Elements Of The Sea Flashcards

1
Q

Electrolysis is?

A

Breaking a substance down using electricity.

If you pass an electric current though an IONIC substance that is molten or in a solution, it breaks down into the elements that it’s made of. This is electrolysis.

It requires a liquid to conduct the electricity, called the electrolyte. The electrolyte contains free ions, which is usually molten or dissolved ionic substance.

In either case it’s the free ions which conduct the electricity.

For the circuit to be complete, there just be a flow of electrons:

Negative ions (anions) go to the positive electrode (the anode), and lose electrons.

Positive ions (cations) go to the negative electrode (the cathode), and gain electrons.

The electrodes must be able to conduct electricity, so they much have free electrons - graphite and platinum are used. They must also be inert so they don’t react.

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2
Q

Anode and cathode formation?

A

Negative ions are formed at the anode.
E.g. 2Br- —> Br2 + 2e-

Positive ions are formed at the cathode.
E.g. Pb2+ + 2e- —> Pb

Cathode - reduction occurs. Ions are accepted.
Anode - oxidation occurs. Ions are donated.

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3
Q

Procedure for carrying out the electrolysis of an aqueous solution?

A
  1. Use wires and clips to connect each electrode to the power supply. The electrode connected to the positive pole will be the anode, and the electrode connected to the negative pole will be the cathode.
  2. Use inert electrodes (such as platinum or carbon electrodes) so that they don’t start reacting and interesting with the electrolysis.
  3. Place the electrodes into a beaker containing the electrolyte, making sure the electrodes are not touching.
  4. Turn the power supply on.
  5. Depending on what electrolyte your using, the products will form as metals (thin layer on surface of the cathode, called ‘plating’) or as gases (bubbles at the cathode or anode).

Look at notes for diagram.

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4
Q

The half equations for anodes and cathodes will show?

A

Half equations for anodes will show negative ions losing electrons to form atoms.

Half equations for cathodes will show positive ions gaining electrons for form atoms.

E.g. molten zinc chloride (ZnCl2):

Anode: 2Cl-(l) —> Cl2 (g) + 2e-
Cathode: Zn2+ (l) + 2e- —> Zn(s)
Atoms and charges must both be balanced.

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5
Q

Electrolysis of a molten salt?

A

A molten salt is just a molten ionic compound.

The product will be elements, because the only ions available are the ones that make up the salt.

For example, PbBr2 is an ionic compound.

Pb is lead - positive. So it is a cation, that will go to the cathode, and gain electrons.
Pb2+ (l) + 2e- —> Pb (s)

Br - negative. So it is an anion, that does to the anode, and loses electrons.

2Br- —> 2e- + Br2 (s)

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6
Q

Electrolysis of an aqueous solution?

A

This is slightly more complicated than ionic compounds.

In aqueous solutions, you have H+ and OH- ions from the water as well as the ions from the ionic compounds.

The products formed at each electrode depend on the reactivity of the ions, as well as the concentration of the salt solution.

There are a few rules to help you work out what will happen at each electrode. They’re on next flashcard.

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7
Q

How to know what is formed in an aqueous electrolysis reaction?

A

The products formed at each electrode depends on the reactivity of ions and the concentration of the salt solution.

Look at reactivity series.

The cathode:

  • if the metal is less reactive than hydrogen, a metal will form.
  • if a metal is less reactive than hydrogen, hydrogen gas is formed (from H+ of water in aqueous solution).

The anode:

  • if the solution doesn’t contain a halide ion (F-, Cl-, etc), then oxygen will form (from OH- in water).
  • if the solution is concentrated, and has a halide ion, then the halogen is formed (Cl2),
  • if it is dilute, oxygen is formed.
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8
Q

Reactivity series?

