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A Level Chemistry > Module 5 - chapter 23 P1 > Flashcards

Module 5 - chapter 23 P1 Flashcards

1
Q

reduction in terms of electrons and oxidation number

A

reduction is the gain of electrons or decrease oxidation number

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2
Q

oxidation in terms of electrons and oxidation number

A

oxidation is the loss of electrons or increase oxidation number

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3
Q

oxidising agent

A

takes electrons from the species being oxidised (the oxidising agent itself gains electrons so contains the species being reduced)

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4
Q

reducing agent

A

adds electrons to the species being reduced (the reducing agent itself loses electrons so contains the species being oxidised)

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5
Q

key points for balancing redox equations

A
  • balance electrons (multiply equations until same e-)
  • balance for charge
  • balance for species (incl H+ and OH-)
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6
Q

how-to balancing ionic equations in acidic conditions

A

1) write out half equations
2) balance half equations for atoms
3) add water on one side to balance for oxygens if necessary. balance the other side for H+ after this.
4) add electrons to balance charges on both sides
5) balance half equations against each other so the same no. e- are required and produced by multiplying/dividing.
6) merge equations and cancel down. NO ELECTRONS SHOULD REMAIN OR YOU’VE BALANCED WRONG
7) check charge balances, if not something’s wrong

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7
Q

how-to balancing ionic equations in alkaline conditions

A

1) write out half equations
2) balance the half equation containing a hydroxide with OH- instead of H2O and H+
3) balance the other half equation for acidic conditions (using H2O and H+)
4) add electrons to balance charges on both sides of both equations
5) balance half equations against each other so the same no. e- are required and produced by multiplying/dividing.
6) neutralise any H+ with the same number of OH- to make the environment alkaline. add the same number of OH- to both sides so remains balanced
7) cancel out H+ and OH- to form water
8) combine equations and cancel down
9) check charges balance and no electrons remain

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8
Q

how-to balance a redox equation using oxidation numbers

A

1) write out the unbalanced equation
2) write out all oxidation numbers

3) identify changes in oxidation numbers and write clearly so you can see how many to balance.
e. g. the element, initial oxidation no., final oxidation no., and the difference between them

4) identify how the oxidation numbers need to balance and what needs to multiply to match them.
5) balance only the atoms which change oxidation state by balancing them within their compounds
6) balance any atoms which have changed due to your balancing

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9
Q

what two redox titrations do you need to know?

A 
potassium manganate (VII) under acidic conditions
and sodium thiosulfate for determination of iodine
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10
Q

key points for manganate (VII) titrations

A
  • potassium manganate (VII) KMnO4 in burette
  • excess of dilute H2SO4 added to conical flask to provide H+ for reduction of MnO4- ions to MN2+
  • reaction is self-indicating
  • manganate decolourises when added
  • as end-point approached, decolourises less rapidly
  • end-point = pale permanent pink colour (excess MnO4- ions)
  • swirl flask so end-point not judged too soon
  • meniscus read from top
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11
Q

why is the meniscus read from the top in manganate titrations?

A

KMnO4 is a deep purple colour, bottom of meniscus difficult to read through intense colour.
titre has same difference between both readings if bottom of meniscus used for initial and final readings

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12
Q

examples of reducing agents analysed with manganate titrations

A 
iron (II) ions
ethanedioic acid (COOH)2
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13
Q

determining percentage purity

A

1) calculate the titre in dm3 and use to find moles of the ion added from the burette (MnO4-)
2) work out the moles of the unknown that reacted, using the previous n
3) scale up for moles in full 250cm3 (if used 25cm3 sample in conical flask eg)
4) use moles to find mass of pure sample
5) find percentage purity by comparing to experimental mass (impure)

