REDOXTransfer of electron at a distanceWhat is means by transfer of electron at a distance?

Look at this diagram :

G

Reducing agent

Oxidizing agent

e e

+─ Positive terminal

Negative terminal

ElectrolyteU-Tube

The reducing agent undergoes oxidation (loss of electron) The oxidizing agent undergoes reduction (gain of electron)

Galvanometer

When oxidizing agent and reducing agent solutions is separated by an electrolyte in a U-Tube, redox reactions occur by transfer of electron using connecting wire. This electron transfer is call “transfer of electron at a distance”.

Solution that can react as oxidizing agent is the substances that undergoes reduction process

SolutionHalf equation (gain of electron) and changes colour of solution

Acidified potassium manganate(VII), KMnO4

MnO4- + 8H+ + 5e →

Mn2+ + 4H2O(purple to colourless)

Acidified potassium dichromate(VI) , K2Cr2O7

Cr2O72- + 14H+ + 6e →

2Cr3+ + 7H2O (orange to green)

Chlorine, Cl2Cl2 + 2e → 2Cl-

(pale yellow to colourless)

Bromine, Br2 Br2 + 2e → 2Br-

(reddish brown to colourless)

Iodine, I2I2 + 2e → 2I-

(brown to colourless)

Give extra attention to half equation for acidified potassium manganate(VII) solution and acidified potassium dicrhomate(VI)solution.

(fisrt time we see this half equation, must memorize)

Solution that can react as reducing agent is undergoes oxidation process

Solution nameHalf equation

(loss of electron) and changes colour of solution

Iron(II) sulphate, FeSO4

Fe2+ → Fe3+ + 1e(green to brown)

Stanum(II) chloride, SnCl2

Sn2+ → Sn4+ + 2e(both of the ions is colourless)

Potassium chloride, KCl

2Cl- → Cl2 + 2e (colourless to pale yellow)

Potassium bromide, KBr

2Br- → Br2 + 2e (colourless to reddish

brown)

Potassium iodide, KI

2I- → I2 + 2e (colourless to brown)

Remember : i. Electron transfer from reducing agent to

oxidizing agent through connecting wire. ii. Carbon/graphite electrode that is immersed in

reducing agent act as negative terminal.iii. Carbon/graphite electrode that is immersed in

oxidising agent act as positive terminal.iv. The deflection of galvanometer needle shows

the electron flowing/moving.

Few reaction that you must know…Example 1:

Can you predict what will happen?

Tips : determine the solution that react as oxidizing agent an reducing agent first.

Oxidizing agent : acidified potassium manganate(VII) solution.

Reducing agent : iron(II) sulphate solution

If you can identify, the rest will be much easier.

G

iron(II) sulphate solution, FeSO4 0.1 mol dm-3

Acidified potassium manganate(VII) solution, KMnO4 0.1 mol dm-3

e e+─Negative

terminal

Sulphuric acid, H2SO4 0.1 mol dm-3 U-Tube

Galvanometer

Positive terminal

[Look at the list of oxidant and reductant that you have prepared above]

Oxidation process ( reducing agent ) Half equation : Fe2+ → Fe3+ + 1e

- 1 iron(II)ion /Fe2+ loss 1 electron to produce 1 iron(III)ion/Fe3+

- Fe2+ is oxidize to Fe3+

- iron(II)ion react as reducing agent- oxidation number of iron is increase from +2

to +3- the colour of the solution is turn from (light)

green to brown-

What will happen to the electrons that has been released?

- The electron will be released through negative terminal (carbon that is immersed in iron(II)sulphate solution), and flow through connecting wire to positive terminal (carbon that is immersed in acidified potassium manganate (VII)

solution), so, the galvanometer needle will deflect.

- Manganate(VII) ion, MnO4- will gain

electron and undergoes the reduction process.

Reduction process (oxidizing agent)

Half equation: MnO4- + 8H+ + 5e → Mn2+ +4H2O

- 1 Manganate (VII) ion, MnO4- gain 5 electron

produce 1 mangan(II) ion, Mn2+ - Manganate(VII) ion is reduced to mangan(II)

ion- Mangnate (VII) ion act as oxidizing agent- oxidation number for mangan is decreasing

from +7 to +2 - the colour of solution is changes from purple

to colourless Can you write the total ionic equation for this reaction?

MnO4- + 8H+ + 5e → Mn2+ + 4H2O

5Fe2+ → 5Fe3+ + 5e

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

How this process is made? Maybe this question will be ask in PAPER2 this year (structure/essay)Example 2:

Can you predict what will happen?