A

(Most reactive)

Potassium
Sodium
Calcium
Magnesium
Aluminium
- Carbon
Zinc
Iron
Tin
Lead
- Hydrogen 
Copper
Silver
Gold
Platinum 

(Least reactive)

please send cows, monkeys and cool zebras into the lovely hot countries, signed general penguin

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9
Q

Purification of copper?

A

The purification of copper is an electrolysis reaction, takes place in an aqueous solution.

Diagram on notes.

At the anode, electrons are lost. This means the copper that makes up the electrode looses electrons. This forms Cu2+ ions.

The ions are attracted to the cathode. The cathode gains electrons, and so a plate of Cu2+ ions forms on the copper cathode.

The anode breaks down and loses mass whilst the cathode builds and gains mass.

Cathode: Cu2+ +2e- —> Cu
Anode: Cu —> Cu2+ + 2e-

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10
Q

Rules for electrolysis of an aqueous solution: at a cathode?

A

Cathode:
1. If the metal is less reactive than hydrogen (e.g. silver or copper), than the metal will be formed.

  1. If the metal is more reactive than hydrogen (e.g. all group 1 and group 2 metals, and aluminium), hydrogen gas will be formed (from the hydrogen ions in the water).
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11
Q

Rules for the electrolysis of an aqueous solution: at an anode?

A

Anode:
1. If the solution doesn’t contain a halide, oxygen will be formed (from hydrogen ions in water).

  1. If the solution is concentrated and contains a halide, then the halogen will be formed.
  2. If the solution contains a halide but is dilute, oxygen will be formed again (from hydrogen atoms in water).
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12
Q

Half equation for the production of oxygen at an anode in an aqueous solution?

A

If the solution doesn’t contain a halide, oxygen will be formed at the anode in an aqueous solution.

Half:

4OH- (aq) —> O2 (g) + 2H2O (l) + 4e-

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13
Q

Group 7?

A

The halogens.

They exist as diatomic molecules (covalent). Usually Cl2, F2, Br2, etc.

In compounds with metals, they exist as 1- ions.
We call these halides: Cl- is chloride, F- is fluoride, Br- is bromide, I- is iodide, As- is astatide.

The halides are ionic.

Melting and boiling points increase down the group.

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14
Q

Physical properties of fluorine?

A

F2

At room temperature: pale yellow gas

Melting point: 53K

Boiling point: 85K

Solubility at 298K/grams per 100g of water: reacts with water.

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15
Q

Physical properties of chlorine?

A

Cl2

At room temperature: green gas

Melting point: 172K

Boiling point: 239K

Solubility at 298K/grams per 100g of water: 0.6g

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15
Q

Physical properties of bromine?

A

Br2

At room temperature: dark red volatile liquid

Melting point: 266K

Boiling point: 332K

Solubility at 298K/grams per 100g of water: 3.5g

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16
Q

Physical properties of iodine?

A

I2

At room temperature: shiny gray/black solid - sublimes to give a purple vapour on warming

Melting point: 387K

Boiling point: 457K

Solubility at 298K/grams per 100g of water: 0.03g

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18
Q

Dissolution of group 7 elements?

A

Halogens are more soluble in non-polar solvents (like cyclohexane) then they are in polar solvents (like water).

  • Chlorine is pale green when dissolved in water.
  • Bromine is orange/yellow when dissolved in water.
  • Iodine is brown when dissolved in water.
  • Chlorine is pale green when dissolved in cyclohexane.
  • Bromine is orange/brown/red when dissolved in cyclohexane.
  • Iodine is violet when dissolved in cyclohexane.
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18
Q

Redox half equations of halogens?

A

These more clearly show the redox processes occurring:

F2 (aq) + 2 I- (aq) → 2 F- (aq) + I2 (aq)

Half equation for oxidation:
2 I-(aq) → I2 (aq) + 2e (aq)

Half equation for reduction:
F2 (aq) + 2e → 2F- (aq)

Oxidation has the e on the right side of the arrow, reduction has the e on the left side of the arrow.

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19
Q

Boiling point trends of halogens?

