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9. Redox processes > Oxidation and reduction > Flashcards

Oxidation and reduction Flashcards

1
Q

What determines oxidation?

A
  • Gain of oxygen
  • Loss of electrons
  • Loss of hydrogen
    ALO (anode, loss of electrons, gain of O2)
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2
Q

What determines reduction?

A
  • Loss of oxygen
  • Gain of electrons
  • Gain of hydrogen
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3
Q

What is a redox reaction?

A

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

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4
Q

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

A
  • Zn –> Zn2+ + 2e- (oxidation)
  • Cu2+ + 2e- –> Cu (reduction)
  • They make up the original ionic equation
  • The electrons on opposite sides cancel out
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5
Q

What are oxidation numbers? What are the used for?

A
  • 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
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6
Q

What are oxidation states?

A
  • A number which, together with its sign, indicates the gain or loss of electron control of an atom during a reaction
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7
Q

How do you determine the oxidation number of an ion?

A
  • The charge of ion e.g. Na+ = +1
  • A neutral atom e.g. Na = 0
  • Charge (+ or -) always infront of number
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8
Q

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

A
  • 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
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9
Q

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

A
  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
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10
Q

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

A
  1. Fluorine has O.S = -1
  2. The alkaline earth metals (Be, Mg, Ca …) and Zn in compounds have O.S = +2
  3. Hydrogen in compounds has O.S = +1 except in metal hydrides (NaH) O.S = -1
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11
Q

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

A
  1. O2 has O.S = -2 except in peroxides where it is -1 and in F2O it has +2
  2. Cl has O.S = -1 unless combined with O2 or F
  3. Charge on metal ion is same state, Zn2+ = +2
  4. Sum of O.S in polyatomic ion add up to charge (CO3 2-) S.O = -2
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12
Q

How do you determine the oxidation number of metals?

A
  • 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
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13
Q

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

A
  • Mostly negative based on their usual ion
  • The number usually corresponds to the position on periodic table (group)
  • Hydrogen has +1
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14
Q

What is the oxidation number of sulfur?

A
  • 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
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15
Q

What are oxidising agents?

A
  • Oxidising agent: a substance that oxidises other substances. Itself is reduced and gains electrons
  • The oxidation number of this agent decreases
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16
Q

What are reducing agents?

A
  • Reducing agent: a substance that reduces other substances. Itself is oxidised and loses electrons
  • The oxidation number of the reducing agent increases
17
Q

How do you identify oxidising and reducing agents?

A
  • Look at whether their oxidation number is increased (reducing agent)
  • Look at whether their oxidation number is decreased (oxidising agent)
18
Q

How do you name transition metal compounds?

A
  • 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
19
Q

What are disproportionation reactions?

A
  • The same species is oxidised and reduced simultaneously during a reaction to form 2 products
    E.g. catalytic decomposition of hydrogen peroxide
    Check book
20
Q

Half-equations in book

A
21
Q

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

A
  • When metals are ranked in order of reactivity
  • Metals higher in reactivity can displace less reactive metals from their compounds in solutions
22
Q

What does the ACTIVITY series determine?

A
  • 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
23
Q

Explain the chemistry behind displacement reactions.

A
  • 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
24
Q

What is a spectator ion?

A
  • 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
25
Q

What does feasibility in the context of displacement reactions mean?

A
  • 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
26
Q

How are redox reactions involved in titrations?

A
  • 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
27
Q

Explain the redox titration of manganate (VII).

A
  • 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
28
Q

Explain the redox reaction of iodine-thiosulphate titration.

A
  • 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
29
Q

What is the Winkler method and what does it determine?

A
  • 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
30
Q

Explain the chemical aspect of the Winkler method.

A
  • 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)
31
Q

What do the results of the Winkler method determine?

A
  • 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

32
Q

See steps for Winkler method calculations in book.

A

9. Redox processes (2 decks)

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