%purity = mass of pure/mass of impure x100

  • when calculating Mr to work out mass from moles, include Mr of any ions in compound, e.g. H+ may be included to make it neutral (in acidified solution)
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14
Q

key points about iodine-thiosulfate titrations

A
  • S2O3 2- are oxidised to S4O6 2- and iodine (I2) is reduced to 2I-
  • sodium thiosulfate = Na2S2O3 in burette
  • excess of potassium iodide added to oxidising agent in conical flask
  • iodide oxidised to iodine = yellow brown
  • iodine reduced back to I- in titration
  • end-point = pale straw colour and starch added to check
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15
Q

uses of iodine-thiosulfate titrations

A

analysing:
ClO- content of bleach
Cu2+ content in copper (II) compounds
the Cu content in copper alloys

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16
Q

how starch helps determine the end-point in iodine-thiosulfate titrations

A

starch is added once the end-point is being approached and a pale straw colour is observed. adding starch makes it go blue-black because iodine tests for starch.
titrate until it becomes creamy white - when only iodide ions remain

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17
Q

what type of titration are iodine-thiosulfate titrations?

A

back titrations - reacted with a known conc of excess reagent (KI) and then titrated against a second reagent to show how much the initial one was in excess, so the unknown concentration can be calculated

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18
Q

how copper alloys are prepared for iodine-thiosulfate titrations

A

salts are just dissolved in water but insoluble compounds can be reacted with conc acids to form CU2+ ions

copper alloys e.g. brass and bronze dissolved in conc HNO3 then neutralised with sodium carbonate to form ions, neutralises copper nitrate - nitrate ions are oxidising agents so interfere with end-point by forming iodine from iodide

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19
Q

voltaic cell

A

converts chemical energy into electrical energy

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20
Q

half cell

A

contains species present in half equation

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21
Q

Where is equilibrium set up in a half cell?

A

At the phase boundary where the metal is in contact with its ions

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22
Q

In isolated half cell?

A

No net transfer of electrons either unit or out of the metal

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23
Q

What are ion/ion half cells?

A

Contain ions if the same element in different oxidation states, e.g. Fe (II) and Fe (III)

Uses inert platinum electrode

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24
Q

Anode?

A

Negative electrode - most reactive metal which loses electrons and is oxidised

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25
Q

Cathode?

A

Positive electrode - less reactive metal which gains electrons to be reduced

26
Q

Standard electrode potential

A

The voltage/potential difference of the half cell combined with standard hydrogen electrode under standard conditions of 298K, 100KPa, and conc 1moldm-3

27
Q

How do you compare standard electrode potentials?

A

Hydrogen gas half cell with inert platinum electrode in dilute H2SO4 (0.5moldm-3 to give 1moldm-3 H+ as 2H+ produced by H2SO4)

Under standard conditions of 298K , 100KPa, all solutions 1moldm-3

28
Q

Standard electrode potential of hydrogen electrode?

A

0V

29
Q

Example of a salt bridge

A

Filter paper strip soaked in KNO3 (aq) - a solvent

30
Q

More negative E* value

A

More Negative Better Reducing Agent

Loses/produces electrons so is oxidised

31
Q

More positive E* value

A

More Positive Better Oxidising Agent

Gains electrons and is reduced

32
Q

Calculating standard cell potential

A

Most positive - most negative

= bottom - top

33
Q

What is current/ voltage?

A

How many electrons pass a certain point in a second

34
Q

What is the purpose of a salt bridge?

A

Replaces/balances the charges of ions used up on both sides of the cell so it remains neutral and to complete the circuit

Doesn’t matter if ions aren’t the same as long as they have the same charge as the others (e.g. 2+ and 2+) and won’t react with the ions to form a solid precipitate

35
Q

What does a voltmeter measure?

A

The potential difference (basically the difference in electron density of the electrodes)

36
Q

Which electrode do the electrons flow from/to?

A

From anode to cathode as oxidation takes place at anode

37
Q

Drawing electrochemical cells key points

A
  • most negative electrode potential on left because oxidised - becomes anode
  • draw direction arrows of electron flow along wire, always from left of cell to right if most neg on left
  • don’t forget to draw the solutions in
  • always draw standard hydrogen electrode on left even if not the most negative
38
Q

Why is hydrogen uses as the standard electrode potential?

