Chemistry: Elements from the Sea (ES)

a Atom economy:

Test yourself
• Atom economy is a measure of the efficiency of a reaction
• In a balanced symbol equation it is calculated with:
  percentage atom economy = relative mass of desired product(s) relative mass of all reactants × 100
• Less waste is produced with higher atom economy
• If there are multiple products which are useful, they will all be the desired product. So many reactions with 2 or more products still have 100% atom economy

b Extraction of halogens from a solution:

Test yourself
• When a concentrated aqueous solution containing halide ions is electrolysed,
  - the halogen element is released at the anode
  - hydrogen gas is formed at the cathode
    2Br^-(aq) → Br₂(aq) + 2e^- (anode)
    2H⁺(aq) + 2e^- → H₂(g) (cathode)

Extraction of chlorine from brine:

• Brine (e.g. from seawater) has a high content of sodium chloride salt (as well as some bromine and iodine salts), so chlorine can be extracted from brine through electrolysis
• During the electrolysis, the cell needs a constant supply of brine
• At the cathode, two hydrogen ions gain two electrons to become one hydrogen molecule
  2H⁺(aq) + 2e^- → H₂(g)
• At the anode, two chloride ions lose electrons and become one chlorine molecule
  2Cl^-(aq) → Cl₂(g) + 2e^-
• Sodium ions stay in the solution because they're more reactive than hydrogen
• The sodium chloride (NaCl) solution needs to be concentrated, if it were dilute, the chloride ions wouldn't be discharged (they won't lose electrons) so instead, the OH^- will lose electrons to give oxygen and water instead of chlorine (which is the desired product)
  4OH^-(aq) → 4e^- + 2H₂O(l) + O₂(g)

Extraction of bromine/iodine from brine:

• Chlorine is more reactive than bromine and iodine
• Therefore, they are displaced by chlorine
• When chlorine gas is bubbled through brine, the chlorine displaces the bromine as it is more reactive
  2Br^-(aq) + Cl₂(g) → Br₂(g) + 2Cl^-(aq)

c Electrolysis:

Test yourself
• Electrolysis is the passing of an electrical current through an aqueous or ionic substance (molten or in solution), breaking it down into the elements it's made up of
• The electrolyte must be an electrically conductive solution, meaning that it must contain free ions
• There are two electrodes. They are usually made from inert substances such as carbon or platinum so they don't react with anything
• The anode is the positive electrode
• The cathode is the negative electrode

The procedure for the electrolysis of an aqueous solution:

• Connect wires from the power supply; positive to the anode and negative to the cathode
• Place the electrodes into a beaker containing the electrolyte. Make sure the electrodes are not touching each other
• Turn on the power supply

Half-equations:

• The movement of electrons during electrolysis can be shown using half-equations
• In electrolysis, the negative ions (anions) move to the anode and lose electrons. The positive ions (cations) move to the cathode and gain electrons
• Therefore, the half-equation at the cathode shows positive ions gaining electrons to form atoms and the half-equation at the anode shows negative ions losing electrons to form atoms
• The half-equations for molten zinc chloride, ZnCl₂ are:
  At the anode: 2Cl^-(l) → Cl₂(g) + 2e^-      the chloride ion is being oxidised to a chlorine molecule
  At the cathode: Zn²⁺(l) + 2e^- → Zn(s)   the zinc ion is gaining two electrons (therefore being reduced)
• The half-equations can be easily predicted for any salt like ZnCl₂, since the first element is always positive in an ionic molecule

Aqueous electrolysis:

• Aqueous electrolysis is where the electrolyte is an aqueous solution of an ionic compound (e.g. CuSO₄(aq) can be used to purify copper)
• Unlike for molten compounds shown above, there will be OH^- and H⁺ ions present in aqueous electrolysis, as well as the ions from the ionic compound
• At the cathode, the metal will be formed if the metal is less reactive than hydrogen (e.g. copper, silver). Otherwise, hydrogen gas will be formed (from the H⁺ ions)
• Salts which are more reactive than hydrogen include (you need to memorise these three only for the exam):
  - Group 1 metal salts
  - Group 2 metal salts
  - Aluminium salts
• At the anode, oxygen will be formed (from OH^- ions) if the solution doesn't contain a halide; 4OH^-(aq) → O₂(g) + 2H₂O(l) + 4e^-, or if it contains a halide but is dilute. If it contains a halide and the solution is concentrated, the halogen gas will be formed

d Oxidation and reduction:

Test yourself
• Losing electrons is oxidation
• Gaining electrons is reduction
• The above can be remembered with the mnemonic OILRIG (Oxidation Is Loss, Reduction Is Gain)
• If a reaction contains oxygen, the oxidised compound will gain oxygen and the reduced one will lose oxygen (this is not required knowledge, but it's a useful shortcut)

Oxidising and reducing agents:

• An oxidising agent gains electrons from another substance and oxidises it
• A reducing agent transfers electrons to another substance, reducing it
• In other words, the reducing agent is being oxidised, donating electrons.
  The oxidising agent is being reduced, accepting electrons

Redox reactions:

• A reaction is a redox reaction if oxidation and reduction occur simultaneously
• A redox reaction can be written as two half-equations; one for the oxidation and one for the reduction

e Oxidation states:

Test yourself
• Oxidation state shows how many electrons an atom has donated/accepted to form an ion/bond
• For every electron lost, the oxidation state of the atom will increase by 1. For every electron gained, it will decrease by 1
• Usually, metals have positive oxidation states because they donate electrons to form positive ions

Basic oxidation state rules:

• Oxygen always has oxidation state -2, unless it's in a peroxide (containing an O-O single bond) where it is -1, or in fluorides (O in OF₂ is +2 and O in O₂F₂ is +1), or in O₂ where it's 0
• Hydrogen is always +1, except in metal hydrides (MH where M is a metal and ) where it is -1, and in H₂ where it's 0
• All diatomic atoms are 0, eg: O₂, H₂, N₂
• An ion with just one one atom has the same oxidation state as its charge, eg: Na⁺ has an oxidation state of +1
• In ions with more than one atom, the overall oxidation state is the same as the overall charge. Each atom will have its own oxidation state but the sum will be equal to the overall charge on the ion
• The oxidation states for halides and relevant metals are shown below. Obvious exceptions (e.g. F₂, Cl₂) are not shown:

  Element	Oxidation state
  Fluorine	-1
  Chlorine	-1 except when with O or F
  Bromine	-1 except when with O, F or Cl
  Iodine	-1 except when with O, F, Cl or Br
  Group 1	+1
  Group 2	+2
  Aluminium	+3

Determining what has been reduced and oxidised in a redox reaction:

• Calculate the oxidation state for each reactant and product, e.g:
       Cl₂ + 2I^- → 2Cl^- + I₂
  Cl  0                    -1
  I             -1                    0
  
  The Cl is reduced because its oxidation state goes from 0 to -1
  The I is oxidised because its oxidation state goes from -1 to 0

f Iodine-thiosulfate titrations:

Test yourself
• These are used to find the concentration of an oxidising agent, such as iodate(V) (IO₃^-) in potassium iodate(V) (KIO₃)

Stage 1, oxidising iodine:

• Measure 25 cm3 of oxidising agent using a volumetric pipette (rinse with water and potassium iodate(V) first)
• Add this to an excess of acidic potassium iodide solution. The iodide ions in this solution will be oxidised by the oxidising agent
  IO₃^-(aq) + 5I^-(aq) + 6H⁺(aq) → 3I₂(aq) + 3H₂O(l)
• Instead of an acidic iodide solution, an iodide solution and sulfuric acid could also be used

Stage 2, finding the number of moles of iodine produced:

• Titrate the solution from stage 1 with sodium thiosulfate (Na₂S₂O₃):
  I₂(aq) + 2S₂O₃^2-(aq) → 2I^-(aq) + S₄O₆^2-(aq)
• Once the solution reaches a pale straw colour, add a few drops of starch solution, an indicator, which will make the solution blue/black
• Starch indicates iodine. Once the solution contains no more iodine, it will therefore be colourless. This is the end point
• Only add starch once the solution is pale straw yellow. If it is added earlier, the iodine will stick to the starch, not reacting with the thiosulfate

