Aluminum (Al H2SO4) reacts with the concentrated sulfuric acid lớn generate aluminum sulfate Sulfer di-oxide as well as water when the sulfuric acid is very concentrated. The reaction is a Redox reaction. So The reaction is-

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

Click to see what occurs when you use diluted sulfuric acid- Al + H2SO4=?

Is the reaction is Redox:

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

The Al oxidation number in the reactants is zero và Al on the left-hand side is + 3 in the aluminum sulfate.On the other hand, the Sulfer oxidation number in the reactants is + 6 but the oxidation number is lowered to lớn + 4 on the left side in Sulfer di-oxide. That means the reaction is a Redox reaction.

Bạn đang xem: 8 al + 15 h2so4 → 4 al2(so4)3 + 3 h2s + 12 h2o

Sulfuric acid and Aluminum (Al H2SO4):

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

Sulfuric acid has a molecular formula H2SO4 & a mineral acid. It is a colorless, odorless, và syrupy liquid that is soluble in water. In this reaction, it acts as an acidic as well as an oxidizing agent. Its corrosiveness can be attributed primarily to lớn its powerful acidic nature và its dehydration and oxidizing characteristics at an elevated concentration.

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 lớn prof the balance of the equation.

Ion-Electron method

The redox reaction equation between aluminum and sulfuric acid (Al H2SO4) in the skeleton form is –

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

♦The H2SO4 will function as both acid and oxidizing agent in this situation.♦


The Oxidizing agent: H2SO4The 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 & 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

Imagination is everything. It is the preview of life’s coming attractions.

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-Albert Einstein

Click on the reaction lớn learn to balance redox reaction easily by ion-electron method

KMnO4 + H2O2 + H2SO4 = O2 + MnSO4 + K2SO4 + H2O

FeSO4 + KMnO4 + H2SO4 = Fe2(SO4)3 + MnSO4 + H2O + K2SO4

NaCl + KMnO4 + H2SO4 = Cl2 + MnSO4 + Na2SO4 + K2SO4 + H2O

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