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Inorganic Chemistry > Topic 14 - Redox II > Flashcards

Topic 14 - Redox II Flashcards

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1
Q

What happens to an element’s oxidation number when it is oxidised?

A

It increases.

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2
Q

Do s-block elements react by gaining or losing electrons?

A

Losing them

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3
Q

Do p-block elements react by gaining or losing electrons?

A
  • Metals -> Losing electrons

* Non-metals -> Gaining electrons

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4
Q

Do d-block elements react by gaining or losing electrons?

A

It varies, by but they tend to form positive ions with positive oxidation numbers.

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5
Q

Describe the structure of a basic electrochemical cell.

A
  • Anode and cathode
  • Each electrode is dipped in a solution of its own ions
  • Salt bridge between solutions
  • External circuit connecting electrodes, with voltmeter
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6
Q

In an electrochemical cell, what is the salt bridge made of?

A

Filter paper soaked in a salt solution

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7
Q

What are the two processes in an electrochemical cell?

A
  • Oxidation

* Reduction

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8
Q

In an electrochemical cell, which electrode does oxidation always happen at?

A

Anode

Remember: A + O are vowels!

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9
Q

In an electrochemical cell, explain which metal becomes the anode and which becomes the cathode? Where do oxidation and reduction occur?

A

MORE REACTIVE METAL:
• Forms ions more readily, so it loses electrons electrons more easily
• This is oxidation
• So this is the anode (relatively positive)
LESS REACTIVE METAL:
• Forms ions less readily, so it instead accepts electrons and ions from the solution, reforming the metal
• This is reduction
• So this is the cathode (relatively negative)

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10
Q

In an electrochemical cell, which electrode does reduction always happen at?

A

Cathode

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11
Q

Do reactive or unreactive metals form ions more readily?

A

Reactive

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12
Q

In an electrochemical cell, which direction do electrons flow in?

A

From the more reactive metal to the less reactive metal.

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13
Q

What is the symbol for cell potential?

A

E(cell)

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14
Q

What is cell potential?

A

The difference between the two half-cells in an electrochemical cell.

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15
Q

Do half-cells have to involve a metal and solution of metal ions?

A

No, they can also involve:
• A solution of two different aqueous ions of the same element (in the same half-cell) with an inert electrode
• Gas with solution of its own ions and an inert electrode

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16
Q

Give an example of a half-cell with a solution of two aqueous ions of the same element. Describe the structure.

A 
• Solution of Fe³⁺ and Fe²⁺
• Platinum (or graphite) electrode 
• Oxidation or reduction (depending on whether the half-cell has the anode or cathode) occurs on the electrode surface
Possible reactions:
• Fe²⁺(aq) -> Fe³⁺(aq) + e⁻
• Fe³⁺(aq) + e⁻ -> Fe²⁺(aq)
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17
Q

When an inert electrode is required in an electrochemical cell, what elements tend to be used?

A
  • Platinum

* Graphite

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18
Q

Given that zinc is more reactive than copper, explain what happens in the zinc/copper electrochemical cell. Include half-equations.

A
  • Zinc loses electrons more easily, so it is oxidised to the ion, releasing electrons into the circuit
  • Zn(s) -> Zn²⁺(aq) + 2e⁻
  • Copper ions in the solution are reduced using electrons from the circuit, to form copper solid
  • Cu²⁺(aq) + 2e⁻ -> Cu(s)
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19
Q

Describe how a half-cell can be set up with a gas.

A
  • The gas is bubbles over a platinum/graphite catalyst sitting in a solution of its aqueous ions
  • Oxidation or reduction occurs on the surface of the electrode
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20
Q

When drawing electrochemical cells, which half-cell is on the left?

A

The one where oxidation occurs (anode).

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21
Q

Are the reactions at each electrode reversible?

A
  • Yes

* The direction each reaction goes in depends on how easily the metal loses electrons

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22
Q

Which way are half-reactions at each electrode in an electrochemical cell written?

A

The reduction reaction is the forward direction.

e.g. Zn²⁺ + 2e⁻ -> Zn

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23
Q

Describe how you can set up an electrochemical cell involving two metals.

A

1) Get a strip of each of the metals you’re investigating. Clean the surfaces using emery paper (or sandpaper).
2) Clean any grease or oil from the electrodes using some propanone.
3) Place each electrode into a beaker with a solution containing ions of that metal. If any solution contains an oxidising agent that contains oxygen (e.g. MnO₄⁻), add acid too.
4) Create a salt bridge to link the two beaters together. Do this by soaking a piece of filter paper in salt solution.
5) Connect the electrodes to a voltmeter using crocodile clips and wires. There should be a voltmeter reading.

