Redox_equations[1].

Report
Redox equations
•Be able to write redox half equation
•Identify the reducing agent and the
oxidising agent in an equation
Redox – what?
Oxidation and reduction
happen at the same time
There is no net gain or loss of
electrons.
e
You can’t just create
them or destroy them!
Example: the thermit reaction
Fe2O3 + Al  Fe + Al2O3
Fe3+
Al0
Fe0
Al3+
Fe3+ + 3e-  Fe
Reduction
Al  Al2O3 + 3e-
Oxidation
Ionic half-equations
Remember:
OIL - Oxidation Is Loss of electrons
RIG - Reduction Is Gain of electrons
And often... (as a quick and simple way to tell):
Oxidation is gain in oxygen or loss of hydrogen
Reduction is loss of oxygen or gain of hydrogen
Oxidation state
Oxidation
Reduction
Oxidation and
reduction can be
seen as movement
up or down a scale
of oxidation states
+7
+6
+5
+4
+3
+2
+1
0
-1
-2
-3
-4
-5
-6
-7
Rules:
Oxygen= -2
Hydrogen = +1
Group 1= +1
Group 2= +2
Group 7= -1
Group 6= -2
Group 4 & 5 (and transition elements) can
change but groups 1,2,6 & 7 can't!
Element= 0
Redox in presence of acid
Example: a past exam question
a) Identify, as oxidation or reduction,
the formation of NO2 from NO3- in the
presence of H+ and deduce the halfequation for the reaction.
NO3
+5
 NO2
+4
Reduction
Steps to take
1. Write a balanced equation for the species
NO3-  NO2
2. Work out “before and after” oxidation states
+5
3. Balance oxidation states with electrons
4. If all the charges don’t balance,
add H+ ions to one of the sides
to balance them
5. If the equation still doesn’t
balance, add enough water
to one side so it balances
+4
NO3- + e-  NO2
NO3- + e- + 2H+  NO2
NO3- + e- + 2H+  NO2 + H2O
Beautiful!
Give these a go
Balance the half equations
Na

Na+
Fe2+

Fe3+
I2

I¯
C2O42-

CO2
H2O2

O2
H2O2

H2O
NO3-

NO
NO2

NO3-
SO42-

SO2
All good?
Na

Na+ +
e-
Fe2+

Fe3+ +
e-

2I¯
C2O42-

2CO2 +
H2O2

I2
+
2e-
H2O2 + 2H+ + 2e- 
O2
2e-
+ 2H+ + 2e-
2H2O
NO3- + 4H+ + 3e- 
NO +
2H2O

NO3- +
2H+ + e-
SO42- + 4H+ + 2e 
SO2 +
2H2O
NO2 + H2O
Combining half equations
Just a mashing together
of two half-equations!
... followed by some
satisfying cancelling-out.
Back to our exam question
b) Deduce the overall equation for the
reaction of copper with NO3- in acidic
conditions to give Cu2+.
NO3- + e- + 2H+  NO2 + H2O
Cu  Cu2+ + 2e-
Reduction
Oxidation
Now what?
Steps to take to combine equations
1. Multiply the equations so that the number of
electrons in each is the same
2+
Cu  Cu + 2e-
X2
2NO3- + 2e- + 4H+  2NO2 + 2H2O
2. Add the two equations and cancel out the electrons
on either side of the equation
Cu + 2NO3- + 2e- + 4H+  Cu2+ + 2e- + 2NO2 + 2H2O
3. If necessary, cancel out any other species which
appear on both sides of the equation
Cu + 2NO3- + 4H+  Cu2+ + 2NO2 + 2H2O
Give these a go
1. Fe2+ ions are oxidised to Fe3+ ions by ClO3- ions in
acidic conditions. The ClO3- ions are reduced to Clions. Write the overall reaction.
Fe2+  Fe3+ + e-
ClO3- + 6e- + 6H+  Cl- + 3H2O
6Fe2+ + ClO3- + 6H+  Fe3+ + Cl- + 3H2O
2. Write an overall reaction for MnO4- reducing H2O2
to O2 and creating Mn2+ .
2MnO4¯ + 5H2O2 + 6H+  2Mn2+ + 5O2 + 8H2O
Complete the half-equation
Na  Na+
Na  Na++ e-
Complete the half-equation
Pb4+  Pb2+
Pb4+ + 2e-  Pb2+
Complete the half-equation
H2  H+
H2  2H+ + 2e-
Complete the half-equation
Cr2O72-  Cr3+
Cr2O72- + 6e-  2Cr3+
What’s wrong with this equation?
Ce3+ + e-  Ce4+
Electron is on the wrong side!
What’s wrong with this equation?
Mg + 2H+ + e-  Mg+ + H2 + e-
Should be Mg2+
Electrons should be cancelled out

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