2}(aq) + \overset{0}{H_2}(g) \qquad \text{(7.39)}$

These three reactions (7.37 to 7.39) are widely used for laboratory preparation of dihydrogen gas. The speed at which $H_2$ bubbles appear tells you about each metal’s reactivity: magnesium produces gas the fastest (strongest reducing agent of the three), zinc is moderate, and iron is the slowest (weakest reducing agent of the three).

Metals that sit at the very bottom of the activity series, such as silver ($Ag$) and gold ($Au$), are so unreactive that they will not displace hydrogen even from hydrochloric acid. Their extreme stability is why they occur as free, uncombined elements in nature.

Halogen Displacement: The Non-Metal Activity Series

Just as metals have an activity series based on their ability to lose electrons (reducing power), the halogens (Group 17) have their own ranking based on their ability to gain electrons (oxidising power). This power decreases as you go down the group:

$F_2 > Cl_2 > Br_2 > I_2$

A higher halogen can always displace a lower halogen from its salt solution. Let us see how this plays out.

Fluorine: The Strongest Oxidiser

Fluorine sits at the top of the halogen activity series. It is so aggressively reactive that it can even displace oxygen from water, something no other halogen can do:

$\overset{+1 \quad -2}{2H_2O}(l) + \overset{0}{2F_2}(g) \rightarrow \overset{+1 \quad -1}{4HF}(aq) + \overset{0}{O_2}(g) \qquad \text{(7.40)}$

• Fluorine drops from $0$ to $-1$ (reduced); it acts as the oxidising agent.
• Oxygen rises from $-2$ to $0$ (oxidised); water’s oxygen is literally forced out as $O_2$ gas.

Because fluorine attacks the water solvent itself, displacement reactions of chlorine, bromine, or iodine using fluorine are not carried out in aqueous solution. The solvent would be destroyed before the intended displacement could happen.

Chlorine Displacing Bromide and Iodide

Chlorine is a strong enough oxidiser to push out both bromide ($Br^-$) and iodide ($I^-$) from their solutions:

$\overset{0}{Cl_2}(g) + \overset{+1 \quad -1}{2KBr}(aq) \rightarrow \overset{+1 \quad -1}{2KCl}(aq) + \overset{0}{Br_2}(l) \qquad \text{(7.41)}$

$\overset{0}{Cl_2}(g) + \overset{+1 \quad -1}{2KI}(aq) \rightarrow \overset{+1 \quad -1}{2KCl}(aq) + \overset{0}{I_2}(s) \qquad \text{(7.42)}$

Both $Br_2$ and $I_2$ are coloured substances. When they dissolve in carbon tetrachloride ($CCl_4$), they produce distinct colours that are easy to spot. This colour appearance is the basis of a well-known laboratory identification method called the Layer Test used to confirm the presence of $Br^-$ and $I^-$ ions in solution.

Writing these same reactions in ionic form makes the electron transfer crystal clear, since the potassium and chloride spectator ions drop out:

$\overset{0}{Cl_2}(g) + \overset{-1}{2Br^-}(aq) \rightarrow \overset{-1}{2Cl^-}(aq) + \overset{0}{Br_2}(l) \qquad \text{(7.41a)}$

$\overset{0}{Cl_2}(g) + \overset{-1}{2I^-}(aq) \rightarrow \overset{-1}{2Cl^-}(aq) + \overset{0}{I_2}(s) \qquad \text{(7.42b)}$

In each case, $Cl_2$ gains electrons (reduced from $0$ to $-1$) and the halide ion loses electrons (oxidised from $-1$ to $0$).

Bromine Displacing Iodide

Following the same logic, bromine is a stronger oxidiser than iodine, so it can displace iodide from solution:

$\overset{0}{Br_2}(l) + \overset{-1}{2I^-}(aq) \rightarrow \overset{-1}{2Br^-}(aq) + \overset{0}{I_2}(s) \qquad \text{(7.43)}$

However, iodine cannot displace chloride or bromide, and bromine cannot displace chloride. Each halogen can only displace those below it in the series.

Why Fluorine Cannot Be Made by Chemical Means

Recovering any halogen from its halide ions requires oxidising the halide:

$2X^- \rightarrow X_2 + 2e^- \qquad \text{(7.44)}$

where $X$ represents a halogen. For $Cl^-$, $Br^-$, and $I^-$, chemical oxidising agents exist that can accomplish this. But for $F^-$, the situation is unique: fluorine itself is the strongest oxidising agent in all of chemistry. No chemical substance can pull electrons away from $F^-$ to form $F_2$. The only route to fluorine gas is electrolysis (using electrical energy to force the oxidation), which you will study in detail later.