In most chemical reactions, the "charge" for any given element remains constant

ie.)Na₂SO₄ + FeCl₃ → Fe₂(SO₄)₃ + NaCl...charges remain constant:

Na⁺¹→ Na⁺¹

SO₄^-2→ SO₄^-2

Fe⁺³→ Fe⁺³

Cl^-1→Cl^-1

However, in some synthesis or single displacement reactions, the charges may change, also known as changes in oxidation

Indicates [hypothetically] the number of electrons lost, gained or shared as a result of chemical bonding. The change in oxidation state of a species (the change in charge) lets you know if it has undergone oxidation or reduction

1. The oxidation number of any uncombined element is zero.

Ex.) H₂, O₂, Fe, etc. = 0 for the oxidation #

2. The oxidation number of monoatomic ions equal the charge of the ion:

Group IA: +1

Group 2A: +2

Flourine (& most Halogens): -1

Oxygen: -2

Hydrogen: +1 (when bonded to a non-metal)

Hydrogen: -1 (when bonded to a metal)

3. Conservation of Charge

-neutral compounds - net charge= 0

      hydrogen  oxygen  sulfur

Ex 1.)H₂SO₄:_subscript→ 2(+1) + 4(-2) + 1(x) =0

                         _charge↑

seperate the compound, H₂SO₄ into the monatomic ions, hydrogen, oxygen and sulfur. According to the above boldface section, the KNOWN charges of the ions are hydrogen and oxygen. Set the equation equal to zero as completed above and find the charge of sulfur_.

2 + -8 + x = 0...x=+6...sulfur has a +6 charge!

hydrogen  oxygen  sulfur

Ex. 2)H₂S₂O₅:      2(+1) + 5(-2) + 2(x) = 0

2 + -10 + 2x = 0... x= +4...sulfur has a +4 charge!

Ex. 3)(NH₄)₂S: The charges of Nitrogen(N) AND Sulfur(S) are unknown. Find the oxidation name for Sulfur by "criss-crossing" the charges first: (NH₄)₂⁺¹S^-2. Now, find Nitrogen's charge:

2(x) + 8(+1) + 1(-2)= 0

2x + 8 + -2 = 0...Nitrogen has a +3 charge while Hydrogen has a +1 charge and Sulfur has a -2 charge

 -Polyatomic ions - net charge= ion charge

nitrogen  oxygen

Ex. 1)NO₂^-1:                  x + 2(-2) = -1

x + -4 = -1...x = +3!

Phosphorus  Oxygen

Ex. 2)PO₄^-3:                  1(x) + 4(-2) = -3

x + -8 = -3...x = +5;

phosphorus has a charge of +5 while oxygen has a charge of -2

Oxidation Vs. Reduction

Oxidizing Agent & Reducing Element

1.Oxidation: the process by which electrons are removed from atoms or ions

(oxidation # increases)

OIL: Oxidation Is Loss (a loss of electrons when oxidation occurs, thus the ion becomes more positive)

2.Reduction: the process by which electrons are added to atoms or ions

(oxidation #decreases)

RIG: Reduction Is Gain

3.Oxidizing Agent: the element that undergoes a decrease in oxidation number (element is BEING reduced, thus its gaining electrons)

4.Reducing Agent: the element that undergoes an increase in oxidation number (element BEING oxidized)

*In Short: the element(1) being oxidized is called the reducing agent because as it loses/gives up its electrons, it in turn helps the other element(2) in the problem become reduced as those given up electrons are gained by that element(2). This element(2) that is being reduced because it gains these electrons is called the oxidizing agent because it helps to take the electrons that the oxidizing element(1) gave.

(The element being oxidized is the reducing agent while the element being reduced is the oxidizing agent)

*Coefficients are IMPORTANT - the equation must be balanced!!

*You must find the Oxidizing 1/2 rxn and the Reducing 1/2 rxn in order to show how the elements are being oxidized and reduced.

Ex. 1) 2Mg + O₂ → 2MgO

...2MgO can be interpreted as 2[Mg⁺²][O^-2]. The reactant Magnesium is oxidized as the product of magnesium loses 2 electrons and thus has a +2 charge. Oxygen is reduced as it gains 2 electrons and thus has a -2 charge.

-Ox. 1/2 rxn: Magnesium is being oxidized as it loses electrons because oxygen, the oxidizing agent, helps it to take them away

...2Mg →2Mg⁺²...oxidation; electrons are lost. There are 2 magnesium ions giving up 2 electrons each. 4 electrons are lost total.

2Mg → 2Mg⁺² + 4e^-! ( 0 = 4 + -4)

-Red. 1/2 rxn: Oxygen is reduced as it gains electrons because magnesium, the reducing agent, helps it to gain them. O₂+ 4e^-→ 2O^-2! (0 + -4 = -4)

_{oxygen takes electrons(4e^-) to become that product(2O^-2)}

Ex. 2) 2Zn + 2HCl → ZnCl₂ + H₂

_{charges:                    0        +1 -1      +2 -1       0}

2H⁺¹ in HCl (the reactant) is being reduced to H⁺⁰ in H₂ (the product). 2Zn⁺⁰ (the reactant) is being oxidized to  Zn⁺² in ZnCl₂ (the product). The oxidizing agent is Hydrogen as it helps to oxidize Zinc. The reducing agent is Zinc as it helps to reduce Hydrogen.

