Download Redox

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Metalloprotein wikipedia, lookup

Evolution of metal ions in biological systems wikipedia, lookup

Oxidation state wikipedia, lookup

Transcript
1.1.4 Redox
Answers
Oxidation numbers
We use the concept of oxidation numbers to help us analyse redox chemistry
at AS/A2. The oxidation number for a species is not the same as ionic
charge; the numbers are assigned to species within covalent compounds, and
ionic compounds. Oxidation numbers are given Roman Numerals (or regular
numbers with the charge before the number, for example +2).
1.
2.
3.
4.
5.
6.
7.
Here are some simple rules to follow when asked to assign oxidation
numbers:
The oxidation number of an uncombined element is always zero;
The oxidation number of a simple ion is the same as the charge on the ion;
The oxidation numbers of all the atoms in a molecule or complex ion must
add up to the total charge on that species;
For non metallic elements in an ion or molecule, we pretend that the most
electronegative element ‘takes’ the electrons to give it the minus charge;
Hydrogen usually has an oxidation number of +1 in compounds (except in
metal hydrides when it is –1);
Oxygen usually has an oxidation number of –2 in compounds (except in
peroxides when it is –1);
Fluorine is -1 in a compound, but the other elements in group 7 have
variable oxidation numbers.
Student activity
Work out the oxidation numbers of the elements shown in the given formulae:
Species
a) Cl2
b) NaCl
c) MgCl2
d) Mg
e) H2O
f) H2O2
g) NaH
h) OHi) MnO4j) MnO2
k) CO2
l) CO32m) SO42n) HSO4o) NH4+
p) NO3q) SO32-
Oxidation
number
Cl=0
Na=+1
Mg=+2
Mg=0
H=+1
H=+1
Na=+1
H=+1
Mn=+7
Mn=+4
C=+4
C=+4
S=+6
S=+6
N=-3
N=+5
S=+6
Oxidation
number
Oxidation
number
Cl=-1
Cl=-1
O=-2
O=-1
H=-1
O=-2
O=-2
O=-2
O=-2
O=-2
O=-2
O=-2
H=+1
O=-2
O=-2
-1-
H=+1
Identifying oxidation and reduction within a reaction
There are many ways of thinking about reduction and oxidation, but remember
they are opposites and they take place together. It is not possible to oxidise
one species without reducing another.
For most inorganic chemistry at this level, we will use the following definitions:
oxidation = loss of electrons
or
increase in oxidation number
reduction = gain of electrons
or
decrease in oxidation number
A good way to remember it is the mnemonic OILRIG (oxidation is loss,
reduction is gain). If a species loses electrons, then it will become less
negative (or more positive), hence the oxidation number increases.
When analysing organic reactions it may still be useful to use the following
definitions:
oxidation = loss of hydrogen
or
gain of oxygen
reduction = gain of hydrogen
or
loss of oxygen
Student activity
In the following equations, write the oxidation number of each species under
the equations; note that for CuO you need a number underneath the Cu and
another number underneath the O. Your teacher will show you how to do the
first one.
→ 2Fe
Fe2O3 + 3C
+ 3CO
In the exam always write oxidation number on the appropriate line.
Never write the multiple
1.
2Mg +
oxidation number 0
O2
0
→ 2MgO
+2 -2
2.
CuO +
+2 -2
H2
0
→ H2O
+1 -2
3.
Zn
0
H2SO4
+1+6-2
→ ZnSO4 +
+2+6-2
4.
CO +
+2 -2
½ O2
0
→ CO2
+4 -2
5.
2NO +
+2 -2
2CO
+2 -2
→ N2
0
+
-2-
+ Cu
0
H2
0
+ 2CO2
+4 -2
Student activity
a) For the same equations, draw arrows showing which change is the
oxidation and which is the reduction. Your teacher will show you how to do the
first one.
Fe2O3 + 3C
→ 2Fe
+ 3CO
1.
2Mg +
Mg oxidised
O2
→ 2MgO
O2 reduced
2.
CuO +
H2 oxidised
H2
→ H2O
Cu reduced
3.
Zn +
Zn oxidised
H2SO4
→ ZnSO4 + H2
H reduced
4.
CO +
C oxidised
½ O2
→ CO2
O2 reduced
5.
2NO +
C oxidised
2CO
→ N2
N reduced
+ Cu
+ 2CO2
b) In a redox reaction, there is a reducing agent (which becomes oxidised)
and an oxidising agent (which becomes reduced). Sometimes we call these
the reductant and oxidant. Identify the oxidising and reducing agents, by
writing R under the reducing agent and O under the oxidising agent (in the
examples above).
Student activity
Complete the following table.
What happens in terms of
electron transfer?
(does it lose or gain
electrons?)
What happens to the
oxidation number of this
agent?
(does the oxidation number
increase or decrease)
Oxidising agent
The oxidising agent will
gain electrons
The oxidation number will
decrease
Reducing agent
The reducing agent will
lose electrons
The oxidation number will
increase
-3-
Using Roman numerals to indicate the magnitude of the oxidation state
within a species
Some compounds have the same name but contain species with different
oxidation states. For example:
sulphuric (IV) acid
H2SO3
sulphuric (VI) acid
H2SO4
Here the sulphur has an oxidation state of +4 (IV) in H2SO3 and +6 (VI) in
H2SO4.
Student activity
Name the following species.
a) MnO2
Manganese (iv) oxide
b) MnO4Manganate (vii)
c) ClO3Chlorate (v)
d) SO32Sulphate (iv)
e) HSO4Hydrogen sulphate (vi)
How do metal and non-metal elements behave when taking part in
reactions?
Look at the equations below.
1.
a) Zn
+
H2SO4
→ ZnSO4
b) 2Fe
+
3Cl2
→ 2FeCl3
c) Mg
+
2HCl
→ MgCl2
d) 3Mg
+
N2
→ Mg3N2
e) Mg
+
I2
→ MgI2
+
H2
+
H2
a) What has happened to the oxidation numbers of the metal
elements? they always increase
b) Have they lost or gained electrons? lost electrons
2.
a) What has happened to the oxidation numbers of the non-metal
elements? they always decrease
b) Have they lost or gained electrons? gained
-4-
-5-