Principles and application

Name: Principles and application
Uploaded: 2017-07-16T20:49:20+00:00
Duration: PTM39S15
Channel: Justina Mason
Description: Principles and application

Principles and application
Redox Principles and application

Course outline Oxidation and reduction Redox reactions
explain oxidation and reduction as an electron transfer process calculate oxidation numbers identify and name oxidants and reductants in equations identify oxidation-reduction reactions using oxidation numbers Oxidation and reduction Redox reactions describe, write equations for and interpret observations for: metal displacement reactions halogen displacement reactions write balanced simple redox equations (metal/metal ion, metal/hydrogen ion and halogen/halide ion)

Course outline Oxidation and reduction
cathode processes Redox Applications – Electrolytic Cells role of the electrolyte direction of ion migration describe and explain how an electric current is conducted in an electrolytic cell direction of electron flow in external circuit electrode product prediction for molten metal halides only describe and explain the following during the operation of an electrolytic cell: predict and name the electrode products for the electrolysis of molten metal halides only. anode processes

OXIDATION Used to be viewed as Gaining oxygen 2 Mg + O2 → 2 MgO
Magnesium gains oxygen Losing hydrogen CH S → C H2S Carbon loses hydrogen

REDUCTION Used to be viewed as Losing oxygen FeO + C → Fe + CO
Iron loses oxygen (Note: carbon also gains oxygen = oxidation) Gaining hydrogen CH O2 → CO H2O Oxygen gains hydrogen (Note: carbon also loses hydrogen and gains oxygen = oxidation)

It was observed that very often when one substance was being oxidised another was being reduced at the same time.

Modern view of oxidation and reduction
Oxidation is defined as losing of electrons LEO Reduction is defined as gaining of electrons GER

Modern view of oxidation and reduction
Oxidation Is Loss of electrons Reduction Is Gain OIL RIG

A redox reaction is an electron transfer process

REDOX = electron transfer
Reducing agent Oxidising agent Loses electrons 2 Fe (ON=0) 4e- + 2 Fe2+O2 – 4e- 4e- O2 (ON=0) Gains electrons 4e- 4e- 4e- 4e-

REDOX = electron transfer
Reducing agent Oxidising agent Loses electrons 2 Na (ON=0) 2e- + 2 Na+Cl – 2e- 2e- Cl2 (ON=0) Gains electrons 2e- 2e- 2e- 2e-

An oxidising agent Is also called the oxidant
Oxidises another substance by taking electrons from it Is also called the oxidant Gains electrons, and therefore Becomes reduced

A reducing agent Is also called the reductant
Reduces another substance by donating electrons to it Is also called the reductant Loses electrons, and therefore Becomes oxidized

A redox equation analysed
Electron transfer S gains electrons Is reduced Is the oxidant Mg + S → Mg S2– +2 –2 Mg loses electrons Is oxidized Is the reductant

What is the ON of P in PH3 ONP + 3 ONH = 0 (neutral) ONP + 3 (+1) = 0

ON of P in H3PO4 3H + P + 4O = 0 3(+1) + P + 4(–2) = 0 (+3) + P + (–8) = 0 P + (–5) = 0 P = +5

3H + P + 3O = 0 3(+1) + P + 3(–2) = 0 (+3) + P + (–6) = 0 P + (–3) = 0
ON of P in H3PO3 3H P + 3O = 0 3(+1) + P + 3(–2) = 0 (+3) P + (–6) = 0 P + (–3) = 0 P = +3

ON of Cr in CrO3 Cr + 3O = 0 Cr + 3(– 2) = 0 Cr + (– 6) = 0 Cr = + 6

ON of Cr in Cr2O72- 2Cr + 7O = – 2 2Cr + 7(– 2) = – 2 2Cr = 14 – 2

Cr + 4O = – 2 Cr + 4(– 2) = – 2 Cr = 8 – 2 = + 6
ON of Cr in CrO42- Cr + 4O = – 2 Cr + 4(– 2) = – 2 Cr = 8 – 2 = + 6

ON of Cr in (NH4)2CrO4 2NH4 + Cr + 4O = 0 2(+1) + Cr + 4(–2) = 0 Cr +2 – 8 = 0 Cr = + 6

ON of C in CH4 C + 4H = 0 C + 4(+1) = 0 C = – 4

C + 2H + O = 0 C + 2(+1) +(– 2) = 0 C + 2 + (– 2) = 0 C = 0
ON of C in CH2O C + 2H + O = 0 C + 2(+1) +(– 2) = 0 C (– 2) = 0 C = 0

