Honors Chemistry

Chapter 20 Notes – Oxidation-Reduction Reactions

(Student edition)

Chapter 20 problem set: 25*, 34, 40, 45, 51, 63, 65, 67, 70

20.1The Meaning of Oxidation and Reduction

The old definition of oxidation was “when a compound reacts with ”

Rusting:4 Fe+ 3 O22 Fe2O3

Combustion:CH4+2 O2CO2+2 H2O

The old definition of reduction was “when a compound loses ” (see reverse of above rxns.)

20.2Oxidation Numbers: the apparent charge assigned to an atom to show the relative

distribution of electrons

Rules for assigning oxidation #’s:

Free elements =

Oxidation #’s of ions =

F =

0 = (peroxides of oxygen = )

H = (metal hydrides, H = )

more electronegative atom gets a charge

Oxidation #’s add up to in compounds

Oxidation #’s = the charge in

Ex1FeO Iron II Oxide orFerrous Oxide

Ex2Fe2O3Iron III Oxide orFerric Oxide

Ex3H2SO4Hydrogen Sulfateor Sulfuric Acid

Ex4H2SO3Hydrogen SulfiteorSulfurous Acid

Ex5H2Cr2O7Hydrogen Dichromate or Dichromic Acid

Ex6NO3–1Nitrate

Ex7NO2-1Nitrite

Identifying Oxidation/Reduction (Redox) Reactions:

The new definition of oxidation: of electrons.

The new definition of reduction: the of electrons.

Na + Cl2  NaCl Na is and Cl is

Oxidizing agent: the molecule/element that causes the oxidation (the loss of an electron) of

another element. The oxidizing agent is the molecule/element that

gets reduced (it gains an electron).

Reducing agent: the molecule/element that causes the reduction (the gain of an electron) of

another element. The oxidizing agent is the molecule/element that

gets oxidized (it loses an electron).

So, in example #1, is the reducing agent and is the oxidizing agent

Zn + HNO3  Zn(NO3)2 + NO2 + H2O

So, lost electrons and gained electrons.

So, is oxidized and is reduced.

So, is the reducing agent and is the oxidizing agent.

But… the oxidizing agent doesn’t always contain oxygen…

Zn+CuCl2ZnCl2+Cu

So, lost electrons and gained electrons.

So, is oxidized and is reduced.

So, is the reducing agent and is the oxidizing agent.

20.3 Balancing Redox Equations with Oxidation Numbers

Some reactions can be balanced by trial and error – others cannot

FeCl3 + Zn  ZnCl2 + Fe

Fe goes from a charge to a charge. This is a of e-.

Zn goes from a charge to a charge. This is a of e-.

The goal is to balance the gain and the loss. Because the least common multiple

is , multiply the by (place a permanent coefficient) and

the by (place as a permanent coefficient). So, …

FeCl3 + Zn  ZnCl2 + Fe

Zn + HNO3  Zn(NO3)2 + NO2 + H2O

Zn + HNO3  Zn(NO3)2 + NO2 + H2O

However… this still doesn’t work….

Zn + HNO3  Zn(NO3)2 + NO2 + H2O

Now, balance the normal way:

Zn + HNO3  Zn(NO3)2 + NO2 + 2 H2O

K2Cr2O7 + H2O + S  SO2 + KOH + Cr2O3

K2Cr2O7 + H2O + S  SO2 + KOH + Cr2O3

Now, balance the normal way:

K2Cr2O7 + H2O + S  SO2 + KOH + Cr2O3

Balancing Redox Equations – Half Reaction Method:

This method allows for showing all ions involved in a reaction:

In acid solution, NO3-1 will react with I2 to produce IO3-1 + NO2. Balance the

equation.

Step 1: write the equation:

Step 2: write a half reaction for the first reactant:

Add water to balance the “O’s” in the reactants:

Add H+ to balance the “H’s” in the water:

Add e- to balance the charge in the above reaction:

Step 3: Write the half reaction for the second reactant:

Now, balance the “I’s” in this half reaction:

Add water to balance the “O’s” in the reactants:

Add H+ to balance the “H’s” in the water:

Add e- to balance the charge in the above reaction:

Step 4: Multiply equations so electrons balance out (need 10 electrons on both sides):

Now, add up the two reactions (canceling appropriately):

CrO2-1 can be changed into the polyatomic ion chromate if combined with the

polyatomic ion hypochlorite (chloride ion is also a product). Balance the equation

in basic solution.

Step 1: Write the half reaction for hypochromite to chromate:

Balance the “O’s” with hydroxides; however, anticipate the amount of

oxygen present in the amount of water that will be needed to

balance out the hydrogens in the hydroxides:

Add water to finish balancing the oxygens and hydrogens:

Add electrons to balance the charge in the above reaction:

Step 2: Write the half reaction for hypochlorite to chloride:

Balance the “O’s” with hydroxides; however, anticipate the amount of

oxygen present in the amount of water that will be needed to

balance out the hydrogens in the hydroxides:

Add water to finish balancing the oxygens and hydrogens:

Add electrons to balance the charge in the above reaction:

Step 3: Multiply equations so electrons balance out (need 6 electrons on both sides):

Now, add up the two reactions (canceling appropriately):

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