1. Redox reagents, equations, titrations, and electrolysis.
Redox Reactions
Redox reagents, equations, titrations, and electrolysis.

2. Index Redox reactions Electrochemical series Writing Redox equations
Redox Titrations
Electrolysis

3. Redox Redox reactions include reactions which involve the loss
or gain of electrons.
The reactant giving away (donating) electrons is called
the reducing agent (which is oxidised)
The reactant taking (accepting) electrons is called
the oxidising agent (which is reduced)
Both oxidation and reduction happen simultaneously,
however each is considered separately using
ion-electron reactions.
O.I.L. R.I.G.
Oxidation is loss Reduction is gain of electrons

4. eg. ½O2 + 2e-  O2- ½Cl2 + e-  Cl- eg. Mg  Mg2+ + 2e-
Note that, in general,
½Cl2 + e-  Cl-
eg.
Mg  Mg2+ + 2e-
Al  Al e-
Metals on the LHS of the Periodic Table ionise by
electron loss and are called reducing agents
Non-metals on the RHS of the Periodic Table ionise
by electron gain and are called oxidising agents

5. Cells and redox A metal lower A metal higher in the series
Ion
bridge
A metal higher
in the series
A metal lower
in the series
ions of metal higher in ECS
ions of metal lower in ECS
Metal atoms will be oxidised.
Metal atoms are the reducing agent.
Metal ions in solution will be reduced,
Metal ions are the oxidising agent.
E.g. Mg  Mg e
E.g. Cu e  Cu
Overall redox equation
Mg (s) + Cu 2+ (aq) => Mg 2+ (aq) + Cu (s)

6. Redox and the electrochemical series
Eo/V Oxidising agents
-3.03v
-2.71v
-2.37v
-0.13v
0.00v
+0.34v
+0.80v
Li+(aq) + e  Li(s)
Na+(aq) + e  Na(s)
Mg2+(aq) + 2e  Mg(s)
Pb2+(aq) + 2e  Pb(s)
2H+(aq) +2e  H2(s)
Cu2+(aq) + 2e  Cu(s)
Ag+(aq) + e  Ag(s)
Increasing powerful reducing agent
(write the reaction backwards)
Hydrogen reference
Increasing powerful oxidising agent
(write the reaction as it appears)
Considering the two ion-equations,
Mg 2+ (aq)  Mg (s) + 2e and Ag + (aq) + e  Ag ,
Mg, being higher up the electrochemical series, would act as the reducing agent. (i.e. the ion-electron equation would be written backwards).
While Ag would be written as it appears in the electrochemical series.

7. Redox The reducing agent in this reaction is the Mg as it will
magnesium(s) + silver nitrate(aq)  magnesium nitrate(aq) + silver(s).
The reducing agent in this reaction is the Mg as it will
donate electrons to the silver ions .
The oxidising agent is the Ag+ ions as they accept electrons
from the Mg
Oxidation:
Mg(s)  Mg2+(aq) e-
Half equations
or ion-equations
Reduction:
2Ag+(aq) + 2e-  2Ag(s)
Mg(s) + 2Ag+(aq)  Mg2+(aq) + 2Ag(s)
Redox equation, electrons cancel out

8. Na(s) + H2O(l)  NaOH(aq) + ½H2(g)
Writing REDOX equations
Consider the reaction between sodium and water:
Na(s) + H2O(l)  NaOH(aq) + ½H2(g)
Consider how the ions are formed in this reaction
Na(s)  Na+(aq) + e-
H2O(l) + e-  OH-(aq) + ½H2(g)

9. H2O(l) + e-  OH-(aq) + ½H2(g)
Na(s)  Na+(aq) + e-
A sodium atom loses an electron
H2O(l) + e-  OH-(aq) + ½H2(g)
and, we could say that a water molecule must be accepting the electron

10. Na(s)  Na+(aq) + e- OIL RIG H2O(l) + e-  OH-(aq) + ½H2(g)
These are called ion-electron equations
(or ionic half equations).
The equation which shows electron loss is called OXDATION.
The equation which shows electron gain is called REDUCTION.

11. H2O(l) + e-  OH-(aq) + ½H2(g)
Na(s)  Na+(aq) + e-
H2O(l) + e-  OH-(aq) + ½H2(g)
Reduction and oxidation occur simultaneously. Adding the two equations together gives us the overall equation for a reaction.

12. Na(s)  Na+(aq) + e-
H2O(l) + e-  OH-(aq) + ½H2(g)
Electrons cancel!

13. Na(s) + H2O(l)  NaOH(aq) + ½H2(g)
Na(s)  Na+(aq)
H2O(l)  OH-(aq) + ½H2(g)
+ e-
+ e-
Na(s) + H2O(l)  NaOH(aq) + ½H2(g)
and we can add the two equations!