O H H is less electronegative than C O is more electronegative than C
How can combined C have an ON of zero? O is more electronegative than C H is less electronegative than C C has ‘lost’ these 2 electrons C e O H C has ‘gained’ these 2 electrons Carbon has ‘lost’ 2 electrons to Oxygen but ‘gained’ 2 from Hydrogen

ON of C in CH3OH C + 4H + O = 0 C + 4(+1) + (– 2) = 0

ON of C in C6H6 6C H = 0 6C (+1) = 0 6C = 0 6C = – 6 C = – 1

ON of Cl in HClO4 H + Cl + 4O = 0 (+1) + Cl + 4(– 2) = 0

ON of Se in H2SeO3 2H + Se + 3O = 0 2(+1) + Se + 3(– 2) = 0 (+2) + Se + (– 6) = 0 Se + (– 4) = 0 Se = +4

Se + 3O = 0 Se + 3(– 2) = 0 Se + (– 6) = 0 Se = +6
ON of Se in SeO3 Se + 3O = 0 Se + 3(– 2) = 0 Se + (– 6) = 0 Se = +6

ON of Mn in MnO4 - Mn + 4O = – 1 Mn + 4(– 2) = – 1 Mn + (– 8) = – 1

Mn + 4O = – 2 Mn + 4(– 2) = – 2 Mn + (– 8) = – 2 Mn = +6
ON of Mn in MnO4 2 – Mn + 4O = – 2 Mn + 4(– 2) = – 2 Mn + (– 8) = – 2 Mn = +6

Mn + 2O = 0 Mn + 2(– 2) = 0 Mn + (– 4) = 0 Mn = +4
ON of Mn in MnO2 Mn + 2O = 0 Mn + 2(– 2) = 0 Mn + (– 4) = 0 Mn = +4

ON of Mn in Mn2O3 2Mn + 3O = 0 2Mn + 3(– 2) = 0 2Mn + (– 6) = 0 2Mn = +6 Mn = +3

With the aid of an electron-dot diagram,
In HClO4 the ON of Cl is +7 ON of O is – 2 ON of H is +1 H + Cl + 4O = 0 +1 + Cl – 8 = 0 Cl = 8 – 1 = +7 With the aid of an electron-dot diagram, show this is consistent with the electronegativities of the atoms in each bond. EN: O = Cl = H = 2.2 In HClO4 the ON of Cl is +7 ON of O is – 2 ON of H is +1 With the aid of an electron-dot diagram, show this is consistent with the electronegativities of the atoms in each bond.

O H e e e Cl H has ‘lost’ 1 electron ON = +1
Cl has ‘lost’ these 2 electrons And also these 2 electrons H has ‘lost’ 1 electron ON = +1 As well these 2 electrons And this 1 electron Each O has ‘gained’ 2 electrons ON = – 2 Cl has ‘lost’ 7 electrons ON = +7

With the aid of an electron-dot diagram,
In H2SO4 the ON of S is +6 ON of O is – 2 ON of H is +1 2H + S + 4O = 0 S – 8 = 0 S = 8 – 2 = +6 With the aid of an electron-dot diagram, show this is consistent with the electronegativities of the atoms in each bond. EN: O = S = H = 2.2 In HClO4 the ON of Cl is +7 ON of O is – 2 ON of H is +1 With the aid of an electron-dot diagram, show this is consistent with the electronegativities of the atoms in each bond.

O H e e e S Each H has ‘lost’ 1 electron ON = +1
S has ‘lost’ these 2 electrons And also these 2 electrons Each H has ‘lost’ 1 electron ON = +1 As well as this 1 electron And this 1 electron Each O has ‘gained’ 2 electrons ON = – 2 S has ‘lost’ 6 electrons ON = +6

Write the half equation for the conversion of IO3– to I2
1. Write the ‘skeleton’ equation including the reactant and product species. Determine the oxidation state of each species, to decide which element is being oxidized/reduced. IO3– → I2 I + 3(–2) = –1 (+5)(–2) (0) so I = (+5) I is reduced

2. Balance the numbers of atoms of the element that is being oxidized/reduced IO3– → I2 2 IO3– → I2

3. Balance the number of oxygen atoms by adding H2O to the side that needs more O IO3– → I2 2 IO3– → I2 2 IO3– → I H2O

4. Balance the number of hydrogen atoms by adding H+ to the side that needs more H IO3– → I2 2 IO3– → I2 2 IO3– → I H2O 2 IO3– H+ → I H2O

