topic 4 - chemical changes Flashcards

reactivity of metals: OILRIG, reactivity series, reaction of acids: with metals, pH scale, neutralisation of acids, salt production, electrolysis: process, electrolysis of molten + aqueous compounds/solutions, aluminium (bauxite), half equations

1
Q

what is the reactivity series

A

potassium
sodium
calcium
magnesium
aluminium
carbon
zinc
iron
tin
lead
hydrogen
copper
silver
gold
platinum

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2
Q

what is OILRIG

A

OILRIG stands for:

Oxidation Is Loss Reduction Is Gain
(of electrons)

at a cathode (-) it is always reduction
at an anode (+) it is always oxidation

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3
Q

what are the acid + metal reactions (3)

A

acid + metal oxide –> metal salt + water
acid + metal hydroxide –> metal salt + water
acid + metal carbonate –> metal salt + carbon dioxide + water

e.g.

hydrochloric acid + copper oxide –> copper chloride + water

hydrochloric acid + sodium hydroxide –> sodium chloride + water

nitric acid + calcium carbonate –> calcium nitrate + carbon dioxide + water

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4
Q

what is the pH scale

A

the pH scale measures how acidic an object is
objects that are not very acidic are called basic
the scale has values ranging from zero (the most acidic) to 14 (the most basic)
pure water has a pH value of 7

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5
Q

electrolysis process

A

investigate what happens when aqueous solutions are electrolysed using inert electrodes

equipment
- copper chloride solution
- copper sulfate solution
- sodium chloride solution
- sodium sulfate solution
- 100cm3 beaker
- petri dish lid
- two carbon rod electrodes
- two crocodile/4 mm plug leads
- low voltage power supply
- blue litmus paper
- tweezers

method

  • pour copper chloride solution into the beaker to about 50cm3
  • add the lid and insert carbon rods through the holes the rods must not touch each other
    attach crocodile leads to the rods and connect the rods to the dc (red and black) terminals of a low voltage power supply
  • select 4v on the power supply and switch on
  • observe both electrodes; is there bubbling at neither, on or both electrodes
  • with tweezers hold a piece of blue litmus paper in the solution next to the positive electrode (one connected to the red terminal)
    *if the gas produced at the +ve electrode doesn’t bleach the litmus paper then it’s oxygen
    *gas produced at the -ve electrode is hydrogen
  • after no more than 5 minutes, switch off the power supply
    –> observe the -ve terminal, is there metal coating on it - this is pure copper
  • clean the equipment carefully and repeat the steps for copper sulfate, sodium chloride, sodium sulfate

findings

copper chloride
- +ve electrode (anode) –> bubbled, chlorine
- -ve electrode (cathode) –> copper formed

copper sulfate
- +ve electrode (anode) –> bubbled, oxygen
- -ve electrode (cathode) –> copper formed

sodium chloride
- +ve electrode (anode) –> bubbled less, chlorine
- -ve electrode (cathode) –> bubbled lots, hydrogen

sodium sulfate
- +ve electrode (anode) –> bubbled, oxygen
- -ve electrode (cathode) –> bubbled more, hydrogen

safety
wear goggles throughout

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6
Q

rules of electrolysis of aqueous solutions

A
  1. if solution contains metal ions from a metal less reactive than hydrogen, we get that metal discharged from the solution
  2. if the metal ions in a solution are more reactive than hydrogen, we get hydrogen gas discharged
  3. if solution contains any chloride (Cl-) bromide (Br-), iodide (I-) ions we get Cl₂, Br₂, I₂
    *if none of these are present you get O₂ (oxygen)
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7
Q

aluminium manufacture

A

aluminium ore is called Bauxite
it contains the compounds Aluminium oxide (Al₂O₃) which has a very high melting point and is insoluble

cryolite is added to Al₂O₃ to lower melting point so it is less expensive to melt (less energy is used)

cathode
Al(+3) + 3e(-) –> Al

anode
2O(-2) - 4e(-) –> O₂

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8
Q

why is the positive electrode in the cell used up during the process of aluminium manufacture

A

positive electrodes are made out of carbon
carbon reacts with oxygen to make CO₂
CO₂ burns away the anode and needs to be replaced frequently

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9
Q

half equations - rule of oxidation and reduction

A

electrons on the right of the equation (after the arrow), show that they are being lost - this means OXIDATION

electrons on the left of the equation (before the arrow), show that they are being gained - this means REDUCTION

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10
Q

how to do half equations

A

check you have the same number of atoms on the left and the right side if not, balance before you do anything

then look at the charges and write them out (no charge = 0, + = 1, +2 = +2 etc)

afterwards see which side has the lowest/negative charge and make the other side equal that –> you can only do this by adding electrons

add the correct amount of electrons

done//

for oxidation

Cu –> Cu(2+)

0 ——–> 2+
0 is the lowest number so make 2+ become zero by adding electrons (e-)

Cu –> Cu(2+) + 2e-
(2+ + 2- = 0)

//

for reduction

Cl₂ –> Cl(-)
Cl₂ –> 2Cl(-) –> balanced

0 ——> (2x-1) -2
-2 is the lowest number so make the 0 become -2 by adding electrons (e-)

2e- + Cl₂ –> 2Cl(-)
(2- + 0 = -2)

//

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