Half-equations separate the processes of oxidation and reduction, showing the loss or gain of electrons.
Syllabus reference R3.2.2Reactivity 3.2.2 - Half-equations separate the processes of oxidation and reduction, showing the loss or gain of electrons.
- Deduce redox half-equations and equations in acidic or neutral solutions.
Guidance
Tools and links
- Tool 1, Inquiry 2 - Why are some redox titrations described as “self-indicating”?
Half-equations
These show the two processes that happen during a redox reaction. There is an oxidation reaction that involves loss of electrons, and a reduction reaction that involves gaining electrons.
Redox equation
2Mg(s) + O2(g) → 2MgO(s)
Oxidation half-equation
Mg(s) → Mg2+ + 2e-
Reduction half-equation
O2(g) + 4e- → 2O2-(s)
The sum of the oxidation half-equation and the reduction half equation must give the stoichiometric equation for the redox reaction. Before the two half equations are added together, the number of electrons in each half-equation must be made the same by judicious multiplication of one (or both) of the equations.
Oxidation half-equation x 2
2Mg(s) → 2Mg2+ + 4e-
Adding up both half-equations
2Mg(s) → 2Mg2+ + 4e-
O2(g) + 4e- → 2O2-(s)
add ----------------------------------------
2Mg(s) + O2(g) → 2MgO(s)
Notice that the electrons now cancel out.
Identifying oxidised and reduced species
We know that there is one reduction and one oxidation reaction involved. To identify the species that gets oxidised we must identify the oxidation state of each atom in the equation. The species that gets oxidised contains the element that increases its oxidation state, i.e. gets more positive or less negative.
The same argument can be applied to the reduced species, only this time it contains the element whose oxidation state (number) decreases (gets more negative, less positive)
Identify the half-equations in the redox equation shown below.
2Mg(s) + O2(g) → 2MgO(s)
Oxidation states:
- Mg = 0
- O (in O2) = 0
- Mg in MgO = +2
- O in MgO = -2
We can see that the magnesium's oxidation state changes from zero to +2, hence it is oxidised. We can see that the oxygen atoms change oxidation state from zero to -2, they are reduced.
Constructing half-equations
The rules for constructing half-equations in acidic or neutral solution are simple:
- We can add hydrogen ions, H+
- We can add water molecules, H2O
- We can add electrons
Once the oxidised (or reduced) species is identified, these three rules are followed to balance the half-equations for atoms and charge.
Construct half-equations for the redox reaction shown below:
Cu(s) + 4HNO3(l) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
Oxidation states left hand side:
- Cu = 0
- H (in HNO3) = +1
- N (in HNO3) = +5
- O (in HNO3) = -2
Oxidation states right hand side:
- Cu (in Cu(NO3)2) = +2
- N (in Cu(NO3)2) = +5
- O (in Cu(NO3)2) = -2
- N (in NO2) = +4
- O (in NO2) = -2
- H (in H2O) = +1
- O (in H2O) = -2
On inspection we can see that the copper changes from oxidation state zero to +2, i.e. it is oxidised. On inspection, we can see that the nitrogen atoms in nitric acid change from +5 to +4 in nitrogen(IV) oxide, NO2. Otherwise, there are no other electron transfers.
We can now construct two half-equations, one for the copper and the other for nitric acid to nitrogen(IV) oxide.
Oxidation of copper
Cu → Cu2+ + 2e-
Reduction of nitric acid
HNO3 → NO2
We can see that this half-equation is not balanced for either atoms or charge. So we must balance the oxygen atoms by adding water to the right hand side, and then balance the hydrogen atoms by adding hydrogen ions to the left hand side:
HNO3 + H+ → NO2 + H2O
Finally, the charge must be balanced by adding an electron to the left hand side:
Nitric acid half-equation
HNO3 + H+ + 1e- → NO2 + H2O
Constructing ionic equations
The full equation for a reaction can be constructed from the two half-equations by adding them together. However, before two half-equations can be added together the number of electrons in each half-equation must be made the same.
