IB Colourful Solutions in Chemistry

Standard level

Half-equations separate the processes of oxidation and reduction, showing the loss or gain of electrons.

Reactivity 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
Identifying oxidised and reduced species
Constructing half-equations
Constructing ionic equations

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^- → 2O^2-(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^- → 2O^2-(s)
add ----------------------------------------
2Mg(s) + O₂(g) → 2MgO(s)

Notice that the electrons now cancel out.

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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.

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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

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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.

half-equation 1: Cu(s) → Cu²⁺(aq) + 2e

half-equation 2: Ag⁺(aq) + 1e → Ag(s)

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:

half-equation 2: 2Ag⁺(aq) + 2e → 2Ag(s)

The half-equations can now be added to give:

2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq)

This is now the balanced ionic redox reaction.

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

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)

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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.

Example: Construct the half-equation for the reduction of the dichromate(VI) ion Cr₂O₇^2- 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.

Cr₂O₇^2- + 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:

Cr₂O₇^2- + 14H+ + 6e → 2Cr³⁺ + 7H₂O

The equation is now balanced electronically.

LHS = -2 + 14 + (-6) = +6
RHS = 2 x (+3) = +6

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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.

SO₃^2- + H₂O → SO₄^2- + 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.

SO₃^2- + H₂O → SO₄^2- + 2H⁺ +2e

Therefore sulfate(IV) ions are a reducing agent.

Balancing redox half-equations

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Worked examples

Q822-01 When the following oxidation reduction equation is balanced, what is the coefficient for H⁺(aq)?

Ag(s) + NO₃^-(aq) + H⁺(aq) → Ag⁺(aq) + NO(g) + H₂O(l)

Answer

The half-equation for silver is:

Ag(s) → Ag⁺(aq) + 1e

The half-equation for the nitrate ion requires four hydrogen ions to absorb two oxygen atoms. Three electrons are needed to equalise the charge on both sides:

NO₃^-(aq) + 4H⁺(aq) + 3e → NO(g) + 2H₂O(l)

To combine these two half-equations, the silver half-equation must first be multiplied by 3:

3Ag(s) → 3Ag⁺(aq) + 3e
NO₃^-(aq) + 4H⁺(aq) + 3e → NO(g) + 2H₂O(l)
------------------------------------------------------------
3Ag(s) + NO₃^-(aq) + 4H⁺(aq) → 3Ag⁺(aq) NO(g) + 2H₂O(l)
--------------------------------------------------------------------------------

Hence the coefficient for H⁺ = 4

Q822-02 What is the coefficient for H⁺ when the equation below is balanced?

_Pb(s) + _NO₃^-(aq) + H⁺(aq) → _Pb²⁺(aq) + _NO(g) + _H₂O(l)

Answer

The species that loses oxygen is the NO₃^-(aq) ion. This produces NO(g). The half-equation needs 4 hydrogen ions to absorb 2 oxygen atoms:

NO₃^-(aq) + 4H⁺(aq) + 3e → NO(g) + 2H₂O(l)

Q822-03 The unbalanced equation for the conversion of sulfur dioxide to sulfuric acid is given below:

SO₂ + H₂O → H₂SO₄

Which other species are used and on which side of the equation to balance it?

H⁺ and e^- on the left
H⁺ on the left and e^- on the right
H⁺ on the right and e^- on the left
H⁺ and e^- on the right

Answer

The species sulfur(IV) oxide turns into sulfate ions.

To do so it adds two oxygen atoms from the water molecules, leaving behind 4 hydrogen ions on the right hand side. To balance the equation electronically 2 electrons are also needed on the right hand side: Response D

SO₂ + 2H₂O → SO₄^2-(aq) + 4H⁺(aq) + 2e

Q822-04 When the following equation is balanced what is the coefficient for Ce⁴⁺?

SO₃^2- + H₂O + Ce⁴⁺ → SO₄^2- + H⁺ + Ce³⁺

Answer

The sulfate(IV) to sulfate(VI) half-equation:

SO₃^2- + H₂O → SO₄^2-(aq) + 2H⁺(aq) + 2e

The cerium half-equation:

Ce⁴⁺ + 1e → Ce³⁺

To match the electrons on both sides the cerium half-equation must be multiplied by 2 before the equations are added:

2Ce⁴⁺ + 2e → 2Ce³⁺
SO₃^2- + H₂O → SO₄^2-(aq) + 2H⁺(aq) + 2e
-----------------------------------------------------------
2Ce⁴⁺ + SO₃^2- + H₂O → 2Ce³⁺ + SO₄^2-(aq) + 2H⁺(aq)
------------------------------------------------------------------------------

The coefficient for Ce⁴⁺ = 2

Q822-05 The element M forms an oxyanion MO₃^- that is readily converted into M²⁺ in acid solution. Calculate the oxidation number of M in the MO₃^- ion. State and explain whether MO₃^- → M²⁺ is an oxidation or a reduction process. Write a balanced half-equation for this reaction.
Answer

the oxidation number of M in the MO₃^- ion = 5+, as each oxygen atom is (2-) and there are three of them = (6-), but the ion only shows (1-) therefore five of the negative charges have been cancelled out by 5+ charges.

