Topic 9 Oxidation and Reduction

• Introduction• Oxidation numbers• Redox equations• Reactivity• Voltaic cells• Electrolytic cells

9.1 Introduction to oxidation and reduction

Definition: Oxidation: Particle that loses electron(s).

Na Na+ + e-

Reduction: Particle that gains electron(s). Cl2 + 2e- 2 Cl- or

• If something is oxidised, something else must be

reduced in a chemical reaction.

OILRIG

Oxidation Is Loss, Reduction Is Gain… of electrons

Half reactions

• Consider the redox reaction: 2 Na + Cl2 2 NaCl

• This reaction can be split into two “half reactions”:– Oxidation reaction: 2 Na 2 Na+ + 2 e-

– Reduction reaction: Cl2 + 2e- 2 Cl-

Oxidation number

• It’s sometimes difficult to see if it is a redox reaction or if a particle has been oxidised/reduced.

• Then you have to find out the oxidation numbers or oxidation state of the atoms in the compound.

• The oxidation number is placed above the atom symbol often in roman figures but not in IB => Use normal figures e.g +2 (Charge: 2+)

Rules for determining oxidation number:

• Elements in their element state: 0• The total oxidation state = the charge of the

compoundH2O total oxidation state = 0

H3O+ total oxidation state = 1

• “Atomic ions”: the same oxidation number as its charge

Cl- oxidation number = -1 elementNa+ oxidation number = +1Al3+ oxidation number = +3

Some elements often have the same oxidation number

• Fluorine: -1

• Hydrogen: +1 (except in hydrides = -1)

• Oxygen: -2 (except in peroxides = -1, bonded to F= pos.)

• The halogens: -1 (except when bonded to oxygen or a more electronegative halogen.)

Find the oxidation number in following compounds

• Cl2 0 => element

• Cu2+ +2 => atomic ions: the same oxidation number as its charge

• CH4 H: +1• C: -4 => total oxidation state = the charge of the compound

• H2SO3 H: +1 => total +2O: -2 => total –6S: +4 ( 2 + 4 - 6 = 0)

• NO3- O: -2=> total -6

N: 5 (5 – 6 = -1)

9.2 In a chemical reaction

• Oxidation: the atom that has an increase in oxidation number.

• Reduction: the atom that has a decrease in oxidation number.

• No change in oxidation number for any atom = no redox reaction.

Is following compounds being oxidised or reduced?

• NO NO3- N: +2 N: +5 => oxidation

• N2O NH3 N: +1 N: -3 => reduction

• HNO2 NO2- N: +3 N: +3 => no redox

Naming compounds:

• Oxidation number is also often used in naming compounds:

• FeCl2: Iron(II)chloride (2 = Ox. no. of Fe)• CuO Copper(II)oxide• Cu2O Copper(I)oxide

Redox reactions

• Balance and put together half-reactions: by calculating electrons

Silver ions react with magnesium metal:– Oxidation reaction: Mg Mg2+ + 2 e-

– Reduction reaction: 2 Ag+ + 2 e- 2 Ag

Total rection: Mg + 2 Ag+ Mg2+ + 2 Ag

Reducing agent• An element like sodium, Na, is eager to become an

ion through oxidation: Na Na+ + e-

• Then some other particle, X, must be reduced (X + e- X-)

• Sodium is then said to be a reducing agent • A reducing agent reduces a compound by self being

oxidised

Oxidising agent • An element like chlorine, Cl, is eager to become

an ion through reduction: Cl + e- Cl-

• Then some other particle must become oxidised• Chlorine is then said to be a oxidizing agent• An oxidising agent oxidises a compound by self

being reduced

Balance redox reactions

• An easy example to start with:Cu + Ag+ Cu2+ +Ag

Divide into Half equations1. Balance one of the reactants; Cu Cu2+ + 2e-

2. Balance the other; Ag+ + e- Ag Have to be multiplied by 2 to get the same number of electrons.

3. Add the reactions; Cu + 2 Ag+ Cu2+ + 2 Ag4. Check atoms and charge.

Balance redox reactions in acidic solutions Cu + HNO3 +…. Cu(NO3)2 + NO + …..1. Half reactions:(1) Cu Cu2+ + 2e- (2) HNO3

NO

2. Check and balance half reactions for O and HO not balanced in (2) => add H2O so oxygen balances

(2) HNO3 NO + 2 H2O ;

