Redox And Electrode Potentials

Question 1

What is oxidising and reducing agent

Answer 1

Oxidising agent takes electrons from species being oxidised, oxidising agent contains species which is reduced
Reducing agent is reverse

Question 2

How to construct a redox equation using half equation

Answer 2

See card 1

Question 3

How to construct redox equation using oxidation numbers

Answer 3

Use this method when hydrogen and oxygen appear in more than one product
See card 2

Question 4

How to predict products of redox reactions

Answer 4

In aq redox reactions H2O often formed, or H+ or OH- ions

As a final check ensure both sides of equation are balanced by charge

Question 5

How it write half equation

Answer 5

See card 3

Question 6

The two common redox reaction

Answer 6

Potassium manganate (VII) (KMnO4(aq)) under acidic conditions 
Sodium thiosulfate (Na2S2O3(aq)) for determination of iodine (I2(aq))

Question 7

Manganate titrations

Answer 7

Standard solution of KMnO4 added to burette.
Using pipette add measured vol of solution being analysed to conical flask, an excess of dilute sulfuric acid is also added to provide H+ ions required for reduction of MnO4- ions. No indicator needed, reaction is self-indicating.
During titration manganate solution reacts and is decolourised as it is being added, endpoint of titration is judged by first permanent pink colour, indicating when there is excess of MnO4- ions present. In titrations this endpt is one of easiest to judge.
Repeat titration until u obtain concordant titres (+-0.1 cm3)

Question 8

How to read meniscus

Answer 8

KMnO4 is deep purple, difficult to see bottom of meniscus, burette readings read from top rather than bottom of meniscus.
Titre is difference between two readings

Question 9

Manganate titrations can be used for analysis of many different reducing agents e.g.

Answer 9

Iron ions, Fe2+ 
Ethanedioic acid (COOH)2

Question 10

How to analyse percentage purity

Answer 10

See card 4

Question 11

Determination of a formula

Answer 11

See card 5

Question 12

What can be analysed

Answer 12

Manganate titrations can be used to analyse reducing agents that reduce MnO4- to Mn2+.
KMnO4 can be replaced with other oxidising agents, e.g. acidified dichromate

Question 13

Iodine/ thiosulfate titrations

Answer 13

See card 6

Question 14

Iodine thiosulfate titrations can be used to determine:

Answer 14

The ClO- content in bleach
The Cu2+ content in copper (II) compounds
Th Cu content in copper alloys

Question 15

Procedure of iodine thiosulfate titrations

Answer 15

1. Add standard solution of Na2S2O3 to burette
2. Prepare solution of oxidising agent to be analysed, using a pipettes add this solution to conical flask, then add excess of potassium iodide, oxidising agent reacts with iodide ions to produce iodine, which turns solution yellow brown
3. Titrate solution with Na2S2O3, during titrations iodine reduced back to I- ions and brown colour fades making it difficult to decide endpoint, so use starch indicator, when end point is being approached iodine colour has faded enough to become pale straw colour
4. Starch indicator added….

Question 16

Why is starch used for endpoint

Answer 16

When endpoint being approached and iodine colour faded enough to become pale straw colour small amount of starch indicator added. A deep blue black colour forms to assist with identification of endpoint. As more sodium thiosulfate added blue black colour fades, at endpoint all iodine reacted and blue black colour disappears

Question 17

Active ingredient of bleach

Answer 17

Chlorate (I) ions ClO- also known as hypochlorite

Bleach is NaClO

Question 18

Titration of ClO- with thiosulfate equation

Answer 18

See card 7

Question 19

How to find conc of ClO- in bleach

Answer 19

See card 8

Question 20

Iodine thiosulfate titrations can be used to determine

Answer 20

Copper content of copper (II) salts or alloys 
For copper (II) salts Cu2+ ion produced simply by dissolving compound in water 
Insoluble copper (II) compounds can be reacted with acids to form Cu2+ ions

Question 21

Analysis of copper

Answer 21

1. For copper alloys eg brass or bronze alloy is reacted and dissolved in concentrated nitric acid followed by neutralisation to form Cu2+ ions.
2. Cu2+ ions react with I- to form solution I2 and white precipitate of copper (I) iodide, CuI (s). Mix appears as brown colour
3. Iodine in brown mix then titrations with standard solution of sodium thiosulfate
   See card 9

Question 22

What is a voltaic cell

Answer 22

Converts chemical energy into electrical energy

Question 23

What is a half cell

Answer 23

Contains Chemical species present in a redox half equation.

