Redox processes

In the industrial production of nitric acid the following changes take place to the nitrogen.

a. Give the oxidation number of the nitrogen atom in each molecule.

b. For each stage, state whether oxidation or reduction has taken place. In each case explain your answer.

c. Give the full systematic name for NO₂.

a.

The oxidation number of N₂ is 0

The oxidation number of NH₃ is -3

The oxidation number of NO is +2

The oxidation number of NO₂ is +4

The oxidation number of HNO₃ is +5

b.

For the stage 1, the reduction has taken place because the oxidation number decreases

For the stage 2 - 4, the oxidation has taken place because the oxidation number increases

c.

The full systematic name for NO₂ is nitrogen(IV) oxide

i. Nitric acid, HNO3, reacts with red phosphorus.

`P + 5HNO_3 -> H_3PO_4 + 5NO_2 + H_2O`

By referring to oxidation number changes, explain why this is a redox reaction.

ii. Explain why nitric acid can be described as an oxidising agent in this reaction.

i.

P is element from reactants with oxidation number of 0 and P in H₃PO₄ from products with oxidation number of +5 while N in HNO₃ has oxidation number of +5 and in NO₂, it is +4 oxidation number. Thus, oxidation and reduction occur together, making this reaction a redox.

ii.

Because nitric acid increases oxidation number P from 0 to +5 after reaction

Calcium reacts with cold water to form calcium hydroxide, Ca(OH)₂, and hydrogen, H₂. 

a. State the oxidation number of calcium in: 

i. calcium metal

ii. calcium hydroxide

b. State the oxidation number of hydrogen in: 

i. water

ii. hydrogen gas

c. Write two half-equations for the reaction between water and calcium hydroxide to show: 

i. the change from calcium to calcium ions 

ii. the change from water to hydroxide ions and hydrogen.

d. In which one of the half-equations in part c is a reduction occurring? Give a reason for your answer.

a. State the oxidation number of calcium

i. Ca: 0

ii. Ca(OH)₂: +2

b. State the oxidation number of hydrogen

i. H₂O: +1

ii. H₂: 0

c.

i. `Ca -> Ca^(2+) + 2e^-`

ii. `2H_2O + 2e^(-) -> 2OH^(-) + H_2`

d.

`2H_2O + 2e^(-) -> 2OH^(-) + H_2` is a reduction occuring because oxidation of hydrogen decreases from +1 to 0

The unbalanced equation for the reaction of sulfur dioxide with bromine is shown below.

`SO_2 + Br_2 + H_2O →  SO""_4^(2–) + Br^– + H^+`

a. State the oxidation number of sulfur in:

i. `SO_2`

ii. `SO""_4^(2-`

b. State the oxidation number of bromine in:

i. `Br_2`

ii. `Br^-`

c. Identify the reducing agent in this reaction. Give a reason for your answer.

d. State the change in oxidation number for:

i. Each S atom

ii. Each bromine atom

e. Construct a balanced equation for this reaction.

a. State the oxidation number of sulfur

i. `SO_2`: +4

ii. `SO""_4^(2-`: +6

b. State the oxidation number of bromine

i. `Br_2`: 0

ii. `Br^-`: -1

c. `SO_2` is a reducing agent because it decreases the oxidation number of bromine from 0 to -1.

d. State the change in oxidation number

i. each S atom: +2

ii. each bromine atom: -1

e. `SO_2 + Br_2 + 2H_2O →  SO""_4^(2–) + 2Br^– + 4H^+`

`SO_2 + Br_2 → SO""_4^(2−) + 2Br^−`

Aluminium reacts with hydrochloric acid to form aluminium chloride, AlCl₃, and hydrogen. This is a redox reaction.

a. Explain in term of electrons, what is meant by a redox reaction.

b.

i. Write a half-equation to show aluminium changing to aluminium ions.

ii.Write a second half-equation to show what happens to the hydrogen ions from the acid.

iii. What is the change in oxidation number when a hydrogen ion turns into a hydrogen atom?

c. Construct a balanced ionic equation for the reaction between aluminium atoms and hydrogen ions.

a. One reactant is oxidised by loss of electrons, and electrons are gained by another reactant. Both processes take place together.

b.

