Asked by Erin
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)
Al (s) + 3AgNO3 (aq) --> Al (NO3)3 (aq) + 3 Ag (s)
Answers
Answered by
DrBob222
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.
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.
Answered by
Doc48
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
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
Answered by
DrBob222
I have head "Leo the lion goes Grr."
L = loss
e = electrons
o = oxidation
L = loss
e = electrons
o = oxidation