5.3

Question 1

5.3.1 a) Explain the rules for determining the electron configuration of both atoms and ions of the d-block elements of Period 4 (Sc-Zn). Use Fe, Fe2+, Fe3+, Cr and Cu as example.

Answer 1

Electrons fill orbitals starting from the lowest energy level (n=1) to the highest energy level. The 3d sub-shell has the highest energy of the Period 4 d-block elements. - this is because the 4s subshell is at a slightly lower energy level than the 3d subshell. So the 4s orbital will fill before any of the 3d orbitals.
e.g. Fe = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6

Chromium and copper do not follow this principle.
1. Chromium has one electron in the 4s orbital, and one electron in each of the five 3d orbitals (i.e. these orbitals are all singly occupied) - this provides more stability:
Cr = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5
Copper has one electron in the 4s orbital, but two electrons in each of the five 3d orbitals (so the 4s subshell is singly occupied, but the 3d subshell is completely full) - again, it is more stable this way:
Cu = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10

Transition elements lose electrons to form positive ions. These electrons are lost from the highest energy level first. However, in this case, it isn’t the 3d sub-shell - this is because once the orbitals are occupied by electrons, the 4s sub-shell is actually at a higher energy level than the 3d sub-shell. So electrons are lost from the 4s orbital first.

e. g. Fe2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d6
e. g Fe3+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d5

Question 2

5.3.1 b) Why are the elements Ti-Cu classified as transition metals? Why aren’t the other Period 4 d-block elements considered to be transition metals?

Answer 2

A transition element is a d-block element that can form
at least one stable ion with an incomplete d-sub-shell.
Scandium and zinc, the first and last members of the Period 4 d-block elements, are not classified as transition metals: they can’t form stable ions with incomplete d sub-shells.
Sc = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d1
Sc3+ = 1s2, 2s2, 2p6, 3s2, 3p6
Zn = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2
Zn2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10
They can each form one ion only: Sc3+ ions have empty d-orbitals and Zn2+ ions have full d-orbitals.

Question 3

5. 3.1 c) i) Explain the existence of more than one oxidation state for each transition element in its compounds.
   ii) Why do transition elements form coloured ions?

Answer 3

The transition metals all form ions with more than one stable oxidation state. The elements show variable oxidation states because the energy levels of the 4s and the 3d sub-shells are very close to one another. So different number of electrons can be gained or lost using fairly similar amounts of energy (e.g. the 3d electron can also be lost when an atom forms a stable ion).
Compounds and ions of transition metals are frequently coloured - in fact, the colour of a transition metal often varies with different oxidation states. This is due to the partially filled d-orbitals of the transition metal ion.
e.g. Fe2+ is pale green and Fe3+ is yellow
e.g. Cr3+ is green and Cr6+ is orange
Scandium and Zinc aren’t transition metals, and don’t form coloured ions: Sc3+ (an empty d sub-shell) and Zn2+ (a full d sub-shell) become colourless in aqueous solution.

Question 4

5.3.1 c) iii) Explain why transition elements and their compounds are very effective catalysts.

Answer 4

A catalyst is a substance that increases the rate of a chemical reaction without itself changing. Transition metals, and their compounds, are important catalysts:
Heterogeneous catalysts include:
- Iron in the Haber process (i.e. where nitrogen and hydrogen react to produce ammonia)
- Nickel in the hydrogenation of alkenes
Homogeneous catalysts include:
- Iron(II) in the reaction between iodide ions and peroxodisulfate (S2O8(2-))

The catalytic behaviour of transition elements is because:

1. Transition metals provide a surface on which a reaction can take place. Reactants are adsorbed onto the surface of the metal and held in place while a reaction occurs.
2. Transition metal ions have the ability to change their oxidation states by gaining or losing electrons. They then bind to reactants, forming intermediates as part of a chemical pathway - this often has a lower activation energy, which in turn increases the reaction of reaction.

Question 5

5.3.1 c) iii) Discuss the importance in the manufacture of chemicals by industry, against the possible risks of transition metals.