-Ox. 1/2 rxn: 2Zn → Zn⁺² + 2e^-! (0 = 2 + -2)

-Red. 1/2 rxn: 2H⁺¹ + 2e^- → H₂! (2 + -2 = 0)

(chlorine is neither being oxidized or reduced because its charges remain the same throughout the equation, thus it is not included here)

Ex. 3) H₂ + Br₂ → 2HBr

_{charges:                           0        0        +1 -1}

H₂⁰ is being oxidized to 2H⁺¹ while Br₂⁰ is being reduced to 2Br^-1. Bromine is the oxidation agent as it oxidizes hydrogen by taking away electrons and making hydrogen more positive. Hydrogen is the reducing agent as it reduces bromine by giving it electrons and making bromine more negative.

-Ox. 1/2 rxn: H₂⁰ → 2H⁺¹ +2e^-! (0 = 2 + -2)

-Red. 1/2 rxn: Br₂⁰ + 2e^- → 2Br^-1! (0 + -2 = -2)

Ex.4) 2FeBr₃ + 3Cl₂ → 2FeCl₃ + 3Br₂

_{charges:               +3 -1          0          +3 -1         0}

^*6Br^-1 (from 2FeBr₃, the reactant) is being oxidized into 3Br₂ while 3Cl₂ is being reduced into ^*6Cl₃^-1 (from 2FeCl_3, the product). Fe's charge doesn't change. Bromine is the reduction agent as it helps to reduce chlorine(by 1) and make it more negative by giving Chlorine 1 electron while Chlorine is the oxidation agent as it helps to oxidize Bromine(by 1) and make it more postive by taking away 1 electron.

[^*Note: Br is a diatomic element so Br₂ exists. However, Br₃ doesn't exist. The 3 on Br is Fe's charge. So, multiply the coefficient(2) with the subscript(3) to get 6Br which does exist. This occurs with 6Cl₃ as well. This is basically like getting your ionic equation back in Unit 9: 2FeBr₃ = 2Fe⁺³ + 6Br^-1]

-Ox. 1/2 rxn: 6Br^-1 → 3Br₂ + 6e^- ! (-6 = 0 + -6)

-Red. 1/2 rxn: 3Cl₂ + 6e^- → 6Cl^-1 ! (0 + -6 = -6)

Ex.5) Zn  CuCl₂ → Cu + ZnCl₂

_{charges:                     0    +2 -1         0     +2 -1}

Zn⁰ oxidizes into Zn⁺² and Cu⁺² reduces into Cu⁰. The charge of Cl₂ doesn't change. Cu is the oxidizing agent as it helps Zn to become more positive(by 2) by giving up its 2 electrons while Zn is the reducing agent as it helps Cu become more negative(by 2) by giving up its(Zn) 2 electrons.

-Ox. 1/2 rxn: Zn → Zn⁺² + 2e^- ! (0 = 2 + -2)

-Red. 1/2 rxn: Cu⁺² + 2e^- → Cu ! (2 + -2 = 0)

Ex.6) 2Al + 3Cl₂ → 2AlCl₃

_{charges:                         0        0          +3 -1}

Oxidized: Aluminum

Reduced: Chlorine

Oxidizing Agent: Chlorine

Reducing Agent: Almuminum

-Ox. 1/2 rxn: 2Al → 2Al⁺³ + 6e^- (0 = 6 + -6)

-Red. 1/2 rxn: 3Cl₂ + 6e^-→ 6Cl^-1 (0 + -6 = -6)

Ex. 7) 2Na + FeCl₂ → 2NaCl + Fe

_{charge:                       0      +2 -1        +1 -1       0}

Oxidized and reducing agent: Na

Reduced and oxidizing agnet: Fe

-Ox. 1/2 rxn: 2Na  → 2Na⁺¹ + 2e^-

-Red. 1/2 rxn: Fe⁺² + 2e^- → Fe

Ex. 8) 2H₂ + O₂ → 2H₂O

_{charge:                              0       0        +1 -2}

Oxidized and reducing agent: H

Reduced and oxidizing agnet: O

-Ox. 1/2 rxn: 2H₂ → 4H⁺¹ + 4e^-

-Red. 1/2 rxn:O₂ + 4e^- → 2O^-2

Ex. 9) 2Cu + 2HNO₃ → 2CuNO₃ + H₂

_{charge:                    0      +1 +5 -2      +1 +5 -2      0}

Oxidized and reducing agent: Cu

Reduced and oxidizing agent: H

-Ox. 1/2 rxn: 2Cu → 2Cu⁺¹ + 2e^-

-Red. 1/2 rxn: 2H⁺¹ + 2e^- → H₂

Ex. 10) AgNO₃ + Cu → CuNO₃ + Ag

_{charge:                   +1 +5 -2      0      +1 +5 -2      0}

Oxidized and reducing agent: Cu

Reduced and oxidizing agent: Ag

-Ox. 1/2 rxn: Cu → Cu⁺¹ + e^-

-Red. 1/2 rxn: Ag⁺¹ + e^- → Ag