Oxidation & Reduction
Oxidation numbers
• Put the sign in front of the number (ion charge
other way)
• Oxygen always -2 (except in H2O2: -1)
• Hydrogen always +1 (except hydride: -1)
• Sum of ON in compounds = 0
• Elements = 0
• Monoatomic ion ON = charge
• Sum of polyatomic ON = charge
• if you get stuck, assume grp1 = +1, grp2 = +2
• With complex formulae, use polyatomic ion
together
+1 -2
1.ClO+6 -2
2.CrO42+4 -2
3.CO2
+4 -2
4.SO32+1 -2
5.H2S
-3 +1
6.NH3
0
ON Practice
+1+1 -2
11.HClO
+1 +3 -2
12.HClO2
+1+5 -2
13.HClO3
+1 +7 -2
14.HClO4
+1 -2
15.Cl2O
+2 -2 +1
16.Fe(OH)2
+1+5 -2
7.N2
17.KBrO3
8.NO
18.NaOCl
+2 -2
+1 +3 -2
+1 -2 +1
+1 -1
9.HNO2
19.H2O2
10.HNO3
20.MnO4-
+1 +5 -2
+7 -2
Identifying oxidants and reductants
The easiest way to do this is work out all ON then
compare atoms before and after the arrow.
If the ON increases, it is being oxidised and is
then therefore a reductant
If the ON decreases (is reduced), it is being
reduced and is then therefore an oxidant
If the ON does not change, it is a spectator ion
If the atom is within a
compound, give that compound
as the species that is being
reduced/oxidised
Appearance & state of reductants & oxidants
Reduced form
Oxidised Form
Formula
Appearance
Formula
Appearance
Cu
Brown solid
Cu2+
Blue ion
SO2
Colourless gas
SO42-
Colourless ion
Mn2+
Colourless ion
MnO4-
Purple ion
H2O2
Colourless liquid
O2
Colourless gas
H2O
Colourless liquid
H2O2
Colourless liquid
Cr3+
Green ion
Cr2O72-
Orange ion
Fe2+
Green ion
Fe3+
Orange ion
Cl-
Colourless ion
Cl2
Pale green gas
I-
Colourless ion
I2(aq)
Brown solution
H2
Colourless gas
H+
Colourless ion
Zn
Grey solid
Zn2+
Colourless ion
Br-
Colourless ion
Br2(aq)
Orange solution
Write balanced redox equations
It doesn’t matter how simple or complex the half
equations are you can follow the same steps:
• Balance atoms that aren’t O or H
• Balance oxygens by adding waters
• Balance hydrogens by adding H+
• Add electrons to the more positive side
To combine the 2 half equations, multiply them up
so the number of electrons are the same. Cancel
like terms
Remember, these are ionic
equations so you must identify
the ions involved in the reaction
from the description
Write balanced redox equations
• Balance atoms that aren’t O or H
• Balance oxygens by adding waters
• Balance hydrogens by adding H+
• Add electrons to the more positive side
To combine the 2 half equations, multiply them up
so the number of electrons are the same. Cancel
like terms
This method is used when the
reaction takes place in acid or
neutral conditions
Write balanced redox equations
• Balance atoms that aren’t O or H
• Balance oxygens by adding waters
• Balance hydrogens by adding H+
• Cancel out H+ by adding OH- to both sides
• Cancel out extra waters etc.
• Add electrons to the more positive side
To combine the 2 half equations, multiply them up
so the number of electrons are the same. Cancel
like terms
Balancing in alkaline conditions
alters the method. Look for
the clues in the question
Titration Recap
n = cV
n = moles (mol)
c = concentration (molL-1)
V = volume (L)
m = cV
m = mass (g)
V = volume (L)
n = m/M
m = mass (g)
n = moles (mol)
c = concentration (gL-1)
M = molar mass (gmol-1)
To convert gL-1
to molL-1=
gL-1/M = molL-1
To convert molL-1
to gL-1=
molL-1 X M = gL-1
Steps in titration calculations
Calculate conc. Of HCl if 20.0 mL reacts with
21.7 mL of 0.0492 molL-1 sodium carbonate.
1.Write balanced equation
Summarise what
you have.
Convert mL to L
Na2CO3 + 2HCl  2NaCl + CO2 + H2O
2.Calculate the amount of known
substance
n(Na2CO3) = 0.0492 X 21.7X10-3
n = 1.068 X 10-3 mol
Write down 1
more s.f. then
you need but leave
all numbers in
calculator
3.Use equation to work out unknown
Use numbers
substance
n(HCl)
= 2
n(Na2CO3)
1
n(HCl) = 2 X n(Na2CO3)
= 2.136 X 10-3 mol
from
equation:
unknown/known
then X this by the
n of known
4.Calculate concentration/mass of
C(HCl) = n/V
unknown
-3
-3
-1
= 2.136X10
/ 20.0X10
= 0.107 molL
Balanced equation:
Known:
Unknown:
Ratio:
Volume:
Volume:
Conc:
Conc:
Amount:
Amount:
Titration expectations
Achieved
Merit
Excellence
At least 2
titres fall
within 0.6mL.