A

Boiling points increases as you go down the group.

This is because there are more instantaneous dipole-induced dipole bonds.

There are more of these bonds because the relative mass of the atoms increases down the group.

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20
Q

Electronegativity of halogens?

A

Electronegativity decreases down the group.

Electronegativity is the ability of an atom to attract electrons towards itself in a covalent bond.

This is because the atoms get larger and so there is an increased distance between the nucleus and the outer electrons as you go down the group.

This means there is more shielding (more shells between nucleus and outer electrons).

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21
Q

Displacement reactions of halogens?

A

In a displacement reaction, hexane (or any organic solvent) is added to the halogen to show the colour change during the reaction.

The halogen will dissolved readily in the organic solvent. This forms a layer above another layer in a test tube. This top layer is called the organic layer. The bottom layer is called the aqueous layer.

The organic and aqueous layer changes colour if there is a displacement reaction.

Displacement reactions PROVE the reactivity and oxidising trends of halogens.

———
More reactive halogens (e.g. KBR) will displaced less reactive halide ions (e.g. Br-)
———

22
Q

Reactivity trend down group 7?

A

Reactive decreases down group 7.

For a reaction to occur, an electron must be gained.

Atoms with a smaller radius have less shielding, and therefore attract electrons more readily than atoms with a larger radius.

F- is most reactive.
I- is less reactive.

23
Q

Halogens as oxidising agents?

A

Halogens are oxidising agents - they remove an electron from another substance.

For example, Ca + F2 → CaF2

  • Here the calcium metal becomes Ca2+, it is oxidised.
  • The fluorine is reduced to fluoride, F-.

The oxidising ability of halogens decreases down a group, for the same reason that reactivity decreases down the group (radius size). Therefore, fluorine is the strongest oxidising agent.

We show this oxidising trend by reacting halogens with halide ions (in displacement reactions using hexane, like on previous flashcard).

A halogen will displace a halide from a solution if the halide is lower in the periodic table.

24
Q

Displacement reaction of KCl?

A

We add hexane (or any oxidising agent), a potassium-halogen and chlorine/bromine/iodine water to a test tube - for the halide ions.

These are the halogens:
Chlorine water (Cl2) produces Cl-. It is almost colourless. 
Bromine water (Br2) produces Br-. It is orange. 
Iodine water (I2) produces I-. It is brown.

The halide ions are in solutions of potassium:
KCl is colourless.

KCl + Cl-: there is no reaction. The aqueous layer and organic layer remain colourless.

KCl + Br-: there is no reaction. The aqueous layer is yellow and the organic layer is orange.

KCl + Br-: there is no reaction. The aqueous layer is brown and the organic layer is purple.

25
Q

Displacement reaction of KBr?

A

KBr + Cl-: the aqueous layer turns yellow and the organic layer turns orange. This is because the halogen is more reactive than the halide and so a displacement occurs.
Equation: Cl2 + 2Br- —> 2Cl- + Br2

KBr + Br-: no reaction. The aqueous layer is yellow and the organic layer is orange.

KBr + I–: no reaction. The aqueous layer is brown and the organic layer is purple.

26
Q

Displacement reaction of KI?

A

KI + Cl-: Cl2 + 2I- —> 2Cl- + I2
The aqueous layer (bottom) is brown and the organic layer is purple.
- Chlorine is higher than iodide in the periodic table, and so iodide is being displaced from its potassium iodide solution by the halogen (Cl2).

KI + Br-: Br2 + 2I- —> 2Br + I2
The aqueous layer is brown and the organic layer is purple.

KI + I-: no reaction. The aqueous layer is brown and the organic layer is purple.

This shows how iodine is the least oxidising, because it is displaced by more reactive halide ions.

27
Q

Hydrogen halides?

A

A molecule that has a hydrogen and a halogen (HX).
X - the halogen.

Formed by adding a concentrated acid to an ionic halide compound (like a salt).