MO₃^- → M²⁺ is a reduction, as the oxidation state of the M atom changes from +5 to +2, i.e. electrons are added.

To absorb three oxygen atoms requires 6 hydrogen ions. Three electrons are also needed on the left hand side to balance the equation's charges:

MO₃^- + 6H+ +3e → M²⁺ + 3H2O

Q822-06 What is the total of all of the coefficients in the balanced half-equation shown below?

H⁺(aq) + MnO₄^-(aq) + e → Mn²⁺(aq) + H₂O(l)

Answer

To absorb four oxygen atoms requires 8 hydrogen ions.

8H⁺(aq) + MnO₄^-(aq) + 5e → Mn²⁺(aq) + 4H₂O(l)

Five electrons are also needed on the left hand side to balance the equation's charges:

Q822-07 Iron in food in the form of Fe³⁺ reacts with ascorbic acid (vitamin C) C₆H₈O₆ to form dehydroascorbic acid, C₆H₆O₆
Write an ionic half-equation to show the conversion of ascorbic acid to dehydroascorbic acid in aqueous solution.
In the other half-equation the Fe³⁺(aq) ions are converted to Fe²⁺(aq) ions. Deduce the overall equation for the reaction between C₆H₈O₆ and Fe³⁺(aq) ions.
Answer

C₆H₈O₆ forms dehydroascorbic acid, C₆H₆O₆, in doing so it loses two hydrogen ions:

C₆H₈O₆(aq) → C₆H₆O₆(aq) + 2H⁺(aq) + 2e

Two electrons are also needed on the right hand side to balance the charges.

The second half-equation for iron is:

Fe³⁺(aq) + 1e → Fe²⁺(aq)

This second equation must be multiplied by 2, to equalise the electrons on both sides before it can be added to the first.

2Fe³⁺(aq) + 2e → 2Fe²⁺(aq)
C₆H₈O₆(aq) → C₆H₆O₆(aq) + 2H⁺(aq) + 2e
-------------------------------------------------------------
2Fe³⁺(aq) + C₆H₈O₆(aq) → 2Fe²⁺(aq) + C₆H₆O₆(aq) + 2H⁺(aq)
---------------------------------------------------------------------------------------

Q822-08 Redox equations may be balanced using changes in oxidation number. For the following redox equation calculate the oxidation number of manganese and carbon. Use these values to balance the equation.

MnO₄^-(aq) + C₂O₄^2-(aq) + H⁺(aq) → Mn²⁺(aq) + CO₂(g) +H₂O(l)

Answer

MnO₄^-(aq) has manganese in the oxidation state (VII).
Mn²⁺(aq) has manganese in the oxidation state (II)
C₂O₄^2-(aq) has carbon in the oxidation state (2 x C) + (4 x O) = -1. ∴ Carbon = +III
CO₂(g) has carbon in the oxidation state (IV)

Hence manganese changes from VII to II, i.e. it gains 5 electrons, and carbon changes from III to IV, i.e. it loses 1 electron per carbon atom.

C₂O₄^2-(aq) + 2e → 2CO₂(g)

MnO₄^-(aq) + 8H⁺(aq) → Mn²⁺(aq) + 4H₂O(l)

To balance the two half-equations the carbon half-equation must be multiplied by 5 and the manganese half-equation by 2.

2MnO₄^-(aq) + 5C₂O₄^2-(aq) + 16H⁺(aq) → 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l)

Q822-09 Hydrogen peroxide H₂O₂ can behave as both an oxidising agent and a reducing agent in acidic solution. When it behaves as an oxidising agent it forms hydroxide ions only. Write a balanced half-equation for this reaction.
Answer

From the information given: H₂O₂ → 2OH^-

We can see here that the atoms on both sides are already balanced, so it only remains to balance the electronic charges. On the left hand side there are two less negative charges than on the right hand side, therefore we must add two electrons to the left hand side: H₂O₂ + 2e → 2OH^-

Q822-10 Chlorate ions ClO₃^-, are oxidising agents in acid medium, forming chloride ions in the process. Write the equation for the half -reaction when chlorate ions behave in this way.
Answer

Write out the reactant and product ions:

ClO₃^- → Cl^-

We need to absorb the 3 oxygen atoms from the chlorate ion so this requires 6 hydrogen ions:

ClO₃^- + 6H+ → Cl^-

Now count up the charges on both sides. There are 5 plus charges on the left and 1 minus charge on the right. The difference, then, is 6 electrons that must be added to the left hand side to balance the half-equation electronically.

ClO₃^- + 6H⁺ + 6e^- → Cl^-

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