At least 3
titres fall
within 0.4mL.
At least 3
titres fall
within 0.2mL.
Average titre
within 0.6mL of
expected
Average titre
within 0.4mL of
expected
Average titre
within 0.2mL of
expected
Only titres
within 0.6mL
range used.
Only titres
within 0.4mL
range used.
Only titres
within 0.2mL
range used.
Minor error in
calculation
allowed.
Composition
correctly
determined.
Composition
correct with
units and s.f
Titration calculations
25.0 mL samples of HCl solution were titrated against standard
sodium carbonate solution of concentration 0.9955 molL-1.
Titres of 29.3 mL, 27.6 mL, 27.8 mL and 27.8 mL were
obtained. Calculate the concentration of the HCl solution
Balanced equation:
Na2CO3 + 2HCl  2NaCl + CO2 + H2O
Known:
Unknown:
Na2CO3
HCl
Ratio:
Volume:
-3 L
27.73 X10n(HCl)
1: 2
Volume:
= 2
25.0 X10-3 L
n(Na2CO3)
1
Conc: n(HCl) = 2 X n(Na Conc:
2CO3)
0.9955 molL-1
Amount:
2.761 X10-2 mol
2.21 molL-1
Amount:
5.521 X10-2 mol
Titration calculations
Jaime titrated 10.0 mL aliquots of sulfuric acid against 0.01122
molL-1 sodium hydroxide solution. She got titres of: 19.82 mL,
19.62 mL, 18.65 mL, 19.68 mL, 19.65 mL. Calculate the
concentration of sulfuric acid.
Balanced equation:
H2SO4 + 2NaOH  Na2SO4 + 2H2O
Known:
Unknown:
NaOH
H2SO4
n(H2SO4)
= 1
Ratio:
2: 1 2
n(NaOH)
n(H2SO4) = 0.5 X Volume:
n(NaOH)
Volume:
19.65 X10-3 L
10.0 X10-3 L
Conc:
0.0112 molL-1
Amount:
2.201 X10-4 mol
Conc:
0.0110 molL-1
Amount:
1.100 X10-4 mol
More complex titration problems
Often used to calculate purity of a substance or % composition
A 5.026g sample of an ore of iron was dissolved in 50mL of
dilute sulfuric acid. The iron was converted to Fe2+(aq). The
resulting solution was titrated against 0.064 02molL-1 KMnO4
solution and required 30.68mL to oxidise all the iron. Calculate
the mass of iron in the ore hence the percentage of iron in the
ore.
C(MnO4-)= 0.06402molL-1
V(MnO4-)=30.68 X 10-3L
n = cv
n(MnO4-)= 1.96413 X 10-3 mol
Fe2+  Fe3+ + eMnO4- + 8H+ + 5e-  Mn2+ +4H2O
MnO4- + 8H+ + 5Fe2+  Mn2+ +4H2O + 5Fe3+
K
U
You need to pick
numbers
outKeep
the values
youin
Write yourWrite
2 half
calculator.
can work
and 4-)
U/Kwith
X n(MnO
equations
to
work
6
figures
in
figure out the ionic
out the full
ionic
standard
form.
equation
5equation
X n(MnO4-)
Give -----------final answer
to 3 s.f.
1
Mole ratios
25.0 mL of diluted hydrogen peroxide solution reacts
with 31.1 mL of 0.0184 mol L–1 MnO4–solution.
2MnO4- + 5H2O2 + 6H+  2Mn2+ + 5O2 + 8H2O
20.0 mL of 0.114 mol L–1 oxalic acid reacted with
23.7 mL of MnO4– solution.
5C2O42- + 2MnO4- + 16H+  10CO2 + 2Mn2+ +
8H2O
Titration problems with 2 equations
A student prepared a 0.01636molL-1 solution of KBrO3, then took
25.0mL samples of this solution, added 1g of KI crystals and 15mL
dilute sulfuric acid, and titrated the liberated iodine against sodium
thiosulfate solution. An average of 27.32mL of thiosulfate were
required. What is the concentration of the thiosulfate?
BrO3- + 6I- + 6H+  Br- + 3H2O + 3I2
K
I2 + 2S2O32- 2I- + S4O62- U
K
U
C(BrO3-)= 0.01636molL-1
V(BrO3-)=25 X 10-3L
n = cv
n(BrO3-)= 4.09000 X 10-4 mol
n(I2) U X U
n(S2O32-)
n(BrO3-K)
Kn(I2)
You
Means
neediodine
to figure
is
out
made
theinknown
the first
and
unknown
reaction in
that
each
is
thenequation
used in the
second reaction
3 X 2
1
n(S2O32-) = 6 X n(BrO3-)
1
Mole ratios
20.0 mL of a 0.0175 mol L–1 KBrO3 solution is reacted
with KI and the iodine liberated reacted with 28.9 mL of
thiosulfate solution.
