the reaction of aluminum metal (Al) with oxygen (O2) forms Al2O3. What is a balanced equation for this redox reaction. What are two half reactions to show how many electrons are gained or lost by each species?
The above response from DrBob222 is almost correct, but they forgot one important thing: we MUST maintain Conservation of Charge!
Aluminum is LOSING three electrons, while oxygen is GAINING four electrons, but Conservation of Charge tells us that we need to be gaining the same amount of electrons as we are losing.
Now that we've identified the number of electrons being lost is 3, and being gained is 4, we have our two numbers we can use to figure out the Least Common Multiple (LCM). The LCM of 3 and 4 is 12, so that is the amount of electrons that need to be lost and gained.
In order to achieve this, we multiply the aluminum half rxn by 4:
4(Al ==> Al^+3 + 3e-)
And we multiply the oxygen half rxn by 3:
3(O2 + 4e- ==> 2O^-2)
Our two new half reactions now look like this:
4Al ==> 4Al^+3 + 12e-
3O2 + 12e- ==> 6O^-2
These two half reactions properly show the losing and gaining of 12 electrons.
Explain the effect of a decrease in pressure on the solubility of solute in a liquid solvent, gas and solid?
To balance the equation for the reaction between aluminum (Al) and oxygen (O2), where aluminum oxide (Al2O3) is formed, follow these steps:
Step 1: Write down the unbalanced equation:
Al + O2 → Al2O3
Step 2: Separate the reaction into two half-reactions, one for the oxidation half and one for the reduction half. In the oxidation half, aluminum loses electrons to form aluminum ions, and in the reduction half, oxygen gains electrons to form oxide ions. Let's balance these half-reactions separately:
Oxidation half-reaction:
Al → Al3+ [Al loses 3 electrons]
Reduction half-reaction:
O2 + 6e- → 2O2- [O2 gains 6 electrons]
Step 3: Balance the number of electrons in both half-reactions by multiplying each half-reaction by the appropriate factor:
Oxidation half-reaction (multiply by 2):
2 Al → 2 Al3+ + 6e- [Now, 6e- lost on both sides]
Reduction half-reaction (multiply by 3):
3 O2 + 6e- → 6 O2- [Now, 6e- gained on both sides]
Step 4: Add the balanced half-reactions together and simplify if possible:
2 Al + 3 O2 → 2 Al3+ + 3 O2-
Now, cancel out common terms on both sides:
2 Al + 3 O2 → 2 Al3+ + 3 O2- [The balanced equation]
In summary, the balanced equation for the reaction of aluminum metal (Al) with oxygen (O2) to form aluminum oxide (Al2O3) is:
2 Al + 3 O2 → 2 Al3+ + 3 O2-
The two half-reactions are:
Oxidation half-reaction: 2 Al → 2 Al3+ + 6e- [Al loses 3 electrons]
Reduction half-reaction: 3 O2 + 6e- → 6 O2- [O2 gains 6 electrons]
4Al + 3O2 ==> 2Al2O3
Al ==> Al^+3 + 3e
O2 + 4e ==> 2O^-2
Check my work.