Redox reactions of complex ions

In a nutshell

When a transition metal ion changes oxidation number a change in colour is usually observed. Simple tests can be carried out to identify the ligands.

Redox reactions

Transition metal ions can lose or gain electrons in redox reactions and will therefore have different oxidation numbers. A colour change is usually observed when a complex ion in solution changes its oxidation number.

Oxidation of iron

Acidified potassium manganate $(VII)$ solution $(KMnO_4)$ can be used to oxidise $Fe^{2+}$ ions to $Fe^{3+}$ ions. The manganese ions are reduced. The solution changes from pale green to yellow.

$MnO_4 \ ^- \space + \space 8H^+ \space + \space 5e^- \rightarrow Mn^{2+}\space + \space 4H_2O \\Fe^{2+} \rightarrow Fe^{3+} \space + \space e^-$

The second half equation is multiplied by five to get a balanced full equation.

$MnO_4 \,^- \space + \space 8H^+ \space + \space 5e^- \rightarrow Mn^{2+}\space + \space 4H_2O \\ Fe^{2+} \rightarrow Fe^{3+} \space + \space e^-$

$MnO_4 \ ^- \space + \space 8H^+ \space + \space 5Fe^{2+} \rightarrow Mn^{2+}\space + \space 4H_2O \space + \space 5Fe^{3+}$

Reduction of iron

Iodide ions can be used to reduce $Fe^{3+}$ ions to $Fe^{2+}$ ions. The iodide ions are oxidised. The solution changes from orange-brown to brown.

$2I^- \rightarrow I_2 \space + \space 2e^- \\Fe^{3+} \space + \space e^- \rightarrow Fe^{2+}$

The second half equation is multiplied by two to get a balanced full equation.

$2I^- \space + \space 2Fe^{3+} \rightarrow 2Fe^{2+} \space + \space I_2$

Oxidation of chromium

Hydrogen peroxide is added to $(Cr(OH)_6\ ^{3-})$ in alkaline conditions to oxidise the $Cr^{3+}$ ions to chromate ions $(CrO_4 \ ^{2-})$ and reduce oxygen. The mixture requires warming up. The solution goes from dark green to yellow.

$H_2O_2 \space + \space 2e^- \rightarrow 2OH^- \\Cr (OH)_6 \^{3-} \space + \space 2OH^- \rightarrow CrO_4 \ ^{2-} + 4H_2O + 3e^-$

The first half equation shows a transfer of two electrons. The second half equation shows a transfer of three electrons. The first half equation is multiplied by three and the second half equation is multiplied by two to get a balanced full equation.

$3H_2O_2 \space + \space 2Cr (OH)_6 \^{3-} \rightarrow 2OH^- \space + \space 2CrO_4 \ ^{2-} + 8H_2O$

Dilute sulfuric acid can be added to chromate $(VI)$ solution to form an orange dichromate $(VI)$ solution.

$2CrO_4 \ ^{2-} + 2H^+ \rightarrow Cr_2O_7 \ ^{2-} \space + \space H_2O$

Reduction of chromium

Acidified zinc can be used to reduce dichromate ions to chromium $(III)$ ions.

$Zn \rightarrow Zn^{2+} \space + \space 2e^- \\ Cr_2O_7 \ ^{2-} \space + \space 14H^+ \space +6e^- \rightarrow 2Cr^{3+} \space + \space 7H_2O$

The second half equation is multiplied by two to get a full balanced equation.

$Cr_2O_7 \ ^{2-} \space + \space 14H^+ \space + \space 3Zn\rightarrow 2Cr^{3+} \space + \space 7H_2O \space + \space 3Zn^{2+}$

Reduction and disproportionation of copper ions

Iodide ions can be used to reduce copper $(II)$ to copper $(I)$ iodide. The starting copper ion solution is pale blue. Copper iodide is a white precipitate, however the solution will appear brown at the end of the reaction because of the dissolved iodine.

$2Cu^{2+}(aq) \space +\space 4I^-(aq) \rightarrow 2CuI(s) \space + \space I_2(aq)$
$Cr2O7 2− + 14H+ + 3Zn→2Cr3+ + 7H2O + 3Zn2+Cr_2O_7 \, ^{2-} \space + \space 14H^+ \space + \space 3Zn\rightarrow 2Cr^{3+} \space + \space 7H_2O \space + \space 3Zn^{2+}$

$Cr_2O_7 \, ^{2-} \space + \space 14H^+ \space + \space 3Zn\rightarrow 2Cr^{3+} \space + \space 7H_2O \space + \space 3Zn^{2+$

$2Cu^{2+}(aq) \space +\space 4I'^-(aq) \rightarrow 2CuI(s) \space + \space I_2(aq)$

$Cu^+$ ions are unstable, therefore, they spontaneously disproportionate. The $Cu^+$ ions oxidise and reduce themselves spontaneously to form $Cu^{2+}(aq)$ and $Cu(s)$ .

Identifying transition metal ions

When aqueous sodium hydroxide is added to transition metals, coloured precipitates usually form.

Aqueous sodium hydroxide can be added drop-wise using a pipette to a solution containing an unknown transition metal ion. The colour of the precipitate formed can be recorded and used to identify the transition metal ion.

The table below shows some of the ions that can be identified using aqueous sodium hydroxide.

Ion	Colour of precipitate
$\bold{Cr^{3+}}$	Dark green
$\bold{Mn^{2+}}$ $Mn^{2+}$ $Cr^{3+}$ $Cr^{3+$	Pale brown
$\bold{Fe^{2+}}$	Pale green
$\bold{Fe^{3+}}$	Orange-brown
$\bold{Cu^{2+}}$	Pale blue

Identifying ligands

Some common ligands can be identified using simple tests. The table below outlines some of the tests used to identify ligands and observations made.

ion	Test	Observation if ligand is present
$\bold{Cl^-}$	Add silver nitrate.	White precipitate forms.
$\bold{Br^-}$	Add silver nitrate.	Cream precipitate forms.
$\bold{I^-}$	Add silver nitrate.	Pale yellow precipitate forms.
$\bold{CO_3 \ ^{2-}}$	Add nitric acid. Gas given off is bubbled through limewater.	Carbon dioxide gas forms. Limewater changes from clear to cloudy.
$\bold{SO_4 \ ^{2-}}$	Add barium nitrate/barium chloride.	White precipitate forms.
$\bold{NH_4 \ ^+}$	Add cold sodium hydroxide to unknown sample and warm. Hold damp, red litmus paper above the unknown solution.	Red litmus paper turns blue.