H not balanced in (2) => add H as H+ (acidic solution, remember…)(2) HNO3

+ 3 H+ NO + 2 H2O

Balance redox reactions in acidic solutions (II)(1) Cu Cu2+ + 2e- (2) HNO3

+ 3 H+ NO + 2 H2O

3. Check and balance half reactions for chargesIf charges not balanced =>add e-

(2) HNO3 + 3 H+ + 3 e- NO + 2 H2O

4. Check and balance half reactions so they use the same no of e-

Multiply (1) with 3 and (2) with 2 => 6 e- produced and consumed in each half-reaction:

(1) 3 Cu 3 Cu2+ + 6 e- (2) 2 HNO3

+ 6 H+ + 6 e- 2 NO + 4 H2O

5. Add the reactions: 3 Cu +2 HNO3 + 6 H+ + 6 e- 3 Cu2+ + 6 e- + 2 NO + 4 H2O

Check atoms and e- : Equation balanced

Balance redox reactions in acidic solutions (III)6. Remove electrons.3 Cu +2 HNO3 + 6 H+ 3 Cu2+ + 2 NO + 4 H2O

7. Fix the reaction (if needed). Add 6 NO3- on both side:

3 Cu +2 HNO3 + 6 H+ + 6 NO3- 3 Cu2+ + 2 NO + 4 H2O + 6 NO3

-

8. Simplify:3 Cu +8 HNO3 3 Cu(NO3)2 + 2 NO + 4 H2O

Check atoms and charges.

9.3 Reactivity

• Redox couple: A species that gain or lose electron(s), e.g. Na+ + e-→ Na

• If a compound is a good reducing agent (easily

oxidized/lose e-), the “other form” will be a bad oxidising agent and vice versa

• Redox couples can be arranged in a reactivity series

Redox reactivity series

Redox coupleOxidised form Reduced form

Na+ + e- Na Good red. agent

Mg2+ + 2e- Mg Fe2+ + 2e- Fe 2 H+ + 2e- H2

Cu2+ + 2e- Cu

I2 + 2e- 2 I-

Br2 + 2e- 2 Br-

Good ox. agent F2 + 2e- 2 F -

• In the upper part of the redox reactivity series the redox couple prefer to be on the left side (oxidised form)

• In the lower part of the table the redox couple prefer to be on the right side (reduced form)

• Eg. sodium is rather Na+ than Na, flourine is rather F2 than F-

You can use the redox reactivity series to predict if a reaction is possible or not by comparing redox couples

• If the redox couple standing above in the table is going to the left and the redox couple standing below is going to the right, then a reaction will occur.– Mg2+ + 2e- Mg– Br2 + 2e- 2 Br-

=> Mg + Br2 MgBr2 is possible

• If the above couple going to the right and the pair below to the left no reaction will occur.

• E.g. F2 + 2I- 2F- + I2 OK but I2 + F- no reaction

9.4 Voltaic cell

• The power of oxidising and reduction can be given in volts (V)

• You can measure the potential between two metals and their ions in a galvanic cell

• In a galvanic/voltaic cell chemical energy is converted into electrical energy

• It’s a spontaneous reaction

• The less noble metal (Zn) will oxidise: Zn Zn2+ +2e-. This will be the negative pole.

• The electrons will pass the voltmeter and reach the copper metal and copper ions will be reduced:

Cu2+ +2e- Cu. This will be the positive pole.

• In the salt bridge ions will go from Copper half cell to Zinc half cell. It balances the charges and will give you a closed circuit.

Cell diagram

- Zn (s) Zn2+ (aq) Cu2+ (aq) Cu (s) +

Negative electrode, anode Positive electrode, cathode

Standard electrode potentials• In the table “standard electrode potential” you find

potentials of different redox couples compared to a “standard hydrogen electrode”.

• If you want to calculate the potential, E, you take the difference between the positive half cell and the negative half cell: E = E+ -E-

• Use of Voltaic cells: Batteries

9.5 Electrolytic cell

• In electrolysis electrical energy is converted to chemical energy. It’s a non-spontaneous reaction.

• Electrolysis can be done in ionic aqueous solution (= electrolyte) or in molten salt (= electrolyte).

Electrolysis of CuBr2

+ pole/electrode = Anode: 2 Br- Br2 + 2e- Oxidation

Reduction -pole/electrod = Cathode: Cu2+ + 2e- Cu

Electrolysis of molten salt: e.g. NaCl(l)

Use of electrolysis: Electroplating, analysis, chargeable batteries, purification of metals

Differences between Voltaic cells and Electrolysis

Voltaic cell• Spontaneous• Chemical energy Electrical

energy• + electrode: Reduction

(Cathode)• - electrode: Oxidation

(Anode)

Electrolysis• Non-spontaneous• Electrical energy Chemical

energy• + electrode: Oxidation

(Anode)• - electrode: Reduction

(Cathode)