5. Balance the charge by adding electrons to the side that has more positive charge IO3– → I2 2 IO3– → I2 2 IO3– → I H2O 2 IO3– H+ → I H2O 2 IO3– H e – → I H2O

Now amend the equation for this reaction in BASIC solution Add 12 OH– to both sides of the half equation. 2 IO3– H e – → I H2O 12 OH– 12 OH–

Now amend the equation for this reaction in BASIC solution The OH– and H+ neutralise 2IO3–+ 12H2O + 10e –→ I2+6H2O +12 OH–

Now amend the equation for this reaction in BASIC solution Cancel out excess H2O 6 2IO3–+ 12H2O + 10e –→ I2+6H2O +12 OH– 2IO3–+ 6H2O e –→ I OH–

Write the half equation for the conversion of Mn2+ to MnO4–
1. Write the ‘skeleton’ equation including the reactant and product species. Determine the oxidation state of each species, to decide which element is being oxidized/reduced. Mn2+ → MnO4– Mn + 4(–2) = –1 (+2) (+7)(–2) So Mn = +7 Mn is oxidized

2. Balance the numbers of atoms of the element that is being oxidized/reduced Mn2+ → MnO4– Mn is already balanced

3. Balance the number of oxygen atoms by adding H2O to the side that needs more O Mn2+ → MnO4– Mn H2O → MnO4–

4. Balance the number of hydrogen atoms by adding H+ to the side that needs more H Mn2+ → MnO4– Mn H2O → MnO4– Mn H2O → MnO4– H+

5. Balance the charge by adding electrons to the side that has more positive charge Mn2+ → MnO4– Mn H2O → MnO4– Mn H2O → MnO4– H+ Mn H2O → MnO4– H+ + 5e –

Now amend the equation for this reaction in BASIC solution Add 8 OH– to both sides of the half equation. Mn H2O → MnO4– H+ + 5e – 8 OH– 8 OH–

Now amend the equation for this reaction in BASIC solution The OH– and H+ neutralise Mn2++ 4 H2O + 8 OH– → MnO4– + 8 H2O + 5e –

Now amend the equation for this reaction in BASIC solution Cancel out excess H2O 4 Mn2++ 4 H2O + 8 OH– → MnO4– + 8 H2O + 5e – Mn OH– → MnO4– H2O + 5e –

The two half reactions are: Cu+ → Cu2+ and Cu+ → Cu
Write the two half equations and the net redox equation for the disproportionation (self-redox) of Cu+ to Cu2+ and Cu The two half reactions are: Cu+ → Cu2+ and Cu+ → Cu

Disproportionation (self-redox) of Cu+ to Cu2+ and Cu
The two half reactions are: Cu+ → Cu2+ and Cu+ → Cu These equations need only to be balanced for charge Cu+ → Cu2+ + e– Cu+ + e– → Cu

Disproportionation (self-redox) of Cu+ to Cu2+ and Cu
Now combine the two half equations Each equation has one electron, so simply add all reactants and all products Cu → Cu2+ + e– Cu+ + e– → Cu 2 Cu → Cu2+ + Cu

Write the two half equations and the net redox equation for the disproportionation (self-redox) of ClO2 to ClO3– and ClO2– in basic solution Some ClO2 molecules are converted to ClO3– and other ClO2 molecules are converted to ClO2– Use the rules to write the two half equations for acidic solution, and then for basic solution.

Disproportionation (self-redox) of ClO2 to ClO3– and ClO2–
The two half reactions are: ClO2 → ClO3– and ClO2 → ClO2– (+4) (+5) (+4) (+3) Use the rules to complete each half equation

Balance each for O by adding H2O ClO2 + H2O → ClO3– ClO → ClO2–

Balance each for H by adding H+ ClO2 + H2O → ClO3– + 2H+ ClO → ClO2–

Balance charges by adding electrons ClO2 + H2O → ClO3– + 2H+ + e – ClO e – → ClO2–

Now add the two half equations ClO2 + H2O → ClO3– + 2H+ + e – ClO2 + e – → ClO2– 2ClO2 + H2O → ClO3– + ClO2– + 2H+