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24
Q

What things are used to clean the electrodes when preparing an electrochemical cell?

A
  • Emery paper (or sandpaper) -> Cleans surface

* Propanone -> Cleans off any grease or oil

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25
Q

What solution could be used in a half-cell of copper metal?

A

CuSO₄ (for example)

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26
Q

What should you do if the solution in a half-cell contains an oxidising agent that contains oxygen (e.g. MnO₄⁻)?

A

Add acid too

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27
Q

Describe what a half-cell electrode potential essentially is.

A

How easily the substance in the half-cell is oxidised (i.e. loses electrons).

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28
Q

Why does a potential difference build up in a half-cell?

A

There is a difference between the charge of the electrode and the ions in the solution.

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29
Q

When there are two half-cell potentials, how can you tell which reaction goes forwards and which goes backwards?

A
  • The half-reaction with the MORE positive electrode potential goes forward.
  • The half-reaction with the MORE negative electrode potential goes backwards.
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30
Q

Zn²⁺ + 2e⁻ -> Zn (-0.76V)
Cu²⁺ + 2e⁻ -> Cu (+0.34V)

Which reaction goes forwards and which goes backwards?

Write the ionic equation for the reaction.

A
  • The zinc reaction is more negative -> It goes backwards
  • The copper reaction is more positive -> It goes forwards

Cu²⁺(aq) + Zn(s) -> Cu(s) + Zn²⁺(aq)

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31
Q

What does the little ° next to the E show?

A

It is the STANDARD electrode potential.

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32
Q

Zn²⁺ + 2e⁻ -> Zn (-0.76V)
Cu²⁺ + 2e⁻ -> Cu (+0.34V)

Which is oxidised and which is reduced?

A
  • Oxidised -> Zinc

* Reduced -> Copper

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33
Q

What is the symbol for standard electrode potential?

A

E°

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34
Q

Define standard electrode potential.

A

The voltage measured under standard conditions when the half-cell is connected to a standard hydrogen electrode.

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35
Q

What are the standard conditions for a standard electrode potential?

A
  • Solutions of the ions you’re interested in must have a concentration of 1.0 mol/dm³
  • 298K
  • 100kPa
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36
Q

Describe the structure of a standard hydrogen half-cell.

A
  • Hydrogen is pumped into an upturned vessel in a 1.00 mol/dm³ H⁺ solution
  • Platinum electrode is in the upturned vessel
  • At 298K and 100kPa
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37
Q

When measuring standard electrode potential, what is the equation at the hydrogen electrode?

A

2H⁺(aq) + 2e⁻ -> H₂(g)
OR
H₂(g) -> 2H⁺(aq) + 2e⁻

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38
Q

In a setup to calculate the standard electrode potential for a half-cell, is the hydrogen half-cell where oxidation or reduction happens?

A

It can be either.

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39
Q

When drawing out the setup to calculate the standard electrode potential for a half-cell, where is the hydrogen half-cell drawn?

A

On the left (regardless of whether it is reduced or oxidised).

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40
Q

Describe how a standard hydrogen half-cell can be used to work out the half-cell electrode potential of an different half-cell.

A
  • Other half-cell is connected on the right of the hydrogen half-cell in order to make a full cell
  • The voltage reading is the half-cell potential of the other half-cell
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41
Q

What is the electrode potential of a standard hydrogen electrode?

A

0.00V

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42
Q

What is the equation for the cell potential of an electrochemical cell?

A

E°(cell) = E°(red) - E°(oxid)

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43
Q

What can be said about the value of E°(cell) for a working electrochemical cell and why?

A
  • It is always positive

* Because the more negative E° value is being subtracted from the more positive E° value

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44
Q

Calculate the cell potential of a magnesium-bromine electrochemical cell:
Br₂ + Mg -> Mg²⁺ + 2Br⁻

Mg²⁺(aq) + 2e⁻ -> Mg(s) E° = -2.37V
1/2Br₂(aq) + e⁻ -> Br⁻(aq) E° = +1.09V

A
  • E°(cell) = E°(red) - E°(oxid)

* E°(cell) = +1.09 - (-2.37) = +3.46V

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45
Q

When doing E°(cell) calculations, do you have to account for the number of electrons transferred at each electrode?

A

No, just use the E° values.

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46
Q

Why are electrode potentials quoted under standard conditions?

A

The temperature, pressure and concentration may affect the equilibrium position, which affects the cell potential.

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47
Q

What is the name for the shorthand method of drawing electrochemical cells?

A

Conventional representation

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48
Q

What are the rules for drawing an electrochemical cell in shorthand?

