Redox

Difficult but necessary

Obviously: Oxidation is adding oxygen

2H2 + O2 2H2O

Reduction is removing oxygen 2FeO + C 2Fe + CO2

But also oxidation is removal of hydrogen And reduction is adding hydrogen

And Oilrig Oxidation is loss of electrons: Cu – 2eˉ Cu2+

Notice the charge increases Reduction is gain: Cu2+ + 2eˉ Cu

Notice the charge has reduced Both often happen in one reaction, so it

is a redox reaction

More…

A redox reaction: 2Al2O3 4Al + 3O2

aluminium reduced +3 to 0, oxygen oxidised, -2 to 0

Not a redox reaction: PbCl2 + 2NaI PbI2 + 2NaCl Pb stays at +2, Cl stays at -1,

Na stays at +1, I stays at -1

e.g.

Mg + CuO MgO + Cu As ions:

Mg + Cu2+ + O2ˉ Mg2+ + Cu + O2ˉ Note the oxygen ions are spectator ions, they

aren’t actually involved, so we get: Mg + Cu2+ Mg2+ + Cu

So the magnesium has reduced the copper And the copper has oxidised the magnesium

Oxidation numbers / states Represent charges where there aren’t

any They are an “accounting trick” to keep

track of how atoms have control over electrons

Apply to ions and covalently bonded atoms

The oxidation numbers of elements are zero e.g.. Fe(s), and even O2

Working them out Rules for assigning: (these rarely change) F is always -1 O is -2, except in OF2

Group 7 are -1, except with O or F Group 1 metals are +1 Group 2 metals are +2 H is +1, except in hydrides, e.g. NaH Al is +3 The total for an ion is its charge (e.g. -1 for CN-) More electronegative atoms get negative numbers The total for a compound is 0, even in O2, Cl2 etc.

Note: the allocation of a high oxidation number does not necessarily mean that electrons have been lent and borrowed.

E.g. in CrO42ˉ the oxidation number of

chromium is +6, yet it is covalently bonded to the oxygens and the energy required to remove 6 electrons would be prohibitive.

All the +6 tells us is that the electrons probably spend more time near the oxygens.

Working out

Overall charge on a compound ion is the sum of the oxidation states:

E.g. for MnO4ˉ in KMnO4

Oxidation state of Mn is +7 because overall charge is -1 and oxygens are -8 (-2 x 4)

So -1 = +7 – 8 Write MnO4ˉ as manganese(VII) oxide or

manganate(VII) Manganate(VII) compounds are

common oxidising agents

e.g. oxidation states in CaSO4

Ca is +2 O is -2 X 4 =-8 Uncharged compound so total oxidation

number is 0 So sulphur is +6 (0 = +2 -8 +6)

Ca O4 S

Call it calcium sulphate(VI)

e.g. the thiosulphate anion S2O32ˉ

This is a common reducing agent, it donates electrons to reduce other chemicals

Overall = oxygens + sulphurs -2 = (3 x -2) + (2 x

sulphur) -2 = (-6) + 4 2 x sulphur = +4 sulphur = +2 This is the sulphur(II) oxide (or

thiosulphate) anion

Practice:

What is the oxidation state of: Chromium in CrO4

2-

Hydrogen and magnesium in MgH2

Both elements in water H2O Chlorine in HClO Sodium and chlorine in Na2ClO3

Carbon in carbonate CO32-

Iron in Fe3O4

Naming: If there is any doubt about the oxidation

state, usually transition metals, it must be given:

CuCl2 Copper(II) chloride CuCl3 Copper(III) chloride NaNO3 Sodium nitrate(V)

in NO3ˉ we count nitrogen using -6 for 3 oxygens, to make -1 for the negative charge, so N is +5

From -1=+5-6 (remember, overall charge is the total of

oxidation states)

Redox or not?

Cl2 + 2KBr 2KCl + Br2

Cl: 0 to -1, Br: -1 to 0 Cl reduced, Br oxidised MnO2 + 4HCl MnCl2 + Cl2 + 2H2O Mn from +4 to +2, some Cl from -1 to 0 Mn reduced, Cl oxidised 2CrO4

2ˉ +2H+ Cr2O72ˉ + H2O

Cr is +6 before and after, nothing else changes either – not redox

Balancing Just when you thought you had got it.... Consider this redox change: MnO4

-(aq) Mn2+(aq)

Continued

Continued.... MnO4

-(aq) Mn2+(aq)

In water Add oxygen in H2O to balance....

Giving MnO4-(aq) Mn2+(aq) + 4H2O(l)

Assume an acidic solution to balance H.... Giving MnO4

-(aq) + 8H+ Mn2+(aq) + 4H2O(l)

Sort-out electrons for charge and redox.... MnO4

-(aq) + 8H+ + 5e- Mn2+(aq) + 4H2O(l)

+7 +2 In fact we’ve always done this, but it was easy examples...

Try:

VO43-(aq) V2+(aq)

MnO4-(aq) MnO2(s)

CrO42-(aq) Cr2+(aq)

SO42-(aq) S8(s)

VO43-(aq) + 8H+(aq) + 3e- V2+(aq) + 4H2O

MnO4-(aq) + 4H+(aq) + 3e- MnO2(s) + 2H2O

CrO42-(aq) + 8H+(aq) + 4e- Cr2+(aq) + 4H2O

8SO42-(aq) + 64H+(aq) + 48e- S8(s) + 32H2O

SO42-(aq) + 8H+(aq) + 6e- S(s) + 4H2O

Some specific half-equations of oxidising agents: Oxygen plus metal: O2 + 4e- 2O2-

chlorine plus metal: Cl2 + 2e- 2Cl-

Sulphur plus metal: S + 2e- S2-

In hydrogen peroxide, oxygen is in a -1 state. Is this likely to be a stable compound?

H2O2 + 2H+ +2e- 2H2O

More....

Concentrated sulphuric acid: 2H2SO4 + Cu CuSO4 + 2H2O + SO2

½ equation: SO42- + 2e-+4H+ SO2 +2H2O

Conc. nitric acid: Cu + 4HNO3 Cu(NO3)2 + 2H2O + 2NO2

Some specific half-equations of reducing agents: Oxygen plus metal: O2 + 4e- 2O2-

chlorine plus metal: Cl2 + 2e- 2Cl-

Sulphur plus metal: S + 2e- S2-

In hydrogen peroxide, oxygen is in a -1 state. Is this likely to be a stable compound?

H2O2 + 2H+ +2e- 2H2O