1.1 - Oxidation and Reduction

Redox Reactions

• A redox reaction is a chemical reaction in which oxidation and reduction occur simultaneously.
• The processes can be recognised in four equivalent ways:

Oxidation	Reduction
gain of oxygen	loss of oxygen
loss of hydrogen	gain of hydrogen
loss of electrons	gain of electrons
increase in oxidation number	decrease in oxidation number

• An oxidising agent oxidises another substance and is itself reduced.

• A reducing agent reduces another substance and is itself oxidised.

• Example: $2Mg+{\mathrm{O}}_2\to 2MgO$. Magnesium loses electrons and is oxidised; oxygen gains electrons and is reduced.

• Oxygen transfer: $2Mg+\mathrm{C}{\mathrm{O}}_2\to 2MgO+\mathrm{C}$. Magnesium gains oxygen and reduces carbon dioxide; carbon dioxide loses oxygen and oxidises magnesium.

• Hydrogen transfer: ${\mathrm{H}}_2\mathrm{S}+\mathrm{C}{\mathrm{l}}_2\to \mathrm{S}+2HCl$. Hydrogen sulphide loses hydrogen and is the reducing agent; chlorine gains hydrogen and is the oxidising agent.

• In magnesium combustion, the half-equations are $\mathrm{Mg}\to \mathrm{M}{\mathrm{g}}^{2+}+2{\mathrm{e}}^{-}$ and ${\mathrm{O}}_2+4{\mathrm{e}}^{-}\to 2{\mathrm{O}}^{2-}$.

• Other examples: carbon reduces copper(II) oxide; ammonia reduces copper(II) oxide in $2N{\mathrm{H}}_3+3CuO\to {\mathrm{N}}_2+3Cu+3{\mathrm{H}}_2\mathrm{O}$.

Electron Transfer at a Distance

• Oxidation and reduction can occur in separate regions when electrons travel through a connecting wire.
• In the iron(II)/acidified manganate(VII) system:
  • oxidation: $\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}$;
  • reduction: $\mathrm{Mn}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to \mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$;
  • overall: $5F{\mathrm{e}}^{2+}+\mathrm{Mn}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}\to 5F{\mathrm{e}}^{3+}+\mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$.
• Electrons flow from the reducing agent at the negative terminal to the oxidising agent at the positive terminal. The electrolyte completes the circuit by allowing ions to move.

Investigation of Electron Transfer

• Aim: Investigate oxidation and reduction in terms of electron transfer at a distance.
• Materials: $1.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ sulphuric acid, freshly prepared $0.2\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ iron(II) sulphate, $0.1\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ acidified potassium manganate(VII), and $0.5\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ sodium hydroxide.
• Apparatus: U-tube, retort stand, carbon electrodes, galvanometer, connecting wires with crocodile clips, dropper and test tube.
• Procedure:
  1. Half-fill a vertically clamped U-tube with sulphuric acid. Carefully add a $3\mathrm{cm}$ layer of iron(II) sulphate in arm X and a $3\mathrm{cm}$ layer of acidified potassium manganate(VII) in arm Y.
  2. Connect the carbon electrodes through the galvanometer and immerse each electrode in its reactant layer. The electrodes must not touch the sulphuric acid layer.
  3. Leave for 30 minutes, noting galvanometer deflection and solution colour changes.
  4. Transfer a little iron(II) sulphate solution to a test tube and add sodium hydroxide to verify the iron(III) product.
• Results: Iron(II) solution changes from green to brown; acidified manganate(VII) changes from purple to colourless. Galvanometer deflection indicates electron flow.
• Conclusion: Electrons move through the wire from iron(II) ions, the reducing agent, to the oxidising-agent side; sulphuric acid is the separating electrolyte.
• To combine half-equations, balance atoms, balance charge with electrons, equalise electrons lost and gained, then add and cancel the electrons.

Common Oxidising and Reducing Agents

• Oxidising agents accept electrons:

Agent	Reduction half-equation
Acidified potassium manganate(VII)	$\mathrm{Mn}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to \mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$
Acidified potassium dichromate(VI)	$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$
Chlorine water	$\mathrm{C}{\mathrm{l}}_2+2{\mathrm{e}}^{-}\to 2C{\mathrm{l}}^{-}$
Bromine water	$\mathrm{B}{\mathrm{r}}_2+2{\mathrm{e}}^{-}\to 2B{\mathrm{r}}^{-}$
Iron(III) chloride	$\mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}\to \mathrm{F}{\mathrm{e}}^{2+}$
Acidified hydrogen peroxide	${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}$

• Reducing agents donate electrons:

Agent	Oxidation half-equation
Potassium iodide	$2{\mathrm{I}}^{-}\to {\mathrm{I}}_2+2{\mathrm{e}}^{-}$
Potassium bromide	$2B{\mathrm{r}}^{-}\to \mathrm{B}{\mathrm{r}}_2+2{\mathrm{e}}^{-}$
Iron(II) sulphate	$\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}$
Reactive metals, such as zinc	$\mathrm{Zn}\to \mathrm{Z}{\mathrm{n}}^{2+}+2{\mathrm{e}}^{-}$
Sulphur dioxide	$\mathrm{S}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{S}{\mathrm{O}}_4^{2-}+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$

Oxidation Number

• Oxidation numbers extend the description of redox reactions to reactions involving covalent compounds. In $\mathrm{C}+{\mathrm{O}}_2\to \mathrm{C}{\mathrm{O}}_2$, carbon changes from $0$ to $+4$ and is oxidised; oxygen changes from $0$ to $-2$ and is reduced.
• The oxidation number or oxidation state is the charge assigned to an element in a compound if bonding electrons are treated as transferred completely.
• Main rules:
  • a free element has oxidation number $0$;
  • a monatomic ion has an oxidation number equal to its charge;
  • elements of Groups 1, 2 and 13 in their compounds have oxidation numbers $+1$, $+2$ and $+3$ respectively;
  • the sum in a neutral compound is $0$; the sum in a polyatomic ion equals the ion charge;
  • fluorine is always $-1$;
  • chlorine, bromine and iodine are normally $-1$, except when bonded to a more electronegative element such as oxygen;
  • hydrogen is normally $+1$, but is $-1$ in metal hydrides;
  • oxygen is normally $-2$, but is $-1$ in peroxides.
• Example: in $\mathrm{Mn}{\mathrm{O}}_4^{-}$, $x+4(-2)=-1$, so Mn is $+7$.
• Roman numerals in IUPAC names show the oxidation number of a metal that has more than one oxidation state, for example iron(II) nitrate, $\mathrm{Fe}(\mathrm{N}{\mathrm{O}}_3{)}_2$, and iron(III) nitrate, $\mathrm{Fe}(\mathrm{N}{\mathrm{O}}_3{)}_3$.
• Further redox examples are $\mathrm{CuO}+{\mathrm{H}}_2\to \mathrm{Cu}+{\mathrm{H}}_2\mathrm{O}$ and $\mathrm{Mg}+2HCl\to \mathrm{MgC}{\mathrm{l}}_2+{\mathrm{H}}_2$.
• In $\mathrm{S}{\mathrm{O}}_3$, $x+3(-2)=0$, so S is $+6$; in $\mathrm{C}{\mathrm{O}}_3^{2-}$, $x+3(-2)=-2$, so C is $+4$.
• Metals with only one oxidation number omit Roman numerals: $\mathrm{Mg}(\mathrm{N}{\mathrm{O}}_3{)}_2$ is magnesium nitrate.
• Other IUPAC names: $\mathrm{Mn}{\mathrm{O}}_2$ = manganese(IV) oxide; $\mathrm{KMn}{\mathrm{O}}_4$ = potassium manganate(VII); $\mathrm{CrC}{\mathrm{l}}_3$ = chromium(III) chloride; ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$ = potassium dichromate(VI).

Interconversion of Iron(II) and Iron(III) Ions

• $\mathrm{F}{\mathrm{e}}^{2+}$ is oxidised to $\mathrm{F}{\mathrm{e}}^{3+}$ by an oxidising agent: $\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}$.

• $\mathrm{F}{\mathrm{e}}^{3+}$ is reduced to $\mathrm{F}{\mathrm{e}}^{2+}$ by a reducing agent: $\mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}\to \mathrm{F}{\mathrm{e}}^{2+}$.

• Iron(II) solution is green; iron(III) solution is brown. Oxidation changes the oxidation number from $+2$ to $+3$; reduction changes it from $+3$ to $+2$.

• Iron(II) gluconate, $\mathrm{Fe}({\mathrm{C}}_6{\mathrm{H}}_{11}{\mathrm{O}}_7{)}_2$, is used in iron pills. Iron ions can be verified with sodium hydroxide, aqueous ammonia, potassium hexacyanoferrate(II), potassium hexacyanoferrate(III) or potassium thiocyanate.

