Redox Reactions

Question 1

Oxidation

Answer 1

Gain O2
Lose H
Lose e-
Increase in oxidation number

Question 2

Reduction

Answer 2

Lose O2
Gain H
Gain e-
Decrease in oxidation number

Question 3

Oxidising agents

Answer 3

Non-metals

More reactive non-metals = more likely to accept e-s, so more positive electrode potentials

Question 4

Reducing agents

Answer 4

Metals, reactive ones have greatest tendency to donate e-s, therefore their half-cells have most -ve electrode potentials

Question 5

Uncombined element oxidation number

Answer 5

0

Question 6

Combined oxygen oxidation number

Answer 6

-2

Question 7

Combined hydrogen oxidation number

Answer 7

+1

Question 8

Simple ion oxidation number

Answer 8

Ion’s charge

Question 9

Combined fluorine oxidation number

Answer 9

-1

Question 10

Standard electrode potential

Answer 10

E.m.f of half- cell compared to an H half-cell

Question 11

H half-cell e.m.f

Answer 11

0V

Question 12

Overall cell reaction calculation

Answer 12

1) calculate difference between half cell values
2) reverse more negative half-equation
3) add half-equations together

Question 13

Redox reaction predictions

Answer 13

• stronger reducing agent with more negative half-cell is on right side of reaction
• stronger oxidising agent with more positive half-cell is on left side of reaction

Question 14

Ionic concentration changes

Answer 14

Still affects equilibrium as per Le Chatelier’s principle

Question 15

Reaction likeliness?

Answer 15

Depends on:
S.E.P must be in aqueous equilibria
Larger difference between half-cells, more likely reaction
If difference is <0.4V, reaction is unlikely to be completed

Question 16

Entropy and equilibrium effect on likeliness

Answer 16

Reaction likely if:
E(cell) is positive
∆Stotal is positive
K>1

Question 17

O2 exceptions

Answer 17

Peroxides (O₂)2- = -1

Oxygen difluoride (F₂O) = +2

Superoxide (K+, rubidium, caesium) MO₂ = 1/2

Question 18

Disproportionation

Answer 18

Same element is both oxidised and reduced

Question 19

Metals bathed in their own ionic solution equilibrium

Answer 19

Mⁿ+(aq) + ne- ⇋M(s)

Question 20

e.m.f

Answer 20

Occurs when no current flows

Cell’s max potential difference

Calculate by: RH - LH
-e.g 0.34-(-0.76) = +1.10 V

Question 21

Half-cell equations

Answer 21

More +ve electrode is written on RHS

More +ve electrode is RED
More -ve is OX

Question 22

Le Chatelier’s principle

Answer 22

Forward reaction = exothermic, +ve Ecell

Question 23

Spontaneous reaction

Answer 23

If Ecell >0.4 V

Question 24

Corrosion (rusting)

Answer 24

Metal corrosion = oxidation

Question 25

Storage cells

Answer 25

Must produce current

Must not leak

-ve pole must produce e-s, +ve pole must accept