A

Half-cell with more negative potential goes on the left:
• Far left -> Reduced form (usually the uncharged metal)
• Dashed line
• Middle left -> Oxidised form (usually the charged ion)
Double vertical dashed lines show the salt bridge linking to the half-cell with the more positive potential:
• Middle right -> Oxidised form (usually the charged ion)
• Dashed line
• Far right -> Reduced form (usually the uncharged metal)

  • Commas separate species that are in the same half-cell and the same physical state
  • If there is a standard hydrogen half-cell, it should be on the left
  • Show inert electrodes on the outside of the diagram (separated by a dashed line)
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49
Q

What do the double vertical dashed lines show on an electrochemical cell shorthand diagram?

A

Salt bridge

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50
Q

What do the single vertical dashed lines show on an electrochemical cell shorthand diagram?

A

The show the boundary between species in different physical states.

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51
Q

How are species in the same half-cell and in the same physical state separated in an electrochemical cell shorthand diagram?

A

Commas

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52
Q

How are inert electrodes shown on an electrochemical cell shorthand diagram?

A

They are on the outside of the diagram, separated by a vertical dashed line.

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53
Q

Draw the conventional representation of the electrochemical cell formed between magnesium and the standard hydrogen half-cell.

A

Pt | H₂(g) | 2H⁺(aq) || Mg²⁺(aq) | Mg(s)

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54
Q

With a half-cell on the left of a standard electrochemical diagram, what can be assumed?

A
  • It is the anode
  • So oxidation happens there -> Electrons are lost
  • So the half-cell electrode potential must be more negative (than the other half-cell), since the half-cell equations are written with reduction in the forward direction

Anode, Oxidation, Negative E°

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55
Q

How does a metal’s reactivity affect how easily it forms ions?

A

The most reactive it is, the more easily it loses electrons to form a positive ion.

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56
Q

How does a non-metal’s reactivity affect how easily it gains electrons?

A

The more reactive it is, the more easily it gains electrons to form a negative ion.

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57
Q

What is an electrochemical series show?

A

How reactive metals and non-metals are based on their standard electrode potentials.

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58
Q

Describe how you can tell how reactive an element is from it’s standard electrode potential.

A
  • See if it is a metal or non-metal
  • If it is a metal -> The more negative the standard electrode potential, the more reactive it is
  • If it is a non-metal -> The more positive the standard electrode potential, the more reactive it is
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59
Q

How can you work out the feasibility of a reaction of a metal with the aqueous ions of another metal using electrode potentials?

A

1) Write the equation as the half-equations
2) Look at the standard electrode potential of each half-equation (in the normal reduction direction)
3) E°(cell) = E°(red) - E°(oxid)
4) If this gives a positive value, the reaction is feasible

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60
Q

When using electrode potentials to work out whether a reaction is feasible, what shows that the reaction is feasible?

A

E°(cell) is positive (calculated by E°(red) - E°(oxid))

61
Q

Predict whether zinc metal reacts with aqueous copper ions.

Zn²⁺(aq) + 2e⁻ -> Zn(s) E° = -0.76V
Cu²⁺(aq) + 2e⁻ -> Cu(s) E° = +0.34

A
  • Zn(s) + Cu²⁺(aq) -> Zn²⁺(aq) + Cu(s)
  • E°(cell) = E°(red) - E°(oxid) = 0.34 -(-0.76) = +1.10V
  • This is positive, so the reaction will happen.
62
Q

What is the name for a species being simultaneously oxidised and reduced?

A

Disproportionation

63
Q

Can electrode potentials be used to predict whether a disproportionation reaction is feasible?

A

Yes

64
Q

Use the following equations to predict whether or not Ag⁺ ions will disproportionate on solution.

Ag⁺(aq) + e⁻ -> Ag(s) E° = +0.80V
Ag²⁺(aq) + e⁻ -> Ag⁺(aq) E° = +2.00V

A
  • 2Ag⁺(aq) -> Ag(s) + Ag²⁺ (aq)
  • E°(cell) = E°(red) - E°(oxid) = 0.80 - (2.00) = -1.20V
  • This is negative, so the reaction will not happen.
65
Q

Remember to check with teacher whether the number of each species in an equation affects the E°(cell) value (when predicting whether w reaction will happen).

A

Do it.

66
Q

When might the prediction of the feasibility of a reaction based on electrode potentials be wrong?

A
  • The conditions are not standard

* The reaction kinetics are not favourable

67
Q

Zn²⁺(aq) + 2e⁻ -> Zn(s) E° = -0.76V
Cu²⁺(aq) + 2e⁻ -> Cu(s) E° = +0.34
Zn(s) + Cu²⁺(aq) -> Zn²⁺(aq) + Cu(s) E°(cell) = +1.10V

What happens to E°(cell) when the concentration of Zn²⁺ is increased?