Answer 5

By employing catalysts, lower temperatures can be used, reducing energy demand from combustion of fossil fuels. This results in lower production costs, more product in a shorter amount of time, and a reduction in CO2 emissions. Catalysts can also reduce waste by allowing a different reaction to be used with a better atom economy. However, the benefits to the environment of improved sustainability should be weighed against the toxicity of some catalysts, and the economic importance that catalysts have can be impacted by how certain catalysts are often specific to only one reaction.

Question 6

5.3.1 d) Define the following terms: ligand and coordinate bond.

Answer 6

A ligand is an atom, ion or molecule that donates a pair of electrons to a central metal ion, forming a coordinate bond (or dative covalent bond). Some ligands are neutral and carry no charge (H2O: / :NH3), while others are negatively charged (:Cl- / :CN-).
A coordinate bond is a covalent bond in which both electrons in the shared pair come from the same atom, i.e. one of the bonded atoms provides both of the electrons for the shared pair.

Question 7

5.3.1 d) Explain the different types ligands, giving examples of each.

Answer 7

• A monodentate ligand can donate just one pair of electrons to the central metal ion (forming one coordinate bond) e.g. H2O: / :NH3 / :Cl- / :CN-.
• A bidentate ligand has two pairs of electrons (on two different atoms) to donate to the central metal ion (forming two coordinate bonds) e.g. ethane-1,2-diamine (‘en’), :NH2CH2CH2:NH2, where each nitrogen donates its lone pair to the metal ion.
• Multidentate ligands have two or more lone pairs of electrons, forming two or more coordinate bonds (e.g. a hexadentate ligand has six lone pairs of electrons available to form 6 coordinate bonds).

Question 8

5.3.1 e) Define the following terms: complex ion and coordination number. How can you work out the overall charge of a complex ion?

Answer 8

A complex ion consists of a central metal ion, surrounded by coordinately bonded ligands.
The coordination number is the number of coordinate bonds that are formed with the central metal ion - this is usually 4 or 6.
The overall charge of the complex ion is a sum of the individual charge of the traditional metal ion and those of the ligands present in the complex.

Question 9

5.3.1 e) i) Explain the shape associated with a six-fold coordination.

Answer 9

Octahedral complexes have six coordinate bonds attached to the central metal ion. Examples include hexaaqua complexes, such as [Cu(H2O)6]2+ and [Fe(H2O)6]3+. Six ligands can fit around a central metal ion only if they’re small enough (H2O: / :NH3 / :CN-).
Octahedral complexes can also occur with multidentate ligands, such as [Ni(NH2CH2CH2NH2)3]2+ (or [Ni(‘en’)3]2+ ).
Four of the ligands are in the same plane, with one ligand above the plane and one ligand below the plane - these two ligands can be represented by straight lines (with a bond angle of 180° between them). The other four ligands experience bond angles of 90° only: two hatched wedges represent the upper-middle ligands and two solid wedges represent the lower-middle ligands.

Question 10

5.3.1 e) ii) Explain the shape associated with a four-fold coordination.

Answer 10

Some ligands are so large that only four can fit around a transition metal ion. This results in either a tetrahedral shape or a square planar shape.
The tetrahedral shape is the more common of the two (with bond angles of 109.5° around the central metal ion). Chloride ligands (:Cl-), for example, can form tetrahedral complexes (tetrachloro) of [CuCl4]2- and [CoCl4]2-. In a tetrahedral complex, there is one straight line going up to a ligand, and three spread out near the bottom: one straight line (left), one solid wedge (middle) and one hatched wedge (right).
A rare configuration is a square planar complex (with bond angles of 90° around the central metal ion). This is when the four monodentate ligands arrange themselves in the same plane as the metal atom. They can be represented by two hatched wedges (at the top) and two solid wedges (at the bottom). Complexes of Platinum can be square planar: consider [Pt(NH3)Cl2] (platin) for example.

Question 11

5.3.1 f) i) Explain how cis-trans isomerism can occur in complexes.

