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Chemistry > Unit 3.2 : Redox Reactions > Flashcards

Unit 3.2 : Redox Reactions Flashcards

1
Q

How are redox titrations carried out?

A

In the same way as acid-base titrations

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2
Q

What’s difference between acid-base titrations and redox titrations?

A

Redox titrations involve redox reactions

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3
Q

2 most important redox reagents in titrations

A

Potassium manganate (VII)
Sodium thiosulfate

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4
Q

Which important redox reagent isn’t used much in titrations anymore and why?

A

Potassium dichromate (VI), it’s carsinogenical

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5
Q

Half equation for the reduction of manganate (VII) ions in acidic solution

A

MnO4- + 8H+ + 5e- —> Mn2+ + 4H2O

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6
Q

What can manganate VII be described as?

A

An oxidising agent

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7
Q

What is potassium manganate (VII) used for in redox titrations?

A

To estimate the concentration of iron (II) ions in aqueous solution

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8
Q

Half equation for the oxidation of iron (II) ions

A

Fe2+ —> Fe3+ + e-

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9
Q

How do we form a full redox equation from half equations?

A

Balance the electrons

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10
Q

Full redox equation of potassium manganate (VII) and iron (II) ions

A

MnO4- + 5Fe2+ + 8H+ ——> Mn2+ + 5Fe3+ + 4H2O

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11
Q

Ratio of the number of moles of MnO4 to the number of moles of Fe2+

A

1:5

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12
Q

How is the titration between potassium manganate (VII) and iron (II) ions set up?

A

Potassium manganate (VII) in the burette, iron (II) complex in flask below

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13
Q

What’s the end point of the redox titration between potassium manganate (VII) and iron (II) ions?

A

Persisting pale pink

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14
Q

Briefly explain how the redox titration between potassium manganate (VII) and iron (II) ions would be completed

A

The potassium manganate (VII) solution is added from a burette to the iron (II) solution in the flask and the reaction is complete when the 1st permanent pink tinge is seen

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15
Q

What is the indicator in the redox reaction between potassium manganate (VII) and iron (II) ions?

A

There isn’t one - it’s self indicating

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16
Q

Why does potassium manganate (VII) not need an indicator?

A

Self-indicating

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17
Q

What colour does potassium manganate (VII) change to and from when the redox titration is complete?

A

Colourless to purple

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18
Q

What needs to be done to potassium manganate (VII) for a redox reaction to proceed satisfactorily?

A

Must be acidified

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19
Q

What do iron (II) do in air?

A

Readily oxidise

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20
Q

What needs to be done as iron (II) ions readily oxidise in air?

A

The solutions need to be acidified with dilute sulfuric acid to prevent oxidation

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21
Q

How does acid force the redox reaction to complete when added to the solution?

A

Removed the oxygen from the oxidising agent

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22
Q

What is the oxidising agent in the redox titration between potassium manganate and iron (II) ions?

A

Manganate (VII) ions

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23
Q

How much sulfuric acid are iron (II) ions acidified with to prevent oxidation?

A

1moldm-3

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24
Q

Why is sulfuric acid chosen as the acid to acidify potassium manganate (VII)?

A

It will not react with the manganate (VII) ions

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25
Q

Is potassium manganate (VII) an oxidising or a reducing agent?

A

An extremely powerful oxidising agent

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26
Q

How powerful of an oxidising agent is potassium manganate (VII)?

A

The most powerful oxidising agent which is reasonably stable in aqueous solution

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27
Q

What does potassium manganate do slowly in aqueous solution?

A

Decompose

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28
Q

What needs to happen to potassium manganate before it can be used in a redox titration reaction?

A

Needs to be standardised

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29
Q

What is usually used to standardise potassium manganate (VII) solution

A

Ammonium iron (II) sulfate

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30
Q

Why is ammonium iron (II) sulfate normally used to standardise potassium manganate (VII)?

A

It does not oxidise readily in solution and can be obtained to a high degree of purity

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31
Q

Why can’t iron (II) salts be used to standardise potassium manganate?

A

They cannot normal be obtained to a high degree of purity

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32
Q

Why would a brown precipitate of manganese (IV) oxide be formed during titration?

A

A neutral/alkaline solution must have been used

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33
Q

What would using a neutral/alkaline solution during the titration of potassium manganate (VII) result in

A

A brown precipitate of manganese (IV) oxide as a different reaction as taken place

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34
Q

What is the concentration of potassium manganate (VII) normally used during its redox titration?

A

0.02moldm-3

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35
Q

What does the concentration of the iron (II) ions need to be in it’s redox titration and how is this worked out?

A

About 0.100moldm-3 as the concentration of potassium manganate (VII) normally used is 0.02moldm-3 and they’re in a 1:5 ratio

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36
Q

What would adding iron (II) ions of concentration 0.100moldm-3 to potassium manganate (VII) of concentration 0.02moldm-3 do?