A
  • The half-cell equilibrium shifts to the right, reducing the ease of electron loss of Zn.
  • The electrode potential of Zn/Zn²⁺ becomes less negative.
  • The whole cell potential will be lower.
68
Q

Zn²⁺(aq) + 2e⁻ -> Zn(s) E° = -0.76V
Cu²⁺(aq) + 2e⁻ -> Cu(s) E° = +0.34
Zn(s) + Cu²⁺(aq) -> Zn²⁺(aq) + Cu(s) E°(cell) = +1.10V

What happens to E°(cell) when the concentration of Cu²⁺ is increased?

A
  • The half-cell equilibrium shifts to the right, increasing the ease of electron gain of Cu²⁺.
  • The electrode potential of Cu²⁺/Cu becomes more positive.
  • The whole cell potential will be higher.
69
Q

Explain why the feasibility prediction for a reaction based on electrode potentials might not be correct due to reaction kinetics.

A
  • The rate of reaction might be so slow that is it might not appear to happen
  • The activation energy may be too high for the reaction to happen spontaneously
70
Q

What two factors is cell potential related to?

A
  • Entropy change

* Equilibrium constant

71
Q

Describe how cell potential is related to entropy change.

A

E° ∝ ΔS(total)

72
Q

Describe how cell potential is related to the equilibrium constant.

A

E° ∝ lnK

73
Q

Where does E° ∝ lnK come from?

A
  • E° ∝ ΔS(total)

* Entropy and the equilibrium constant are linked

74
Q

How can you tell a reactive metal and non-metal from its standard electrode potential?

A
  • The more reactive a metal, the more negative the standard electrode potential.
  • The more reactive a non-metal, the more positive the standard electrode potential.
75
Q

What are energy storage cells?

A
  • They are essentially batteries

* They work just like electrochemical cells

76
Q

How can you work out the voltage produced by an energy storage cell?

A

Using the electrode potentials of the substances used in the cell.

(i.e. Just like with a normal electrochemical cell)

77
Q

The nickel-iron cell has a nickel oxide hydroxide (NiO(OH)) cathode and an iron (Fe) anode with potassium hydroxide as the electrolyte.

Fe(OH)₂ + 2e⁻ -> Fe + 2OH⁻ E°=-0.89V
NiO(OH) + H₂O + e⁻ -> Ni(OH)₂ + OH⁻ E°=+0.49V

a) Write our the full equation for the reaction.
b) Calculate the cell voltage produced by the nickel-iron cell.

A

a) Combine the two reactions in the feasible reaction (so that E° is positive):

2NiO(OH) + 2H₂O + Fe -> 2Ni(OH)₂ + Fe(OH)₂

b) E°(cell) = E°(red) - E°(oxid)

E°(cell) = +0.49 -(-0.89) = 1.38V

78
Q

What are fuel cells?

A

Electrochemical cells that have fuel stored outside the cell and fed in when electricity is required.

79
Q

Give an example of a fuel cell.

A

Hydrogen-oxygen fuel cell

80
Q

What are the two types of hydrogen-oxygen fuel cells?

A
  • Acidic

* Alkaline

81
Q

What can hydrogen-oxygen cells be used for?

A

Powering electric vehicles

82
Q

Describe the structure of an alkaline hydrogen-oxygen fuel cell.

A
  • H₂ gas is fed next to negative platinum electrode -> H₂O flows away
  • O₂ gas is fed next to positive platinum electrode
  • Electrons flow through circuit from negative hydrogen electrode to positive oxygen electrode
  • Electrodes are separated by an anion-exchange membrane that allows OH⁻ and H₂O to pass through it, but not H₂ and O₂
  • Electrolyte between membranes is KOH, in which OH⁻ ions flow from oxygen to hydrogen side
83
Q

In a hydrogen-oxygen fuel cell, what are the electrodes made of?

A

Platinum

84
Q

In an alkaline hydrogen-oxygen fuel cell, what are the membranes called?

A

Anion-exchange membranes

85
Q

In an alkaline hydrogen-oxygen fuel cell, what are the membranes used for?

A
  • Allow anions (OH⁻) and water to pass through

* Don’t allow H₂ and O₂ to pass through

86
Q

In an alkaline hydrogen-oxygen fuel cell, what is the electrolyte?

A

KOH

87
Q

In a fuel cell, how can you tell which is the positive and negative electrode?

A

Electrons flow away from the negative electrode.

88
Q

What is the anode?

A

The electrode which:
• Anions flow to
• Electrons flow away from

89
Q

What is the cathode?

A

The electrode which:
• Cations flow to
• Electrons flow to

90
Q

Describe the direction of flow of electrons in an alkaline hydrogen-oxygen fuel cell.