Answer 11

One type of stereoisomerism shown by complexes is cis-trans isomerism. In a cis isomer, two of the same ligands are adjacent to each other (their coordinate bonds are 90° apart), and in a trans isomer, two of the same ligands are opposite each other (their coordinate bonds are 180° apart).
1) Cis-trans isomerism is possible in some six-coordinate complexes with an octahedral shape:
- four of one type of (monodentate) ligand, and two of another type of (monodentate) ligand (e.g. [Co(NH3)4(Cl2]+)
- two monodentate ligands and two bidentate ligands (e.g. [Cr(C2O4)2(H2O)2)]-, [Co(NH2CH2CH2NH2)2Cl2]+)
2) Cis-trans isomerism is possible in some four-coordinate complexes with a square planar shape (the complex must have four ligands of two different types: two of one ligand and two of another) e.g.[Pt(NH3)2Cl2], [NiCl2(NH3)2].
[Note: In organic chemistry, cis-trans stereoisomerism requires the presence of a C=C double bond. In complex ions, this isn’t necessary]

Question 12

5.3.1 f) ii) Explain how optical isomerism can occur in complexes.

Answer 12

The requirements for optical isomerism (a type of stereoisomerism that only occurs in octahedral complexes) are:
- a complex with three bidentate ligand (e.g. [Ni(en)3]2+)
- a complex with two bidentate ligands, and two monodentate ligands (e.g. [Co(en)2Cl2])
- a complex with one hexadentate ligand (e.g. [Cu(EDTA)]2-)
Optical isomers are non-superimposable mirror images of each other. They rotate plane-polarised light differently - one of the isomers rotates the light clockwise and the other rotates the light anticlockwise. A mixture containing equal amounts of two optical isomers has no effect on plane-polarised light because the rotations cancel out - this is called a racemic mixture.
[Note: while some cis isomers (e.g cis-[Co(NH2CH2CH2NH2)2Cl2]+) also exist as two optical isomers, trans-isomers cannot form optical isomers: mirror image is exactly the same, and can be superimposes]

Question 13

5.1.3 g) How is cis-platin used in medicine?

Answer 13

Cis-platin (the cis-isomer of a platinum complex, e.g.[PtCl2(NH3)2], is used as an anti-cancer drug. The two chloride ligands are very easy to displace. So the cis-platin loses them, and binds to two nitrogen atoms on the DNA molecule inside a cancerous cell instead. This prevents the cancerous cell from reproducing by cell division - the cell will die, since it is unable to repair the damage. Unfortunately cis-platin has some unpleasant side effects, and can lead to kidney damage.

Question 14

5.1.3 h) i) How is [Cu(NH3)4(H2O)2]2+ formed in a ligand substitution reaction?

Answer 14

An aqueous solution of copper(II) ions contains the pale blue complex [Cu(H2O)6]2+.

When an excess of aqueous ammonia solution is added to [Cu(H2O)6]2+ (aq), the pale blue solution changes colour to form a deep blue solution containing [Cu(NH3)4(H2O)2]2+ (aq) complex ions:
[Cu(H2O)6]2+(aq) + 4NH3(aq) ⇄ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)

If you carry out this experiment in the laboratory (and add the ammonia dropwise), the reaction will take place in two distinct steps. First, when a small amount of aqueous ammonia solution is initially added, a blue precipitate of [Cu(H2O)4(OH)2] (or Cu(OH)2) is formed:
[Cu(H2O)6]2+(aq) + 2NH3(aq) → Cu(H2O)4(OH)2 + 2NH4+(aq)
Then (as more ammonia is added) the Cu(OH)2 precipitate will dissolve in excess ammonia to form the deep blue solution of [Cu(NH3)4(H2O)2]2+ shown above:
[Cu(H2O)6]2+(aq) + 4NH3(aq) ⇄ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)

Question 15

5.1.3 h) i) How is [CuCl4]2- formed in a ligand substitution reaction?

Answer 15

An aqueous solution of copper(II) ions contains [Cu(H2O)6]2+, and is a pale blue colour. When concentrated hydrochloric acid is added, a yellow solution of [CuCl4]2- forms:
[Cu(H2O)6]2+(aq) + 4Cl-(aq) ⇄ [CuCl4]2-(aq) + 6H2O(l)
An intermediate green solution also forms: this is not a new species, but a result of the yellow solution mixing with the blue solution as the reaction proceeds.
[Note: the chloride ligands are larger than the water ligands and experience stronger repulsions. Fewer chloride ligands can fit around the central metal ion - this explains the change in coordination number and shape]
[Note: this reaction exists in equilibrium and is reversible - adding water to the yellow solution can return it to a pale blue colour]

Question 16

5.1.3 h) ii) How is [Cr(NH3)6] formed in a ligand substitution reaction?

