Module 5: Chapter 23 (Redox and Electrode Potentials) Flashcards Preview

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Flashcards in Module 5: Chapter 23 (Redox and Electrode Potentials) Deck (56)
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1

What is oxidation?

loss of electrons or increase in oxidation number

2

What is reduction?

loss of electrons or decrease in oxidation number

3

What is an oxidising agent?

- takes electrons from the species that is being oxidised
- contains the species that is reduced
- for example: oxygen, dichromate, manganate (VII)

4

What is a reducing agent?

- adds electrons to the species being oxidised
- contains the species that is being oxidised
- for example: magnesium, zinc, sulfur dioxide, hydrogen and Fe2+

5

How do you write a redox equation from half-equations?

1. balance the electrons
2. add the equations and cancel out the electrons
3. cancel any species that is on both sides of the equation

6

Write a full equation:
H2O2 + 2e- -> 2OH-
Cr^3+ + 8OH- -> CrO4^2- + 4H2O + 3e-

3H2O2 + 2Cr^3+ + 10OH- -> 2CrO4^2- + 8H2O

7

What are the rules for oxidation numbers?

1. Group one elements = +1
2. group two elements = +2
3. F = -1
4. H = +1 (unless in a metal hydride = -1)
5. O = -2
6. Cl (and descending halides) = -1

8

How to write a redox equation from oxidation numbers?

1. summarise any info provided into an equation
2. assign oxidation numbers
3. balance only the species that contain elements that have changed oxidation numbers
4. balance any remaining atoms (H+ or H2O may have to be added to balance)

9

Balance:
S + HNO3 -> HSO4 + NO2 + H2O

S + 6HNO3 -> HSO4 + 6NO2 + 2H2O

10

Balance:
BrO3^- + Br- + H+ -> Br2 + H2O

BrO3^- + 5Br- + 6H+ -> 3Br2 + 3H2O

11

Balance (in acidic solution):
MnO4^- + SO3^2- -> SO4^2- + Mn^2+

6H+ + 2MnO4^- + 5SO3^2- -> 5SO4^2- + 2Mn^2+ + 3H2O

12

Balance (in acidic solution):
Cr2O7^2- + Sn^2+ -> Sn^4+ + Cr^3+

Cr2O7^2- + 3Sn^2+ + 14H+ -> 3Sn^4+ + 2Cr^3+ + 7H2O

13

What happens overall in a manganate (VII) titration?

- carried out under acidic conditions
- MnO4- ions are reduced to Mn2+ and the other chemical is oxidised

14

What is the general method for manganate (VII) titrations?

1. standard solution of potassium manganate (VII) fills the burette
2. the other solution is pipetted into a conical flask, along with an excess of dilute sulfuric acid (acid provides the H+ ions)
3. during the titration the manganate (VII) solution reacts and is decolourised. (MnO4- is pink/purple, Mn2+ is very pale/pink and often appears colourless)
4. the end point is the first permanent pink colour (no indicator required)
5. Repeat until concordant titres are obtained

15

What is the equation for the overall manganate equation?

5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O
OR
5(COOH)2 + 2MnO4^- -> 16H+ -> 10CO2 + 10H+ Mn^2+ + 8H2O

16

Calculating the percentage purity of a product from manganate titration (e.g. FeSO4•7H2O)

(use a full balanced equation as reference)
1. Calculate the amount of MnO4- that reacted using mean titre in n = c x V
2. Determine the amount of Fe2+ that reacted using the balanced equation
3. Find out unknown information (mol used in the titration scaling up to what has been used in the standard solution, then find the mass of the substance in the

17

Question:
A 0.1203g sample of (COOH)2•xH2O was dissolved in 25.0cm^3 of 1.0 moldm^-3 H2SO4 in a conical flask. The contents were heated to 60ºC and then titrated against 0.0200 moldm^-3 MnO4-. The mean titre was 19.10cm^3. Determine the value of x.
Half equations:
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O and (COOH)2 -> 2CO2 + 2H+ + 2e-

(COOH)2•2H2O (x = 2)

18

What is the equation for the iodine/thiosulfate equation?

2S2O3^2- + I2 -> S4O6^2- + 2I-

19

What the solution colours in an iodine/thiosulfate equation?

I2 = yellow-brown
I- = colourless

20

What are the 2 reducing agents manganate titrations can be used to analyse for?

- Iron (II) iona, Fe^2+(aq)
- Ethanedioic acid, (COOH)2(aq)

21

What can the iodine/thiosulfate titration be used to determine?

- the ClO- content in household bleach
- the Cu2+ content in copper (II) compounds
- the Cu content in copper alloys

22

What is the analysis of chlorate (I) ions?

ClO- + 2I- + 2H+ -> Cl- + I2 + H2O

- ClO- is from NaClO (the active ingredient in bleach)

- I2 produce reacts with S2O3^2- ions so therefore 1 mol of ClO- produces one mol of I2, which reacts with 2 moles of S2O3^2-
- 1 mol ClO- = 2 moles of S2O3^2-

23

What is the analysis of copper (II) ions?

2Cu2+ + 4I- -> 2CuI + I2

- CuI is a white ppt
- I2 produced is titrated with thiosulfate
- 1 mol of Cu2+ is equivalent to one mol of S2O3^2-

24

What is a half-cell?

Half-cells contain all the chemical species present in a redox half-equation

25

What is a voltaic cell?

These cells are made by connecting 2 different half-cells which then allows electrons to flow
- these cells convert chemical energy to electrical energy (therefore a type of electrochemical cell)

26

What should chemicals in 2 half-cells be separated?

If they were not kept apart electrons would flow in an uncontrolled way and heat energy would be released instead of electrical energy

27

What is electrical energy?

Electrical energy results from the movement of electrons and chemical energy is required to transfer electrons from one species to another

28

What does the vertical like "|" in notations mean?

| represents the phase boundary between the solid and the solution or ions for example

29

What are metal/metal ion half-cells?

- These half-cells consist of a metal rod dipped into a solution of its aqueous metal ions, e.g. Zn2+(aq) | Zn(s)
- at the phase boundary, where the metal is in contact with its ions, an equilibrium is set up
- by convention the forwards reaction shows reduction and the backwards reaction shows oxidation (Zn2+ + 2e- ⇌ Zn)

30

What happens in an isolated half-cell?

There is no net transfer of electrons into or out of the metal