i. `Al → Al^(3+) + 3e^−`

ii. `2H^+ + 2e^− → H_2`

iii. The change in oxidation number when a hydrogen ion turns into a hydrogen atom is -1

c. `2Al + 6H^+ → 2Al^(3+) + 3H _2`

Iodine, `I_2`, reacts with thiosulfate ions, `S_2O""_3^(2–` to form iodide ions, `I^–`, and tetrathionate ions, `S_4O""_6^(2–`

`I_2 + 2S_2O""_3^(2–) → 2I^– + S_4O""_6^(2–)`

a. State the oxidation number of each sulfur atom in:

i. a `S_2O""_3^(2–` ion

ii. a `S_4O""_6^(2–` ion

b. Explain in terms of electron transfer why the conversion of iodine to iodide ions is a reduction reaction

c. When a salt containing iodide ions is warmed with concentrated sulfuric acid and `MnO_2`, iodine is evolved.

`2I^– + MnO_2 + 6H^+ + 2SO""_4^(2–) -> I_2 + Mn^(2+) + 2HSO""_4^(–) + 2H_2O`

i. State the systematic name for `MnO_2`

ii. What is the oxidation number of S in the `SO""_4^(2–)` ion?

iii. Which reactant gets oxidised in this reaction? Explain your answer by using oxidation numbers.

iv. Which substance is the oxidising agent? Explain your answer.

a. State the oxidation number of each sulfur atom

i. +2

ii. +5/2

b. Because the oxidation number of iodine atoms decreases from 0 to -1 meaning that they gain electrons

c.

i. manganese(IV) oxide

ii. The oxidation number of S in the `SO""_4^(2-` ion is +6

iii. `I^-` ions get oxidised in this reaction because their oxidation number increases from -1 to 0

iv. The oxidising agent is `MnO_2` because it increases the oxidation number of iodine from -1 to 0 or we can also say that the oxidation number of the Mn has decreased

The compound KBrO3 decomposes when heated.

`2KBrO_3 →  2KBr + 3O_2`

a. State the oxidation numbers of bromine in:

i. KBrO₃

ii. KBr

b. Explain using oxidation numbers why this reaction is a redox reaction.

c. State the systematic name of KBrO₃.

a. State the oxidation numbers of bromine

i. +5

ii. -1

b. The oxidation number of Br decreases from +5 to -1, whereas the oxidation number of Oxygen increases from -2 to 0. Simultaneously, both processes are taking place.

c. The systematic name of KBrO₃ is potassium bromate(V)

When KBrO₃ reacts with hydrazine, N₂H₄, nitrogen gas is evolved.

`2KBrO_3 + 3N_2H_4 → 2KBr + 3N_2 + 6H_2O`

i. What is the oxidation number change of the bromine atom when KBrO₃ is converted to KBr?

ii. What is the oxidation number change for each nitrogen atom when N₂H₄ is converted to N₂?

iii. Use your answers to i and ii to explain why 2 moles of KBrO₃ react with 3 moles of N₂H₄.

i. The oxidation number change of the bromine atom when KBrO₃ is converted to KBr is -6

ii. The oxidation number change for each nitrogen atom when N₂H₄ is converted to N₂ is +2 because the oxidation number of N in N₂H₄ is -2 and that of N in N₂ is 0

iii. Regarding definition, oxidation number changes must balance before and after reactions. Thus, the total oxidation number of N in N₂H₄ is 3 x (2x +2) = +12, leading the moles of KBrO₃ to be 2 because the following total oxidation number of bromine will be 2 x(-6) = - 12.

The heptahydrate mineral of magnesium sulfate is used in Epsom Salts for preparing therapeutic baths. Magnesium sulfate is also used in some medical applications. 

What is the oxidation state of MgSO₄?

A. -2

B. +2

C. +7

D. 0

The answer is D.

For MgSO₄, the oxidation number of Mg is +2; oxygen is -2 and 2 is +6. Thus, they are added up together giving: (+2) + (+6) + 4 x (-2) = 0

In which of the following reactions is the change in oxidation number the smallest for nitrogen?

A. `N_2 + 3H_2 → 2NH_3`

B. `3Mg + N_2 → Mg_3N_2`

C. `4NH_3 + 5O_2 → 4NO + 6H_2O`

D. `2NO + O_2 → 2NO_2`

The answer is D.

The oxidation number of nitrogen in reaction D changes from +2 to +4, as such, an overall change is +2

A and B are incorrect because in both reactions, the oxidation number of nitrogen changes from 0 to -3

C is incorrect because the oxidation number of nitrogen changes from -3 to +2