We make HCl by adding concentrated phosphoric acid.

E.g. NaCl + H3PO4 —> HCl + NaH2PO4 (HX)

28
Q

Halide ions as reducing agents?

A

Halide ions are reducing agents because they lose an electron in a reaction.

The ionic radius increases as you go down the group (distance between the nucleus and the outer electrons becomes larger). This means there is also more shielding.

This means the attractive forces are weaker as you go down the group and so the outer electrons are more readily lost.

I- is the most powerful reducing agent.

There are two tests to prove this trend:

  1. Reaction with sulfuric acid.
  2. Reaction with silver nitrate.
29
Q

Testing hydrogen halides as reducing agents?

A

In this test, we are examining the ability of the hydrogen halides to reduce sulfur in sulfuric acid (the powerfullness of the hydrogen halide as a reducing agent).

NaHSO4 - This is what you get when reacti a hydrogen halide with sulfuric acid (will make more sense on next card). The oxidation state of NaHSO4 is +6.

The oxidation state of the products:
SO2 = +4
S = 0
H2S = -2

We can tell when sulfuric acid is reduced because the oxidation state will decrease (e.g. go from 0 to -2) in the reaction equation.

30
Q

Reacting Cl- with sulfuric acid?

A

This produces the hydrogen halide, HCl.
Sulfuric acid - H2SO4

H2SO4 + NaCl —> NaHSO4 + HCl

This is not a redox reaction.

Sulfur has not been reduced here. The oxidation state of NaSO4 is +6.

White, misty fumes are produced.
This shows that Cl- (from NaCl) is not a reducing agent for sulfur.

31
Q

Reacting Br- with sulfuric acid?

A

Br- and I- undergo step A of the reaction with sulfuric acid (white, Misty fumes are produced).

H2SO4 + NaBr —> NaSO4 + HBr (hydrogen halide).

They further undergo step B: reducing sulfur to form sulfur dioxide (SO2 - which has an oxidation state of +4).

H2SO4 + 2HBr —> Br2 + SO2 + 2H2O

The oxidation state of sulfur has went from +6 to +4, showing how it has been reduced.

An orange vapour is produced at step B.

32
Q

Reacting I- with sulfuric acid?

A

Also undergoes step A and B.

Iodine further reduces sulfur in sulfuric acid to S - sulfur.

H2SO4 + 6HI —> 3I2 + S + 4H2O

The oxidation state of H2SO4 is +4 and the oxidation state of S is 0. This is has been reduced (step C).

A yellow solid is produced here.

Further reduction:
Then, sulfur is further reduced to hydrogen sulphide gas (H2S - oxidation state of -2).

H2SO4 + 8HI —> 4I2 + H2S + 4H2O

This is step D - a rotten egg smell of H2S is produced. This is a toxic gas.

33
Q

Stability if hydrogen halides?

A

Hydrogen halide stability decreases down group 7.

Hydrogen fluoride and hydrogen chloride are stable when heated. They do not split into H+ and F-/Cl- ions when heated.

Hydrogen bromide partially splits.

Hydrogen iodide will split more easily.

Why?
As you go down group 7, the atom gets larger. The bonding electrons are further away from the nucleus (more shielding) and this weakens the attractive forces, which weakens the bond enthalpy.

H-F = 565KjMol-1

H-Cl = 431

H-Br = 366

H-I = 299

Bond enthalpy decreases down the group, meaning that the forces are weaker, making the atoms less stable.

34
Q

Acidity of hydrogen halides?

A

Hydrogen halides are acidic.

They’re gases - dissolve in water to form acidic solutions.

They react with water to form a white, misty fume.

They dissociate (break apart) when dissolved in water.

HCl + H2O —> H3O+ + Cl-

HCl —(in water)—> hydrochloride acid
HBr —> hydrobromic acid
HI —> hydroiodoic acid

HF doesn’t dissociate enough to form a strong acid in water.

