# Writing redox and half-reactions ```Writing redox and half-reactions
Redox reactions and half-reactions
Remember from Grade 11 that oxidation and reduction occur simultaneously in
a redox reaction. The reactions taking place in electrochemical cells are redox reactions.
Two questions should be asked to determine if a reaction is a redox reaction:

Is there a compound or atom being oxidised?

Is there a compound or atom being reduced?
If the answer to both of these questions is yes, then the reaction is a redox reaction. For
example, the reaction given below is a redox reaction:
2Fe3+(aq)+Sn2+(aq) → 2Fe2+(aq)+Sn4+(aq)
You can write a redox reaction as two half-reactions, one showing the reduction process,
and one showing the oxidation process. Fe3+ is gaining an electron to become Fe2+. Iron is
therefore being reduced and tin is the reducing agent (causing iron to be reduced).
The reduction half-reaction is:
Fe3+(aq) + e−→Fe2+(aq)
Sn2+ is losing two electrons to become Sn4+. Tin is therefore being oxidised and iron is
the oxidising agent (causing tin to be oxidised). The oxidation half-reaction is:
Sn2+(aq)→Sn4+(aq) + 2e−
Notice that in the overall reaction the reduction half-reaction is multiplied by two. This is so
that the number of electrons gained in the reduction half-reaction match the number of
electrons lost in the oxidation half-reaction.
Balancing redox reactions
Half-reactions can be used to balance redox reactions. We are going to use some worked
examples to help explain the method.
Example 1: Balancing redox reactions
Question
Chlorine gas oxidises Fe2+ ions to Fe3+ ions. In the process, chlorine is reduced to chloride
ions. Write a balanced equation for this reaction.
Write down the unbalanced oxidation half-reaction
Fe2+(aq) → Fe3+(aq)
Balance the number of atoms on both sides of the equation
There is one iron atom on the left and one on the right, so no additional atoms need to be
Once the atoms are balanced, check that the charges balance
The charge on the left of the equation is +2, but the charge on the right is +3. Therefore,
one electron must be added to the right hand side so that the charges balance. The halfreaction is now:
Fe2+(aq) → Fe3+(aq)+e−
Repeat steps 1 - 3 with the reduction half-reaction
The unbalanced reduction half-reaction is:
Cl2(g) → Cl−(aq)
The atoms don't balance, so we need to multiply the right hand side by two to fix this.
Cl2(g) → 2Cl−(aq)
Two electrons must be added to the left hand side to balance the charges.
Cl2(g)+2e− → 2Cl−(aq)
Compare the number of electrons in each equation
Multiply each half-reaction by a suitable number so that the number of electrons released
(oxidation) is equal to the number of electrons accepted (reduction).
oxidation half-reaction: &times;2: 2Fe2+(aq) → 2Fe3+(aq)+2e−
reduction half-reaction: &times;1: Cl2(g)+ 2e−→ 2Cl−(aq)
Combine the two half-reactions to get a final equation for the overall reaction
2Fe2+(aq)+Cl2(g) → 2Fe3+(aq)+2Cl−(aq)
Do a final check to make sure that the equation is balanced
We check the number of atoms and the charges and find that the equation is balanced.
Example 2: Balancing redox reactions in an acid medium
Question
The following reaction takes place in an acid medium:
Cr2O2−7(aq)+H2S(g) → Cr3+(aq)+S(s)
Write a balanced equation for this reaction.
Write down the unbalanced reduction half-reaction
In Cr2O2−7 chromium exists as Cr6+. It becomes Cr3+. Therefore electrons are gained, this
is the reduction half-reaction:
Cr2O2−7(aq) → Cr3+(aq)
Balance the number of atoms on both sides of the equation
We need to multiply the right side by two so that the number of Cr atoms will balance. In an
acid medium there are water molecules and H+ ions in the solution, so these can be used
to balance the equation.
To balance the oxygen atoms, we will need to add water molecules to the right hand side:
Cr2O2−7(aq) → 2Cr3+(aq)+7H2O(ℓ)
Now the oxygen atoms balance but the hydrogens don't. Because the reaction takes place
in an acid medium, we can add hydrogen ions to the left side.
Cr2O2−7(aq)+14H+(aq) → 2Cr3+(aq)+7H2O(ℓ)
Once the atoms are balanced, check that the charges balance
The charge on the left of the equation is (−2 + 14) = +12, but the charge on the right is +6.
