Unit 11: Redox

U11L1: Intro to Redox

●

●

●

●

RedOx = Reduction and Oxidation

Reduction - loss of electrons by an element during a reaction

Oxidation - gain of electrons by an element during a reaction

○

LEO the lion says GER

■ Lose Electrons Oxidation

■ Gain Electrons Reduction

The electrons lost by one element are gained by a different element -

they don’t just magically disappear! This happens simultaneously. This

is how the conservation of charge is maintained.

U11L2: Assigning Oxidation Numbers

●

Some rules for assigning oxidation numbers

○

*All singular elements have an oxidation state of 0

■ Cu: ox. # = 0

○

○

○

○

○

○

○

○

■ O2: ox# = 0

■ Ag: ox # = 0

■ F2: ox # = 0

*The oxidation state of an ion is the same as the charge of the ion

■ Ag+1: ox. # = +1

■ Cu+2: ox. # = +2

■ Cl-1: ox. # = -1

■ O-2: ox. # = -2

*H when combined with a METAL: oxidation state = -1

*H when combined with a NONMETAL: oxidation state = +1

Group 1 metals: oxidation state = +1 and Group 2 metals: oxidation state = +2

Oxygen: oxidation state = -2 (except in peroxide X2O2 and the oxidation state = -1)

Fluorine: oxidation state = -1

The first 4 rules should be memorized - everything can be determined using the Periodic Table

Some elements have multiple oxidation states - you can use what you know to determine the missing

oxidation state.

●

Elements with multiple oxidation states - use the chart method to help you

○

The sum of all oxidation states in a compound add up to 0

K S Na S O

O K2SO4 Ox # Na2S2O3

Sub. 2 1 4 Sub 2 2 3

Ox # +1 +6 -2 +1 +2 -2

Total +2 ? (+6) -8 0 Total +2 ? (+4) -6 0

+2 + X -8 = 0

X-6 = 0

X = +6

+2 + X - 6 = 0

X - 4 = 0

X = +2

Note: The two S’s together have a +4 charge - so

each individual S has a +2 charge.

○

The sum al all oxidation states in a polyatomic ion

P O (PO3)3- Cl O (ClO3)1-

Sub. 1 3 Sub 1 3

Ox # +3 -2 Ox # +5 -2

Total ? (+3) -6 -3 Total ? (+5) -6 -1

X - 6 = -3

X = +3

X - 6 = -1

X = +5

○

If you see a polyatomic in a compound - you can use the charge from Table E to help solve for unknown

oxidation numbers

Cu S O CuSO4 Sub. 1 1 4

Ox # +2 +6 -2

Total +2 ? (+6) -8 0

+2 + X -8 = 0

X-6 = 0

X = +6

Cu can have the ox. # +1 or +2

S can have the ox. #-2, +4, or +6

How do you solve for two unknowns?

SO4 is polyatomic ion on Table E w/ a -2 charge

The formula says there is one Cu - so to cancel out the

-2 from the SO4, Cu must be +2.

Now you can continue normally to determine the ox. #

for S.

●

Determining if a reaction is a redox reaction

○

If the oxidation numbers of two or more elements change - YES

■ You can determine the oxidation state for each element but that might take a while

■ If a reaction is a SINGLE REPLACEMENT - it is automatically a redox reaction

U11L3: Half Reactions

●

Half reactions show half of what is going on in a redox reaction

●

●

●

Electrons are always written on the more p o s i t i v e side of the reaction

Conservation of Charge - net charge of reactants = net charge of products

Reduction Half Reactions

●

Oxidation Half Reactions

○

Electrons GAINED/written as a reactant

○

○

Cu2+ + 2 e-

→ Cu0

○

Electrons LOST/written as a product

Ni0 → Ni3+ + 3 e-

●

●

Balancing Half Reactions - electrons lost must equal electrons gained

○

(Cu2+ + 2 e-

→ Cu) 3 - multiply the whole equation by 3 to get 6 electrons: 3 Cu2+ + 6 e-

→3 Cu

○

(Ni → Ni3+ + 3 e-) 2 - multiply the whole equation by 2 to get 6 electrons: 2 Ni → 2 Ni3+ + 6 e

Overall Balanced Redox Reaction - combine both half reactions - don’t write things that are written as a reactant

AND a product

○

○

Unsimplified Balanced Redox Reaction: 3 Cu2+ + 2 Ni + 6 e-

Simplified Balanced Redox Reaction: 3 Cu2+ + 2 Ni →3 Cu + 2 Ni3+ + 6 e

+ →3 Cu + 2 Ni3+

U11L4: Voltaic Cells

●

Table J - Activity Series

○

The higher the element is - the more active - more

likely to lose electrons - it is oxidized - is anode

○

The lower the element is - the less active - less

●

likely to lose electrons - it is oxidized - is anode

Electrons flow ANODE → CATHODE (ALWAYS) through wire

●

Spontaneous

●

Converts CHEMICAL ENERGY → ELECTRICAL ENERGY

●

Labeling and analyzing a voltaic cell

○

Cu(s) is the anode (Higher on Table J)

■ Oxidation occurs at anode (AN

○

Ag(s) is the cathode (Lower on Table J)

○

Reduction occurs at cathode (RED

OX)...always

■ Loss of electrons = oxidation (LEO)

○

CAT)...always

Gain of electrons = reduction (GER)

■ Mass of anode decreases

■ Half Reaction: Cu0

(s) → Cu2+

○

Mass of cathode increases

(aq) + 2e-

○

Half Reaction: Ag+

(aq) + e-

→ Ag0

(s)

U11L5: Electrolytic Cells

Uses for electroplating/decomposing compounds

Nonspontaneous - requires the input of energy to function

Converts ELECTRICAL ENERGY → CHEMICAL ENERGY

The weird object = cathode - site of reduction

The other metal = anode - site of oxidation

Electrons flow ANODE → CATHODE (ALWAYS) through the wire

Unit 11: Redox

Voltaic (Galvanic) vs. Electrolytic Cells

Voltaic/Galvanic Cell Electrolytic Cell

Flow of e- (Spontaneous or Forced) Spontaneous Forced

Energy Conversion Chemical → Electrical Electrical → Chemical

(+) Electrode Cathode Anode

(-) Electrode Anode Cathode

Direction of e- Flow Anode to Cathode Anode to Cathode

Reduction Half Cell Cathode Cathode

Oxidation Half Cell Anode Anode