Two of these connected together form a voltaic cell, allowing electrons to flow

Question 24

In the cell why are the chemicals in the two half cells kept apart

Answer 24

If allowed to mix electrons would flow in an uncontrollable way and heat energy would be released rather than electrical energy

Question 25

Metal/ metal ion half cells

Answer 25

Metal rod dipped in solution of its aqueous metal ion.
Represented using vertical line for phase boundary between aqueous solution and metal
See card 10

Question 26

What is equilibrium in a half cell

Answer 26

At the phase boundary, where the metal is in contact with its ions.
It is written so that forward reaction shows reduction and reverse reaction shows oxidation
See card 11

Question 27

In an isolated half cell

Answer 27

No net transfer of electrons in or out of metal

Question 28

When two half cells are connected direction of electron flow

Answer 28

Depends on relative tendency of each electrode to release electrons

Question 29

Ion/ion half cells

Answer 29

Contains ions of same element in different oxidation states
E.g. see card 12
In this half cell there is no metal to transport electrons into or out of half cell so insert metal electrode made of platinum is used

Question 30

Which electrode has greater tendency to lose electrons

Answer 30

In a cell with two metal/ metal ion half cells connected, the more reactive metal releases electrons more readily and is oxidised

Question 31

In an operating cells which metal loses electrons

Answer 31

The more reactive metal loses electrons and is oxidised, this is negative electrode
Reverse for less reactive metal

Question 32

Standard electrode potential

Answer 32

Tendency to be reduced and gain electrons.
Standard chosen is a half cell containing hydrogen gas H2(g) and a solution containing H+ (aq) ions.
Insert platinum electrode used to allow electrons in and out of half cell

Question 33

Sign of standard electrode potential shows sigh of

Answer 33

Half cell connected to standard hydrogen electrode and shows the relative tendency to gain electrons compared with hydrogen half cell

Question 34

How to measure a standard electrode potential

Answer 34

Half cell connected to a standard hydrogen electrode

See card 13

Question 35

The more negative the E value

Answer 35

The greater the tendency to lose electrons and undergo oxidation.
Th greater the reactivity of metal in losing electrons

Question 36

More positive the E value

Answer 36

The greater the tendency to gain electrons and undergo reduction.
Greater reactivity of non metal in gaining electrons

Question 37

Metals tend to have … E values

Answer 37

Negative.
They lose electrons.
Reverse for non metals

Question 38

emf measured is then a

Answer 38

Cell potential

Question 39

How to measure a standard cell potential (the experiment)

Answer 39

Prepare 2 standard half cells(temp 298K): for metal half cell conc of metal ion: 1 Moldm-3, for ion cell metal ions must have same conc and inert electrode eg platinum present, for half cell contains gases gas must be at 100 kPa in contact with solution conc 1 moldm-3 and inert electrode.
Connect metal electrodes to voltmeter with wires.
Prep salt bridge, by soaking filter paper in aq KNO3.
Connect solutions with salt bridge.
Record standard cell potential from voltmeter

Question 40

How to calculate standard cell potential from standard electrode potentials

Answer 40

See card 13

Question 41

Limitations of predictions using E values

Answer 41

Reactions that have very large activation energy result in very slow rate, electrode potentials indicate feasibility of reaction but not rate of reaction.
Non standard conditions will affect the electrode potentials and therefore the overall cell potential.
Standard electrode potentials apply to aqueous equilibria, many reactions that take place are not aqueous.

Question 42

Primary cells

Answer 42

Non rechargeable.
Used once only.
Chemicals eventually used up

Question 43

Secondary cells

Answer 43

Rechargeable.
Unlike primary cells Cell reaction producing electrical energy can be reversed during recharging.
Chemicals in cell then regenerated.

Question 44

Fuel cells

Answer 44

Uses energy from reaction of fuel with oxygen to create voltage.
Fuel and oxygen flow into fuel cell, products flow out, electrolyte remains in cell.
Can operate continuously when fuel and oxygen supplied into cell.
Do not have to be recharged.
Hydrogen fuel cells produces no CO2 only water as combustion product

Question 45

Alkali hydrogen fuel cell

Answer 45

See card 14

Question 46

Acid hydrogen fuel cell

Answer 46

See card 15