Synthesis of Alum

Experiment 2 Online Tutorial >> Synthesis of Alum >> Tutor 1

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Experiment 2 - Synthesis of Alum

Tutor 1 - Balancing Redox Reactions

Reactions in which electrons are transferred from one reactant to another are called oxidation-reduction reactions. The reaction of Al(s) in aqueous KOH is an example of an oxidation-reduction or redox reaction. The Al metal is oxidized to aluminum ion with an oxidation number of +3 and the hydrogen in KOH or in water is reduced from an oxidation number of +1 to zero in hydrogen gas.

??Al_(s) + ??KOH_(aq) + ??H₂O_(l) => ??Al(OH)₄^-_(aq) + ??K⁺_(aq) + ??H₂_(g)

or, omitting the K⁺, spectator ions:

??Al_(s) + ??OH^-_(aq) + ??H₂O_(l) => ??Al(OH)₄^-_(aq) + ??H₂_(g)

One method that can be used to balance redox reactions is called the Half-Reaction method.

The Half Reaction Method

Redox reactions often occur in aqueous acid or base solutions; our reaction occurs in an aqueous base solution. Below is the Half-Reaction method for balancing oxidation-reduction reactions occuring in basic solutions.

To balance a reaction involving oxidation and reduction, we must first identify which species is oxidized and which is reduced. We then break the net reaction into two imaginary half reactions, one of which involves only oxidation and the other only reduction, and apply the Half-Reaction method to each half reaction. A reaction is balanced when the number of atoms of each element is the same on both sides, and the net charge is the same on both sides.

Here are the steps we will follow in balancing this redox reaction:

Step 1:

In the first step, we break the reaction into half reactions, one involving oxidation and the other reduction. Please identify which reaction below is the oxidation and which is the reduction:

Hint

Al → Al(OH)4-

Select

Oxidation half reaction

Reduction half reaction

OH- → H2

Select

Oxidation half reaction

Reduction half reaction

Good Job!

That's not quite right.

Hint:

In this half reaction, the Al loses three electons to form an atom with a formal charge of 3. This loss of electrons is an oxidation.

Hint:

In this half reaction, the OH- gains electons producing H2. This gain of electrons is a reduction.

Step 2:

In the next step, we balance the number of atoms on each side of the half reaction. For each half reaction below, balance the number of atoms that are oxidized or reduced:

Hint

Select

→

Select

Al(OH)4-

Select

OH-

→

Select

Good Job!

That's not quite right.

Hint:

To balance this half reaction, we first must determine which atom is oxidized?

get next hint

Hint:

The Al is oxidized in this half reaction. To balance the number of Al atoms, how many Al atoms appear on each side of the reaction?

get previous hint

get next hint

Hint:

Since there is only one Al on each side of the equation, the atoms are already balanced. Enter one for Al.

get previous hint

Hint:

The Al is oxidized in this half reaction. To balance the number of Al atoms, how many Al atoms appear on each side of the reaction?

get next hint

Hint:

Since there is only one Al on each side of the equation, the atoms are already balanced. Enter one for Al(OH)4-.

get previous hint

Hint:

To balance this half reaction, we first must determine which atom is reduced?

get next hint

Hint:

H is reduced in this half reaction. To balance the number of H atoms on the left, look at how many H atoms appear on the right side of the reaction.

get previous hint

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Hint:

Since there are two H atoms on the right, enter 2 for the number of OH- 's on the left. This will balance the H atoms in this half reaction.

get previous hint

Hint:

H is reduced in this half reaction. To balance the number of H atoms on the right, look at how many H atoms appear on the left side of the reaction.

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Hint:

Since 2 was entered for the coefficient of OH^- atoms and we are starting with 2 H atoms on the right, enter 1 to balance the H atoms in this half reaction.

get previous hint

Step 3:

In the next step, we balance the oxygen atoms by adding H2O to the half reactions. Please select the number of H2O molecules needed to balance oxygen atoms in the half reactions:

Hint

Al +

Select

H2O → Al(OH)4-

2 OH- →

Select

H2O + H2

Good Job!

That's not quite right.

Hint:

To balance the number of oxgen atoms in this half reaction, we will add one water molecule for each O atom on the right side of the reaction. How many oxygen atoms appear on the right side of the reaction?

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Hint:

There are 4 O atoms on the right side of the reaction, so 4 water molecules must be added to balance the half reaction.

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Hint:

To balance the number of oxgen atoms in the second half reaction, we will add one water molecule for each O atom on the left side of the reaction. How many oxygen atoms appear on the left side of the reaction?

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Hint:

There are 2 O atoms on the left side of the reaction, so 2 water molecules must be added to balance the half reaction.

get previous hint

Step 4:

In the fourth step, we balance the H atoms by adding H+ to one side of half reaction. Please select the number of H⁺ needed to balance H atoms in the half reactions:

Hint

Al + 4H2O → Al(OH)4- +

Select

H+

2OH- +

Select

H+ → H2 + 2H2O

Good Job!

That's not quite right.

Hint:

To balance the number of H atoms in this half reaction, we will add additional H+ to the right side of the reaction. How many H atoms appear on the left and right sides of the reaction?

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Hint:

There are 8 H atoms on the left and 4 on the right side of the reaction, so 4 H+ will need be added on the right to balance the H atoms.

get previous hint

Hint:

To balance the number of H atoms in this half reaction, we will add additional H+ to the left side of the reaction. How many H atoms appear on the left and right sides of the reaction?