G

Bromine water

Potassium iodide solution, KI 0.1 mol dm-3

Sulphuric acid, H2SO4 0.1 mol dm-3

U-tube

Galvanometer

Tips : recognize the solution that react as oxidizing agent and reducing agent first.

Oxidising agent : bromine water

Reducing agent : potassium iodide solution

Negative terminal: The carbon that immersed in potassium iodide solution

Positive terminal: The carbon that immersed in bromine solution/water

After you have recognize it, the others will be easier. Look at the table that you have prepare above,

Oxidation process / (Reducing agent)Half equation : 2I- → I2 + 2e

- 2 iodide ions loss 2 electron to produce 1 molecule of iodine

- Iodide ion is oxidize to iodine- Iodide ion react as reducing agent- Oxidation number for iodine increase from -1 to

0. thus oxidation occur- Colourless solution change to brownish yellow

What will occur to this releasing electron?- This electron will be released through negative

terminal - graphite electrode that is immersed in potassium

iodide solution, and flow through connecting wire to positive terminal

- graphite electrode that is immersed in bromine solution),

- Thus, galvanometer needle will deflect.- bromine will gain electron and reduction process

will occur.

Reduction processHalf equation for reduction : Br2 + 2e → 2Br-

- 1 molecule bromine gains 2 electron to produce 2 bromide ion

- Bromine is reduce to bromide ion.- Bromine react as oxidizing agent- Oxidation number for bromine decreasing from 0

to -1- Brownish-yellow solution change to colourless

Can you write the total ionic equation for this reaction?

Br2 + 2I- 2Br- + I2

example 3: (cont 5.1 & 5.2 :28.7.2008)

Tips : recognize the solution that react as oxidizing agent and reducing agent first.

Oxidizing agent : acidified potassium dicrhomate(VI)

Terminal positive: Carbon electrod that is immersed into acidified potassium dichromate(VI)

G

stanum(II) chloride solution, SnCl2 0.1 mol dm-3

Acidified potassium dichromate solution, K2Cr2O7

Sulphuric acid, H2SO4 0.1 mol dm-3

U-Tube

Galvanometer

Reducing agent : stanum(II) chloride solution

Terminal negative: carbon immersed into stanum(II) chloride solution

Oxidation process (reducing agent) Cont: 5.1Half equation : Sn2+ → Sn4+ + 2e [observation: no changes in colour of solution]

Cl- Cl2 + 2e[observation: The colour changes from colourless to pale yellow]

(depend on the question)

- 1 stanum(II) ion loss 2 electrons to produce 1 stanum(IV) ion

- Stanum(II) ion is oxidize to stanum(IV) ion- Stanum(II) ion react as reducing agent- Oxidation number for stanum is increasing from

+2 to +4

What will occur to this releasing electron?

- This electron will be released through negative terminal

- graphite electrode that is immersed in stanum(II) chloride solution) is negative terminal, and flow through connecting wire to positive terminal

- electrode that is immersed in acidified potassium dichromate(VI) solution is positive terminal,

- Galvanometer needle will deflect.- Dicrhomate(VI)ion will gain electron and

reducing process will occur

Reduction process / (oxidising agent)half equation for reduction:

Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O

- 1 dichromate(VI) ion gain 6 electron to produce 2 chromium(III)ion- dichromate(VI) ion is reduced to 2 chromium(III) ion- dichromate(VI) ion react as oxidizing agent.- Oxidation number for chromium decreasing from

+6 to +3- Orange solution turn to green

Can you write the total ionic equation for this reaction?

Can you answer the following question;i. State the electrode that react as positive terminal

Carbon electrode that immersed in acidified K2Cr2O7 solution

ii. State the electrode that react as negative terminalCarbon electrode that immersed in SnCl2 solution

iii. Mark the direction of the flow of current at the wire

iv. Name the substance that undergoes oxidation processstanum(II) Ion

v. Name the substance that undergoes reduction processdichromate(VI) Ion

vi. Name the oxidizing agentAcidified potassium dicrhomate(VI) solution /dikromat(VI)ion

vii. Name the reduction agentstanum(II)chloride solution / stanum(II)ion

viii.State the observation at the end of this reactionColour of solution for acidified potassium dicrhomate(VI)solution change from purple to green

ix. Write the half equation for the both electrodeSn2+ ----> Sn4+ + 2eCr2O7

2- + 14H+ + 6e ---> 2Cr2+ + 7H2O x. Write the total ionic equation for this reaction

3Sn2+ + Cr2O72- + 14H+ ---> 3Sn4+ + 2Cr2+ +

7H2O