A

Complete reaction after the addition of 25cm3 of manganate (VII) solution to 25cm3 of the iron (II) solution

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37
Q

What tend to be self indicating in titrations?

A

Transition metals

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38
Q

How do we know exactly how much of a salt has been added in a reaction?

A

Mass of solid
Mass of solid + container
Mass of container

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39
Q

What happens to the solution that’s left unreacted at the end of a reaction?

A

It’s still present

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40
Q

Dichromate ions

A

Cr2O7^2-

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41
Q

Oxyanions

A

Negative ions that contain oxygen atoms bonded to another element

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42
Q

Negative ions that contain oxygen atoms bonded to another element

A

Oxyanions

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43
Q

Cr2O7^2- ion

A

Dichromate

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44
Q

Under which conditions can half equations for Oxyanions occur?

A

Either acidic or alkaline

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45
Q

Steps to constructing half equations for Oxyanions under acidic conditions

A

1.) balance the non-oxygen atoms
2.) balance the oxygen atoms by adding water molecules to the opposite side of the equation from the Oxyanion
3.) balance the hydrogen atoms from the water molecules by adding hydrogen atoms to the opposite side of the equation from the water molecules (i.e 0 the same side as the oxyanion)
4.) balance the overall charges on both sides of the equation by adding electrons to the side with the most positive overall charge (usually the same side a the hydrogen ions)

46
Q

Steps to constructing half equations for Oxyanions under alkaline conditions

A

1.) balance the non-oxygen atoms
2.) for every oxygen atom in the oxyanion add a water molecule to the same side of the equation as the oxyanion
3.) balance the oxygen atoms by adding hydroxide ions to the opposite side of the equation from the water molecules and the oxyanion
4.) balance the overall charges on both sides of the equation by adding electrons to the side with the most positive overall charge (usually the opposite side from the hydroxide ions)

47
Q

Which side of an equation usually has the most positive overall charge?

A

The side with the hydrogen ions, or the opposite side from the hydroxide ions

48
Q

Ion/electron half equation for the reduction of acidified Cr2O7^2- to Cr3+

A

Cr2O7^2- + 14H+ + 6e- —> 2Cr3+ + 7H2O

49
Q

Ion/electron half equation for the reduction of acidified MnO4- to Mn2+

A

MnO4 + 8H+ + 5e- —> Mn2+ + 4H2O

50
Q

Ion/electron half equation for the oxidation of S2O3^2- to S4O6^2-

A

2S2O3^2- —> S4O6^2- + 2e-

51
Q

2 extra ways to confirm if an oxyanion half equation is correct

A

If you have an oxidising agent, you need hydrogen ions
For every oxygen atom, we need twice the hydrogen atoms (as oxygen is -2)

52
Q

What do you need if you have an oxidising agent?

A

Hydrogen ions

53
Q

How do we combine half equations?

A

Balance he electrons by multiplying them out
Confirm that charges balance at the end

54
Q

What do we do if we have hydrogen atoms on both sides of combined half equations?

A

We can simplify by subtracting the smaller amount from the other side

55
Q

When can we simplify the combination of half equations and how?

A

When we have hydrogen atoms on both sides of the equation - by subtracting the smaller amount from the other side

56
Q

Manganate (VII) ions

A

MnO4-

57
Q

Are big numbers included in overall charge calculations?

A

Yes

58
Q

What can the same simplifying method as used with hydrogen atoms when combining half equations be used for?

A

Waters

59
Q

Thiosulfate ions

A

S2O3^2-

60
Q

S2O3^2-

A

Thiosulfate ions

61
Q

What is done as it isn’t straightforward to analyse a solution for copper (II) ions directly?

A

An indirect route is used

62
Q

How is iodine formed during the titration involving copper (II) ions?

A

Addition of a colourless solution containing iodide ions, such as potassium iodide, to a blue solution containing copper (II) ions leads to the formation of a cloudy brown solution due to the formation of iodine

63
Q

What type of solution contains iodide ions for the copper (II) titration?

A

A colourless solution, such as potassium iodide

64
Q

What type of solution contains the copper (II) ions for their titration?

A

Blue

65
Q

What colour does the solution turn upon adding a colourless solution containing iodide ions to a blue solution containing copper (II) ions to each other and why?

A

Brown due to the formation of iodine

66
Q

Why is iodine formatted during the copper (II) ions titration?

A

Cu2+ ions in solution react with iodide ions to generate a brown solution of iodine

67
Q

As well as reacting with iodide ions to generate iodine, what else happens to Cu2+ ions during their titration?

A

Reduced to copper (I) in a precipitate of CuI

68
Q

Colour of precipitate formed by CuI

A

White

69
Q

Colour of solution formed by iodine

A

Brown

70
Q

Equation for the reaction between Cu2+ ions and iodide ions

A

2Cu^2+ + 4I- —> 2CuI + I2

71
Q

What type of compounds are all reducing agents?

A

Halide ions

72
Q

What are all halide ions?

