Question

Consider the reaction between copper(II) chloride and aluminum.Write the general skeleton reaction.Write the balanced reduction half reaction.Write the balanced oxidation half reaction.Write the balanced net ionic equation.Write the full balanced redox reaction.

Answer 1

The students' chemical reaction between aluminum and copper(II) chloride is a redox reaction involving the transfer of electrons, resulting in aluminum being oxidized and copper(II) chloride being reduced. After identifying, balancing, and adding the oxidation and reduction half-reactions, the full balanced redox reaction is 2Al(s) + 3Cu^2+(aq) → 2Al^3+(aq) + 3Cu(s).

Step-by-step explanation:

The chemical reaction between copper(II) chloride and aluminum is an example of a redox reaction, which involves the transfer of electrons between substances. In this reaction, aluminum is oxidized, and copper(II) chloride is reduced.

Steps to Balance the Redox Reaction

Identify and write the oxidation and reduction half-reactions separately.

Balance all elements in each half-reaction except for hydrogen and oxygen.

Balance oxygen atoms by adding H2O.

Balance hydrogen atoms by adding H+.

Balance the charges in each half-reaction by adding electrons (e−).

If necessary, multiply one or both half-reactions to have the same number of electrons transferred.

Add the half-reactions together and cancel out anything that is the same on both sides to get the net ionic equation.

Oxidation Half-Reaction:

Al(s) → Al3+(aq) + 3e−

Reduction Half-Reaction:

Cu2+(aq) + 2e− → Cu(s)

For the complete balanced redox reaction between aluminum and copper(II) chloride, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. After balancing the atoms and charges, we combine the half-reactions to get the fully balanced equation:

2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

Answer 2

• General skeleton reaction:

• Balanced reduction half reaction:

• Balanced oxidation half reaction:

• Balanced net ionic equation:

• Full balanced redox reaction:

`3CuCl_2+2Al\rightarrow2AlCl_3+3Cu`

Explanation:

1st) It is necessary to write the skeleton reaction (unbalanced reaction):

`CuCl_2+Al\rightarrow AlCl_3+Cu`

In the reaction, copper (II) chloride react with aluminum to produce aluminum chloride and copper.

2nd) We need to know the oxidation number of each atom in the reaction, then we can find the element that oxidized and the element that id reduced:

Oxidation numbers:

+2 -1 0 +3 -1 0

With the oxidation number we can see that Al goes from 0 to +3, so the aluminum atom oxidizes. And, the copper atom goes from +2 to 0, so it is reduced.

3rd) To write the balanced reduction half reaction, it is necessary to balance all elements except oxygen and hydrogen (in this case, there is no oxygen or hydrogen atoms).

`Cu^(+2)+2e^-\rightarrow Cu^0`

Then, it is important to balance the electrons in each half reaction, so in the Cu half reaction we have to multiply everything by 3:

4th) To write the balanced oxidation half reaction, we have to proceed like in the previous step:

`Al^0\rightarrow Al^(+3)+3e^-`

We have to balance the electrons here too, but in this case we have to multiply by 2:

Note: Electron balance is done to cancel out the electrons in both reactions, so we need to have the same number of electrons on each side of the half reactions.

5th) Now, we can write the balanced net ionic equation, without the electrons:

6th) Finally, we can write the full balanced redox reaction, including the other elements, in this case Cl: