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Aqa a level chemistry > Topic 7: Oxidation and reduction (redox reactions) (physical chem) > Flashcards

Topic 7: Oxidation and reduction (redox reactions) (physical chem) Flashcards

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1
Q

define oxidation

A

loss of electrons

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2
Q

define reduction

A

gain of electrons

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3
Q

When does a redox reaction occur

A

when oxidation and reduction occur simultaneously (one species looses electrons which are then donated and gained by the other species)

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4
Q

What is an oxidation number

A

Gives the oxidation state of an element or ionic substance
-e.g hydrogen = +1
-oxygen = -2
Group 7 = -1
Group 1 = +1

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5
Q

Oxidising agent

A

-accepts electrons from the species that is being oxidised so it gains electrons and is reduced (oxidation no. becomes more negative)

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6
Q

Reducing agent

A

-donates electrons to the species being reduced
-looses electrons and is oxidised (positive oxidation no.)

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7
Q

half equations

A

-used to show separate oxidation and reduction reactions that occur

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8
Q

state in terms of electrons the meaning of the term oxidising agent

A

electron accepter

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9
Q

equation for oxidation of iron

A

Fe2+ –> Fe3+ + e-

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10
Q

when is oxidation number the same as charge

A

only for monotomic elements

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11
Q

Cl2 (aq) + 2Br (aq) –> 2Cl- (aq) + Br2 (aq) –> what is being oxidised and what is being reduced

A

Cl2 = reduced (neutral to negative)
Br = oxidised (negative to neutral)

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12
Q

negative to neutral

A

oxidisation

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13
Q

reduction vs oxidation

A

-Elements can either be oxidised or reduced (transfer of electrons between elements and the system)

-reduction = gain of electrons

-oxidation = loss of electrons

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14
Q

Al3+ + 3e- –> Al what is happening in this equation

A

aluminum is reduced (gains electrons)

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15
Q

2Cl- –> Cl2 + 2e- what is happening here

A

chlorine is oxidsed (looses electron)

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16
Q

what side of the equation has oxidised and what side is reduction

A

oxidised = right hand side
reduction = left hand side

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17
Q

steps to calculating half equations

A

1) molecule –> compound or compound –> molecule

2) Balance atoms

3) See charges to tell whether it is reduced or oxidised

4) Add electrons

5) Balance equation

18
Q

solid vs aqueous

A

Solid = no charge

Aqueous = charge

19
Q

Zn + CuSO4 –> ZnSO4 + Cu
write the two half equations

A

Zn –> Zn2+ + 2e- (oxidised)

Cu2+ + 2e- –> Cu (reduction)

20
Q

molecule to compound

A

reduction

21
Q

calculate the oxidation number in Fe2O3

A

Fe = +3
O = -2

22
Q

state in terms of electrons why iodine is classified as an oxidising agent in this reaction

A

-electron acceptor
-iodine gains electrons and becomes reduced

23
Q

what is oxidised and reduced in this equation using oxidation numbers

H2S + Cl2 –> S + 2HCl

A

S has been oxidised from -2 in H2S to 0 in S

Cl has been reduced from 0 in Cl2 to -1 in 2HCl

24
Q

define oxidation number

A

-Oxidation number shows us how many electrons are gained or lost by an element when atoms turn into ions

-Oxidation number –> the charge that an ion has or the charge it would have if the bonding electrons are completely transferred

25
Q

more positive vs negative oxidation number

A

The more electrons removed (positive oxidation number) = oxidation

The more electrons added (negative oxidation number) = reduction

26
Q

rules for oxidation numbers

A

-uncombined element = zero

-ions of just one atom = charge of the ion

-neutral compound = zero

-sum of oxidation numbers for an ion = charge of ion

-Flourine in its compound = -1

-Group 1 metals = +1

-H = +1 except in metal hydrides which would be –1

-Oxygen = -2 expect when bonded to flourine or H2O2

27
Q

oxidising agent

A

oxidises another element and gets reduced itself by receiving electrons from the element becoming oxidised. (element reduced)

28
Q

reducing agent

A

Reducing agent –> reduces another element and gets oxidised by transferring its electrons to the elements being reduced (element oxidised)

29
Q

reducing and oxidation number in equation

Mg + CuSO4 –> MgSO4 + Cu

A

-> Mg is oxidised (0 –> +2)

–> Cu is reduced (+2 –> 0)

–> Cu in CuSO4 is oxidising agent

—> Mg is the reducing agent

30
Q

equation for Cr3+ –> Cr2O7

A

2Cr3+ + 7H2O –> Cr2O7^2- + 14H+ + 6e-

31
Q

disproportionation

A

-Disproportionation is where an element is simultaneously oxidised and reduced in the same reaction

-e.g Cl2 can be both oxidised and reduced in two separate compounds so has experienced disproportionation

32
Q

balancing redox equations

A

-chemical equation –> ionic equation –> half equation –> redox

1) Write half equations for reduction and oxidation

2) Add H2O to balance oxygen

3) Add H+ to balance hydrogen atoms

4) Add electrons to balance the charges (must be the same on both sides e.g -1 on both sides)

33
Q

combining half equations

A

1) Write both half equations balanced

2) Multiply so number of electrons on both equations are equal

3) Add half equations together and cancel out electrons, H+ ions and common species

34
Q

Explain in terms of oxidation states why magnesium is the reducing agent

A

Mg changes oxidation state from 0 to +2 so electrons are lost

35
Q

oxidation

A

ion –> molecule + electron

36
Q

A solution of sodium chlorate was added to a colourless solution of potassium iodidie. Suggest what happens and why

A

-solution goes brown
-bc of iodine
-iodine is oxidised

37
Q

oxidiation vs reduction

A

oxidation = increase in oxidation number
reduction = decrease in oxidation number

38
Q

reducing agents are

A

electron donors
(species that is oxidised)

39
Q

write a half equation for the reduction of chlorate ions to form chlorine in acidic conditions

A

2ClO- + 2e- + 4H+ —> Cl2 + 2H2O

40
Q

true or false - oxidation state must equal charge of compound

e.g SO4^2- –> charge of s is 6 instead of 8

A

true

Aqa a level chemistry (19 decks)

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