Al H2SO4

Al + H2SO4 = Al2(SO4)3 + SO2 + H2O

When the Sulfuric acid is very concentrated the Aluminium (Al H2SO4) reacts with the concentrated Sulfuric acid to produce Aluminium sulfate Sulfer di-oxide as well as Water. The reaction is a Redox reaction. So The reaction is-

Al + H2SO4 (conc) = Al2(SO4)3 + SO2 + H2O

Click to know what happens when the Sulfuric acid is diluted- Al + H2SO4 (Dill) = ?

Is the reaction is Redox:

Yes, The above reaction is a Oxidation-Reduction reaction because-

  1. The Oxidation number of the Al in the reactants is zero and Al in the Aluminium sulfate at left side is +3.
  2. On the other hand the Oxidation number of the Sulfer in the reactants is +6 but in the left side in Sulfer di-oxide the Oxidation number is reduced to +4. That means the reaction is a Redox reaction.

Aluminium and Sulfuric Acid (Al H2SO4):

Aluminium or aluminum is a chemical element with symbol Al and atomic number 13. It is a metal and very useful to our everyday life. It is a silvery-white, soft and nonmagnetic metal.

Sulfuric acid is a mineral acid with molecular formula H₂SO₄. It is a colorless, odorless, and syrupy liquid that is soluble in water. In this reaction it acts as an acidic as well as oxidizing agent. Its corrosiveness can be mainly ascribed to its strong acidic nature, and if at a high concentration its dehydrating and oxidizing properties.

Balancing the equation of the reaction:

After balancing the equation of the reaction we will get-

2 Al + 6 H2SO4 (conc) = Al2(SO4)3 + 3 SO2 + 6 H2O

As the reaction is a Redox reaction, we can use the ion-electron method to prof the balance of the equation.

Al H2SO4

Ion-Electron method

The skeleton reaction equation of the redox reaction between Aluminium and Sulfuric acid (Al H2SO4) is-

 Al + H2SO4 (conc) = Al2(SO4)3 + SO2 + H2O

In this case the H2SO4 will act as both acid as well as Oxidizing agent.


The Oxidizing agent: H2SO4
The Reducing agent: Al

Oxidation Half Reaction:

Al – 3e = Al3+ … … … … (1)

Reduction Half Reaction:

4H++ SO42- + 2e = SO2 + 2H2O … … … … (2)

Now, equation (1)x2 + (2)x3,

2Al – 6e = 2Al3+

12H+ 3SO42- + 6e = 3SO2 + 6H2O

2Al  +  12H+ 3SO42- = 2Al3+ + 3SO2 + 6H2O

Now adding necessary ions and radicals we get,

2Al  +  (12H+ 3SO42-) + 3SO42- = Al2(SO4)3+ 3SO2 + 6H2O

2Al  +  (12H2 + 6SO42-)= Al2(SO4)3+ 3SO2 + 6H2O

2 Al + 6 H2SO4(conc) = Al2(SO4)3 + 3 SO2 + 6 H2O

Click on the reaction to learn balancing redox reaction easily by ion electron method

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