Balancing Equations for Reactions Occurring in Basic Solutions:

I have the answer for it and I've gone and redone it twice and I can't match the answer...I was just seeing if anyone could show the steps so that I could see where I messed up...

NO2-(aq)+Al(s)-->NH3(aq)+Al(OH)4-(aq)
[[on NO2-, the two is a subscript and the - is the charge and on the Al(OH)4-, the 4 is a subscript and the - is the charge]]

Here is how I do the first one. It isn't the fastest way nor the easiest way; however, I use it because it makes the student go through the oxidation state.

Step 1. Copy the half equation.
NO2^- ==> NH3

Step 2.
Determine the oxidation states of the elements. I'll assume you know how to do that. N on the left is +3 and on the right is -3.

Step 3.
Determine the change in oxidation state and add electrons to the appropriate side to balance. Going from +3 to -3 means 6e must have been added, so
NO2^- + 6e ==> NH3

Step 4.
Count up the charge on each side and add
a. OH^- if in basic solution to balance the charge or.
b. H^+ if in acid solution to balance the charge. Note here that this is a basic solution; however, this procedure works the same way for acid solution so if you understand how to do this one in basic solution you will also understand how to do all of the acidic ones.
The charge on the left side is -7 and on the right side is 0, therefore, add 7 OH^- to the right.
NO2^- + 6e ==> NH3 + 7OH^-

Step 5.
Now count the H on both sides and balance with water. I see 10 H atoms on the right and none on the left; therefore, I add 5H2O to the left.
5H2O + NO2^- + 6e ==> NH3 + 7OH^-

Step 6.
That SHOULD balance the oxygen atoms. MOST of the time it will do that. The final step is to check everything; i.e., (a)atoms (b)charge (c) change in oxidation state.
(a)I see 10 H on the left and right.
I see 7 O on the left and right.
I see 1 N on the left and right.
(a) is ok.

(b)I see 7 minus charges on the left and 7 on the right.
(b) is ok.

(c) Change in oxidation state is +3 to -3 which is a change of 6e and those were added in the previous step
(c) is ok.

Therefore, all three balance and the equation is balanced. Viola!

NO2^- + 6e + 5H2O ==> NH3 + 7OH^-

and
Al + 4OH^- ==> Al(OH^-) + 3e.
=============================
Multiply equation 1 by 1 and equation 2 by 2, add, then remove atoms/ions/charges that are common to both sides.

okay then I had the +6e- for the first and +3e- for the second...but I just didn't get the H2Os and OHs and I don't understand how I messed that up...:S

tks

To balance equations for reactions occurring in basic solutions, you need to follow these steps:

Step 1: Break down the equation into half-reactions:
- Write two separate half-reactions, one for the reduction and one for the oxidation.

In this case, we have:
Half-reaction for reduction:
NO2-(aq) → NH3(aq)
Half-reaction for oxidation:
Al(s) → Al(OH)4-(aq)

Step 2: Balance the atoms:
- Start by balancing atoms other than hydrogen and oxygen.

Balancing the reduction half-reaction:
NO2- → NH3
Since there are no nitrogen atoms on the right side, we need to add NH3 on the left side. Adding a coefficient of 4 will balance the nitrogen atoms.
4NO2- → NH3

Balancing the oxidation half-reaction:
Al → Al(OH)4-
The aluminum is already balanced, but the oxygen atoms are not. There are four oxygen atoms on the right side, so we need to add four water (H2O) molecules on the left side.
Al + 4H2O → Al(OH)4-

Step 3: Balance the charge:
- Add electrons (e-) to one side of each half-reaction to balance the charges.

Balancing the reduction half-reaction:
NO2- + 4e- → NH3

Balancing the oxidation half-reaction:
Al → Al(OH)4- + 3e-

Step 4: Balance the number of electrons:
- Multiply each half-reaction by a factor to make the number of electrons equal in both half-reactions.

In this case, we need to multiply the reduction half-reaction by 3 and the oxidation half-reaction by 4 to balance the number of electrons.

Balancing the reduction half-reaction:
3NO2- + 12e- → 3NH3

Balancing the oxidation half-reaction:
4Al → 4Al(OH)4- + 12e-

Step 5: Combine the half-reactions:
- Add the two balanced half-reactions together.

Multiplying the reduction half-reaction by 4 and the oxidation half-reaction by 3 allows us to cancel out the electrons and have equal numbers on both sides.

Final balanced equation for the reaction occurring in basic solution:
12NO2-(aq) + 4Al(s) + 3H2O(l) → 4NH3(aq) + 4Al(OH)4-(aq)

Make sure to double-check that all the atoms and charges are balanced in the final equation.