OILRIG

• Oxidation is loss of electrons.• Reduction is gaining electrons.• REDOX – oxidation and reduction happening together.• Example• Mg(s) —> Mg 2+ (aq) + 2e Oxidation• Cu 2+ (aq) + 2 e —> Cu (s) Reduction• Cl 2(g) + 2e —> 2Cl- (aq) Reduction

REDOX

• CuSO4(aq) + Zn —> ZnSO4 (aq) + Cu (s)• The Cu ions gain electrons to form Cu atoms• Cu 2+ (aq) + 2e —> Cu (s) Reduction• The Zn atoms loose electrons forming Zn ions• Zn (s) —> Zn 2+ (aq) + 2e• REDOX ( we combine the oxidation and Reduction

equations)• Cu 2+ (aq) + 2e + Zn (s) —> Cu (s) + Zn 2+ (aq) + 2e• We rewrite eq without electrons since they cancel each

other out.

• An oxidising agent is a substance that accepts electrons.

• In the previous example the Cu 2+ ions are the oxidising agent.

• A reducing agent is one that donates electrons.• The Zn atoms act as a reducing agent.

More complex ion electron equations!

• Example ( data book p 11)• Permanganate ions reducing to form Manganese ions.• 1. Write formula of each ion.• MnO4

-(aq) —> Mn 2+ (aq)• 2. Balance – both equal in this case.• 3. Balance O by adding water to the other side.• MnO4

-(aq) —> Mn 2+ (aq) + 4H2O (l)• 4. Balance H by adding H+ ions to other side• MnO4

- (Aq) + 8 H+ (aq) —> Mn 2+ (aq) + 4 H2O (l)

• 5. Balance charge by adding electrons• Mn04

- (aq) + 8 H+ (aq)+ 5e —> Mn 2+ (aq) + 4H2O (l) +

• Total charge on LHS Total charge on RHS• 1 - + 8 + = 7+ 2+• So if we add 5e to LHS both side now have a

charge of 2+

Volumetric Titrations

• This is when we work out the volume or concentration of an acid required to neutralise a fixed conc. / volume of an alkali.

• C1xV1 = C2xV2• Example • What is the conc. of HCl if 25 cm3 neutralises 20 cm3 of 2

mol/l NaOH?• C1xV1= C2xV2• C1 = C2xV2/V1• = 2 x 0.02/ 0.025 = 1.6 mol/l

Redox Titrations

• We use redox titration to find out the volume or concentration of a reducing agent.

• We need to know:• Accurate volumes of reactants.• Concentration of oxidising agent.• Balanced redox equation.• Recognisable end pioint – e.g.colour change!

Example of Redox Titration

• Calculate the concentration of Iron(ii) sulphate solution if 20 cm3 of it react with 24 cm3 of Potassium permanganate – concentration 0.02 mol/l.

• 1. Get redox equation ( p11)• MnO4

- (aq)+ 8H+ (aq) + 5e —> Mn 2+ (aq) + 4 H2O(l) • (purple) ( colourless)

Fe 2+ (aq) —> Fe 3+ (aq) + e

• 2. Balanced redox equation• MnO4

- + 8H+ + 5 Fe 2+ —> 5 Fe 3+ + Mn 2+ + 4 H2O

• 3.Mole ratio• MnO4:Fe• 1:5• 4.Work out number of moles of MNO4 used• M = C xV = 0.02 x 0.024 = 0.00048 ( 4.8 x 10 –4)

• 5. Use mole ratio to work out number of moles of Iron used.

• MnO4:Fe 2+

• 1:5• So if MnO4 = 0.00048 moles, Fe 2+ will be;• 0.00048 x 5 = 0.0024 moles• We can now find the concentration of the Fe 2+

solution;• C = M/V = 0.0024/ 0.02 = 0.12 mol/l

End Point

• We can identify the end point using the colour change – when all the iron ions have been used up there will be excess permanganate ions and so we will see a purple colour.

• We call this a self indicating reaction – using one of the reactants to indicate end point.

Quantative Electrolysis

• Electrolysis – breaking down a compound using electricity.

• Oxidation occurs at the at the positive electrode.• Reduction occurs at the negative electrode.• In the redox reaction – the total number of electrons lost

must be the same as the total number of electrons gained.

• 1 mole of electrons requires 96,500 C of charge.• ( This is also known as 1 Faraday = 1F)

• We can calculate the quantity of electricity required to produce 1 mole of a product – in electrolysis.

• Q = I xT • Q = quantity of electricity ( coulombs)• I = current( amps)• T = time ( s)

Worked Example

• Calculate the mass of Sodium produced in the electrolysis of NaCl – using a current of 10 A for 32 mins.

• Q = IT= 10 x ( 32 x 60)10 x 1920= 19200 C

Na + (aq) + e —> Na(s)To produce 1 mole of Sodium( 23g) we need 1 mole of electrons

– 1 F = 96,500C.

• Therefore if we have 19200C ( from Q=IT)• We will produce 19200/96500 x 23 = 4.57g of Na• Example 2• Calculate the mass of Cu produced in the electrolysis of

Cu Cl2 if a current of 20A is passed for 25 minutes.• Q = IT

= 20 x ( 25 x 60 ) = 20 x 1500. = 30000C

• Cu 2+ (aq) + 2e —> Cu(s)• 1 mole of Cu ( 63g) requires 2 moles of electrons = 2F =

2 x 96,500C = 193,000C.• Therefore – 30000C ( from Q=IT) will produce:• 30000/193,000 x 63 = 9.79g of Cu• Example 3• Calculate the volume of Hydrogen gas produced( molar

volume 24l) when a current of 15A is passed through a solution of sulphuric acid for 25 minutes.

• Q = IT = 15 x ( 25 x 60 ) = 15 x 1500

= 22500C

2H +(aq) + 2e —> H2 (g)

We need 2 F ( 2 x 96,500C = 193, 000C) to produce 1 mole of H2 gas – 24 l

• Therefore - 22500C will produce :• 22500/193,000 x 24 = 2.79litres.• Example 4• How long will it take to produce 6.3 g of Cu at

the positive electrode in the electrolysis of CuCl2

using a current of 10A?• Rearrange Q = IT to find T• T = Q/I

• I mole of Cu requires 2 mole of electrons = 2F = 2 x 96,500C. ( 193,00C)

• Cu 2+ (aq) + 2e —> Cu (s)• To find 1 mole = 63g of Cu :• T = Q/I

= 193,000/10 = 19300s.For 6.3 g = 6.3/63 = 0.1 moles.0.1 x 19300 = 1930 seconds.