AS Chemistry OXIDATION STATES, HALF EQUATIONS and REDOX REACTIONS

AS Chemistry

OXIDATION STATES, HALF EQUATIONS and

REDOX REACTIONS

OXIDATIONOXIDATION REDUCTIONREDUCTION

= OXYGEN= OXYGEN

= HYDROGEN= HYDROGEN

= ELECTRON= ELECTRON

= OXIDATION STATE= OXIDATION STATE

REDOX REACTIONS = reactions involving REDuction and OXidation

Definitions:

GAINGAIN LOSSLOSS

LOSSLOSS GAINGAIN

LOSSLOSS GAINGAIN

INCREASEINCREASE DECREASEDECREASE

Remember “OILRIG” : Oxidation Is Loss ; Reduction Is Gain (of electrons)

Oxidation states (also called oxidation numbers) are numbers assigned to EACH ATOM that takes part in a reaction.

Oxidation states are assigned using a set of International rules.

Rules for deciding Oxidation States (Numbers) :

5. In a BINARY (2 elements) COMPOUND

1. In all UNCOMBINED ELEMENTS, atom’s ox. no. = 0 .

2. In all COMPOUNDS, sum of ox. no.’s equals zero.

3. In all IONS, sum of ox. no.’s equals ion charge.

4. In all COMPOUNDS : Gp 1 elements + 1

+ 2

- 1

+ 3

the more electronegative atom given NEGATIVE ox. no. and the less electronegative atom given POSITIVE ox. no.

In most COMPOUNDS,

6. H = + 1 except when bonded to a metal- metal must have the positive ox. no. (Rule 5)

7. O = - 2 except when bonded to F or in peroxides, e.g. Na2O2- F must have the negative ox. no. (Rule 4)

LEA

RN

and PR

AC

TIS

E

Gp 2 elements

Gp 3 elementsFluorine

ASSIGN AN OXIDATION NUMBER / STATE TO EACH ATOM IN :

Cl2 CO32-

Ca2+ SO32-

Al3+ ClO-

H2O IO4-

CO2 CH4

ClF MnO4-

NO3- Na2S4O6

CuCl CuBr2

N2 C2O4

2-

BrF5 Mn2O3

SF6 CO

S2- BrF

VCl2 Na2S

NO2- BrO3

-

NH4+ H2SO4

SO42- I-

S2O32- NH3

CCl4 Cr2O7

2-

Cl(0)

Ca(+2)

Al(+3)

H(+1) O(-2)

O(-2) C(+4)

F(-1) Cl(+1)

O(-2) N(+5)

Cl(-1) Cu(+1)

N(0)

F(-1) Br(+5)

F(-1) S(+6)

S(-2)

Cl(-1) V(+2)

O(-2) N(+3)

H(+1) N(-3)

O(-2) S(+6)

O(-2) S(+2)

Cl(-1) C(+4)

O(-2) C(+4)

O(-2) S(+4)

O(-2) Cl(+1)

O(-2) I(+7)

H(+1) C(-4)

O(-2) Mn(+7)

Na(+1) & O(-2) S(+2.5)

Br(-1) Cu(+2)

O(-2) C(+3)

O(-2) Mn(+3)

O(-2) C(+2)

F(-1) Br(+1)

Na(+1) S(-2)

O(-2) Br(+5)

O(-2) & H(+1) S(+6)

I(-1)

H(+1) N(-3)

O(-2) Cr(+6)

Rem. OXIDATION REDUCTION

= OXIDATION No. INCREASE DECREASE

Work out the oxidation number change for each of the following process and use it to decide whether it is an OXIDATION or a REDUCTION.

PROCESSPROCESS Ox. No.’sOx. No.’s OxidationOxidation ReductionReduction

ClCl2 2 Cl Cl--

CaCa Ca Ca2+2+

NONO2 2 NO NO

33--

MnOMnO44- - Mn Mn2+2+

SOSO22 SO SO

442-2-

IOIO44- - I I

22

HH22SOSO4 4 S S2-2-

BrBr2 2 BrO BrO--

NHNH44+ + NH NH

33

CrCr22OO77

2- 2- Cr Cr3+3+

Cl(0) (-1)

Ca(0) (+2)

N(+4) (+5)

Mn(+7) (+2)

S(+4) (+6)

I(+7) (0)

S(+6) (-2)

Br(0) (+1)

N(-3) (-3)