Therefore, six electrons must be added to the left hand side so that the charges balance.
This makes sense as electrons are gained in the reduction half-reaction. The half-reaction is
now:
Cr2O2−7(aq)+14H+(aq)+6e− → 2Cr3+(aq)+7H2O(ℓ)
Repeat steps 1 - 3 with the oxidation half-reaction
The unbalanced oxidation half-reaction is:
H2S(g) → S(s)
However, you can ignore the H+ in H2S as they are accounted for by the acid medium and
the reduction half-reaction.
S2−(aq) → S(s)
The atoms balance, however the charges do not. Two electrons must be added to the right
hand side of the equation.
S2−(aq) → S(s)+2e−
Compare the number of electrons in each equation
Multiply each half-reaction by a suitable number so that the number of electrons released
(oxidation) is equal to the number of electrons accepted (reduction).
oxidation half-reaction: &times;3: 3S2−(aq) → 3S(s)+ 6e−
reduction half-reaction: &times;1: Cr2O2−7(aq)+14H+(aq)+ 6e−→ 2Cr3+(aq)+7H2O(ℓ)
Combine the two half-reactions to get a final equation for the overall reaction
Cr2O2−7(aq)+14H+(aq)+3S2−(aq) → 3S(s)+2Cr3+(aq)+7H2O(ℓ)
However, H2S(g) is the reactant, so it would be better to write:
Cr2O2−7(aq)+8H+(aq)+3H2S(g) → 3S(s)+2Cr3+(aq)+7H2O(ℓ)
Do a final check to make sure that the equation is balanced
We check the number of atoms and the charges and find that the equation is balanced.
Example 3: Balancing redox reactions in an alkaline medium
Question
The complex ion hexaamminecobalt(II) (Co(NH3)2+6) is oxidised by hydrogen peroxide to
form the hexaamminecobalt(III) ion (Co(NH3)3+6). Write a balanced equation for this
reaction.
Write down the unbalanced oxidation half-reaction
Ammonia (NH3) has an oxidation number of 0. Therefore, in Co(NH3)2+6 cobalt exists
as Co2+. In Co(NH3)3+6 cobalt exists as Co3+. Electrons are lost and this is the oxidation
half-reaction:
Co2+(aq) → Co3+(aq)
Balance the number of atoms on both sides of the equation
The number of atoms are the same on both sides.
Once the atoms are balanced, check that the charges balance
The charge on the left of the equation is +2, but the charge on the right is +3. One electron
must be added to the right hand side to balance the charges in the equation:
Co2+ → Co3++e−
Repeat steps 1 - 3 with the reduction half-reaction
Cobalt is oxidised by hydrogen peroxide, therefore hydrogen peroxide is reduced.
Reduction means a gain of electrons. The product of the reduction of H2O2 in an alkaline
medium is OH−:
H2O2(ℓ) → OH−(aq)
Next you need to balance the atoms:
H2O2(ℓ) → 2OH−(aq)
Then you need to balance the charges:
H2O2(ℓ)+2e− → 2OH−(aq)
Compare the number of electrons in each equation
Multiply each half-reaction by a suitable number so that the number of electrons released
(oxidation) is equal to the number of electrons accepted (reduction):
oxidation half-reaction: &times;2: 2Co2+(aq) →2Co3+(aq)+ 2e−
reduction half-reaction: &times;1: H2O2(ℓ)+ 2e−→ 2OH−(aq)
Combine the two half-reactions, and add in the spectator ions, to get a final equation
for the overall reaction
2Co(NH3)2+6(aq)+H2O2(ℓ) → 2Co(NH3)3+6(aq)+2OH−(aq)
Do a final check to make sure that the equation is balanced
We check the number of atoms and the charges and find that the equation is balanced.
Exercise 1: Balancing redox reactions
Problem 1:
Balance the following equations:
1. HNO3(ℓ)+PbS(s) → PbSO4(s)+NO2(g)+H2O(ℓ)
2. NaI(aq)+Fe2(SO4)3(aq) → I2(s)+FeSO4(aq)+Na2SO4(aq)
Practise more questions like this
1. Pb2+ is a spectator ion.
The unbalanced reduction half-reaction is:
NO−3(aq) → NO2(g)
Add water molecules to the right and H+ ions to the left (acid medium) to balance the
oxygen and hydrogen atoms:
NO−3(aq)+2H+(aq) → NO2(g)+H2O(ℓ)
Balance the charge by adding an electron to the left (this makes sense as this is the
reduction half-reaction, and N5+ → N4+):
NO−3(aq)+2H+(aq)+e− → NO2(g)+H2O(ℓ)
The unbalanced oxidation half-reaction is:
S2−(aq) → SO2−4(aq)
Add water molecules to the left and H+ ions to the right to balance the oxygen and
hydrogen atoms:
S2−(aq)+4H2O(ℓ) → SO2−4(aq)+8H+(aq)
Balance the charge by adding eight electrons to the right (this makes sense as this is
the oxidation half-reaction, and S2− → S6+):
S2−(aq)+4H2O(ℓ) → SO2−4(aq)+8H+(aq)+8e−
We multiply the reduction half-reaction reaction by 8 to balance the number of
electrons in both equations:
8NO−3(aq)+16H+(aq)+8e− → 8NO2(g)+8H2O(ℓ)
Adding the two equations together gives the balanced equation (electrons are equal on
both sides and can be removed):
8NO−3(aq)+16H+(aq)+S2−(aq)+4H2O(ℓ) → SO2−4(aq)+8H+(aq)+8NO2(g)+8H2O(
ℓ)
Putting the spectator ions back into the equation and removing any extra H+ ions and
water molecules we get:
8HNO3(ℓ)+PbS(s) → PbSO4(s)+8NO2(g)+4H2O(ℓ)
2. Na+ and SO2−4 are spectator ions.
Fe has the oxidation number +3 in Fe2(SO4)3, and +2 in FeSO4. This must be the
reduction half-reaction:
Fe3+(aq) → Fe2+(aq)
The atoms are balanced. Balance the charge by adding an electron to the left:
Fe3+(aq)+e− → Fe2+(aq)
The unbalanced oxidation half-reaction is:
I−(aq) → I2(s)
There must be 2 I− to balance the atoms:
2I−(aq) → I2(s)
Balance the charge by adding two electrons to the right:
2I−(aq) → I2(s)+2e−
We multiply the reduction half-reaction by 2 to balance the number of electrons in both
equations:
2Fe3+(aq)+2e− → 2Fe2+(aq)
Adding the two equations together and removing the electrons gives:
2I−(aq)+2Fe3+(aq) → I2(s)+2Fe2+(aq)
Putting the spectator ions back into the equation we get:
2NaI(aq)+Fe2(SO4)3(aq) → I2(s)+2FeSO4(aq)+Na2SO4
Problem 2:
Permanganate(VII) ions ( MnO−4 ) oxidise hydrogen peroxide ( H2O2 ) to oxygen gas. The
reaction is done in an acid medium. During the reaction, the permanganate(VII) ions are
reduced to manganese(II) ions (Mn2+). Write a balanced equation for the reaction.
Practise more questions like this
The unbalanced reduction half-reaction is:
MnO−4(aq) → Mn2+(aq)
As this is in an acid medium, we can add water molecules to the right and H+ ions to the left
to balance the oxygen and hydrogen atoms:
MnO−4(aq)+8H+ → Mn2+(aq)+4H2O(ℓ)
Balance the charge by adding five electrons to the left (this makes sense as this is the
reduction half-reaction, and Mn7+ → Mn2+):
MnO−4(aq)+8H+(aq)+5e− → Mn2+(aq)+4H2O(ℓ)
The unbalanced oxidation half-reaction is:
H2O2(ℓ) → O2(g)
Add H+ ions to the right to balance the hydrogen atoms:
H2O2(ℓ) → O2(g)+2H+(aq)
Balance the charge by adding two electrons to the right (this makes sense as this is the
oxidation half-reaction, and 2O− → O2):
H2O2(ℓ) → O2(g)+2H+(aq)+2e−
We multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5 to balance
the number of electrons in both equations:
2MnO−4(aq)+16H+(aq)+10e− → 2Mn2+(aq)+8H2O(ℓ)
5H2O2(ℓ) → 5O2(g)+10H+(aq)+10e−
Adding the two equations together gives the balanced equation (electrons are equal on both
sides and can be removed):
2MnO−4(aq)+16H+(aq)+5H2O2(ℓ) → 5O2(g)+10H+(aq)+2Mn2+(aq)+8H2O(ℓ)
Removing any extra H+ ions we get:
2MnO−4(aq)+6H+(aq)+5H2O2(ℓ) → 5O2(g)+2Mn2+(aq)+8H2O(ℓ)
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