Hydrogen halides react with ammonia gas to make white fumes of ammonia halides.

NH3 (g) + HCl (g) —> NH4Cl (s)

35
Q

Halide ions react with silver nitrate?

A

This is the second test.

Test for halides: add dilute nitric acid (HNO3) and silver nitrate solution (AgNO3) to the halide.

The colour of precipitate helps identify the ions.

White precipitate - silver chloride
Ag+ (aq) + Cl- (aq) —> AgCl (s)

Cream precipitate- silver bromide
(Same equation)

Yellow precipitate - silver iodide
(Same equation)

We add nitric acid to react with any anions (other than halides) which also form a precipitate with silver nitrate solution. This would give a false positive.

The precipitates are difficult to distinguish between, so we had ammonia solution to further distinguish between the ions.
NH3 - ammonia solution.

Cl-: the precipitate will dissolve in dilute ammonia solution. It will be clear when dissolved.

Br-: the precipitate dissolves in concentrated ammonia solution.

I-: the precipitate is insoluble in concentrated ammonia solution. It will not dissolve.

F- does not form a precipitate because it is insoluble in AgNO3 in the first place.

36
Q

Uses of chlorine?

A
  • Sterilises drinking water (rids of bacteria and cholera).

- Used to make bleach, for sanitising and making paper.

37
Q

Risks of chlorine?

A
  • Toxic and erosive (keep away from skin and eyes).
  • Oxidising agent (keep away from flammable material).

It is transported under pressure, as a liquid in small containers. This means more can be transported at once. It is usually a gas at room temp.

38
Q

Atom economy?

A

% atom economy =

molecular mass of desired product / sum of molecular mass of all products

x100

Make sure equation is balanced first.

E.g. “Fe2O3 + 1/2C —> 2Fe + 1/2CO2 Work out the atom economy of iron”.

Fe = 55.8 (x2 because there is 2 moles in equation) = 111.6

CO2 = 44. 44 x 1.5 (because 1/2 from equation) = 66.
111.6 + 66 = 177.6

111.6 / 177.6 x100 = 62.8%

39
Q

Benefits of atom economy?

A

Atom economy tells us how efficient a reaction is.

Company’s will try to use reactions that are as close to 100% atom economy.

  • High atom economy’s mean that raw materials are used more efficiently. This is more sustainable.
  • High atom economy’s mean that we have less waste and this benefits the environment.
  • High atom economy’s means less by-products so less time and money is used separating the by-products from the desirable products.
40
Q

How to make hydrogen iodide and hydrogen boride?

A

Use phosphoric acid,

We use sulphuric acid or phosphoric acid for hydrogen bromide.

Hydrogen chloride forms in water, but can be made using sulfuric acid too.

41
Q

What do you require to find out if a reaction is endothermic or exothermic?

A

Enthalpy change of formation.

42
Q

What is an oxidising agent compared to a reducing agent?

A

Oxidising agent = species that removes electrons from another.

Reducing agent = species that gives electrons to another.

43
Q

Reversible reactions?

A

Reactions that go forward and backward. They’re shown with the double arrows.

E.g. A + B <> C + D

The forward reaction is the formation of the products. The forward reaction happens quickly (rate is high) initially, but then slows (rate is slow) as the concentration falls. This is because a higher concentration means more collisions, but as concentration lowers, collisions reduce.

The backward reaction is the reformation of the reactants. The reactants are initially formed slowly (rate is slow) but as concentration increases, form quicker (rate is fast).

Eventually, when the rate is the same for both forward and backward reaction, a dynamic equilibrium is reached. Here, the concentration of all substances is constant.

Dynamic equilibrium can only be reached in a closed system.

44
Q

How to write kc?

A

Kc is equilibrium constant.

If 2A + B <> 2C + D, then:
Kc = (C)2 (D) / (A)2 (B)

Products are always over the reactants.