Answer 16

An aqueous solution of chromium(III) ions contains the dark green complex [Cr(H2O)6]3+. [Note: Some textbooks describe the [Cr(H2O)6]3+ complex as a violet colour. However, when it is produced during a reaction in a test tube, it is often green - you don’t need to know the chemistry of why it’s green]

When an excess of aqueous ammonia solution is added to [Cr(H2O)6]3+ (aq), the dark green solution changes colour to form a purple solution containing [Cr(NH3)6]3+ (aq) complex ions:
[Cr(H2O)6]3+(aq) + 6NH3(aq) ⇄ [Cr(NH3)6]3+(aq) + 6H2O(l)

If you carry out this experiment in the laboratory (and add the ammonia dropwise), the reaction will take place in two distinct steps. First, when a small amount of aqueous ammonia solution is initially added, a gray-green precipitate of [Cr(H2O)3(OH)3] (or Cr(OH)3) is formed:
[Cr(H2O)6]3+(aq) + 3NH3(aq) → Cr(H2O)3(OH)3 + 3NH4+(aq)
Then (as more ammonia is added) the Cr(OH)3 precipitate will dissolves in excess ammonia to form the purple solution of [Cr(NH3)6]3+ shown above:
[Cr(H2O)6]3+(aq) + 6NH3(aq) ⇄ [Cr(NH3)6]3+(aq) + 6H2O(l)

Question 17

5.1.3 i) Explain the biochemical importance of iron in haemoglobin, including any ligand substitutions.

Answer 17

Haemoglobin is composed of four polypeptide chains. Each of these polypeptide chains contains four non-protein components called haem groups. Each haem group has a Fe2+ central metal ion. It is the Fe2+ ion which bonds to oxygen, allowing it to be transported in blood:
The Fe2+ ion forms 6 coordinate bonds. Four of the lone pairs come from four nitrogen atoms in the haem structure. A fifth lone pair comes from a nitrogen atom on the protein globin. The last lone pair comes from a water ligand, which can be substituted with an oxygen molecule (forming oxyhaemoglobin).
If carbon monoxide is inhaled, the haemoglobin substitutes it’s water ligand/oxygen ligand for a carbon monoxide ligand instead (forming carboxyhaemoglobin). Carbon monoxide binds to haemoglobin more strongly than oxygen, and the reaction isn’t reversible: the haemoglobin can’t transport oxygen any longer.

Question 18

5.3.1 j) Give the ionic equation, and the accompanying colour change of aqueous Cu2+ with aqueous sodium hydroxide and aqueous ammonia.

Answer 18

An aqueous solution of copper(II) ions contains [Cu(H2O)6]2+, and is a pale blue colour.

1. When aqueous sodium hydroxide is added, a blue precipitate is formed:
   [Cu(H2O)6]2+(aq) + 2OH-(aq) → Cu(H2O)4(OH)2 + 2H2O(l)
   or Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)
   The precipitate is insoluble in excess sodium hydroxide.
2. When a small amount of aqueous ammonia solution is added, a blue precipitate is formed:
   [Cu(H2O)6]2+(aq) + 2NH3(aq) → Cu(H2O)4(OH)2 + 2NH4+(aq)

When an excess of aqueous ammonia solution is added, a deep blue solution is formed:
[Cu(H2O)6]2+(aq) + 4NH3(aq) ⇄ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)

Question 19

5.3.1 j) Give the ionic equation, and the accompanying colour change of aqueous Fe2+ with aqueous sodium hydroxide and aqueous ammonia.

Answer 19

An aqueous solution of iron(II) ions contains [Fe(H2O)6]2+, and is a pale green colour.

1. When aqueous sodium hydroxide is added, a green precipitate is formed (which darkens on standing):
   [Fe(H2O)6]2+(aq) + 2OH-(aq) → Fe(H2O)4(OH)2 + 2H2O(l)
   or Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
   The precipitate is insoluble in excess sodium hydroxide, but turns a rusty brown colour on standing in air - this is because iron(II) is oxidised to iron(III):
   Fe(OH)2 (s) → Fe(OH)3 (s)
2. When a small amount of aqueous ammonia solution is added, a green precipitate is formed (which darkens on standing):
   [Fe(H2O)6]2+(aq) + 2NH3(aq) → Fe(H2O)4(OH)2 + 2NH4+(aq)
   The precipitate is insoluble in excess ammonia.

