Al+NO3^-1-->Al(OH)4^-1+NH3

I am supposed to balance this reaction and label the reduction/oxidation reactions and oxidizing/reducting agents

I have my specific steps to solving this but when I get to step 2 I do not see how I can split these into 2 half reactions

To me it seems that Al-->Al but NO3-1 gets split up into NO3-1-->OH4^-+NH3

is this right? what would NO'3 agent be?

To balance the given chemical equation, you need to follow a stepwise approach. Let's break it down:

Step 1: Write down the unbalanced equation:
Al + NO3^-1 -> Al(OH)4^-1 + NH3

Step 2: Split the reaction into two half-reactions, one for reduction and one for oxidation.

In this case, aluminum (Al) is being oxidized as it loses electrons, and nitrate ion (NO3^-1) is being reduced as it gains electrons.

So, the oxidation half-reaction can be written as: Al -> Al(OH)4^-1
And the reduction half-reaction can be written as: NO3^-1 -> NH3

Step 3: Begin balancing the half-reactions.

For the oxidation half-reaction:
Since the aluminum (Al) goes from 0 to -1 charge, it loses 1 electron. So, we can write it as:
Al -> Al(OH)4^-1 + 3e^-

For the reduction half-reaction:
Since the nitrate ion (NO3^-1) goes from -1 to 0 charge, it gains 1 electron. So, we can write it as:
NO3^-1 + 6e^- -> NH3

Step 4: Balance the charge by multiplying each half-reaction by a suitable number to equalize the number of electrons gained and lost. In this case, we need to multiply the oxidation half-reaction by 6, and the reduction half-reaction by 2.

Balanced oxidation half-reaction:
6Al -> 6Al(OH)4^-1 + 18e^-

Balanced reduction half-reaction:
2NO3^-1 + 12e^- -> 2NH3

Step 5: Combine the two balanced half-reactions into the overall balanced equation.

12Al + 6NO3^-1 -> 6Al(OH)4^-1 + 2NH3

To determine the oxidizing and reducing agents, you need to look at which species is being reduced and which is being oxidized. In this case:

The reducing agent is Al (it gets oxidized) because it loses electrons.
The oxidizing agent is NO3^-1 (it gets reduced) because it gains electrons.

So, to summarize:
Balanced equation: 12Al + 6NO3^-1 -> 6Al(OH)4^-1 + 2NH3
Oxidation reaction: Al -> Al(OH)4^-1
Reduction reaction: NO3^-1 -> NH3
Reducing agent: Al
Oxidizing agent: NO3^-1

To balance the given reaction and identify the oxidation and reduction reactions, let's break it down into two half-reactions: oxidation and reduction.

Step 1: Write the unbalanced half-reactions:
Oxidation half-reaction: Al --> Al(OH)4^-1
Reduction half-reaction: NO3^-1 --> NH3

Step 2: Balance the atoms other than hydrogen and oxygen in each half-reaction:
Oxidation half-reaction: Al --> Al(OH)4^-1
Reduction half-reaction: 3NO3^-1 --> NH3

Step 3: Balance the oxygen atoms in each half-reaction by adding water molecules (H2O):
Oxidation half-reaction: Al --> Al(OH)4^-1 (balancing by adding 4H2O on the product side)
Reduction half-reaction: 3NO3^-1 + 2H2O --> NH3 (balancing by adding 6H2O on the product side)

Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+):
Oxidation half-reaction: Al --> Al(OH)4^-1 + 4H+ (balancing by adding 4H+ on the reactant side)
Reduction half-reaction: 3NO3^-1 + 2H2O + 4H+ --> NH3 (balancing by adding 4H+ on the reactant side)

Step 5: Balance the charges in each half-reaction:
Oxidation half-reaction: Al --> Al(OH)4^-1 + 4H+ + 4e^-
Reduction half-reaction: 3NO3^-1 + 2H2O + 4H+ + e^- --> NH3

Step 6: Multiply the half-reactions by appropriate coefficients to balance the number of electrons transferred:
Oxidation half-reaction: 4Al --> 4Al(OH)4^-1 + 16H+ + 16e^-
Reduction half-reaction: 3NO3^-1 + 2H2O + 4H+ + e^- --> 3NH3 + 6H2O

Step 7: Combine the half-reactions and cancel out spectator ions on both sides:
4Al + 3NO3^-1 + 2H2O + 4H+ --> 4Al(OH)4^-1 + 3NH3

By balancing the reaction, we see that the oxidation half-reaction involves the aluminum (Al) being oxidized (losing electrons) from 0 to +3 oxidation state, while the reduction half-reaction involves the nitrate ion (NO3^-1) being reduced (gaining electrons) from +5 to -3 oxidation state.

In this reaction, the reducing agent is Al since it loses electrons (undergoes oxidation), while the oxidizing agent is NO3^-1 since it gains electrons (undergoes reduction).