Kc shows molar concentration in a reaction.
The brackets mean concentration in mol dm3

You put this into your calculator to work out kc.

Then work out the units: 
moldm-3 moldm-3 : moldm-3 moldm-3 moldm-3 
Cancel them out 
You get 1/moldm-3 
Invert moldm-3 to make mol-1dm3
45
Q

How does temperature effect the kc value?

A

Kc is only valid for one temperature.

Changing the temperature will change the equilibrium concentrations and so kc will change too.

You can tell if a reaction is exothermic or endothermic by the 🔺H. If it’s negative, it’s exothermic. If it’s positive, it’s endothermic. The 🔺H always tells you the enthalpy change of the forward reaction, so the backward reaction is always the opposite.

If temperature increases, the equilibrium will shift to the endothermic side. Kc decreases.

If temperature decreases, the equilibrium will shift to the exothermic side. Kc increases.

46
Q

How does pressure and temperature effect equilibrium position?

A

Pressure:
- If we increase the pressure then the equilibrium will shift to the side with the fewest number of gas particles (look at the big number in equation). This will reduce the pressure.

  • If we decrease the pressure then the equilibrium will shift to the side with the most gas particles. This increases the pressure.

Temperature:
- If we increase the temperature then the equilibrium will shift to reduce the temperature. So the equilibrium moves to the endothermic side (can see if this is right or left by looking at 🔺H).

  • If we decrease the temperature then the equilibrium will shift to increase the temperature. So the equilibrium moves to the exothermic side (can see if this is right or left by looking at 🔺H).
47
Q

How does concentration affect kc value?

A

Concentration doesn’t affect the kc value.

However, the formula kc = products /reactants can alter the kc.

  • If product concentration increases, kc rises.
  • If product concentration decreases, kc drops.
49
Q

What does the value of kc tell you about the position of equilibrium?

A

If kc is much greater than 1, equilibrium will lie very to the right. There will be many, many more products than reactants.

If kc is just greater than 1, equilibrium will lie to the right. There will me more products than reactants.

If kc is equal to 1, equilibrium will be in the middle. The same number of products and reactants.

If kc is lower than 1, equilibrium will lie to the left. There will be more reactants than products.

If kc is much lower than 1, equilibrium will lie very to the left. There will be many, many more reactants than products.

50
Q

Electrolysis of brine?

A

Brine is highly concentrated solution with salts such as chlorides, bromides and iodides salts.

It’s found in sea water from minerals in rocks dissolving into the water.

Cathode: H+ ions from the water make up hydrogen gas. 2H+ + 2e- —> H2

Anode: Cl- ions from salt make up Cl2 gas.
2Cl- —> Cl2 + 2e-

In the solution:
Sodium ions are more reactive than hydrogen and so remain in the solution. They react with hydroxide ions to make sodium hydroxide (NaOH).

For this to work:
The solution must be concentrated to produce electrolysis.
If the solution is dilute, the chloride ions don’t release any electrons. This means water and oxygen is produced instead.
4OH- —> 2H2O + O2 + 4e-

51
Q

Extraction of bromine?

A

Bromine is extracted using a displacement reaction.

If we use a more reactive halogen, we extract bromine from the brine.

Chlorine is more reactive and bromine is produced which is condensed and purified into a liquid.

2Br- + Cl2 —> Br2 + 2Cl-

52
Q

Extraction of iodine?

A

If we use a more reactive halogen, we extract bromine from the brine.

Chlorine is more reactive and iodine is produced which is condensed and purified into a grey solid.

2I- + Cl2 —> I2 + 2Cl-

53
Q

Reducing agent and oxidising agent?

A

Reducing agents - lose electrons. They are oxidised themselves.

Oxidising agents - gain electrons. They are reduce themselves.

54
Q

Oxidation and reduction?

A

Oxidation - loss of electrons. (E is on the right side of the arrow).

Reduction - gain of electrons. (E is on the left side of the arrow).