14. Balancing Redox equations
Most redox reaction you will come across will occur in
neutral or acidic conditions.
Make sure there are the same number of atoms of the
element that is oxidised or reduce on each side of the
half equation.
2. If there are any oxygen atoms present, balance them by
adding water molecules to the other side of the half-equation.
3. If there are any hydrogen atoms present, balance them
by adding hydrogen ions on the other side of the half-equation.
4. Make sure the half-reactions have the same overall
charge on each side by adding electrons.
For basic solutions H atoms are balanced using H2O and then the same
number of OH- ions to the opposite side to balance the oxygen atoms

15. SO2(g) + 2H2O(l)  SO42-(aq) + 4H+(aq)
1. Write down what you know
sulphur dioxide is oxidised to sulphate ions
SO2(g)  SO42-(aq)
2. Balance the oxygen atoms by adding water
2H2O(l)
SO2(g)  SO42-(aq)
3. Balance the hydrogen atoms by adding hydrogen ions
SO2(g) + 2H2O(l)  SO42-(aq) +
4H+(aq)
4. Balance the charges by adding electrons
SO2(g) + 2H2O(l)  SO42-(aq) + 4H+(aq) +
2e-
charge is zero
4 - and 4 + equals zero

16. Redox Titrations
Titration is a technique for measuring the concentration
of a solution, when one is known and another is not.
Redox titrations involve solutions of reducing and
oxidising agents.
At equivalence-point of a redox titration precisely enough
electrons have been removed to oxidise all of the reducing
agent.

17. RedoxTitration 1 2 3 4 What to do: Carefully fill the burette
with potassium permanganate . 2
Carefully pipette exactly
20 ml of iron II sulphate
into the conical flask.
Then add 20 ml 1 mol l-1 H2SO4 3
Add the permanganate until a
permanent purple colour appears
in the conical flask. 4
A rough titration is done first to give a rough
equivalence-point (end-point), then repeated
more accurately to give concordant results.

18. Redox titrations n = V x C
5 Fe2+ (aq) + 8H+ (aq) + MnO4- (aq)  5 Fe3+ (aq) + Mn2+ (aq) + 4H2O(l)
purple
colourless
Use a standard solution of potassium permanganate to find
out the unknown concentration of an iron (ii) sulphate solution
x = [MnO4- (aq)]
y = [Fe 2+ (aq) ]
n y = 5
n x = 1
V x x C x
n x =
V y x C y
n y Or
C y =
V x x C x
x n y
V y x n x

19. Redox titrations, Vitamin C
I2 (aq) + 2e-  2I - (aq)
C6H8O6 
C6H6O H+ (aq) e-
reduction
oxidation 
Purple (in the presence of starch)
I2 (aq) + C6H8O6
colourless
C6H6O H I- (aq)
Iodine, those concentration is known (in the burette)
acts as an oxidising agent,
Vitamin C, the unknown concentration (in the conical flask)
is a reducing agent.
Starch is added to show when the end-point is reached.
V x x C x
n x =
V y x C y
n y

20. Faraday Faraday was the first person to measure the amount
of electrical charge needed to deposit a certain amount
of substance at an electrode.
Amount of
electrical
charge
(electrons)
electrolysis
Mass of substance
deposited
Electrical charge is the amount of electrons

21. Electroysis Q I x t Quantity of charge = current x time Q = I x t
Current is the flow of an electrical charge
The amount or quantity of charge (Q) is measured in Coulombs (C)
Quantity of charge = current x time
Q = I x t
96,500 coulombs is called 1 Faraday (F).
The number of coulombs required to deposit 1 mole of atoms or molecules of an element is 96,500 x n. (F x n) n being either 1,2,3 or 4.
The multiplying factor n, can be equated to the number of electrons associated with the production of one atom or molecule of the element.

22. Electrolysis
96,500 coulombs = 1 mole of electrons

23. Electrolysis and Hydrogen
2H+ (aq) +  H2 (g)
So to produce 1 mole of H2, 2 moles of electrons
So to produce 1 mole of H2 , x 2 C of charge are needed
So by electrolysing hydrochloric acid, collecting the hydrogen
gas at the anode measuring its volume, and taking the molar
volume as 24 litres). By measuring the current in amps
and timing in seconds how long the current was flowing for
it is possible to confirm that x 2 C of charge are
needed to produce 1 mole of H2 gas.

24. Gases by electrolysis
The mass or volume of an element discharged can be calculated from the quantity of electricity passed and vice-versa.
Example: A solution of HCl is electrolysed. What current is needed
to produce 2.4 litres of H2 gas in 16min 5 sec? Molar volume = 24 l mol-1
Since 2H e  H2 so 2 moles of electrons produce 1 mole of gas.
i.e x 2 C of charge is needed to produce 1 mole of gas
Since 2.4l is 0.1 mole of gas, ( x 2 ) x 0.1 C of charge is needed
Q = I x t
So I = Q/t
( x 2 ) x 0.1 /
(60 * 16) + 5
Ans: 20 A