Add 2 OH– to each side 2ClO2 + H2O → ClO3– + ClO2– + 2H+ 2 OH– 2 OH–

Add 2 OH– to each side Then cancel excess H2O 1 2ClO2 + H2O → ClO3– + ClO2– + 2H2O 2 OH– 2ClO2 + 2OH– → ClO3– + ClO2– + H2O

Disproportionation (self-redox) of MnO42– to MnO4– and MnO2
The two half reactions are: MnO42– → MnO4– (+6) (+7) MnO42– → MnO2 (+6) (+4) Use the rules to complete each half equation

Balance O by adding H2O MnO42– → MnO4– MnO42– → MnO H2O

Balance H by adding H+ MnO42– → MnO4– MnO42– + 4H → MnO H2O

Balance charges by adding electrons MnO42– → MnO4– + e – MnO42– + 4H+ + 2e –→ MnO H2O

Electrons lost = Electrons gained 2MnO42– → 2MnO4– + 2e – MnO42– + 4H+ + 2e –→ MnO H2O

Add the two half equations 2MnO42– → 2MnO4– + 2e – MnO42– + 4H+ + 2e – → MnO2 + 2H2O 3MnO42– + 4H+ → 2MnO4– +MnO2 +2H2O

Chemistry Project Redox Titration

Content Definition Experiment Discussion

Definition Redox reaction
A redox reaction is involving transfer of electrons. Oxidation is a process in which a substance loses electrons. Reduction is a process in which a substance gains electrons. They cannot take place without each other.

Definition Redox reaction
An oxidizing agent is a substance that oxidizes others by accepting electrons. A reducing agent is a substance that reduces others by donating electrons.

Definition Titration Titration is a process of volumetric analysis that is used to determine the amount of solute in a solution. In simple titration, indicator is used to detect the equivalence point of reaction.

Definition Example Iodometric titration
Titration involving oxidation of iron (Ⅲ) Titration involving reduction of iron (Ⅱ) ( In some case,indicators need not to add. e.g. Titration with potassium manganate (Ⅶ) or potassium dichromate (Ⅵ) )

Experiment of iodimetric titrition
Thoery A standard iodine solution is titrated with a thiosulphate solution with unknown molarity. The standardized thiosulphate solution can be used to titrate another iodine solution of unknown concentration.

2)• Since iodine is volatile, we cannot prepare standard solution of iodine directly by accurately weighing a certain amount of iodine solid & dissolving the iodine in water . • Moreover , iodine is only very slightly soluble in water

• so a standard solution of iodine is first prepare by dissolving a known of pure potassium iodate(v) solid into an acidic medium (diluted H2SO4)containing excess iodide. IO3- (aq) +5I- (aq) + 6H+ (aq) > 3 I2 (aq) +3H2O (l) or [IO3- (aq) +8I- (aq) + 6H+ (aq) > 3 I3- (aq) + 3H2O (l) ]

3) 1ST titration • The standard iodine solution is then used to titrate with a thiosulphate solution of unknown concentration . • Since the colour of iodine in iodide cannot be used to accurately detect the end point starch is used as the indicator. (the change in colour ‘brown->yellow->colourless’ is very difficult to observe) Starch + iodine  blue ‘complex’

• Since starch irreversibly combines with iodine at a high concentration of I2(aq), so that the I2 will not be released from starch at the end point . • The starch solution should be added at the later stage of the titration (when the solution just turns from brown to pale yellow ). After the addition of starch ,the mixture turns deep blue . • The end point is shown by the complete decolourisation of the blue colour .

4) 2nd titration • After standardization ,the thiosulphate solution can be used to titrate another solution of unknown concentration of iodine. This method of titration (using iodine as an oxidizing agent in the titration ) is known as the iodimetric titration or iodimetry.

IO3- (aq) +5I- (aq) + 6H+ (aq) > 3 I2 (aq) +3H2O (l) …..(1) or [IO3- (aq) +8I- (aq) + 6H+ (aq) > 3 I3- (aq) + 3H2O (l)] I2 (aq) + 2S2O32- (aq) > S4O62- (aq) + 2I- (aq) ……(2) or [I3- (aq) + 2S2O32- (aq) > S4O62- (aq) +3I- (aq)]

The concentration of thiosulphate solution can be determined as follow: no. of moles of I2 = 3 x no. of moles of IO3- no. of moles of I2 = 1/2 x no. of moles of S2O32- no. of moles of S2O32- = 6 x no. of moles of IO3-

Determination of iron(Ⅲ)
1) Excess NaI (aq) is added to an oxidizing agent (e.g. Fe3+) with unknown molarity. 2) The iodine liberated is then determined by the titration with standard thiosulphate solution . 2Fe3+ (aq) + 2I- (aq) > 2Fe2+ (aq) + I2 (aq) This method of titration (using iodide as reducing agent in the titration )is known as the iodometric titration or iodometery.