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Hint:

There are 6 H atoms on the right and only 2 on the left side of the reaction, so 4 H+ will need to be added on the left to balance the H atoms.

get previous hint

We balanced the equation first with H+. Since we are dealing with a basic solution, the answer must be converted to one in which OH- is used instead. This is done by adding to each side of the equation the number of hydroxide ions equal to the number of H⁺ ions appearing in the equation.

Hint

For the oxidation half reaction:

Al + 4H2O

→ Al(OH)4-

4H+

+ 4OH-

Al + 4H2O

+ 4OH-

→ Al(OH)4-

4H+

+ 4OH-

Realizing that 4H+ + 4 OH- = 4H2O and cancelling H2O on each side gives the result:

Al + 4OH-

→ Al(OH)4-

Add hydroxide ions to complete the reduction half reaction:

2OH- + 4H+

→ H2 + 2H2O

Select

OH-

Select

OH-

2OH- + 4H+ + 4OH-

→ H2 + 2H2O + 4OH-

Combine H+ + OH- to form water molecules and cancel OH- and H2O on each side to simplify the reduction half reaction equation:

Select

2H2O

2OH^-

→

Select

2H2O

2OH^-

Select

2H2O

2OH^-

Good Job!

That's not quite right.

Hint:

To convert the reaction to a basic environment, we must add an OH- ion for each H+. How many H+ ions appear on the left side of the reaction?

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Hint:

There are 4H+ ion on the left side of the reaction, so 4OH- ions must be added.

get previous hint

Hint:

To convert the reaction to a basic environment, we must add to each side of the equation, an OH- ion for each H+ ion appearing in the equation. How many H+ ions appear in the reaction?

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Hint:

We added 4 OH- to the left side of the equation, so we must add the same number of OH- to the right side.

get previous hint

Hint:

As in oxidation half reaction example, to balance the equation start by combining H+ and OH- to make H2O.

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Hint:

Combining 4H+ and 4OH- will yield 4H2O on the left. This will cancel 2H2O on the right leaving 2H2O remaining on the left.

get previous hint

Hint:

The H2 term on the right side of the reaction cannot be further simplified or cancelled. Include this term as is in the final reaction.

Hint:

The 4OH- terms on the right side of the equation can cancel the 2OH- on the left side of the equation, leaving 2OH- remaining.

Step 5:

In the next step, we balance the charge of each half reaction by adding electrons. Select the number of electrons to add to balance the charge of each half reaction.

Hint

Al + 4OH^- → Al(OH)₄^- +

Select

Oxidation half reaction

Reduction half reaction

2H2O +

Select

e- → H2 + 2OH-

Select

Oxidation half reaction

Reduction half reaction

Good Job!

That's not quite right.

Hint:

To balance the charge on the first reaction, start by counting the number of ions on both sides of the reaction.

get next hint

Hint:

There are 4 negative charges on the left side of the reaction from the 4OH- ions and 1 negative charge on the right side from the Al(OH)4- ion. To balance this, we will need to add 3 e- to the right side of the reaction.

get previous hint

Hint:

In this half reaction, the Al loses three electrons to form an atom with a formal charge of 3. This loss of elections is an oxidation.

Hint:

To balance the charge on the second reaction, start by counting the number of ions on both sides of the reaction.

get next hint

Hint:

There are no negative charges on the left side of the reaction and 2 negative charges on the right side from the 2OH- ions. To balance this, we will need to add 2 e- to the left side of the reaction.

get previous hint

Hint:

In this half reaction, the H+ in the water molecule gains electrons producing H2. This gain of electons is a reduction.

Step 6:

In the last step, each half reaction must be completely balanced (the same numbers of atoms and charge on each side). To do this, we will multiply the half reactions by the number of electrons in the other half reaction so that the number of electons on each side of the total reaction will cancel. Then we add the two half reactions together.

Hint

2Al + 8OH-

→

2Al(OH)4- + 6e-

Oxidation half reaction

6H2O + 6e-

→

3H2 + 6OH-

Reduction half reaction

2Al + 8OH- + 6H2O + 6e-

→

2Al(OH)4- + 6e- + 3H2 + 6OH-

Simplify the above equation:

Select

Al +

Select

OH- +

Select

H2O →

Select

Al(OH)4- +

Select

Simplify the above equation:

2Al(s) + 2KOH(aq) + 6H2O(l) → 2AL(OH)4-(aq) + 2K+(aq) + 3H2(g)

Good Job!

That's not quite right.

Hint:

The Al term on the left side cannot be cancelled or further reduced. The numbger of Al atoms in this reaction is 2.

Hint:

OH- appears on both the left and right side of the equation. Can this term be reduced or cancelled?

get next hint

Hint:

6 OH- to the right side of the equation can cancel 6 OH- to the left, leaving 2 OH- remaining.

get previous hint

Hint:

The H2O term on the left cannot be cancelled or further reduced. The number of H2O molecules in this reaction is 6.

Hint:

The Al(OH)4- term on the right cannot be cancelled or further reduced. The number of Al(OH)4- molecules in this reaction is 2.

Hint:

The H2 term on the right side of the equation cannot be further simplified or cancelled. The number of H2 molecules in this reaction is 3.

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