Grade 11 Revision

K Warne

G12 Prior knowledge from 10 & 11

• the use of equations of motion in solving problems dealing with momentum, work, energy and power

• the use of Newton’s first, second and third laws of motion

• conservation of mechanical energy • sound waves and properties of sound• electromagnetism• Stoichiometric calculations • Concentration calculations• Balancing of chemical equations • Use of oxidation numbers • Identification and description of intermolecular forces

(Van der Waals forces and hydrogen bonds)

G12 Prior knowledge from 10 & 11

• the use of equations of motion in solving problems dealing with momentum, work, energy and power

• the use of Newton’s first, second and third laws of motion

• conservation of mechanical energy • sound waves and properties of sound• electromagnetism• Stoichiometric calculations • Concentration calculations• Balancing of chemical equations • Use of oxidation numbers • Identification and description of intermolecular forces

(Van der Waals forces and hydrogen bonds)

G12 Prior knowledge from 10 & 11

• the use of equations of motion in solving problems dealing with momentum, work, energy and power

• the use of Newton’s first, second and third laws of motion

• conservation of mechanical energy • sound waves and properties of sound• electromagnetism• Stoichiometric calculations • Concentration calculations• Balancing of chemical equations • Use of oxidation numbers • Identification and description of intermolecular forces

(Van der Waals forces and hydrogen bonds)

G11 Session 2

• Decimal conversions

• Concentration Calculations

• Calculations map

• No. of particles

• Redox reactions

Decimal Conversions

King Henry Died a miserable death calledmeasles

Kilo

Hecta Decca m(unit) deci centi milli

1000 100 10 1 1/101/100

1/1000

Volume Conversions

1 dm = ……. cm1 dm3 = ……… cm3

1 m3 = ……………… dm3 = ……………………………. cm3

….cm3

…………………………………

…… cm3

…… cm3

……. cm3 ... m3

Volume Conversions

1 dm = 10 cm1 dm3 = 1000 cm3

1 m3 = 1000 dm3 = 1 000 000 cm3

1cm3

1 dm3 (1 litre)

10 cm3

10 cm3

10 cm3 1 m3

Concentration - MolarityThe concentration of a solution is defined as the ………………. of

……………………… per ………………. (dm3) of ………………….

solute

solute

Final volume of ……………..

500cm3

=+

Concentration =Amount of ……… (……….)

Volume of ………………

30g of NaCl

C = n

v

Concentration - MolarityThe concentration of a solution is defined as the AMOUNT of SOLUTE

per LITRE (dm3) of SOLUTION.

c(NaCl) = (m/Mr)x1/v = ((30/(23+35.5))x1/0.5 =

solute

solute

Final volume of solution

500cm3

=+

Concentration =Amount of solute (moles)

Volume of solution

30g of NaCl

n

vC =

The Mole• The mole is defined as, “the amount of matter with

the same number of elementary particles as 12 grams of carbon 12”.

602 300 000 000 000 000 000 000Six hundred and two thousand, three hundred, billion billion !

6.023x1023 particles

12.00 g

CSymbol (L)

Number of particles = no of moles x no. particles in a mole

Particles = n x L

ASKEDGIVEN

Mole Calculations

MOLES MOLES

MASS MASS

VOLUME VOLUME

CONCENTRATIONCONCENTRATION

MOLARRATIO

Number Of

particles

Number Of

particles

Mixed example

Ammonia gas is made by reacting ammonium chloride with calcium hydroxide according to:

NH4Cl + Ca(OH)2 NH3 + CaCl2 + H2O

If 32.1 g of ammonium chloride reacts with 500 cm3 of a 0.75 M calcium hydroxide solution, Show by calculation; which is the limiting reagent, what volume of ammonia is produced at S.T.P in m3 and how many hydroxide ions are left after the reaction?

Redox Reactions• Involve electron transfer• Oxidation is ……….. Reduction is ………… O….R….

Na Na+ + e- Cl2 + 2e- 2Cl-

• Both processes ALWAYS occur ………………...• Oxidation is caused by …………… AGENTS – ……………!!

Na Na+ + e-, Na ……………… .: ………….. agent!! (Good one)

• Reduction is caused by …………… AGENTS – ……………!Cl2 + 2e- 2Cl-, Cl2 ………………. .: …………….. agent!