A

From negative electrode near hydrogen to the positive electrode near oxygen. OH⁻ carry the charge from there through the electrolyte to the negative electrode again.

91
Q

Describe how an alkaline hydrogen-oxygen fuel cell works. Include half-equations.

A

• Hydrogen is fed to the negative electrode. It combines with the OH⁻ travelling through the electrolyte:
2H₂(g) + 4OH⁻(aq) -> 4H₂O(l) + 4e⁻
• Oxygen is fed to the positive electrode. It combines with water in the solution to produce OH⁻:
O₂(g) + 2H₂O(l) + 4e⁻ -> 4OH⁻(aq)
• OH⁻ travel through the solution to the negative electrode, where they are used by the hydrogen.
• Electrons move away from the negative electrode (hydrogen-side) and travel around the circuit to the positive electrode (oxygen-side).

Overall effect:
2H₂(g) + O₂(g) -> 2H₂O(l)

92
Q

What are the half-equations and full equations in the alkaline hydrogen-oxygen fuel cell?

A 
• Negative electrode:
2H₂(g) + 4OH⁻(aq) -> 4H₂O(l) + 4e⁻
• Positive electrode:
O₂(g) + 2H₂O(l) + 4e⁻ -> 4OH⁻(aq)
• Overall:
2H₂(g) + O₂(g) -> 2H₂O(l)
93
Q

What is the half-equation for the negative electrode in the alkaline hydrogen-oxygen fuel cell?

A

2H₂(g) + 4OH⁻(aq) -> 4H₂O(l) + 4e⁻

94
Q

What is the half-equation for the positive electrode in the alkaline hydrogen-oxygen fuel cell?

A

O₂(g) + 2H₂O(l) + 4e⁻ -> 4OH⁻(aq)

95
Q

Remember to practise drawing out the structure of an alkaline hydrogen-oxygen fuel cell.

A

Pg 160 of revision guide

96
Q

Describe the structure of an acidic hydrogen-oxygen fuel cell.

A
  • H₂ gas is fed next to negative electrode
  • O₂ gas is fed next to positive electrode
  • Electrons flow through circuit from negative hydrogen electrode to positive oxygen electrode
  • Electrodes are separated by a polymer electrolyte membrane that allows H⁺ ions to get through, but not electrons, O₂ or H₂
  • Electrolyte is in the polymer electrolyte membrane
97
Q

In an acidic hydrogen-oxygen fuel cell, what are the membranes called?

A

Polymer electrolyte membrane (PEM)

98
Q

In an acidic hydrogen-oxygen fuel cell, how does the PEM work?

A

It allows H⁺ ions to travel across, but not O₂, H₂ or electrons. So electrons are forced around the external circuit.

99
Q

Describe the direction of flow of electrons in an acidic hydrogen-oxygen fuel cell.

A

From negative electrode near hydrogen to the positive electrode near oxygen. H⁺ carry positive charge from the hydrogen side to the oxygen.

100
Q

Describe how an acidic hydrogen-oxygen fuel cell works. Include half-equations.

A

• Hydrogen is fed to the negative electrode. It is forced to split into H⁺ and e⁻:
H₂ -> 2H⁺ + 2e⁻
• Oxygen is fed to the positive electrode. It combines with H⁺ ions travelling across the PEM:
1/2O₂ + 2H⁺ + 2e⁻ -> H₂O
• H⁺ travelled through the polymer electrolyte membrane (PEM) from the hydrogen to the oxygen side.
• Electrons move away from the negative electrode (hydrogen-side) and travel around the circuit to the positive electrode (oxygen-side).

Overall effect:
2H₂(g) + O₂(g) -> 2H₂O(l)

101
Q

What are the half-equations and full equations in the acidic hydrogen-oxygen fuel cell?

A 
• Negative electrode:
H₂ -> 2H⁺ + 2e⁻
• Positive electrode:
1/2O₂ + 2H⁺ + 2e⁻ -> H₂O
• Overall:
2H₂ + O₂ -> 2H₂O
102
Q

What is the half-equation for the negative electrode in the acidic hydrogen-oxygen fuel cell?

A

H₂ -> 2H⁺ + 2e⁻

103
Q

What is the half-equation for the positive electrode in the acidic hydrogen-oxygen fuel cell?

A

1/2O₂ + 2H⁺ + 2e⁻ -> H₂O

104
Q

Remember to practise drawing out the structure of an acidic hydrogen-oxygen fuel cell.

A

Pg 161 of revision guide

105
Q

In an alkaline or acidic hydrogen-oxygen fuel cell, which is the positive and negative electrode, and what are they called?