Determination of iron (Ⅱ)
1)Standard acidified KMnO4 (aq) is added to an reducing agent (e.g. Fe2+) MnO4- (aq) +8H+ (aq) +5Fe2+ (aq)  Mn2+ (aq) +4H2O (l) +5Fe3+ (aq) 2) An indicator is not required as acidified potassium permanganate(Ⅶ) itself serves as an indicator.

Determination of iron (Ⅱ)
3) When manganate (Ⅶ) is added to iron (Ⅱ) in acidic solution , manganate(Ⅶ) is reduced to manganate(Ⅱ) while iron (Ⅱ) is oxidized to iron (Ⅲ). 4) The colour of the reaction mixture at the equivalence point is yellow (due to the presence of FeFe3+ (aq) ) 5) When an extra drop of acidified potassium manganate (Ⅶ)and thus become light purple .

Conclusion Reducing agent can be determined by titration with potassium permanganate Oxidizing agent can be determined by titration with sodium iodide (iodometric titration)

Discussion Error in the idiometric titration
A standard solution of iodine must be used immediately because its molarity changes with time because: iodine is volatile =>I2 can escape from the solution , causing the decrease of [I2] with time iodine can be oxidized by air (promoted by acids, heat & light )=>[I2]increase with time

Discussion An aqueous solution of KMnO4 is unstable .
Error in the titration with KMnO4 An aqueous solution of KMnO4 is unstable . 4MnO4- + light > MnO2 (s) +2 O2 (g) + MnO22- (aq) So the KMnO4 (aq) should be stored in a brown bottle & should be standardized before use.

Making Galvanic Cells No bridge Voltage Points to Deposit Cu – Al 0 V
Cu – Zn 1.1 V Al – Zn 0.9 V Zn 3) The salt bridge allows the cell to continue to function. K+ or NO3– can move into one of the half cells to balance the charge created by + metal ions forming or leaving solution. 4) No. We did not get the same values in the lab

Cu – Zn cell Electron flow Cu (+) Salt bridge Zn (–) Cu2+ Zn2+
Zn  Zn2+ + 2e– (oxidation - LEO) Cu2+ + 2e–  Cu (reduction - GER) Cu2+ + Zn  Cu + Zn2+

Al – Zn cell Electron flow Al (–) Salt bridge Zn (+) Al3+ Zn2+
Zn2+ + 2e–  Zn (reduction - GER) Al  Al3+ + 3e– (oxidation - LEO) 2Al + 3Zn2+  2Al3+ + 3Zn

For more lessons, visit www.chalkbored.com
Making Galvanic Cells Cu-Zn voltage Al-Zn voltage No bridge No bridge 0 sec  0.9 0 sec  0.5 30 sec  0.9 30 sec  0.5 Deposit Cu Deposit Al Points to Cu Points to Al 3) The salt bridge allows the cell to continue to function. K+ or NO3– can move into one of the half cells to balance the charge created by + metal ions forming or leaving solution. 4) No. We did not get the same values in the lab For more lessons, visit

Fe (+) Salt bridge Mg (–) Fe2+ Mg2+ Mg  Mg2+ + 2e– (oxidation - LEO)
Electron flow e- Fe (+) Salt bridge Mg (–) Fe2+ Mg2+ Mg  Mg2+ + 2e– (oxidation - LEO) Fe2+ + 2e–  Fe (reduction - GER) Mg + Fe2+  Mg2+ + Fe

For more lessons, visit www.chalkbored.com
Cu and Al Electron flow e- Cu (+) Salt bridge Al (–) Cu2+ Al3+ Al  Al3+ + 3e– (oxidation - LEO) Cu2+ + 2e–  Cu (reduction - GER) 3Cu2+ + 2Al  3Cu + 2Al3+ For more lessons, visit

Principles and application

Similar presentations

Presentation on theme: "Principles and application"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Principles and application

Similar presentations

Presentation on theme: "Principles and application"— Presentation transcript:

Similar presentations

About project

Feedback