Chapter 22 Oxidation – Reduction Rxns. Section 22.1 The Meaning of Oxidation and Reduction. OBJECTIVES: Define the terms “oxidation” and “reduction” in terms of electron loss or gain. State the characteristics of a RedOx Rxn, and recognize oxidizing and reducing agents.
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OIL RIG = Oxidation Is Loss of Electrons
Reduction Is Gain of Electrons
LEO goes GER =
Loss of Electrons is Oxidation
Gain of Electrons is Reduction
In this example elemental magnesium and sulfur combine to produce the compound magnesium sulfide (a synthesis reaction).
The magnesium atom becomes a magnesium cation by losing two electrons. The two electrons are needed on the product side to balance the charge of this OXIDATION HALF REACTION.
The sulfur atom gains the two electrons lost by the magnesium atom to become a sulfide anion. This is the REDUCTION HALF REACTION.
The over all balanced reaction is the sum of the two half reactions. NO electrons can show, and it must balance by both mass and charge!
The S was reduced, so it is the oxidizing agent.
Loss of electrons
Shift of electrons away from a covalent bond
Gain of Oxygen
Increase in oxidation number (charge)
Loss of hydrogen (biology definition)
Gain of electrons
Shift of electrons toward an atom in a covalent compound
Loss of Oxygen
Decrease in Oxidation Number (charge)
Gain of hydrogen (biology definition)
Ex: In SO2, S = +4 and O = -2
(S) + 2(O) = 0 or (+4) + 2(-2) = 0
Ex: In CO32-, C = +4 and O = -2
(C) + 3(O) = -2 or (+4) + 3(-2) = -2
S = +3, O = -2
2(S) + 3(O) = 0 or 2(+3) + 3(-2) = 0
K = +1, Mn = +7, O = -2
K + Mn + 4(O) = 0 or (+1) + (+7) + 4(-2) = 0
N = 0 (remember, diatomic elements are all zero!)
Cr = +6, O = -2
2(Cr) + 7(O) = -2 or 2(+6) + 7(-2) = -2
2AgNO3(aq) + Cu(s) 2Ag(s) + Cu(NO3)2(aq)
For this single replacement rxn, the balanced complete ionic rxn is:
2Ag1+(aq) + Cu0(s) + 2NO31-(aq) 2Ag0(s) + Cu2+(aq) + 2NO31-(aq)
The spectator ion must be 2NO31-(aq), so that leaves the net ionic rxn as:
2Ag1+(aq) + Cu0(s) 2Ag0(s) + Cu2+(aq)
(Note that this rxn is balanced by both mass and charge!)
The half rxns are:
2Ag1+(aq) + 2e1- 2Ag0(s)Reduction
Cu0(s) Cu2+(aq) + 2e1- Oxidation
Thus, 2Ag1+(aq) is reduced (making it the oxidizing agent), and Cu0(s) is oxidized (making it the reducing agent). Notice that all the answers to a potential NYS Regents question on this rxn are on the reactant (left) side of the reaction!
Li(s) + Al(NO3)3(aq) LiNO3(aq) + Al(s)
Write the complete ionic reaction:
Li(s) + Al3+ (aq) + 3(NO3)1-(aq) Li1+ (aq) + NO31-(aq) + Al(s)
Eliminate the spectator ions. Write the net ionic reaction:
Li(s) + Al3+ (aq) Li1+ (aq) + Al(s)
Now write the two half-rxns:
3e1- + Al3+ (aq) Al(s) Reduction
Li(s) Li1+ (aq) + 1e1- Oxidation
As you can see, the number of electrons is not the same, so we will multiple the oxidation half by 3:
3e1- + Al3+ (aq) Al(s) Reduction
3Li(s) 3Li1+ (aq) + 3e1- Oxidation
Now add the two half-reactions back together to get:
3Li(s) + Al3+ (aq) 3Li1+ (aq) + Al(s)
Put the spectator ions back in and you get:
3Li(s) + Al(NO3)3(aq) 3LiNO3(aq) + Al(s)
Of course you could have done this by simple “inspection”, but the technique we just used can be applied to solve far more difficult problems!