A

Reducing agents

73
Q

Strongest reducing agent from the halide ions

A

Iodide ions

74
Q

What do we titrate the iodine formed during the reaction between Cu2+ and I- ions with?

A

With sodium thiosulfate of known concentration

75
Q

Equation for sodium thiosulfate reacting with aqueous iodine in a redox titration

A

I2 + 2S2O3^2- —> 2I- + S4O6^2-

76
Q

Molar ratio between iodine and thiosulfate

A

1:2
Iodine:thiosulfate

77
Q

Half equation of iodine being reduced

A

I2 + e- —> 2I-

78
Q

Half equation for the oxidation of thiosulfate ions

A

2S2O3^2- —> S4O6^2- + 2e-

79
Q

Method for the redox titration for copper (II) ions

A

1.) add excess iodide ions (e.g - KI) to Cu2+(aq) to ensure that all of the Cu2+ reacts to produce I2. The mixture formed is a cloudy brown solution
2.) titrate the brown I2 (aq) against sodium thiosulfate (Na2S2O3) until the mixture is straw coloured
3.) add starch indicator which goes blue-black, and continue until the colour vanishes. The mixture is often described as flesh coloured

80
Q

Why are excess iodide ions added during the titration for copper (II) ions?

A

To ensure that all of the Cu2+ reacts to produce I2

81
Q

How long do we titrate brown I2 against sodium thiosulfate for?

A

Until the mixture is straw coloured

82
Q

What colour does starch indicator go?

A

Blue-black

83
Q

What goes in the flask during the titration for copper (II) ions?

A

Iodine solution

84
Q

What comes from the burette during the titration for copper (II) ions?

A

Sodium thiosulfate solution

85
Q

What is the indicator during the redox titration for copper (II) ions?

A

Starch solution

86
Q

At which point is starch solution added during the redox titration of copper (II) ions?

A

Close to the end point

87
Q

What is the end point of the redox titration for copper (II) ions?

A

When the blue colouration of the starch in the sodium thiosulfate and iodine solution just disappears

88
Q

What has happened when the blue colour of the starch has gone during the redox titration for copper (II) ions?

A

All of the iodine has been used up

89
Q

What type of sodium thiosulfate is used during the redox titration for copper (II) ions?

A

Standard sodium thiosulfate (may be standardised before use)

90
Q

What is the use of iodine/sodium thiosulfate titrations?

A

To estimate the concentration of copper (II) ions in aqueous solution

91
Q

Explain what is actually happening on a chemical level during the redox titration for copper (II) ions

A

Copper (II) ions undergo a redox reaction with iodide ions in aqueous solution
The iodide ions are oxidised to iodine, which can be titrated against sodium thiosulfate
The copper (II) ions are reduced to copper (I) ion, which precipitate as copper (I) iodide

92
Q

Under which conditions do copper (II) ions undergo a redox reaction with iodide ions?

A

In aqueous solution

93
Q

How do iodide ions turn into iodine that can be titrated against sodium thiosulfate for the redox titration for copper (II) ions?

A

They’re oxidised

94
Q

What happens to the copper (II) ions during their redox titration?

A

Reduced to copper (I) ions, which precipitate a copper (I) iodide

95
Q

How do copper (I) ions precipitate in the copper (II) ions redox titration?

A

As copper (I) iodide

96
Q

How does copper (I) iodide appear?

A

An off-white precipitate

97
Q

Why are excess iodide ions added during the redox titration for copper (II) ions (besides ensuring that all of the Cu2+ reacts)?

A

The iodine solution is complexed with unreacted iodide ions as a triiodide ion, I3-

98
Q

Colour of iodine solution

A

Brown

99
Q

What is the brown iodine solution complexed with?

A

Unreacted iodide ions as a triiodide ion, I3-

100
Q

How does iodine appear at room temperature?

A

A grey solid that sublimes to a purple vapour

101
Q

How could we make the redox titration for copper (II) ions faster?

A

By adding the excess iodide ions in solid form

102
Q

Molar ratio between copper (II) ions and iodine molecules

A

2:1

103
Q

Molar ratio between a molecule of iodine and thiosulfate ions

A

1:2

104
Q

Molar ratio between thiosulfate and copper

A

1:1

105
Q

How do we know that the molar ratio between thiosulfate and copper is 1:1?

A

Copper (II) ions to iodine —> 2:1
Iodine to thiosulfate ions —> 1:2
Combined…
2Cu2+ = 1I2 = 2S2O3^2-

106
Q

Ethanedioate ions

A

C2O4^2-

107
Q

C2O4^2- ions

A

Ethanedioate

108
Q

How do we work out what’s oxidised and what’s reduced for Ecell calculations?

A

Write the full equation (combining the half equation) and work it out from this

109
Q

Hi

A

Ok

110
Q

Why do we have to standardise potassium manganate (VII)?

A

Since it:
- may absorb moisture from the environment
- may not be in its purest form

Chemistry (39 decks)

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