Change-in-oxidation number method to balance equation....?

Question

How do you balance the equation AL + NO3(-) --> AL(OH)4(-) + NH3 using the change in oxidation number method? I got stuck on identifying the oxidation numbers, I couldn't remember how to....any help?

Answer 1

Okay first let's see which elements are changing their oxidation numbers. O is -2 on both sides of the equation. I don't see H on the left, so you might want to double check your equation to see if you copied it down correctly.

N is changing from +5 to -3. Here is why.

In the nitrate ion, NO3(-), the total charge on the ion is -1 and the oxygen atom is almost always going to be -2
The sum of the oxidation numbers must be equal to zero if it's a neutral compound or equal to the overall charge if it's an ion.
You have three O's so 3 x (-2) = -6
Therefore N has to be +5 so that the sum of the oxidation numbers equals the charge on the ion (-1).

In NH3, each H is +1 (H is almost always going to be +1 in compounds) and you have three H's, so the N has to be -3 to make the sum of the oxidation numbers equal to zero (for a neutral compound, NH3)

Since N is changing from +5 to -3, you would need to have 8 electrons on the left side of the equation. This would be classified as a reduction. The overall balanced half-reaction would look something like this, after balancing the charge with OH(-) ions and adding water molecules to balance the H's and the O's

NO3(-) + 8 e(-) + 6 H2O ==> NH3 + 9OH(-)

Al is changing from zero (as an element) to +3

Al ==> Al(3+) + 3 e(-)

Then we have to add the two half-reactions together, making sure that the electrons lost are the same as the electrons gained.

Multiply this reaction by three:
NO3(-) + 8 e(-) + 6 H2O ==> NH3 + 9OH(-)

Multiply this reaction by eight:
Al ==> Al(3+) + 3 e(-)

3 NO3(-) + 24 e(-) + 18 H2O ==> 3 NH3 + 27OH(-)
8 Al ==> 8 Al(3+) + 24 e(-)

Notice that 24 electrons appears on the left and the right, so that when we add these two half-reactions together, the electrons cancel out.

3 NO3(-) + 18 H2O + 8 Al ==> 3 NH3 + 27OH(-) + 8 Al(3+)

Although the above equation is balanced according to chanrge and atoms, I'm a little confused about making
Al(OH)4(-) appear as a product in the reaction. If you have 8 Al ions, you would need 32 OH(-) ions, and you only have 27.

I suppose you could add 5 OH(-) ions to both sides:

3 NO3(-) + 18 H2O + 8 Al + 5 OH(-) => 8 Al(OH)4(-) + 3 NH3

Anyway, that info should help you at some level.

Answer 2

OK...

Al starts with an oxidation number of 0 (all elements in their standard states have an oxidation number of 0).

In Al(OH)4(-) there are several ways to determine the oxidation number, so I'll give you one. Each OH- ion has a net charge of -1 since you have 4 of them, you have a total of -4 charges from those. Since the ion has a net charge of -1, the Al atom must have a charge (oxidation number) of +3. So, Aluminum is being oxidized from 0 to +3, and must be losing 3 electrons.

In a nitrate ion, each oxygen has an oxidation number of -2, and you have 3 of them, so that gives you -6 charges. Since the ion has a net charge of -1, the N atom must have a +5 charge in the nitrate ion. In ammonia (NH3), each H has a +1 charge, so the N must have an oxidation number of -3. So, N is being reduced from +5 to -3, and so must be gaining 8 electrons.

See what you can do from this point on...

Answer 3

Your equation is perplexing. you genuinely did not mean Arcturium (Ai) simply by fact people have not got here upon it yet and its optimal oxidation state is +2. Aluminum looks like the ingredient you meant simply by fact it has an oxidation state of +3 and the "i" is close to the l on the keyboard. regrettably, Al (like Fe) would not dissolve in HNO3 and you won't form the [Al(OH)4]^- and NH3 in an acidic answer. Of Ac, Ag, Am, Ar, As, At, and Au, basically As might undergo this reaction. in case you have not made a typo errors, then your professor is proposing you with a hypothetical ingredient for the purpose of preparation. i'm going to humor you the two and stability it. Ai is oxidized from 0 to +3 jointly as N is decreased from +5 to -3. 0) Ai + NO3- ===> Ai(OH)4- + NH3 a million) Ai(s) + 4H2O(l) > [Ai(OH)4]^-a million(aq) + 4H^+a million(aq) + 3e^- 2) [NO3]^-a million(aq) +9H^+a million(aq) + 8e^- > 3H2O(l) + NH3(g) Multiply equation a million with the aid of 8 and equation 2 with the aid of 3. 3) 8Ai(s) + 32H2O(l) > 8[Ai(OH)4]^-a million(aq) + 32H^+a million(aq) + 24e^- 4) 3[NO3]^-a million(aq) +27H^+a million(aq) + 24e^- > 9H2O(l) + 3NH3(g) upload equations 3 and four at the same time, then cancel all electrons and extra H2O and H^+a million. 5) 8Ai(s) + 3[NO3]^-a million(aq) + 23H2O(l) > 8[Ai(OH)4]^-a million(aq) + 5H^+a million(aq) + 3NH3(g) Voila! there is your complicated, hypothetical reaction.