Understanding Points
Reactivity 3.2.1—Oxidation and reduction can be described in terms of electron transfer, change in oxidation state, oxygen gain/loss or hydrogen loss/gain.
Reactivity 3.2.2—Half-equations separate the processes of oxidation and reduction, showing the
loss or gain of electrons.
Reactivity 3.2.3—The relative ease of oxidation and reduction of an element in a group can be
predicted from its position in the periodic table. The reactions between metals and aqueous metal ions demonstrate the relative ease of oxidation of different metals.
Reactivity 3.2.4—Acids react with reactive metals to release hydrogen.
Reactivity 3.2.5—Oxidation occurs at the anode and reduction occurs at the cathode in
electrochemical cells.
Reactivity 3.2.6—A primary (voltaic) cell is an electrochemical cell that converts energy from
spontaneous redox reactions to electrical energy.
Reactivity 3.2.7—Secondary (rechargeable) cells involve redox reactions that can be reversed using electrical energy.
Reactivity 3.2.8—An electrolytic cell is an electrochemical cell that converts electrical energy to
chemical energy by bringing about non-spontaneous reactions.
Reactivity 3.2.9—Functional groups in organic compounds may undergo oxidation.
Reactivity 3.2.10—Functional groups in organic compounds may undergo reduction.
Reactivity 3.2.11—Reduction of unsaturated compounds by the addition of hydrogen lowers the
degree of unsaturation.
Redox
•
Oxidation: loss of e-, loss of H, gain of O, increase in oxidation state
•
Reduction: gain of e-, gain of H, loss of O, decrease in oxidation state
•
“OILRIG”= Oxidation is Loss of e-, Reduction is Gain of e-
•
Oxidation and reduction always occur together and each take up one half equation
•
Species being oxidized vs reduced can be identified by comparing changes in oxidation states in reactant and product sides
Oxidizing/reducing agent
•
Reducing Agent: Species being oxidized; reduces the other species
•
Oxidising Agent: Species being reduced; oxidizes the other species
Example | Reducing agent | Oxidising agent |
Oxidised by releasing e-s/Hs and reducing the other species | Reduced with e-s and H from oxidised species | |
2Mg(s) + O2(g) → 2MgO(s) | Mg | O2 |
3Mg(s) + N2(g) → Mg3N2(s) | Mg | N2 |
(Iron ore extraction)
Fe2O3(s) + 3C(s) → 2Fe(s) + 3CO(g) | C | Fe |
(acid + metal)
M + 2HA → MA2(aq) + H2(g) | M | HA |
CH3CH2OH → CH3CHO | N/A | K2Cr2O7 → Cr2O3/ KMnO4 → MnO2 |
CH3CHO → CH3CH2OH | LiAlH4 → Li+ + AlH3/ NaBH4 → Na+ + BH3 | N/A |
1.
acid rxn with base and carbonate ≠ redox ˙.˙ proton transfer
2.
Reactant can be both oxidising and reducing agent (similar to amphiprotic)
a.
Depending on ‘redox potential’ of reacting species
b.
E.g. water
3.
2H2O(l) + 2Na(s) → 2NaOH(aq) + H2(g)
4.
2H2O(l) + 2F2(g) → 4HF(aq) + O2(g)
Half-equations
Oxidation | Reduction |
e- and charged form on product side | e- and charged form on reactant side |
Mg(s) → Mg2+(s) + 2e– | Ag+ + e- → Ag(s) |
1.
Balance no. of e-s in the two half-equations
2.
Combine two half equations to cancel out e-s
3.
For 3 species reaction,
a.
Identify non-oxidised/reduced species as ‘spectator’ ion
b.
Compounds are either oxidising or reducing agent depending on reduced or oxidised component element
c.
Write out oxidation and reduction half equations
d.
Combine 2 half-equations and add ‘spectator’ ion
e.
E.g. Cl2(aq) + 2KI(aq) → 2KCl(aq) + I2(aq)
4.
For 4 species rxn with 2 products, reduced species may be distributed b/w 2 products, e.g. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
5.