A
  • Hydrogen side -> Anode -> Negative

* Oxygen side -> Cathode -> Positive

106
Q

Describe the differences in an alkaline and acidic hydrogen-oxygen fuel cell.

A

Electrons flow in the same direction and the overall reaction is the same, except:
• Alkaline cell uses KOH electrolyte and anion-exchange membranes, while acidic cell uses polymer electrolyte membrane -> So in an alkaline cell OH⁻ moves from positive to negative electrode, while in an acidic cell H⁺ moves from positive to negative electrode
• So there are different half-equations at each electrode

107
Q

What other types of fuel can fuel cells use except hydrogen?

A

Hydrogen-rich fuels, including methanol and ethanol.

108
Q

How can fuel cells use hydrogen-rich field, such as ethanol and methanol?

A
  • Traditionally, they can be converted into H₂ in the car by a reformer -> They are then used
  • Alternatively, there is a new generation of fuel cells that can use alcohols directly without having to reform them into hydrogen first
109
Q

Describe how new generation fuel cells use alcohols (hydrogen-rich fuels) without reforming them into hydrogen first. Include half-equations, using methanol as an example.

A

• Alcohol is oxidised at the anode in the presence of water:
CH₃OH + H₂O -> CO₂ + 6e⁻ +6H⁺
• The H⁺ ions pass through the electrolyte and are oxidised to water.
6H⁺ + 6e⁻ + 3/2O₂ -> 3H₂O

110
Q

What moves through the electrolyte in new generation fuel cells that use alcohols (hydrogen-rich fuels) without reforming them into hydrogen first?

A

H⁺ moves through the electrolyte from the anode to the cathode.

111
Q

What are the two types of titration?

A
  • Acid-base titrations

* Redox titrations

112
Q

Explain simply how an acid-base titration works.

A
  • You work out how much acid is needed to neutralise a given volume of alkali of known concentration (or vice versa)
  • This is used to work out the concentration of the unknown acid/alkali
113
Q

What equipment is required to carry out an acid-base titration?

A
  • Pipette
  • Burette
  • Conical flask
  • Acid/Alkali of known concentration
  • Acid/Alkali of unknown concentration
114
Q

What are redox titrations used for and on what principle do they work?

A
  • Used to work out the concentration of an oxidising or reducing agent
  • It works by titrating a reducing agent against an oxidising agent of a known concentration (or vice versa)
  • From this, the concentration can be worked out
115
Q

What are redox titrations usually carried out with? Why?

A

Transition element ions, because they are good at changing oxidation number. This makes them good oxidising and reducing agents.

116
Q

In a redox titration, which solution is added to which?

A

The solution of known concentration is added to the solution of unknown concentration.

117
Q

Describe how to carry out a redox titration to find out how many manganate(VII) ions (MnO₄⁻) are required to react with a reducing agent of unknown concentration.

A

1) Pipette a quantity of a reducing agent (e.g. aqueous Fe²⁺) into a conical flask.
2) Add an excess of dilute sulfuric acid (this is ensure that the oxidising agent can easily be reduced).
3) Do a rough titration by adding MnO₄⁻ using a burette until the colour changes from colourless to purple.
4) Repeat and average to get an accurate titration volume.
5) Do the right calculations to work out the concentration.

118
Q

In a redox titration, is the reducing agent added to the oxidising agent, or vice versa?

A

This is irrelevant. The solution of known concentration is added to the solution of unknown concentration.

119
Q

In a redox titration, what is added aside from the reducing and oxidising agent? Why?

A
  • An excess of dilute sulfuric acid is added

* This is to make sure there are plenty of H⁺ ions to allow the oxidising agent to be reduced

120
Q

Do redox titrations use an indicator?

A

Usually no, because the transition metals tend to change colour when they change oxidation state.

121
Q

Is MnO₄⁻ a reducing or oxidising agent?

A

Oxidising agent

122
Q

What are MnO₄⁻ ions reduced to in redox titrations?

A

Mn²⁺

123
Q

What colour is MnO₄⁻ and Mn²⁺?

A
  • MnO₄⁻ -> Purple

* Mn²⁺ -> Colourless

124
Q

Give the half-equations and full equation for the oxidation of Fe²⁺ to Fe³⁺ by manganate(VII) ions.

Describe the colour change.

A

• MnO₄⁻ + 8H⁺ + 5e⁻ -> Mn²⁺ + 4H₂O
• 5Fe²⁺ -> 5Fe³⁺ + 5e⁻
Full equation:
• MnO₄⁻ + 8H⁺ + 5Fe²⁺ -> Mn²⁺ + 4H₂O + 5Fe³⁺

From purple to colourless.

125
Q

Is Cr₂O₇²⁻ a reducing or oxidising agent?

