Redox and
electrode
potentials
, Redox reactions
o Reduction- gain of electron or decrease in oxidation number
o Oxidation- loss of electrons or increase in oxidation number
The oxidising agent contains the species being reduced and
takes electrons from the species being oxidised
the reducing agent contains the species that is oxidised and
adds electrons to the species being reduced.
, Redox equations using half equations
You can write equations for a redox reaction from the half-
equations for the reduction and oxidation reactions. The procedure
ensures that the number of electrons transferred are balanced.
Example
The redox reaction between hydrogen peroxide, H₂O₂, and alkaline
Cr3+ ions has the following half equations. Write the overall redox
equation.
H₂O₂ + 2e- 2OH- 1
Cr3+ + 8OH- CrO₄2- + 4H₂O + 3e- 2
1.Balance the electrons
3x 1 3H₂O₂ + 6e- 6OH-
2x 2 2Cr3+ + 16OH- 2CrO₄2- + 8H₂O + 6e-
2. Add and cancel electrons
3H₂O₂ + 6e- + 2Cr3+ + 16OH- 6OH- + 2CrO₄2- + 8H₂O + 6e-
3. Cancel any species
3H₂O₂ + 2Cr3+ + 16OH-- 6OH- + 2CrO₄2- + 8H₂O
10OH
3H₂O₂ + 2Cr3+ + 10OH- 2CrO₄2- + 8H₂O
, Construction redox equations using oxidation numbers
Oxidation numbers can be used to write & balance a redox equation.
Example
Sulfur, S, reacts with concentrated nitric acid, HNO₃, to form sulfuric
acid, H₂SO₄, nitrogen dioxide, NO₂, and water
Construct the overall equation for this redox reaction
1:Summarise the information provided
S + HNO₃ H₂SO₄ + NO₂ + H₂O
2:assign oxidation numbers to identify the atoms that change their
oxidation number
S + HNO₃ H₂SO₄ + NO₂ + H₂O
0 +1 +5 -2 +1 +6 -2 +4 -2 +1 -2
+6 -1
3:Balance only the species that contain the elements that have changed
oxidation number
to match an increase of +6 for sulfur, you need a total decrease of -6 from
nitrogen. So HNO₃ and NO₂ also need to be multiplied by 6 to give 6x-1=-6
S + 6HNO₃ H₂SO₄ + 6NO₂ + H₂O
4:balance any remaining atoms
As the oxidation numbers are now balanced, the hydrogen and oxygen
atoms are balanced by adding 2 in front of the H₂O
S + 6HNO₃ H₂SO₄ + 6NO₂ + 2H₂O
, Writing a half equation
Examples
Acidified MnO₄- ions are reduced to Mn2+ ions
MnO₄- + H+ Mn2+
Write the half equation for this reduction
Step 1: assign the oxidation numbers & the change in oxidation number
MnO₄- + H+ Mn2+
+7 -2 +1 +2
-5
Step 2: balance the electrons
A decrease in oxidation number of 5 requires 5e- on the LHS. If it was an
oxidation reaction, and so there was an increase in oxidation number of 5,
e- would be required on the RHS.
MnO₄- + 5e- + H+ + Mn2+
Step 3: balance any remaining atoms & predict any remaining species
There are 4 oxygen atoms & H+ on the LHS but not hydrogen or
oxygen atoms on the RHS. A likely product is H₂O. Add H₂O on the
RHS & balance.
MnO₄- + 5e- + H+ + Mn2+ + H₂O
electrode
potentials
, Redox reactions
o Reduction- gain of electron or decrease in oxidation number
o Oxidation- loss of electrons or increase in oxidation number
The oxidising agent contains the species being reduced and
takes electrons from the species being oxidised
the reducing agent contains the species that is oxidised and
adds electrons to the species being reduced.
, Redox equations using half equations
You can write equations for a redox reaction from the half-
equations for the reduction and oxidation reactions. The procedure
ensures that the number of electrons transferred are balanced.
Example
The redox reaction between hydrogen peroxide, H₂O₂, and alkaline
Cr3+ ions has the following half equations. Write the overall redox
equation.
H₂O₂ + 2e- 2OH- 1
Cr3+ + 8OH- CrO₄2- + 4H₂O + 3e- 2
1.Balance the electrons
3x 1 3H₂O₂ + 6e- 6OH-
2x 2 2Cr3+ + 16OH- 2CrO₄2- + 8H₂O + 6e-
2. Add and cancel electrons
3H₂O₂ + 6e- + 2Cr3+ + 16OH- 6OH- + 2CrO₄2- + 8H₂O + 6e-
3. Cancel any species
3H₂O₂ + 2Cr3+ + 16OH-- 6OH- + 2CrO₄2- + 8H₂O
10OH
3H₂O₂ + 2Cr3+ + 10OH- 2CrO₄2- + 8H₂O
, Construction redox equations using oxidation numbers
Oxidation numbers can be used to write & balance a redox equation.
Example
Sulfur, S, reacts with concentrated nitric acid, HNO₃, to form sulfuric
acid, H₂SO₄, nitrogen dioxide, NO₂, and water
Construct the overall equation for this redox reaction
1:Summarise the information provided
S + HNO₃ H₂SO₄ + NO₂ + H₂O
2:assign oxidation numbers to identify the atoms that change their
oxidation number
S + HNO₃ H₂SO₄ + NO₂ + H₂O
0 +1 +5 -2 +1 +6 -2 +4 -2 +1 -2
+6 -1
3:Balance only the species that contain the elements that have changed
oxidation number
to match an increase of +6 for sulfur, you need a total decrease of -6 from
nitrogen. So HNO₃ and NO₂ also need to be multiplied by 6 to give 6x-1=-6
S + 6HNO₃ H₂SO₄ + 6NO₂ + H₂O
4:balance any remaining atoms
As the oxidation numbers are now balanced, the hydrogen and oxygen
atoms are balanced by adding 2 in front of the H₂O
S + 6HNO₃ H₂SO₄ + 6NO₂ + 2H₂O
, Writing a half equation
Examples
Acidified MnO₄- ions are reduced to Mn2+ ions
MnO₄- + H+ Mn2+
Write the half equation for this reduction
Step 1: assign the oxidation numbers & the change in oxidation number
MnO₄- + H+ Mn2+
+7 -2 +1 +2
-5
Step 2: balance the electrons
A decrease in oxidation number of 5 requires 5e- on the LHS. If it was an
oxidation reaction, and so there was an increase in oxidation number of 5,
e- would be required on the RHS.
MnO₄- + 5e- + H+ + Mn2+
Step 3: balance any remaining atoms & predict any remaining species
There are 4 oxygen atoms & H+ on the LHS but not hydrogen or
oxygen atoms on the RHS. A likely product is H₂O. Add H₂O on the
RHS & balance.
MnO₄- + 5e- + H+ + Mn2+ + H₂O