Question 20

5.3.1 j) Give the ionic equation, and the accompanying colour change of aqueous Fe3+ with aqueous sodium hydroxide and aqueous ammonia.

Answer 20

An aqueous solution of iron(III) ions contains [Fe(H2O)6]3+, and is a yellow colour.

1. When aqueous sodium hydroxide is added, an orange/brown precipitate is formed:
   [Fe(H2O)6]3+(aq) + 3OH-(aq) → Fe(H2O)3(OH)3 + 2H2O(l)
   or Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s)
   The precipitate is insoluble in excess sodium hydroxide.
2. When a small amount of aqueous ammonia solution is added, an orange/brown precipitate is formed:
   [Fe(H2O)6]3+(aq) + 3NH3(aq) → Fe(H2O)3(OH)3 + 3NH4+(aq)
   The precipitate is insoluble in excess ammonia.

Question 21

5.3.1 j) Give the ionic equation, and the accompanying colour change of aqueous Mn2+ with aqueous sodium hydroxide and aqueous ammonia.

Answer 21

An aqueous solution of manganese(II) ions contains [Mn(H2O)6]2+, and is a pale pink colour.

1. When aqueous sodium hydroxide is added, a pink precipitate is formed (which darkens on standing):
   [Mn(H2O)6]2+(aq) + 2OH-(aq) → Mn(H2O)4(OH)2 + 2H2O(l)
   or Mn2+(aq) + 2OH-(aq) → Mn(OH)2(s)
   The precipitate is insoluble in excess sodium hydroxide.
2. When a small amount of aqueous ammonia solution is added, a pink precipitate is formed (which darkens on standing):
   [Mn(H2O)6]2+(aq) + 2NH3(aq) → Mn(H2O)4(OH)2 + 2NH4+(aq)
   The precipitate is insoluble in excess ammonia.

Question 22

5.3.1 j) Give the ionic equation, and the accompanying colour change of aqueous Cr3+ with aqueous sodium hydroxide and aqueous ammonia.

Answer 22

An aqueous solution of chromium(III) ions contains [Cr(H2O)6]3+, and is a dark green colour.

1) When aqueous sodium hydroxide is added, a grey-green precipitate is formed:
[Cr(H2O)6]3+(aq) + 3OH-(aq) → Cr(H2O)3(OH)3 + 2H2O(l)
or Cr3+(aq) + 3OH-(aq) → Cr(OH)3(s)

When an excess of aqueous sodium hydroxide is added, Cr(OH)3 further reacts (and redissolves) to form [Cr(OH)6]3-, which is a dark green colour:
Cr(OH)3(H2O)3 + 3OH-(aq) → [Cr(OH)6]3-(aq) + 3H2O

2) When a small amount of aqueous ammonia solution is added, a grey-green precipitate is formed:
[Cr(H2O)6]3+(aq) + 3NH3(aq) → Cr(H2O)3(OH)3 + 3NH4+(aq)

When an excess of aqueous ammonia solution is added, a purple solution is formed:
[Cr(H2O)6]2+(aq) + 6NH3(aq) ⇄ [Cr(NH3)6]2+(aq) + 6H2O(l)

Question 23

5.3.1 k) i) Give the redox reaction and accompanying colour change for the conversion of Fe2+ into Fe3+.
Fe2+ can be oxidised with H+/MnO4-.

Answer 23

Fe2+(aq) [pale green] is oxidised to Fe3+(aq) [yellow].

One example for this reaction is the redox titration by acidified potassium manganate(VII), KMNO4(aq).
The pale green Fe2+ (aq) ions are reduced to yellow Fe3+ (aq) ions. Unfortunately, this colour change is obscured by the reduction from purple manganate(VII) ions, MnO4- to colourless manganate(II) ions, Mn2+. The end point is the first trace of permanent pink in the solution (showing a tiny excess of manganate(VII) ions).
Half equations:
MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (l)
{The manganate is reduced}
Fe2+ (aq) → Fe3+ (aq) + e-
{The iron is oxidised}
Full equation:
MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) → Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)

Question 24

5.3.1 k) i) Give the redox reaction and accompanying colour change for the conversion of Fe3+ into Fe2+.
Fe3+ can be reduced with I-.