A

Oxidising agent

126
Q

What are Cr₂O₇²⁻ ions reduced to in redox titrations?

A

Cr³⁺

127
Q

What colour is Cr₂O₇²⁻ and Cr³⁺?

A
  • Cr₂O₇²⁻ -> Orange

* Cr₃⁺ -> Green

128
Q

Give the half-equations and full equation for the oxidation of Zn to Zn²⁺ by dichromate(VI) ions.

Describe the colour change.

A

• Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -> 2Cr³⁺ + 7H₂O
• 3Zn -> 3Zn²⁺ + 6e⁻
Full equation:
• Cr₂O₇²⁻ + 14H⁺ + 3Zn -> 2Cr³⁺ + 7H₂O + 3Zn²⁺

Orange to green.

129
Q

What can make the colour change in a redox reaction easier to spot?

A

Doing it in front of a white surface.

130
Q

How can the concentration of the unknown in a redox titration be calculated?

A

1) Work out the number of moles used of the reactant of known concentration.
2) Use the equation to find the number of moles of the reactant of unknown concentration.
3) Use “conc = moles / vol” to work out the concentration of the reactant of unknown concentration.

131
Q

In a titration, 27.5 cm³ of 0.0200 mol/dm³ aqueous potassium manganate(VII) reacted with 25.0 cm³ of acidified iron(II) sulfate solution.

Give the ionic equation for this and calculate the concentration of Fe²⁺ ions in the solution.

A
  • MnO₄⁻ + 5Fe²⁺ + 8H⁺ -> Mn²⁺ + 5Fe³⁺ + 4H₂O
  • Number of moles of MnO₄⁻ = Conc x Vol = 0.0200 x (27.5 / 1000) = 5.50 x 10⁻⁴ mol
  • Therefore, the number of moles of Fe²⁺ = 5.50 x 10⁻⁴ x 5 = 2.75 x 10⁻³ mol
  • Concentration of Fe²⁺ = Moles / Vol = (2.75 x 10⁻³) / (25.0 x 10⁻³) = 0.110 mol/dm³
132
Q

What practical involves redox titrations?

A

Estimating the percentage of iron in iron tablets.

133
Q

In what form is iron usually found in iron tablets?

A

Iron(II) sulfate

134
Q

A 2.56g iron tablet was dissolved in dilute sulfuric acid to give 250 cm³ of solution. 25.0 cm³ of this solution was found to react with 12.5 cm³ of 0.0250 mol/dm³ potassium manganate(VII) solution.

Give the ionic equation for this and calculate the percentage of iron in the tablet.

A
  • MnO₄⁻ + 8H⁺ + 5Fe²⁺ -> Mn²⁺ + 4H₂O + 5Fe³⁺
  • Moles of MnO₄⁻ = Conc x Vol = 0.0250 x (12.5 x 10⁻³) = 3.125 x 10⁻⁴ mol
  • Therefore moles of Fe²⁺ = 1.5625 x 10⁻³ mol
  • Moles of Fe²⁺ in 250cm³ = 1.5625 x 10⁻³ x 10 = 1.5625 x 10⁻² mol
  • Mass of iron in the tablet = Mr x Moles = 55.8 x 1.5625 x 10⁻² = 0.871g
  • Percentage of iron in tablet = (0.871 / 2.56) x 100 = 34.1%
135
Q

In a redox titration equation, what tends to be on the reactant side and product side (aside from the oxidising and reducing agent)?

A
  • Reactant: H⁺

* Product: H₂O

136
Q

What redox titration in particular can be used to find the concentration of an oxidising agent?

A

Iodine - Sodium thiosulfate

137
Q

What is the formula for the thiosulfate ion?

A

S₂O₃²⁻

138
Q

Describe briefly how iodine - sodium thiosulfate titrations work to calculate the concentration of an oxidising agent.

A

1) Use a sample of the oxidising agent to oxidise as much I⁻ (from KI) as possible
2) Find out how many moles of I₂ have been produced -> By reacting it with sodium thiosulfate until the colour fades and the starch that is added remains blue
3) Calculate the concentration of the oxidising agent from the moles of I₂ produced

139
Q

Describe how a sodium - thiosulfate titration could be used to work out the concentration of potassium iodate(V) solution. Include equations.

