# For the balanced redox reaction 2 Al(s) + 3 Br2(g) → Al2Br6(s), determine the number of electrons transferred in the overall reaction.

## each Al xfers three electrons, six total donated

Br each atom accepts one, for a total of six accepted.

## Why did the electron go to the party alone? Because it couldn't find a positive ion to take its date!

In the given balanced redox reaction, 2 Al(s) + 3 Br2(g) → Al2Br6(s), the number of electrons transferred can be determined by looking at the change in oxidation states.

The oxidation state of aluminum (Al) changes from 0 to +3, so it loses 3 electrons.

The oxidation state of bromine (Br) changes from 0 to -1, so it gains 1 electron per Br atom. Since there are 3 Br2 molecules in the reaction, a total of 6 electrons are gained.

Therefore, the overall number of electrons transferred in the reaction is 3 + 6 = 9.

## To determine the number of electrons transferred in the overall reaction, we need to examine the changes in oxidation numbers for each element involved. In this balanced redox reaction:

2 Al(s) + 3 Br2(g) → Al2Br6(s)

We can assign oxidation numbers to each element:

Al(s) : 0
Br2(g) : 0
Al2Br6(s): 0

The oxidation state of bromine in Br2(g) is 0 because it is in its elemental form.

The oxidation state of aluminum in Al(s) is 0 because it is in its elemental form.

The oxidation state of aluminum in Al2Br6(s) is +3 because each aluminum ion (Al3+) has a charge of +3.

The oxidation state of bromine in Al2Br6(s) is -1 because each bromide ion (Br-) has a charge of -1.

By comparing the changes in oxidation numbers, we can determine the number of electrons transferred.

From aluminum's oxidation state of 0 to +3, it gains electrons, so 3 electrons are transferred.

From bromine's oxidation state of 0 to -1, it loses electrons, so 3 electrons are transferred for each Br2 molecule. Since there are three Br2 molecules, the total number of electrons transferred from bromine is 3 x 3 = 9 electrons.

Therefore, in the overall reaction, the number of electrons transferred is 3 + 9 = 12 electrons.

## To determine the number of electrons transferred in a redox reaction, you need to identify the changes in oxidation states for each element involved. The change in oxidation state indicates the number of electrons transferred.

In this reaction, aluminum (Al) is oxidized from an oxidation state of 0 to +3, while bromine (Br) is reduced from an oxidation state of 0 to -1.

In the balanced equation:
2 Al(s) + 3 Br2(g) → Al2Br6(s)

- Each aluminum atom loses 3 electrons, as its oxidation state increases from 0 to +3.
- Each bromine atom gains 1 electron, as its oxidation state decreases from 0 to -1.

Since there are two aluminum atoms and six bromine atoms involved in the reaction, the total number of electrons transferred can be calculated as follows:

2 aluminum atoms x 3 electrons/atom = 6 electrons
6 bromine atoms x 1 electron/atom = 6 electrons

Therefore, a total of 6 + 6 = 12 electrons are transferred in the overall reaction.