6 species rxn still counts as 1 redox
Balancing redox equations - write using half-equations
•
Completion of redox equation that is not balanced b/w reactants and products in number of H, O atoms and charge (opposite of previously given complete redox equation with need to identify oxidised and reduced)
e.g. NO3– + Cu → NO + Cu2+
1. Assign oxidation states to determine which atoms are being oxidized and which are being reduced: | NO3– + Cu → NO + Cu2+
+5 -2 0 +2 -2 +2 |
2.1 Balance the atoms other than H and O | Cu(s) → Cu2+(aq)
NO3–(aq) → NO(g) |
2.2 Balance each half-equation for O by adding H2O as needed. | NO3–(aq) → NO(g) + 2H2O(l) |
2.3 Balance each half-equation for H by adding H+ as needed. | NO3–(aq) + 4H+(aq) → NO(g) + 2H2O(l) |
2.4 Balance each half-equation for charge by adding electrons to the side with the more positive charge. | Cu(s) → Cu2+(aq) + 2e–
NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O(l) |
3. Equalize the number of electrons | 3Cu(s) → 3Cu2+(aq) + 6e–
2NO3–(aq) + 8H+(aq) + 6e– → 2NO(g) + 4H2O(l) |
4. Add 2 half-equations together | 3Cu(s) + 2NO3–(aq) + 8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l) |
5. Confirm charge on both sides add up to equal value | Both left and right = 6+ |
1.
Product stoichiometry may need to be multiplied (Step 2.1)
a.
e.g. P4 → H3PO4 to P4 → 4H3PO4 + 20e-
2.
H2O may need to be added to both half-equations (Step 2.2)
Activity Series/reactivity series:
•
Orders elements in order of reactivity/reducing or oxidizing agent ability
•
A more reactive element will preferentially be oxidized and displace the other metal in a displacement rxn
Activity | Reducing agent (Reactivity) | Oxidising agent (Reactivity) | ||
Increasing
↑
Decreasing | K
Na
Ca
Mg
Al
Zn
Fe
Pb
Cu
Ag
Au
Pt | More reactive,
Stronger reducing agent
↑
Less Reactive,
Weaker reducing agent | F
Cl
Br
I | More reactive,
Stronger oxidizing agent
↑
Less reactive
Weaker oxidizing agent |
Displacement reaction
Verify strength of reducing/oxidising agents by
1.
reacting ‘test’ element in atomic state with ‘comparing’ element in compound state
2.
check whether ‘test’ element reduces/oxidises ‘comparing’ element
a.
Successful displacement - ‘test’ element is more reactive than ‘comparing element’
b.
Unsuccessful displacement - ‘test’ element is more reactive than ‘comparing element’
3.
E.g. Zn(s) + Cu2+(SO42-)(aq) → Zn2+(SO42-)(aq) + Cu(s)
a.
Visibly: Blue colour fades
b.
Brown solid copper precipitates and deposits
Redox reactions of acids and metals
•
Acid is the oxidizing agent, metal is the reducing agent
•
Produces hydrogen gas
•
e.g. Zn(S) + 2HCl(aq) → ZnCl2(aq) + H2(g)
◦
Zn(s) → Zn2+(aq) + 2e-
◦
2H+(aq) + 2e- → H2(g)
•
Redox titration
BOD: the amount of oxygen (mg) consumed per dm3 of water over a five day period by microbes
•
The higher the demand the less oxygen that will be present in water
•
The less oxygen present in the water, the more anaerobic bacteria will flourish
•
Less oxygen means less oxygen for aquatic plants ; they will die
•
More anaerobic bacteria means a lower pH in the water
Winkler Method
•
Used to measure dissolved oxygen in a solution
•
By titrating with a standard solution of sodium thiosulfate we can measure the amount of iodine formed, which is in turn directly proportional to the amount of dissolved oxygen
•
Step 1: 2Mn2+ (aq) + 4OH- (aq) + O2 (aq) → 2MnO2 (s) + 2H2O (l)
•
Step 2: MnO2 (s) + 2I- (aq) + 4H+ (aq) → Mn2+ (aq) + I2 (aq) + 2H2O (l)
•
Step 3: 2S2O32- (aq) + I2 (aq) → S4O62- (aq) + 2I- (aq)
•
Ratio of O2 to S2O32- is 1:4 → n(O2) = moles of S2 O3-4
•
Primary voltaic cell
1.
Generate electricity from spontaneous redox reactions
2.
Redox reaction is exothermic (stronger bonds are formed)
3.
Harness the heat released to generate electricity
Primary voltaic cells consist of two half cells connected together to allow electrons transferred during a redox reaction to produce energy in the form of electricity.