(Good)

OVERALL: 2Na + Cl2 Na+ + 2Cl-

Redox Reactions• Involve electron transfer• Oxidation is loss Reduction is gain OILRIG

• Na Na+ + e- Cl2 + 2e- 2Cl-

• Both processes ALWAYS occur together.• Oxidation is caused by OXIDIZNG AGENTS – REDUCED!!

Na Na+ + e-, Na oxidised .: reducing agent!! (Good one)

• Reduction is caused by REDUCING AGENTS – OXIDISED!Cl2 + 2e- 2Cl-, Cl2 reduced .: Oxidizing agent!

(Good)

0 0 +1 -1OVERALL: 2Na + Cl2 Na+ + 2Cl-

Reactions written as reductions.

Positive potentials accept electrons are good OXIDISING AGENTS.

Negative potentials donate electrons are good REDUCING AGENTS.

ELECTROCHEMICAL SSERIESequilibrium E° (volts)

1.5

0.8

0.34

0

-0.13

-0.44

-0.76

-1.66

-2.37

-2.71

-2.87

-2.92

-3.03

Electrochemical half-cell Electrochemical half-cell potentials are listed from +ve to potentials are listed from +ve to –ve E–ve Eθθ values. values.

Reactions take place

Reactions written as reductions.

Positive potentials accept electrons are good OXIDISING AGENTS.

Negative potentials donate electrons are good REDUCING AGENTS.

ELECTROCHEMICAL SERIESequilibrium E° (volts)

1.5

0.8

0.34

0

-0.13

-0.44

-0.76

-1.66

-2.37

-2.71

-2.87

-2.92

-3.03Ele

ctro

ns

flow

in e

xter

nal

cir

cuit

.

Electrochemical half-cell Electrochemical half-cell potentials are listed from +ve to potentials are listed from +ve to –ve E–ve Eθθ values. values.

Oxidizing Agents

REDUCINGREDUCING AGENTSAGENTS

Reactions take placeReactions take placeTop LEFT to bottom RIGHTTop LEFT to bottom RIGHT

Oxidation numbersOxidation numbersOxidation numbers can be used to identify

oxidation and reduction as well as balance equations – particularly when it is difficult to identify e- transfer.

Oxidation numbers are ………………………… charges charges that an atom of an element that an atom of an element …………….……………. in a compound, if all bonds were …………..bonds were …………...

OOHH

HH++ --

H = …., O = …..

NH3

H = ….., N = …..

Oxidation numbersOxidation numbersOxidation numbers can be used to identify

oxidation and reduction as well as balance equations – particularly when it is difficult to identify e- transfer.

Oxidation numbers are imaginaryimaginary charges charges that an atom of an element that an atom of an element would havewould have in a compound, if all bonds were ionicbonds were ionic.

OOHH

HH++ --

H = +1, O = -2

NH3

H = +1, N = -3

Oxidation number rulesOxidation number rules • The O.N. of a free element is 0• Hydrogen is +1, (except hydrides -1)• Metals: GI +1, GII +2, GIII +3, Zn +2• Halides -1, Oxides -2, sulphides -2, nitrides

-3• The O.N. of a simple ion is equal to its ionic

charge i.e. Mg2+ O.N. is +2

• In allocation of O.N., charge is conserved i.e. – the sum of the total O.N. of the atoms in a

compound is zero, Na+Cl- +1 = (-1) = 0– while that for a polyatomic ion is equal to ionic

charge on the ion. MnO4-

Mn +7 + (4xO -2) = -1

BALANCING HALF REACTIONS

1. Balance oxygens by adding H2O2. Balance hydrogens by adding H+ ions3. Balance charges by adding electrons

(to most + side)(Balanced M’s & NM’s first)

To combine 1/2 reactions1. Multiply by suitable factors to equal

out the electrons.2. Add the reactions and cancel things

appearing on both sides.3. Add spectator ions.

Redox ExampleAn iron II solution can be standardized using a

permanganate solution according to the reaction:

FeCl2 + KMnO4 FeCl3 + Mn2+

Determine; • the oxidation states of all reacting species• The forumula of the oxidising and reducing

agents • the balanced overall (ionic) reaction• The identity of spectator ions in the given

reaction