It is important to note here that only reduction half-equations can be added to oxidation half-equations and vice versa. It would be impossible to add two reduction half-equations together as we would end up with electrons that have no home to go to.
For the following two half-equations we can construct the full (ionic) equation by simple addition. We add each left hand side together, then we add each right hand side together, as if it were a simple maths sum:
Cu2+(aq) + 2e → Cu(s)
Zn(s) → Zn2+(aq)
+ 2e
-----------------------------------------
Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s)
The electrons that appear on either side of the reaction arrow cancel out, leaving only the ions that react.
The above reaction is uncomplicated, as the number of electrons on each side of the reaction cancel out. The following example shows a situation in which the electrons do not initially cancel out, in which one half-equation has to be manipulated to equalise the electrons.
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half-equation 1: Cu(s)
→ Cu2+(aq)
+ 2e
|
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half-equation 2: Ag+(aq)
+ 1e → Ag(s)
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It should be seen that if we simply add these equations together there will be one extra electron remaining on the right hand side. Before the equations can be added, the second equation must be multiplied through by 2 to give:
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half-equation 2: 2Ag+(aq)
+ 2e → 2Ag(s)
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The half-equations can now be added to give:
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2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)
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This is now the balanced ionic redox reaction.
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Example: Given the two following half-equations, construct the full ionic equation for the redox reaction: Al(s) → Al3+ + 3e Fe3+ + 3e → Fe We can see that the number of electrons is the same on both sides, so the half-equations can be added together directly: Al + Fe3+ → Al3+ + Fe |
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Example: Given the two following half-equations, construct the full ionic equation for the redox reaction: Mg(s) → Mg2+(aq) + 2e Ag+(aq) + 1e → Ag(s) The number of electrons is NOT the same on both sides, so the half-equations can NOT be added together directly. First the silver half-equation must be doubled to make the electrons the same in both equations: 2Ag+(aq) + 2e → 2Ag(s) Now they can be added together and the electrons cancelled out: Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2Ag(s) |
Reactions in acidic solution
When potassium manganate(VII) acts as an oxidising agent, the manganate(VII) ion is reduced in (acidic solution) according to the half-equation:
MnO4- + 8H+ + 5e → Mn2+ + 4H2O
The eight hydrogen ions are needed to absorb the four oxygen atoms from the manganate(VII) ion. Notice that the half-equation is balanced electronically. On the left hand side there are 8 positive charges and 6 negative charges making an overall 2+ chage. On the right hand side, there is also a 2+ charge, the equation is balanced both in terms of atoms and electrical charges.
To find the number hydrogen ions required is usually a matter of seeing how many water molecules must be created from any oxygen in the oxyions.
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Example: Construct the half-equation for the reduction of the dichromate(VI) ion Cr2O72- in acidic solution, if the product is the Cr3+ ion. The dichromate(VI) ion contains 7 oxygen atoms wheras the product chromium (III) ion contains none. This means that all of the oxygen atoms must go to produce water. This requires the use of 14 hydrogen ions. Cr2O72- + 14H+ → 2Cr3+ + 7H2O However, the equation is not balanced electronically. The two chromium atoms in the dichromate ion are in the (VI) oxidation state and they both end up in the (III) oxidation state; they have absorbed between them 6 electrons. These must be added to the equation: Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O The equation is now balanced electronically.
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Use of water
Sulfate(IV) ions can be oxidised to form sulfate(VI) in acidic solution. In this case the sulfate(IV) must pick up an oxygen from the aqueous medium. It used a water molecule.
SO32- + H2O → SO42- + 2H+
The equation needs to be balanced in terms of charges by adding electrons to the appropriate side. One sulfur atom changes from the (IV) oxidation state to the (VI) oxidation state, thus releasing two electrons. The electrons must be added to the right hand side.
SO32- + H2O → SO42- + 2H+ +2e
Therefore sulfate(IV) ions are a reducing agent.
Balancing redox half-equations