I was able to get the balanced molecular equation but I am having a hard time getting the net ionic and oxidation half- reaction. Below is what I got for the balanced molecular equation:
Al (s) + 3AgNO3 (aq) --> Al (NO3)3 (aq) + 3 Ag (s)
Convert the balanced molecular equation you have to an ionic equation.
Al(s) + 3Ag^+(aq) + 3NO3^-(aq) ==> Al^3+(aq) + 3NO3^-(aq) + 3Ag(s)
Now cancel the ions common to both sides. Those are the 3NO3^-(aq). What's left is the net ionic equation; i.e., Al(s) + 3Ag^+(aq) ==> Al^3+(aq) + 3Ag(s)
It should be obvious which is the oxidation half and which is the reduction half.
If you haven't seen this acronym, it may be helpful in identifying oxidation and reduction half-reactions...
OIL RIG …
O => Oxidation
I => Is
L => Loss (of electrons)
R => Reduction
I => Is
G => Gain (of electrons)
Alᵒ(s) + 3AgNO₃(aq) => Al(NO₃)₃(aq) + 3Agᵒ(s)
1(0) 3(+1) 1(+3) 3(0)
Alᵒ(s) => Al⁺³(aq) + 3eˉ => (3 electrons lost from Alᵒ => Oxidation)
3Ag⁺¹(aq) + 3eˉ => 3Agᵒ(s) => (3 electrons gained by each of 3 Ag⁺¹ => Reduction)
Oxidation Is Loss = Reduction Is Gain
I have head "Leo the lion goes Grr."
L = loss
e = electrons
o = oxidation
To obtain the net ionic equation and oxidation half-reaction, let's review the steps:
1. Start with the balanced molecular equation you have provided:
Al (s) + 3AgNO3 (aq) → Al(NO3)3 (aq) + 3Ag (s)
2. Identify the state of each species involved in the reaction (s = solid, aq = aqueous):
Al (s) + 3AgNO3 (aq) → Al(NO3)3 (aq) + 3Ag (s)
3. Split all the reactants and products into their respective ions (if applicable) based on their state:
Al (s) → Al3+ (aq)
AgNO3 (aq) → Ag+ (aq) + NO3- (aq)
Al(NO3)3 (aq) → Al3+ (aq) + 3NO3- (aq)
Ag (s) → Ag (s)
Here, Al (s) remains as a solid, AgNO3 (aq) splits into Ag+ (aq) and NO3- (aq), Al(NO3)3 (aq) splits into Al3+ (aq) and 3NO3- (aq), and Ag (s) remains as a solid.
4. Cancel out any spectator ions. Spectator ions are present on both sides of the equation and do not participate in the reaction. In this case, NO3- ions are spectator ions.
Al (s) + 3Ag+ (aq) → Al3+ (aq) + 3Ag (s)
Now, we have the net ionic equation, which only includes the species actually involved in the reaction.
5. Determine the oxidation numbers for the elements in each species:
In the reactant, Al (s), the oxidation number of Al is 0. In the product, Al3+ (aq), the oxidation number of Al is +3. Thus, Al is oxidized from 0 to +3.
In the reactant, Ag+ (aq), the oxidation number of Ag is +1. In the product, Ag (s), the oxidation number of Ag is 0. Thus, Ag is reduced from +1 to 0.
6. Write the oxidation half-reaction and the reduction half-reaction based on the changes in oxidation numbers:
Oxidation half-reaction:
Al (s) → Al3+ (aq) + 3e-
Reduction half-reaction:
3Ag+ (aq) + 3e- → 3Ag (s)
In the oxidation half-reaction, Al loses 3 electrons, whereas in the reduction half-reaction, each Ag+ ion gains 1 electron.
Remember to balance the number of electrons transferred in each half-reaction to ensure charge conservation.
That explains the process of obtaining the net ionic equation and the oxidation half-reaction for the balanced molecular equation you provided.