Calculations:

• Because number of moles = concentration × volume, multiply the volume of sodium thiosulfate used by its concentration to get the number of moles of thiosulfate. Now divide this by 2 to get the number of moles of iodine produced in stage one (because 1 iodine molecule reacts with 2 thiosulfate ions in the titration)
• Divide the number of moles of iodine by 3 to get the number of moles of potassium iodate(V) (look at the equation in stage 1)
• Finally, use n = c × V again to calculate the concentration of potassium iodate(V) solution - volume is 0.025 dm3 and number of moles was calculated above

Redox equations:

• The equation for a redox reaction can be made by combining the two half-equations
• To balance a redox reaction equation, ensure that the oxidation states are also balanced. Do this by finding the change in oxidation state for each element in the reaction. Now balance the symbol equation as usual and multiply the coefficients of each element by a number to ensure that the oxidation state changes cancel.
  Now, there should be the same number of each element on both sides, and charges should be balanced (summing to 0 on both sides)

Example: balancing Ca + Al³⁺ → Ca²⁺ + Al:

• First, calculate the oxidation states:
  - Ca has oxidation state 0 because it's on its own and has no charge
  - Al³⁺ has oxidation state +3 because of its charge
  - Ca²⁺ has oxidation state +2 because of its charge
  - Al has oxidation state 0 because it's on its own and has no charge
• Now sum them on each side:
  - On the left, the total oxidation state is +3
  - On the right, the total oxidation state is +2
• These are not equal, so the equation is not balanced
• To balance the oxidation states, we need to multiply the Al³⁺ by 2 and the Ca²⁺ by 3. This gives us Ca + 2Al³⁺ → 3Ca²⁺ + Al
• Now the oxidation states are balanced, but the elements are not. So now balance the elements without touching the oxidation states: 3Ca + 2Al³⁺ → 3Ca²⁺ + 2Al

g Naming ions:

Test yourself
• In an ionic compound, the name of the positive ion is in front of the name of the negative ion (e.g. NaCl is sodium chloride because the Na is positive and Cl is negative)
• If a molecule can have multiple oxidation states, it is shown next to the name in Roman numerals (e.g. iron in iron(II) sulfate has oxidation state +2 and formula FeSO₄)
• If a name ends in -ate, it must contain oxygen (e.g. sulfate is sulfur and oxygen)
  - Roman numerals are sometimes needed here (e.g. SO₃^2- is sulfate(IV) because sulfur has oxidation state +4)
• More than one Roman numeral may be needed - e.g. in iron(III) sulfate(VI), where Fe has oxidation state +3 and S is +4, so it must be SO₄^2-. Therefore, this compound is Fe₂(SO₄)₃
• A Roman numeral indicates the oxidation state of a non-oxygen atom before it

h Properties of halogens:

Test yourself
• Some of the properties of the halogens are shown below:
  
• As you go down the group, the halogens:
  - Have higher melting and boiling points
  - Change from gases to liquids to solids at room temperature
  - Become less volatile (harder to vaporise from a liquid to a gas)
     - This is due to stronger instantaneous dipole-induced dipole bonds
  - Become less reactive (all halides will react with water to some extent)
     - F₂ will react with an irreversible reaction
     - Cl₂, Br₂ and I₂ will react with a reversible ⇌ reaction
• There is no trend for solubility in water down the group, but are generally not very soluble because they're covalent and non-polar
• Halogens are more soluble in organic solvents (e.g. hexane) than in water
• Note: a halogen is a molecule made from two group 7 atoms. A halide is an ion of a single group 7 atom

i Reactivity of halogens:

Test yourself
• The halogens are in group 7, so they tend to remove electrons from other elements to complete their outer shell (so they're oxidising agents)
• Fluorine atoms are very small so the attraction between the core and the electrons in the outer shell is very strong
• Chlorine atoms are larger and the outer shell is further from the core so the attraction is weaker. Additionally, there is more electron shielding, where outer electrons are shielded from the charge of the nucleus by inner electrons
• This means that fluorine gains the extra electron more easily than chlorine does
• Therefore fluorine is the most reactive and the strongest oxidising agent

j Reactions of the halogens with halide ions:

Test yourself
• When chlorine solution is added to potassium iodide solution, iodine is produced with a redox reaction:
  Cl₂(aq) + 2KI(aq) → 2KCl(aq) + I₂(aq)
• The K⁺ ions are unchanged so they are not included in the overall ionic equation, which is Cl₂(aq) + 2I^-(aq) → 2Cl^-(aq) + I₂(aq)
• In this reaction, the iodide loses an electron and becomes oxidised, and the chloride ions in the chlorine gain electrons - which makes chlorine the oxidising agent
• This is easier to see from the half-equations:
  - Cl₂(aq) + 2e^- → 2Cl^-(aq)
  - 2I^-(aq) → I₂(aq) + 2e^-
• This reaction also happens with chlorine displacing bromide ions, and bromine displacing iodide ions
• The solution formed will have the colour of the formed halogen in water. For example, after the reaction between chlorine and potassium iodide above, the solution formed will be brown

Aqueous halogen colours:

• Chlorine solution is colourless
• Bromine solution is orange
• Iodine solution is brown

k Reactions of halide ions with silver ions:

Test yourself
• Silver halide precipitates form when a solution of silver ions is added to a solution containing chloride, bromide or iodide ions. Therefore, silver nitrate solution (AgNO₃) is a good test for halides:
  - First add dilute nitric acid to remove ions that may interfere
  TO DO: more detail
  
  - Next, add silver nitrate solution to form the precipitate
• The general equation for this is:
  Ag⁺(aq) + X^-(aq) → AgX(s), where the X represents the halide ion
• The colour of these precipitates is:
  - Silver fluoride (AgF) - soluble
  - Silver chloride (AgCl) - white
  - Silver bromide (AgBr) - cream
  - Silver iodide (AgI) - yellow

Silver halides and ammonia:

• Sometimes it is difficult to identify between the colours of the precipitates so ammonia solution is added. The solubilities are:
  - silver chloride (most soluble)
  - silver bromide
  - silver iodide (least soluble)
• The above facts can be used with ammonia concentrations:
  - silver chloride will dissolve in dilute ammonia solution
  - silver bromide will dissolve in concentrated ammonia solution
  - silver iodide will not dissolve at all

l, m Preparation of hydrogen halides with phosphoric acid:

Test yourself
• Hydrogen halides can be made by adding concentrated phosphoric acid to a solid ionic halide

Preparation of hydrogen halides with sulfuric acid:

• Hydrogen chloride can be made by adding concentrated sulfuric acid (H₂SO₄), an oxidising agent, to the solid ionic halide
• Hydrogen bromide and iodide cannot be made this way with sulfuric acid
• Instead, the sulfuric acid would react with the bromide/iodide to make the halide gas
• This is because bromine and iodine are strong enough reducing agents to reduce the sulfuric acid
• With bromide, Br₂, a sulfate salt, H₂O and SO₂ are formed in this reaction
• With iodide, I₂, H₂S and H₂O are formed. This is different than for bromide ions because iodide ions are stronger reducing agents than bromide ions

m Properties of hydrogen halides:

Test yourself
• A hydrogen halide is one hydrogen covalently bonded to a halide ion - e.g. hydrogen chloride, HCl

Hydrogen halides when heated:

• When heated, hydrogen fluoride and hydrogen chloride are stable and won't break into H⁺ and F^-/Cl^-, whereas hydrogen bromide will slightly dissociate and hydrogen iodide will fully dissociate
• This is due to the strength of the hydrogen-halide bonds; going down group 7, the the bond gets weaker (bond enthalpy decreases down the group)
• As the halogen atom is bigger further down the group, the bonding electrons are farther away from the nucleus, and are more shielded by the inner electrons