•
Half cells are metals immersed in aqueous solution of its own ions e.g. Zn in ZnSO4
•
Half cells are connected by a wire to transfer electrons and a salt bridge to transfer ions
•
The metal higher in the reactivity series is oxidised and the metal lower in the reactivity series is reduced
•
Oxidation occurs at the Anode (negative electrode)
•
Reduction occurs at the Cathode (positive electrode)
•
Notation:
Secondary voltaic cell
•
The redox reaction involved may be reversed using electricity → rechargeable
•
e.g. lithium-ion, lead-acid, nickel-cadmium
Electrolytic cell
•
Converts electrical energy to chemical energy using non-spontaneous redox reactions
•
A battery is used to provide electrical energy for non-spontaneous reactions to occur
•
Electricity is passed through an electrolyte - substance that does not conduct electricity when solid, but does when molten/dissolved in water
•
Oxidation occurs at the Anode (positive electrode)
•
Reduction occurs at the Cathode (negative electrode)
•
mnemonic: red cat and an ox (reduction at the cathode, oxidation at the anode) → for both voltaic and electrolytic cells
•
Moles of e- going through the circuit is given by: I tF (I= current, F=Faraday’s constant)
Due to presence of water, two possible redox rxn at each electrode depending on
1.
relative E⊖ values of the ions
a.
More +ve E⊖ means more likely to be reduced
b.
More -ve E⊖ means more likely to be oxidized
2.
relative concentrations of the ions in the electrolyte
a.
More available the ion (i.e. higher conc.), more preferentially discharged
3.
nature of the electrode
a.
Can be inert or be a part of the redox reaction
i.
Inert: graphite, Pt
ii.
Part of rxn: Cu in CuSO4 electrolysis; +ve electrode itself can release e- (oxidation) and form Cu2+ ions
E.g. Dilute solution of NaCl
Exceptions
•
At the anode: if we use high concentrations (saturated) of a solution then the anion present in the salt will always be oxidised
•
At the cathode: if we make the cathode out of a particular metal, then that metal will always be preferentially discharged
•
Molten solution = no water
Comparing electrochemical cells
Voltaic cell | Electrolytic cell |
Spontaneous | Non-spontaneous (requires work) |
Anode = -ve
Cathode = +ve | Anode =+ve
Cathode = -ve |
Redox of metal electrodes and soln | Redox of soln |
e- flow direction -ve → +ve | e- flow direction +ve → -ve |
e- flow from anode to cathode | |
Reduction at cathode and oxidation at anode |
Oxidation of alcohols
•
Reagent: Cr2O7 2- / H+
•
Reduction of carbonyl compounds
Carbonyl compound | Product | Reducing agent |
Carboxylic acid | Primary alcohol | LiAlH4 (Lithium Aluminium Hydride) in dry ether/heat |
Carboxylic acid | Aldehyde | NaBH4 (Sodium Borohydride) in aqueous alcohol/ heat |
Ketone | Secondary alcohol | NaBH4 (Sodium Borohydride) in aqueous alcohol/ heat |
Aldehyde | Primary alcohol | NaBH4 (Sodium Borohydride) in aqueous alcohol/ heat |
•
LiAlH4 is too strong of a reducing agent to reduce carboxylic acid to aldehyde hence NaBH4 is used
Hydrogenation
•
Compounds containing a double or triple bond can be reduced by addition of hydrogen
•
Requirements: 180℃, nickel catalyst
•
This decreases their degree of unsaturation
•
Index of hydrogen deficiency (IHD)
•
The degree of unsaturation of a compound or index of hydrogen deficiency provides a useful clue to the structure of a molecule once its molecular formula is known
•
The IHD can be used to determine the number of rings or double/triple bonds in a molecule
•
Essentially we are calculating how many molecules of H₂ needs to be added to convert the molecule to a saturated, non-cyclic molecule (i.e. remove double/triple bond and break cyclic structure)
•
IHD equation for hydrocarbons, CxHy :
(2x + 2 - y ) 2
IHD | Bonds present |
0 | Single bonds |
1 | Double bond or cyclic structure |
2 | 1 Triple bond or 2 double bond |
4 | Benzene (“3 double bonds” + cyclic structure) |
Note that:
•
Oxygen and sulfur do not affect the IHD
•
If there are elements such as nitrogen or a halogen in the compound use the unified equation:
(2C+2)+N-X-H2
C = carbon, N = nitrogen, X = halogen, H = hydrogen