Answer 24

Fe3+(aq) [yellow] is reduced to Fe2+(aq) [pale green].

This reaction takes place when a solution of iron(III) reacts with a solution of iodide ions. The yellow Fe3+ (aq) ions are reduced to pale green Fe2+ (aq) ions. Unfortunately, this colour change is obscured by the oxidation of iodide ions, I- (aq), to iodine, I2 (aq) - iodine is has brown colour to it.
Half equations:
2I- (aq) → I2 (aq) + 2e-
{The iodide is oxidised}
Fe3+ (aq) + e- → Fe2+ (aq)
{The iron is reduced}
Full equation:
2I- (aq) + 2Fe3+ (aq) → 2Fe2+ (aq) + I2 (aq)

Question 25

5.3.1 k) ii) Give the redox reaction and accompanying colour change for the conversion of Cr3+ into Cr2O7(2-).
Cr3+ can be oxidised with H2O2/OH-.

Answer 25

Cr3+ (aq) [green] is oxidised to Cr2O7(2-) [orange]. However, the Cr3+ first forms (CrO4)2-, which then forms (Cr2O7)2-.

1. The [dark green] Cr3+ ion in [Cr(OH)6]3- or Cr(OH)6(3+) is oxidised to the [yellow] chromate(VI) solution, (CrO4)2-(aq). This is done by warming it with hydrogen peroxides solution, H2O2(aq), in alkaline conditions.
   Half equations:
   H2O2 (aq) + 2e- → 2OH- (aq)
   {The oxygen is reduced}
   2[Cr(OH)6]3- (aq) + 4OH- (aq) → 2CrO4(2-) (aq) + 8H2O (l) + 6e-
   {The chromium is oxidised}
   Full equation:
   3H2O2 (aq) + 2[Cr(OH)6]3- (aq) → 2OH- (aq) + 2CrO4(2-) (aq) + 8H2O (l)
2. Adding the dilute sulfuric acid to the chromate(VI) solution produces the orange dichromate(VI) solution:
   2CrO4(2-) (aq) + 2H+ (aq) → Cr2O7(2-) (aq) + H2O (l)

Question 26

5.3.1 k) ii) Give the redox reaction and accompanying colour change for the conversion of (Cr2O7)2- into Cr3+.
(Cr2O7)2- can be reduced with Zn/H+.

Answer 26

(Cr2O7)2-(aq) [orange] is reduced to Cr3+(aq) [green] by acidified zinc.
Half equations:
Zn (s) → Zn2+ (aq) + 2e-
{The zinc is oxidised}
(Cr2O7)2- (aq) + 14H+ (aq) + 6e- → 2Cr3+ (aq) + 7H2O (aq)
{The chromium is reduced}
Full equation:
(Cr2O7)2- (aq) + 14H+ (aq) + 3Zn (s) → 2Cr3+ (aq) + 7H2O (l) + 3Zn2+ (aq)

Question 27

5.3.1 k) ii) Give the redox reaction and accompanying colour change for the conversion of Cu2+ into Cu+.
Cu2+ can be reduced with I-.

Answer 27

A [pale blue] solution of Cu2+(aq) is reduced to the [off-white precipitate] copper(I) iodide, CuI, by an excess of iodide ions, I-.
2Cu2+(aq) + 4I-(aq) → 2CuI(s) + I2(aq)
{The iodide is oxidised and the copper is reduced}

Question 28

5.3.1 k) ii) Give the redox reaction and accompanying colour change for the conversion of Cu+ into Cu2+.

Answer 28

In aqueous conditions, Cu+(aq) readily disproportionates to produce Cu2+(aq) and Cu(s).
One Cu+ (aq) ion gains an electron and is reduced to solid Cu (s). The other Cu+ (aq) ion loses an electron and is oxidised to Cu2+ (aq):
2Cu+(aq) → Cu(s) + Cu2+(aq)
[Note: this disproportionation of Cu+ ions is apparent when solid copper(I) oxide, Cu2O, reacts with dilute sulfuric acid to form a brown precipitate of copper and a blue solution of copper(II) sulfate]