BrO3- + 6I- + 6H+  Br- + 3I2 + 3H2O
I2 + 2S2O32-  2I- + S4O62-
10.0 mL of a Cu2+ solution was reacted with KI and the
iodine liberated titrated against 0.025 mol L–1
thiosulfate solution. 26.4 mL of thiosulfate were
required.
2Cu2+ + 4I-  2CuI + I2
I2 + 2S2O32-  2I- + S4O62-
Electrochemistry
Copper
nitrate
Zinc
Electrochemistry
Electrochemistry
Zinc metal disappears, blue
colour fades and copper is
deposited
Zn(s) + Cu(NO3)2(aq) ➞ Zn(NO3)2(aq) + Cu(s)
Zn(s) ➞ Zn2+(aq) + 2eCu2+(aq) + 2e- ➞ Cu(s)
These 2 half-equations can occur in separate
beakers as long as there’s a path for the
electrons to travel (wire) and the ions to travel
(salt bridge)
Electrochemistry
eIf voltage is +ve,
e-Zinc
move L to R.
If(Zn)
voltage is –ve,
this is reversed
Zinc
nitrate
(Zn2+)
V
Remember; RED
CAT
Copper
(Cu)
Copper
nitrate
(Cu2+)
Please note that
electrons move
from –ve to +ve.
- flow
Salt
bridge
allows
e
to
copper
Anode
is
–ve
in
e flow from Zn leaving
ions to move to
electrochemical
beaker
to combine with
Zn2+ in beaker
keep
the
system
cells!! Cu2+ to form
electrically Cu
Zn Zn2+ + 2e2+ + 2e- Cu
Cu
neutral
Oxidation at anode
Reduction at cathode
Electrochemistry
•The voltmeter connecting the 2 half-cells is measuring
the electromotive force (emf or E)
•We can use emf figures to compare the strength of
reductants and oxidants
•To compare you must use the standard half-cell with an
emf of 0 (H+/H2) and standard conditions: 25C (298K),
1.0molL-1, 1.0atm (101.3 kPa)
•In these comparisons, the hydrogen half-cell is on the
right-hand side and connected to the positive terminal of
the voltmeter to that cell
Cell diagrams
/ represents
change of
phase
Zinc
(Zn)
Zinc
nitrate
(Zn2+)
e-
V
//
represents
the salt
bridge
Copper
(Cu)
Copper
nitrate
(Cu2+)
Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
Write left
hand electrode
first as
oxidation
Write right
hand electrode
last as
reduction
Zn(s)/ Zn2+(aq)// Ag+(aq)/ Ag(s)
Pb(s)/ Pb2+(aq)// Fe3+(aq), Fe2+(aq)/ C(s)
Cu(s)/ Cu2+(aq)// MnO4–(aq), Mn2+(aq)/ C(s)
Pt(s)/ Cl-(aq)/ Cl2(g)// BrO3-(aq),Br2/ C(s)
Calculating E(cell)
E(cell) = E(red) – E(ox)
• If E(cell) is positive: cell on left
experiences oxidation and cell on right
experiences reduction. Reaction is
spontaneous.
• If E(cell) is negative: cell on left
experiences reduction and cell on right
experiences oxidation. Reaction is nonspontaneous.
Calculating E(cell)
Calculate the E(cell) for the following cell,Reduction
then
write the overall cell equation.
always on
right
Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
E(Zn2+/Zn) = -0.76V
E(Cu2+/Cu) = +0.34V
E(cell) = E(red) – E(ox)
= +0.34V –(-0.76V)
= +1.10V
- a –
becomes a
+!
E(cell)is positive so occurs as written in the cell
diagram:
Zn  Zn2+ + 2eCu2+ + 2e- Cu
Zn + Cu2+ Zn2+ + Cu
There are many
different methods
but this one always
works!
Calculate the E(cell) for the following cell.
C(s)/C2O42-(aq)/CO2(g)//Cr2O72-(aq),Cr3+(aq)/C(s)
E(CO2/C2O42-) = -0.20V E(Cr2O72-(aq),Cr3+(aq)/C(s) = +1.33V
E(cell) = E(red) – E(ox)
= +1.33V –(-0.20V)
Which
reaction is
reduction?
= +1.53V
positive: spontaneous. The reaction will
occur
Will sulfur precipitate when H2S gas is bubbled
through NiSO4 solution?
E(S/H2S) = +0.17V
E(Ni2+/Ni) = -0.23V
H2S → S
H2S/S//Ni2+/Ni
Ni2+ → Ni
E(cell) = E(red) – E(ox)
= -0.23V –(+0.17V)
= -0.40V
Negative: Non-spontaneous. The reaction
will not occur; sulfur will not precipitate
Strongest & weakest
•Standard reduction potentials (ECell) are written
with reductants on the right
•Strongest reductant is the species on the right
with the most negative E
•Strongest oxidant is the species on the left with
the most positive E
•Weakest oxidant = strongest reductant = most –ve
•Weakest reductant = strongest oxidant = most +ve