Oxidation and reduction

Question 1

What determines oxidation?

Answer 1

• Gain of oxygen
• Loss of electrons
• Loss of hydrogen
  ALO (anode, loss of electrons, gain of O2)

Question 2

What determines reduction?

Answer 2

• Loss of oxygen
• Gain of electrons
• Gain of hydrogen

Question 3

What is a redox reaction?

Answer 3

When either oxidation or reduction occurs, the other one will also take place too

Question 4

How do you write a half-equations of Zn + Cu2+ –> Zn2+ + Cu?

Answer 4

• Zn –> Zn2+ + 2e- (oxidation)
• Cu2+ + 2e- –> Cu (reduction)
• They make up the original ionic equation
• The electrons on opposite sides cancel out

Question 5

What are oxidation numbers? What are the used for?

Answer 5

• The numerical part of oxidation state, given as a roman numeral
• Tell if oxidation or reduction has taken place
• Work out what has been oxidised/reduced
• Construct half equations

Question 6

What are oxidation states?

Answer 6

• A number which, together with its sign, indicates the gain or loss of electron control of an atom during a reaction

Question 7

How do you determine the oxidation number of an ion?

Answer 7

• The charge of ion e.g. Na+ = +1
• A neutral atom e.g. Na = 0
• Charge (+ or -) always infront of number

Question 8

How do you determine the oxidation number of compounds or molecules?

Answer 8

• The sum of oxidation numbers of the atoms is 0
• The more electronegative species will have a negative number
• Electronegativity increases across a period and decreases down a group
• E.g. O further right than C, hence O2 is negative
• Look at their position in the periodic table

Question 9

What are rules 1, 2 & 3 to determine the oxidation states?

Answer 9

1. Free elements e.g. O2, oxidation state = 0
2. The sum of oxidation states of all atoms in a compound equal the net charge on compound (could be 0 or +3 or -2)
3. The group 1 metals have O.S = +1. Group 2 metals have O.S = +2

Question 10

What are rules 4, 5 & 6 to determine the oxidation states?

Answer 10

4. Fluorine has O.S = -1
5. The alkaline earth metals (Be, Mg, Ca …) and Zn in compounds have O.S = +2
6. Hydrogen in compounds has O.S = +1 except in metal hydrides (NaH) O.S = -1

Question 11

What are rules 7, 8, 9 & 10 determine the oxidation states?

Answer 11

7. O2 has O.S = -2 except in peroxides where it is -1 and in F2O it has +2
8. Cl has O.S = -1 unless combined with O2 or F
9. Charge on metal ion is same state, Zn2+ = +2
10. Sum of O.S in polyatomic ion add up to charge (CO3 2-) S.O = -2

Question 12

How do you determine the oxidation number of metals?

Answer 12

• The number usually corresponds to the position on periodic table (group)
• Metals have a positive values in compounds
• Transition metals can have multiple oxidation numbers

Question 13

How do you determine the oxidation number of non-metals?

Answer 13

• Mostly negative based on their usual ion
• The number usually corresponds to the position on periodic table (group)
• Hydrogen has +1

Question 14

What is the oxidation number of sulfur?

Answer 14

• E.g. S4O6 2-
  -2 x 6 = -12
  4x = 10
  x = 2.5, therefore the oxidation number is a fraction
• This is technically not possible, single atoms can only have a whole number oxidation number

Question 15

What are oxidising agents?

Answer 15

• Oxidising agent: a substance that oxidises other substances. Itself is reduced and gains electrons
• The oxidation number of this agent decreases

Question 16

What are reducing agents?

Answer 16

• Reducing agent: a substance that reduces other substances. Itself is oxidised and loses electrons
• The oxidation number of the reducing agent increases

Question 17

How do you identify oxidising and reducing agents?

Answer 17

• Look at whether their oxidation number is increased (reducing agent)
• Look at whether their oxidation number is decreased (oxidising agent)

Question 18

How do you name transition metal compounds?

Answer 18

• Transitions have varying oxidation numbers
• If an element has a variable oxidation number, it is written in Roman numerals (stock notation)
  Check book for specific examples

Question 19

What are disproportionation reactions?

Answer 19

• The same species is oxidised and reduced simultaneously during a reaction to form 2 products
  E.g. catalytic decomposition of hydrogen peroxide
  Check book

Question 20

Half-equations in book

Question 21

What is the reactivity series? What is it used for?

Answer 21

• When metals are ranked in order of reactivity
• Metals higher in reactivity can displace less reactive metals from their compounds in solutions

Question 22

What does the ACTIVITY series determine?

Answer 22

• Metals listed in order of their strength as reducing agents
• In other words the ease with which they undergo oxidation
• The activity series allows metals to be ranked from strongest reducing agents to weakest
• Found in section 25 in data booklet

Question 23

Explain the chemistry behind displacement reactions.

Answer 23

• The more reactive a metal is, the better it is at pushing electrons into less reactive metal ions
• Metals above hydrogen can displace hydrogen ions from solution to produce hydrogen gas

Question 24

What is a spectator ion?

Answer 24

• An ion that appears unchanged on both sides of the complete ionic equation, it is neither oxidised nor reduced
• Not included in net ionic equation

Question 25

What does feasibility in the context of displacement reactions mean?

Answer 25

- The activity series helps predict whether a displacement reaction will take place or not (feasible) - If one of the elements is more reactive than the one in the compound, a reaction will take place - The more reactive metal --> oxidation - The less reactive metal --> reduction

Question 26

How are redox reactions involved in titrations?

Answer 26

- A titration in which the concentration of a solution (in conicle flask) is determined by titrated with solution of known concentration - The oxidising agent is titrated against a reducing agent

Question 27

Explain the redox titration of manganate (VII).

Answer 27

- Reaction between acidified manganate (VII) ions and iron (II) ions (reaction in book) - Manganate (VII) has strong purple colour, disappears at end point, no further indicator needed

Question 28

Explain the redox reaction of iodine-thiosulphate titration.

Answer 28

- Reaction between iodine and thriosulfate ions - Iodine is naturally brown, and becomes colourless when converted to iodine ions - Starch is added before reaction is finished to clarify end point - Solution turns blue/black until iodine reacts, then colourless - Aim is to find concentration of oxidising agent (iodine) - Known quantity of sodium thiosulfate solution

Question 29

What is the Winkler method and what does it determine?

Answer 29

- Technique used to measure the concentration of dissolved oxygen in a water sample - Dissolved O2 used as an indicator of the health of a water body (more O2 little pollution) - Used to determine the Biochemical Oxygen Demand (BOD), amount of O2 used to decompose the organic matter over a period of time - A high BOD, means low level of dissolved O2

Question 30

Explain the chemical aspect of the Winkler method.

Answer 30

- Manganese (II) sulfate is added to a water sample, NaOH is added to make the solution alkaline - In the solution the dissolved O2 will oxidize Mn (II) ions to Mn (IV), (equation in book) - Manganese (IV) oxide (MnO2) appears as a brown precipitate - Then potassium iodide (KI) is added to the solution. Mn (IV) is reduced back to Mn (II) releasing iodine (equation in book) - Thiosulphate is used as an indicator to titrate the liberated iodine (equation in book)

Question 31

What do the results of the Winkler method determine?

Answer 31

- The number of moles of iodine produced (2 moles), we can work out the number of moles of O2 molecules present in the original water sample - Oxygen content presented as mg/dm3 or ppm - Check how the O2 content decreased after 5 days Calculations found in book

Question 32

See steps for Winkler method calculations in book.