Electrochemistry

• Define oxidation and reduction. • Determine oxidation numbers for atoms.

• Identify the oxidizing agent, the reducing agent.

• Distinguish between redox and non-redox reactions.

Burning and corrosion needs oxygen – oxidation.

Oxidation-reduction reactions – (redox)Chemical changes when electrons are transferred

from one reactant to another.

Oxidation - an atom loses one or more electrons.Reduction - an atom gains one or more electrons.

"LEO says GER” Losing Electrons is Oxidation, Gaining Electrons is Reduction

2 Mg (s) + O2 (g) → 2 MgO (s)

Mg – neutralMg2+ ion

O – neutral O2– ion

Magnesium is oxidized. Oxygen is reduced.Mg → Mg2+ 2e- O + 2e- → O2-

Mg (s) + Cl2 (g) → MgCl2 (s)

Written difference between ion and oxidation:Chlorine ion – Cl1- ion charge = 1-

oxidation number = -1

Mg → Mg2+ 2e- Cl + 1e- → Cl1-

Sometimes these numbers are the same (like above) sometimes they are very different – which is why we write

them differently.

Oxidation number represents the charge the atom

would have if every bond were ionic.

1. Assign known numbers first (below). Then calculate the others.

• All uncombined elements (diatomic) – zero.• Monatomic ions in ionic bond - equals ion

charge.

**You assign them to EACH atom**

Neutral compound:Sum of ox.numbers for each atom must be zero.

Polyatomic ion: Sum of ox.numbers must be the charge of that ion.

Assign oxidation numbers to each atom in SO2.

S = +4 O = –2

In compounds:• Alkali metals - always +1.• Earth metals - always +2• Al: +3, F: -1, H: +1*, O: –2*

Assign ox.numbers for each atom in K2Cr2O7

Step 1: Start with atoms which are known.O: –2

K: +1Step 2: Solve for other atoms.All O atoms: –2 × 7 = –14All K atoms: +1 × 2 = +2

The total for the compound must be zero. All Cr atoms: -14 + 2 + ?? = 0 Two Cr atoms - (+12) ÷ 2 = +6 each

K = +1 Cr = +6 O = –2.

Assign ox. numbers for each atom in Fe(NO3)3

Ionic bond between Fe3+ and NO3–

Fe: +3

Sum of ox.numbers for the compound must be 0.

All O atoms: -2 × 9 = –18 +3 (-18) + ?? = 0All N atoms: 15 ÷ 3 =

Fe = +3 N = +5 O = –2.

Use ox.numbers to determine if reaction is a redox reaction.

SO2 + H2O → H2SO3

Ox.numbers do not change – no e- transferred – NOT a redox reaction.

-2+4+1-2+1-2+4

-6+4+2-2+2-4+4

Is the following reaction a redox reaction?

• Cu – oxidized (loss of electrons). • Ag – reduced (gain of electrons).

Oxidation cannot occur without reduction.

0-2+5+1-2+5

0-6+5+1-6+5

+10

+10Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s)

Oxidizing agent - causes the oxidation of another substance. AgNO3 is the oxidizing agent

Reducing agent - causes the reduction of another substance.

Oxidizing agent becomes reduced and the reducing agent becomes oxidized.

Cu is the reducing agent

+1

0

0

+1Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s)

Identify the substance oxidized, the substance reduced, the oxidizing agent and the reducing agent.

2 HNO3(aq) + 3 H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l)

S – oxidizedN – reducedH2S – reducing agentHNO3 – oxidizing agent

0-2+2-2+1

0-2+2-2+2

-2+5

-6+5

+1

+1

-2+1

-2+2

How many electrons are transferred in the reaction below:

Stoichiometry used to determined total electrons transferred.

HNO3(aq) + H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l)

0-2+2-2+1-2+5+1 -2+1

gains 3e-

loses 2e-

S: (3 atoms) x (2e- lost) = 6 electrons lostN: (2 atoms) x (3e- gained) = 6 electrons

gained

2 3

Strong oxidizing agents:Are very reactive – will take from anything

Oxidizing AgentsReaction Products

O2

O2–, H2O, CO2

F2, Cl2, Br2, I2 F–, Cl–, Br–, I–

MnO4–

Mn2+

Cr2O72–

Cr3+

HNO3

NO, NO2

H2O2

O2, H2O

Strong reducing agents:Are very reactive – will give to anything.Metals, substances that burn easily – H2, CxHy