Investigation of Iron-ion Interconversion

• Aim: Study conversion of iron(II) ions to iron(III) ions and vice versa.
• Materials: Freshly prepared $0.5\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ iron(II) sulphate, $0.5\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ iron(III) chloride, bromine water, zinc powder and $2.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ sodium hydroxide.
• Apparatus: Boiling tubes, test tubes, boiling-tube holder, dropper, spatula, filter funnel, filter paper, Bunsen burner and test-tube rack.
• Procedure:
  1. Add bromine water dropwise to $2\mathrm{c}{\mathrm{m}}^3$ iron(II) sulphate while shaking until no further colour change occurs. Heat gently and add sodium hydroxide to verify iron(III) ions.
  2. Add half a spatula of zinc powder to $2\mathrm{c}{\mathrm{m}}^3$ iron(III) chloride. Heat gently, filter, and add sodium hydroxide to the filtrate to verify iron(II) ions.
• Conclusion: Bromine water oxidises iron(II) ions; zinc reduces iron(III) ions.

Displacement as a Redox Reaction

Metals

• A more electropositive metal releases electrons more readily and displaces a less electropositive metal from its salt solution.

• Example: $\mathrm{Mg}+\mathrm{C}{\mathrm{u}}^{2+}\to \mathrm{M}{\mathrm{g}}^{2+}+\mathrm{Cu}$.

  • $\mathrm{Mg}\to \mathrm{M}{\mathrm{g}}^{2+}+2{\mathrm{e}}^{-}$: magnesium is oxidised and is the reducing agent.
  • $\mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$: copper(II) ions are reduced and are the oxidising agent.

• If the solid metal is less electropositive than the metal ion, no displacement occurs.

• Electropositivity is the tendency of an atom to release electrons and form a cation. The electrochemical series arranges metals from the most negative to the most positive $E^0$.

• In the illustrated zinc/copper(II) nitrate reaction, the blue solution becomes paler, brown copper deposits, and zinc becomes thinner: $\mathrm{Zn}+\mathrm{C}{\mathrm{u}}^{2+}\to \mathrm{Z}{\mathrm{n}}^{2+}+\mathrm{Cu}$.

• Metals higher in the series are stronger reducing agents; metal ions lower in the series are stronger oxidising agents.

Metal-displacement investigation: Magnesium, lead and copper are tested with their nitrate solutions; apparatus comprises test tubes, a test-tube rack and sandpaper.

• Use cleaned magnesium, lead and copper with $0.5\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ solutions of their nitrates.
• Aim: Investigate redox in metal displacement. Evidence includes deposits, changes to the metal and changes in solution colour; electron transfer explains reacting pairs.

Halogens

• Halogen atoms accept electrons to form halide ions, so halogens act as oxidising agents.
• Oxidising strength decreases down Group 17: $\mathrm{C}{\mathrm{l}}_2>\mathrm{B}{\mathrm{r}}_2>{\mathrm{I}}_2$.
• A more reactive halogen displaces the halide ion of a less reactive halogen, for example: $\mathrm{C}{\mathrm{l}}_2+2B{\mathrm{r}}^{-}\to 2C{\mathrm{l}}^{-}+\mathrm{B}{\mathrm{r}}_2$
• The organic solvent layer makes the halogen colour easier to identify.

Halogen-displacement Investigation

• Aim: Investigate redox in displacement of a halogen from its halide solution.
• Materials: Chlorine water, bromine water, iodine solution, $0.5\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ potassium chloride, potassium bromide and potassium iodide, and 1,1,1-trichloroethane.
• Apparatus: Test tubes, measuring cylinder and test-tube rack.
• Procedure:
  1. Mix $2\mathrm{c}{\mathrm{m}}^3$ of each halogen solution with $2\mathrm{c}{\mathrm{m}}^3$ organic solvent; shake vigorously and leave for 30 seconds to establish its colour in each layer.
  2. Mix $2\mathrm{c}{\mathrm{m}}^3$ halide solution with $2\mathrm{c}{\mathrm{m}}^3$ another halogen solution, shake, then add $2\mathrm{c}{\mathrm{m}}^3$ organic solvent, shake and leave for 30 seconds.
  3. Compare the aqueous and organic layers for bromine/iodine with chloride, chlorine/iodine with bromide, and chlorine/bromine with iodide.
• Precaution: Chlorine, bromine and iodine are poisonous; handle their solutions carefully.
• Conclusion: Oxidising strength decreases down Group 17; reducing strength of halide ions increases: $\mathrm{C}{\mathrm{l}}^{-}<\mathrm{B}{\mathrm{r}}^{-}<{\mathrm{I}}^{-}$.
• For chlorine and potassium bromide, colourless bromide solution becomes brown as bromine forms; greenish-yellow chlorine becomes colourless as chloride forms.
• Half-equations: $2B{\mathrm{r}}^{-}\to \mathrm{B}{\mathrm{r}}_2+2{\mathrm{e}}^{-}$ and $\mathrm{C}{\mathrm{l}}_2+2{\mathrm{e}}^{-}\to 2C{\mathrm{l}}^{-}$. Bromide is oxidised and chlorine is reduced.