.7 Oxidation, reduction, redox equations Flashcards
Identify whether each reaction is an example of reduction or oxidation:
2FeO + C 🡪 2Fe + CO2
2Mg + O2 🡪 2MgO
Na 🡪 Na+ + e-
2I- - 2e- 🡪
I2 ½ O2 + 2e- 🡪 O2-
Reduction Oxidation Oxidation Oxidation Reduction
Define oxidation and reduction in terms of electrons, what is an oxidation number? Why do we use oxidation states?
Can be expressed in two ways: Oxidation is the gain of oxygen
Reduction is the loss of oxygen
Oxidation is the loss of electrons
Reduction is the gain of electrons
OiL RiG
‘An oxidation number is the charge an atom would have if electrons in its bonds belonged completely to a more electronegative atom’.
∴ The more electronegative atom becomes negative
Tell if oxidation or reduction has taken place
Work out what has been oxidised / reduced
Construct half equations and balance redox equations
Oxidation Number Rules (Lesson + ones from Dr B) (in order of importance) including benchmarks
Lesson:The oxidation number of elements = 0 So Na = 0 and the Cl in Cl2 = 0
Oxidation numbers are written as charge;value and must be placed above the specific element (e.g +2 on top of Mg -2 on top of O in MgO)
Simple ions have the same ionic charge and oxidation number Na+, K+ both have the oxidation number +1 S2-, O2- both have the oxidation number -2
The sum of oxidation numbers in (free elements) natural molecules and compounds= 0
The sum of oxidation numbers in complex ions = the charge of the ion
Metals Group 1 = +1 Group 2 = +2 Group 3 = +3 (mostly, but Al is always +3)
Non metals
Vary but are mostly negative. To avoid ambiguity, oxidation number is often given in the name e.g. sulphur(VI)oxide for SO3
Hydrogen +1 BUT 0 in H2
-1 in metal hydrides e.g. NaH (due to hydride ion: H-)
Oxygen -2 BUT 0 in O2
-1 in peroxides e.g.H2O2
+2 with fluorine, e.g. F2O
Fluorine -1 BUT 0 in F2
In compound, most EN element always has a negative oxidation state
(Check Mr B ones for last rule abt group 16 15 etc)
Practice Questions from powerpoint
(Should’ve done anyway)
Hydrogen burning in air is a common reaction, what is actual happening?
H2(g) + ½ O2(g) → H2O(g)
Oxidation states (H) 0 → +1 Oxidation states 0 → -2
The H has lost electrons, i.e become oxidised and value has increased
O has gained electrons, i.e become reduced and value has decreased.
New Definition of REDOX (in terms of oxidation number)
If an atom is oxidized, its oxidation number increases
Oxidation = Loss of electrons (OIL)
If an atom is reduced, its oxidation number decreases
Reduction = Gain of electrons (RIG)
Calculate the changes that take place in the underlined element in the following reactions, classify as oxidation, reduction or neither (O,R or N)
- NO3- → NO (N underlined)
- HNO3 → N2O
- C2O42- → CO2 (C)
- SO32- → SO42- (S)
- H2O2 → H2O (H)
- H2O2 → O2 (O)
- NO3- → NO : R +5 to +2
- HNO3 → N2O : R +5 to +1
- C2O42- → CO2 : O +3 to +4
- SO32- → SO42- : O +4 to +6
- H2O2 → H2O : R -1 to -2
- H2O2 → O2 : O -1 to 0
Do half equation for
2Mg + O2 🡪 2MgO (overall reaction)
By definition, reduction and oxidation take place simultaneously and we can show an overall equation or two half equations for redox reactions.
1) Mg: Mg – 2e- → Mg2+ (but we don’t show – electrons so I need to draw this with e on RHS)
Mg → Mg2+ + 2e-
2) O: 1/2O2 + 2e- → O2-
3) 2Mg + O2 + 2e- → 2Mg2+ + 2O2- + 2e-
4) Cancel out electrons
5) 2Mg + O2 → 2Mg2+ + 2O2- (combined half equations – ionic equation)
What do Ionic Equations only show? How do you do them?
Only show species in a reaction that are chemically altered (have a change in oxidation state)
Sn(s) + PbCl2(aq) → SnCl2(aq) + Pb(s) (Full equation)
Identify which species in this equation are chemically altered and which are spectator ions (not chemically altered)
Write reduction and oxidation equations for these
Combine the equations
Do ionic equation for
Sn(s) + PbCl2(aq) → SnCl2(aq) + Pb(s)
Sn → Sn2+ + 2e- (Oxidation)
Pb2+ + 2e- → Pb (Reduction)
2Cl- → 2Cl- (no change - spectator ion)
Sn + Pb2+ → Sn2+ + Pb (ionic equation)
In a redox reaction, what is an agent?
What are the agents (and what types) in this reaction
Sn + Pb2+ → Sn2+ + Pb
An MI5 agent works for MI5. An agent causes something to happen.
An oxidising agent makes oxidisation happen. It itself is reduced.
Sn loses electrons. It is oxidised. Therefore, it is the reducing agent.
Pb gains electrons. It is reduced. Therefore, it is the oxidising agent.
For each of the following write two half equations and then combine them to write ionic equations.
Zn + 2AgCl 🡪 ZnCl2 + 2Ag
2K + H2 🡪 2KH
Mg + 2HCl 🡪 MgCl2 + H2
PbO2 + 4HCl 🡪 PbCl2 + Cl2 + 2H2O
1) Zn + 2AgCl →ZnCl2 + 2Ag
Zn → Zn2+ + 2e-
2Ag+ + 2e- → 2Ag
Cl- is a spectator ion
Zn + 2Ag+ → Zn2+ + 2Ag
2) 2K + H2 → 2KH
2K → 2K+ + 2e-
H2 + 2e- → 2H-
2K + H2 → 2H- + 2K+
3)Mg + 2HCl → MgCl2 + H2
Mg → Mg2+ + 2e-
2H+ + 2e- → H2
Cl is a spectator ion
Mg + 2H+ → Mg2+ + H2
4)PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O
Pb4+ + 2e- → Pb2+
2Cl- → Cl2 + 2e-
2O2-, 2Cl- and 4H+ are spectator ions
Pb4+ + 2Cl- 🡪 Pb2+ + Cl2
What is disproportionation?
A reaction where a substance undergoes both oxidation and reduction simultaneously is called a disproportionation reaction.
Balancing redox half equations steps
Practice with MnO4- being reduced to Mn2+ in acidic solution
- Identify the atom being oxidised or reduced, and make sure there are the same number of that atom on both sides (by balancing).
- Work out oxidation state of the element before and after the change
- Add electrons to one side of the equation so that the oxidation states balance.
(Often reactions take place in aqueous conditions, so a number of spectator species are also present, and although neither is oxidised nor reduced, they still exist and need to be balanced).
- Balance O atoms by adding water ; Balance H atoms by adding H+ ions.
i. MnO4- → Mn2+
ii. +7 → +2
iii. MnO4- + 5e- → Mn2+
iv. MnO4- + 5e- → Mn2+ + 4H2O
MnO4- + 5e- + 8H+ → Mn2+ + 4H2O
Full complex eq for
Cr2O72- being reduced to Cr3+ in acidic solution
i. Cr2O72- → 2Cr3+
ii. +6 → +3 (each chromium)
iii. Cr2O72- + 6e- → 2Cr3+
iv. Cr2O72- + 6e- → 2Cr3+ + 7H2O
Cr2O72- + 6e- + 14H+ → 2Cr3+ + 7H2O