Cr(+6) (+3)

NONE

Half Equations= equations showing the SEPARATE oxidation (loss of e-) and

reduction (gain of e-) processes in any redox reaction

e.g. 1 2Ca(s) + O2(g) 2CaO(s)

Ca atoms - Ca oxidised O2 mols - O2 reduced

Oxidation :Reduction :

HALF EQUATIONS :

0 +2

0 -2

Ca Ca2+ + 2e-

O2 + 4e- 2O2-

e.g. 2 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

Na atoms - Na oxidised H2O mols - H2O reduced


HALF EQUATIONS :

0 +1H(+1) H(0)

Na Na+ + e-

2H2O + 2e- 2OH- + H2

General Method for Writing Half Equations

e.g.1 MnO4- Mn2+ (NOT balanced ; occurs in acid)

Mn(+7) (+2) MnO4- is reduced

Number of electrons in half-equation

3. Complete the balance (for atoms and charges) by inserting H2O and H+ or OH- as appropriate

1. Write formulas of “redox” particles and balance “changed” atoms

Reduction : MnO4- Mn2+

CHANGE in oxidation number of “redox” atoms

2. Insert e- on left for reduction, right for oxidation

Reduction : MnO4- + 5e- Mn2+

=

Reduction : MnO4- + 8H+ + 5e- Mn2+ + 4H2O

i.e. 4 O 4H2O 8H+

e.g.2 Cl2 ClO4- (NOT balanced ; occurs in alkali)

Cl(0) (+7) Cl2 is oxidised




Reduction : Cl2 2 ClO4-



Reduction : Cl2 2 ClO4- + 14e-

=

Reduction : Cl2 + 16OH- 2 ClO4- + 14e- + 8H2O

i.e. 8 O 8H2O 16OH-

e.g.3 Cu + HNO3 Cu2+ + NO2 (NOT balanced ; occurs in acid)

Cu(0) (+2)

N(+5) (+4)

Cu oxidised

and HNO3 reduced




Oxidation : Reduction :Cu Cu2+ HNO3 NO2



Oxidation : Reduction :Cu Cu2+ + 2e- HNO3 + e- NO2

=


Cu Cu2+ + 2e-

HNO3 + H+ + e- NO2 + H2O

Write a half-equation for each of the following changes.

1. Cl2 to Cl- 6. Br- to Br2

2. Pb2+ to Pb 7. Al to Al3+

3. H2SO4 to H2S 8. At- to At2

4. HNO3 to NO 9. Fe to Fe2+

5. H2SO4 to SO2 10. Br- to Br2

Combining half-equations to produce the full equation

(a) An oxidation half-equation must be combined with a reduction half-equation

(b) Combine in the ratio which balances out the electrons lost during oxidation with those gained during reduction.

i.e. oxidation number changes must balance

Oxidation :

Reduction :

Cu Cu2+ + 2e-

HNO3 + H+ + e- NO2 + H2O

X 1

X 2

Add :

Cu + 2HNO3 + 2H+ + 2e- Cu2+ + 2e- + 2NO2 + 2H2O

Example 1

Remember(a) An oxidation half-equation must be combined with a

reduction half-equation


Oxidation :

Reduction :

Fe2+ Fe3+ + e-

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

X 5

Add :

5Fe2+ + MnO4- + 8H+ + 5e- 5Fe3+ + 5e- + Mn2+ + 4H2O

Example 2

X 1

Remember(a) An oxidation half-equation must be combined with a

reduction half-equation


Oxidation :

Reduction :

Cl2 + 2OH- ClO- + H2O + e-

Cl2 + 2e- 2Cl-

X 2

Add : 2Cl2 + 4OH- + 2e- 2ClO- + 2e- + 2Cl- + 2H2O

Example 3

X 1

Cl2 + 2OH- ClO- + Cl- + H2OCancel by 2

Note : Chlorine is BOTH oxidised and reduced. Such a reaction is called a DISPROPORTIONATION

Use the half-equations written earlier and combine them to form the overall equation for.

1. Cl2 to Cl- with Br- to Br2

2. Pb2+ to Pb with Al to Al3+

3. H2SO4 to H2S with I- to I2

4. HNO3 to NO with Fe to Fe2+

5. H2SO4 to SO2 with Br- to Br2

The End

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AS Chemistry OXIDATION STATES, HALF EQUATIONS and REDOX REACTIONS