A

STAGE 1:
• Measure out a 25cm³ sample of potassium iodate(V)
• Add this to an excess of acidified potassium iodide solution (KI).
• Some of the iodide ions are oxidised to iodine:
IO₃⁻ + 5I⁻ + 6H⁺ -> 3I₂ + 3H₂O
STAGE 2:
• From a burette, add sodium thiosulfate of known concentration to the solution drop by drop:
I₂ + 2S₂O₃²⁻ -> 2I⁻ + S₄O₆²⁻
• When the colour fades to a pale yellow, add 2cm³ of starch solution to detect the presence of iodine -> This turns the solution dark blue
• Add sodium thiosulphate until the blue colour disappears (end point)
• Using the titre volume, calculate the moles of iodine in the original solution
STAGE 3:
• Using the original equation, work out the moles of the oxidising agent and therefore its concentration

140
Q

Describe how an iodine - sodium thiosulphate titration is carried out.

A

1) Take a flask containing I₂ produced by a redox reaction of an oxidising agent.
2) From a burette, add sodium thiosulfate to the flask, drop by drop.
3) When the iodine fades to a pale yellow colour, add 2cm³ of starch solution (to detect the presence of iodine). This gives a blue-black colour.
4) Continue adding the sodium thiosulfate drop by drop until the blue colour disappears (end point).
5) Use the titre volume and the equation to work out the moles of I₂ in the original solution.

141
Q

Describe the colour changes in an iodine-sodium thiosulfate titration.

A

Yellow -> Pale yellow -> Blue-black (when starch added) -> Colourless

142
Q

Give the ionic equation for the oxidation of acidified iodide ions using potassium iodate(V) oxidising agent.

A

IO₃⁻(aq) + 5I⁻(aq) + 6H⁺(aq) -> 3I₂(aq) + 3H₂O(l)

143
Q

Give the ionic equation for the reaction between iodine and thiosulfate ions.

A

I₂(aq) + 2S₂O₃²⁻(aq) -> 2I⁻(aq) + S₄O₆²⁻(aq)

144
Q

25.0 cm³ of potassium iodate(V) solution is used to react with an excess of acidified potassium iodide. The resulting solution is reacted fully with 11.0 cm³ of 0.120 mol/dm³ sodium thiosulfate solution. Give the equations for each reaction and work out the concentration of the potassium iodate(V).

A

• IO₃⁻ + 5I⁻ + 6H⁺ -> 3I₂ + 3H₂O
• I₂ + 2S₂O₃²⁻ -> 2I⁻ + S₄O₆²⁻
Second reaction:
• Moles of thiosulfate = 0.120 x 11.0 x 10⁻³ = 1.32 x 10⁻³ mol
• Therefore, moles of iodine = 1.32 x 10³ / 2 = 6.60 x 10⁻⁴ mol
First reaction:
• Therefore, moles of iodate = 6.60 x 10⁻⁴ / 3 = 2.20 x 10⁻⁴ mol
• Concentration of potassium iodate(V) = 0.00880 mol/dm³

145
Q

Describe how a sodium - thiosulfate titration could be used to work out the percentage of copper in an alloy. Include equations.

A

STAGE 1:
• Dissolve a weighed amount of the alloy in some concentrated nitric acid
• Pour this mixture into a 250cm³ volumetric flask and make up to 250cm³ using deionised water
• Pipette out 25cm³ and transfer to a flask.
• Slowly added sodium carbonate to neutralise any remaining acid until a precipitate form. Add drops of ethanoic acid to remove it.
• Add an excess of potassium iodide solution:
2Cu²⁺ + 4I⁻ -> 2CuI + I₂
• A white precipitate forms. The copper(II) ions have been reduced to copper(I).
STAGE 2:
• Titrate the mixture against sodium thiosulfate (see other flashcards) to find the number of moles of iodine present.
• I₂ + 2S₂O₃²⁻ -> I⁻ + S₄O₆²⁻
STAGE 3:
• Use the first equation to work out the moles of copper in 25cm³, and therefore 250cm³
• From this, calculate the mass of copper in the whole piece of the alloy
• Work out the percentage by mass of the copper in the alloy

146
Q

Give the half equation for the reaction of copper(II) ions with iodide ions.

A

2Cu²⁺(aq) + 4I⁻(aq) -> 2CuI(s) + I₂(aq)

147
Q

What is the solubility and colour of copper(I) iodide?

A

White solid (insoluble)

148
Q

Remember to practise writing out how you can work out the percentage of copper in an alloy.

A

Pg 166 of revision guide

149
Q

What are some of the sources of error in iodine - sodium thiosulfate titrations?

A

1) The starch indicator has to be added at the right point, when most of the I₂ has reacted, or the blue colour will be very slow to disappear
2) Starch solution needs to be freshly made or it won’t behave as expected
3) In copper alloy calculations, the copper(I) iodide precipitate can make seeing the colour of the solution difficult
4) Iodine produced before the iodine-thiosulfate titration can evaporate, giving a too low percentage or concentration of the initial reactant.

Inorganic Chemistry (21 decks)

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