A

Easy to ensure purity

39
Q

Shorthand for electrochemical cells

A

Metal anode / anode ion// cathode ion/ cathode metal

/ = phase boundary within half cell
//= external circuit

E.g. Zn/Zn2+//Cu2+/Cu

40
Q

Short hand for standard hydrogen cells

A

Pt[H2(g)] / H+(g) // Mg2+ (aq)/ Mg(s)

41
Q

Key points about electrode potential equations to predict feasibility

A

Electrons always on left of equation

Most negative on top and most positive on bottom

Anti-clockwise arrows

Bottom - top = E*

If calculated E* = positive then feasible

42
Q

Why not always feasible despite being positive E*

A

If cell potential very low but positive reaction still may not occur as activation energy may be too high

If not under standard conditions may not occur, e.g if greater concentration of solutions then would shift equilibrium right making the E* positive but if lower would shift left so E* more neg

Rate of reaction might be slow even if feasible

43
Q

Ways to increase cell potential

A

Increase conc of solution which is strongest oxidising agent (most pos E) - shifts equilibrium to right so E is higher and do opposite with other so E* is lower and difference is bigger for bigger cell potential.

Increase SA of electrodes so more metal atoms to be oxidised

44
Q

What are fuel cells

A

Use energy produced when a fuel reacts with oxygen to produce a voltage in a very exothermic reaction

45
Q

How do fuel cells work

A

Fuel and oxygen flow into cell as gases and products (water and electrons) flow out

Can flow continuously if fuel and oxygen continuously supplied into cell -doesn’t require recharging

46
Q

Examples of fuels for fuel cells

A

Hydrogen, hydrocarbons, alcohols, and any hydrogen-rich molecules e.g. methanol

Hydrogen don’t produce CO2 in combustion and only water

47
Q

Electrodes and electrolyte in fuel cells

A

Titanium sponge coated in platinum,

electrolyte is a membrane allowing the movement of ions from one cell to another like a salt bridge

Electrolyte can be acid or alkali

48
Q

Cell voltage produced by fuel cells

A

1.23V in both acid and alkali

Electrode potentials for half cells add up to same in acid and alkali conditions

49
Q

Overall reaction in fuel cells

A

H2(g) + 1/2O2(g) -> H2O(l)

50
Q

Anode of alkali fuel cell

A

Hydrogen oxidised by reaction with hydroxyl ions to produce electrons and water. Water removed from anode

51
Q

Cathode of alkali fuel cell

A

Oxygen reduced by electrons from hydrogen and reacts with water to produce hydroxyl ions

52
Q

At anode of acid fuel cell

A

Hydrogen oxidised by dissociating into ions and electrons

53
Q

Cathode of acid fuel cell

A

Oxygen reduced by the electrons from the hydrogen and reacts with hydrogen ions to produce water. Water removed from cathode

54
Q

At anode of fuel cell

A

Hydrogen enters system and flows over the electrode. Electrons flow from anode to cathode

55
Q

At cathode of fuel cell

A

Oxygen reduced by electrons from hydrogen

56
Q

Storage cells

A

Supply electricity by anode providing electrons to external circuit and cathode receives electrons - primary and secondary cells

57
Q

Primary storage cells

A

Single use batteries bc reaction is irreversible so cannot be recharged

E.g. alkaline battery

58
Q

Secondary storage cells

A

Rechargeable bc reaction is reversible. Recharged by applying external voltage to reverse reactions so cathode reduced and anode oxidised again to produce voltage. when recharged direction of reaction switches

E.g. nickel-cadmium batteries

59
Q

Benefits of electrochemical cells

A
  • hydrogen fuel cells can be used in fuel cell vehicles. Less polluting bc produce less CO2 than petrol as only water is a side product
  • fuel cells more efficient than combustion engines
60
Q

Risks of electrochemical cells

A
  • toxicity - lithium ion batteries release toxic gases when heated, electrolyte contains LiPF6 which releases fluoride and hydrogen fluoride gas
  • fire from Li based cells- Li highly reactive so if coating punctured spark ignites lithium
  • hydrogen fuel cells for vehicles must be stored at low temps as gas v flammable
  • hydrogen explosive so difficulty with storage and transportation

A Level Chemistry (21 decks)

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