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Chemistry · Ch 8 — Redox Reactions

Non-metal displacement

Non-metal displacement

This category includes hydrogen displacement and, more rarely, oxygen displacement.

Hydrogen Displacement from Water

Very strong reducing metals (alkali metals and some alkaline earth metals like Ca, Sr, Ba) displace hydrogen from cold water:

2Na0(s)+2H2+1O−2(l)→2Na+1O−2H+1(aq)+H20(g)(7.33) 2\overset{0}{Na}(s) + 2\overset{+1}{H_2}\overset{-2}{O}(l) \rightarrow 2\overset{+1}{Na}\overset{-2}{O}\overset{+1}{H}(aq) + \overset{0}{H_2}(g) \qquad(7.33)

Ca0(s)+2H2+1O−2(l)→Ca+2(O−2H+1)2(aq)+H20(g)(7.34) \overset{0}{Ca}(s) + 2\overset{+1}{H_2}\overset{-2}{O}(l) \rightarrow \overset{+2}{Ca}(\overset{-2}{O}\overset{+1}{H})_2(aq) + \overset{0}{H_2}(g) \qquad(7.34)

Less active metals like magnesium and iron react with steam (not cold water) to produce hydrogen:

Mg0(s)+2H2+1O−2(l)→ΔMg+2(O−2H+1)2(s)+H20(g)(7.35) \overset{0}{Mg}(s) + 2\overset{+1}{H_2}\overset{-2}{O}(l) \xrightarrow{\Delta} \overset{+2}{Mg}(\overset{-2}{O}\overset{+1}{H})_2(s) + \overset{0}{H_2}(g) \qquad(7.35)

2Fe0(s)+3H2+1O−2(l)→ΔFe2+3O3−2(s)+3H20(g)(7.36) 2\overset{0}{Fe}(s) + 3\overset{+1}{H_2}\overset{-2}{O}(l) \xrightarrow{\Delta} \overset{+3}{Fe_2}\overset{-2}{O_3}(s) + 3\overset{0}{H_2}(g) \qquad(7.36)

Hydrogen Displacement from Acids

Many metals that do not react with water or steam can still displace hydrogen from acids. Even cadmium and tin — which do not react with steam — can produce hydrogen from acids.

Laboratory preparation of dihydrogen gas:

Zn0(s)+2H+1Cl−1(aq)→Zn+2Cl2−1(aq)+H20(g)(7.37) \overset{0}{Zn}(s) + 2\overset{+1}{H}\overset{-1}{Cl}(aq) \rightarrow \overset{+2}{Zn}\overset{-1}{Cl_2}(aq) + \overset{0}{H_2}(g) \qquad(7.37)

Mg0(s)+2H+1Cl−1(aq)→Mg+2Cl2−1(aq)+H20(g)(7.38) \overset{0}{Mg}(s) + 2\overset{+1}{H}\overset{-1}{Cl}(aq) \rightarrow \overset{+2}{Mg}\overset{-1}{Cl_2}(aq) + \overset{0}{H_2}(g) \qquad(7.38)

Fe0(s)+2H+1Cl−1(aq)→Fe+2Cl2−1(aq)+H20(g)(7.39) \overset{0}{Fe}(s) + 2\overset{+1}{H}\overset{-1}{Cl}(aq) \rightarrow \overset{+2}{Fe}\overset{-1}{Cl_2}(aq) + \overset{0}{H_2}(g) \qquad(7.39)

The rate of hydrogen evolution reflects the reactivity of the metal — fastest for Mg, slowest for Fe. Very unreactive metals like silver and gold (which occur in the native state) do not react even with hydrochloric acid.

Tip

The relative reducing power of metals follows the activity series. For Zn, Cu, and Ag, the order of reducing activity is: Zn > Cu > Ag. This means Zn can displace Cu from its compounds, but Cu cannot displace Zn.

Halogen Displacement

Just as metals have an activity series, halogens also have a reactivity order. The oxidising power of halogens decreases down Group 17:

F2>Cl2>Br2>I2\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2

Fluorine is so reactive that it can displace chloride, bromide, and iodide ions from their compounds. In fact, fluorine is so aggressive that it even attacks water and displaces oxygen:

2H2+1O−2(l)+2F20(g)→4H+1F−1(aq)+O20(g)(7.40) 2\overset{+1}{H_2}\overset{-2}{O}(l) + 2\overset{0}{F_2}(g) \rightarrow 4\overset{+1}{H}\overset{-1}{F}(aq) + \overset{0}{O_2}(g) \qquad(7.40)

Because fluorine reacts with water itself, displacement reactions of chlorine, bromine, and iodine using fluorine are not carried out in aqueous solution.

Chlorine can displace bromide and iodide ions in aqueous solution:

Cl20(g)+2K+1Br−1(aq)→2K+1Cl−1(aq)+Br20(l)(7.41) \overset{0}{Cl_2}(g) + 2\overset{+1}{K}\overset{-1}{Br}(aq) \rightarrow 2\overset{+1}{K}\overset{-1}{Cl}(aq) + \overset{0}{Br_2}(l) \qquad(7.41)

Cl20(g)+2K+1I−1(aq)→2K+1Cl−1(aq)+I20(s)(7.42) \overset{0}{Cl_2}(g) + 2\overset{+1}{K}\overset{-1}{I}(aq) \rightarrow 2\overset{+1}{K}\overset{-1}{Cl}(aq) + \overset{0}{I_2}(s) \qquad(7.42)

In ionic form:

Cl20(g)+2Br−−1(aq)→2Cl−−1(aq)+Br20(l)(7.41a) \overset{0}{Cl_2}(g) + 2\overset{-1}{Br^-}(aq) \rightarrow 2\overset{-1}{Cl^-}(aq) + \overset{0}{Br_2}(l) \qquad(7.41a)

Cl20(g)+2I−−1(aq)→2Cl−−1(aq)+I20(s)(7.42b) \overset{0}{Cl_2}(g) + 2\overset{-1}{I^-}(aq) \rightarrow 2\overset{-1}{Cl^-}(aq) + \overset{0}{I_2}(s) \qquad(7.42b)

Bromine can also displace iodide ions:

Br20(l)+2I−−1(aq)→2Br−−1(aq)+I20(s)(7.43) \overset{0}{Br_2}(l) + 2\overset{-1}{I^-}(aq) \rightarrow 2\overset{-1}{Br^-}(aq) + \overset{0}{I_2}(s) \qquad(7.43) …