Question

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)

Answer 1

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.

Answer 2

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

Answer 3

I have head "Leo the lion goes Grr."
L = loss
e = electrons
o = oxidation