Hydrogen halides with water:

• Hydrogen chloride, bromide and iodide all dissolve in water to create a strong acid
• When the hydrogen halide dissolves in water, it dissociates to give a halide ion and a hydrogen ion:
  HCl(g) → H⁺(aq) + Cl^-(aq)
• This applies to HCl, HBr and HI but not to HF, because it doesn't fully dissociate in water (only some molecules dissociate) making it a weak acid

Hydrogen halides and ammonia:

• Ammonia, NH₃, is a base. This means that it can accept a H⁺ from the hydrogen halide to form NH₄⁺
• This ammonium ion can bond with a negatively charged halide ion to form an ammonium halide:
  X^-(aq) + NH₃⁺(aq) → NH₄X(aq) where X is a halide

Hydrogen halides and sulfuric acid:

• Hydrogen fluoride and chloride do not react with sulfuric acid, because they are not strong enough reducing agents to reduce the sulfur
• 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O  HBr reduces H₂SO₄ to SO₂
  8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O     HI reduces H₂SO₄ to H₂S

n Chlorine:

Test yourself
• Chlorine is a toxic gas and is also corrosive. It also irritates the respiratory system
• It is stored in cylinders. Transport is designed so that they are not damaged
• Chlorine can be transported by road or rail, in pressurised tank containers
• It is transported as a liquid, because more liquid can be stored in a fixed volume under pressure than gas
• There is a possibility of the pressure getting too high. This is why a venting mechanism is usually built in
• It is an oxidising agent, and therefore can increase the risk of fires

Uses of chlorine:

• Chlorine is used for sterilising water; when chlorine is added to the water it kills bacteria and other harmful pathogens
• It is also used in household bleach products, to kill bacteria on surfaces or to remove stains from clothes
• The bleach is an oxidising agent, and removes the stains by breaking bonds in coloured chemicals to form colourless products

o Dynamic equilibrium:

Test yourself
• Dynamic equilibrium is when:
  - the concentrations of the reactants and products stay constant
  - the forward and reverse reactions are both occurring
  - the rate of the forward reaction equals the rate of the reverse reaction

p Equilibrium:

Test yourself
• If the equilibrium lies to the left, there'll be more reactants than products
• If the equilibrium lies to the right, there'll be more products than reactants

The equilibrium constant:

• Square brackets around a molecule represents its concentration
• The equilibrium constant ( for a concentration equilibrium) is calculated by dividing the concentrations of each product to the power of its quantity by the concentrations of each reactant to the power of its quantity (use mol dm-3 for the concentrations)
• So for the reaction aA + bB ⇋ cC + dD:
  
• doesn't always have units (they can cancel due to the fraction used to calculate it). Units are not required in the AS exams
• A value is only valid for a particular temperature

Expressing the position of equilibrium:

• means that the equilibrium is very far right - much more products than reactants present
• means that the equilibrium is slightly right - there are slightly more products than reactants
• means that the equilibrium is in the middle - the same concentrations of products and reactants
• means that the equilibrium is slightly left - there are slightly less products than reactants
• means that the equilibrium is very far left - much less products than reactants present

q Effects of changing pressure:

Test yourself
• This is only for equilibria involving gases
• Increasing pressure shifts the equilibrium to the side with the fewest moles of gas
• Decreasing pressure shifts the equilibrium to the opposite side - the one with the most moles of gas

Effects of changing temperature:

• One direction will always be endothermic (positive Δ H), and the opposite direction will be exothermic (negative Δ H)
• Increasing the temperature shifts the equilibrium in the endothermic direction
• Decreasing the temperature shifts the equilibrium in the exothermic direction

Explaining equilibria movements with :

• I'll use the equilibria (from p) as an example
• If the concentration of C is increased, the top of the fraction will have a larger value
• Because the equilibrium constant will stay the same, the bottom of the fraction must increase to compensate
• Therefore, the concentrations of A and B will increase
• This explains what we learnt at GCSE - adding more of the substances on the right-hand side of the ⇌ will increase the concentrations of the left-hand side