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Chemistry HL > Redox Processes (topic 9/19) > Flashcards

Redox Processes (topic 9/19) Flashcards

1
Q

Oxidation number

A

Roman numerals ex. v, vii

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2
Q

Oxidation state

A

+5, +7.
**Start with the plus or minus before the number otherwise there will be no point given.

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3
Q

Oxidation states of elements

A

0

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4
Q

The sum of the oxidation states

A

The sum of the oxidation states is the same as the charge on the ions. No charge = 0

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5
Q

Oxygen in a compound

A

Has an oxidation state of -2 **except in peroxides where it is -1

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6
Q

Hydrogen in a compound

A

Has an oxidation state of +1 **except if bonded to a metal, where it is -1

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7
Q

Flourine in a compound

A

Has an oxidation state of -1

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8
Q

Group 1, 2, and 13 in a compound

A

Have oxidation states of:

Group 1: +1
Group 2: +2
Group 13: +3

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9
Q

Halogens in a compound (group 17)

A

Probably -1 with exceptions. Except flourine which is -1

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10
Q

Tricky ones

A
  1. VNO3 = +1, +5, -2
  2. CuSO4 * 5H2O = +2, +6, -2, +1, -2
  3. Fe(ClO4)2 = +2, ClO4-, ClO4-
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11
Q

Upsetting ones

A
  1. H2CO (methanal) = +1, 0, -2
  2. C3H8 (propane) = -8/3, +1
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12
Q

Oxidation (definitions)

A
  • Adding oxygen
  • Removing hydrogen
  • Loss of electrons
  • Oxidation state increases
  • OIL (oxidation is loss of electrons)
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13
Q

Reduction (definitions)

A
  • Removing oxygen
  • Adding hydrogen
  • Gain of electrons
  • RIG (reduction is gain of electrons)
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14
Q

Oxidizing agents

A

Force other chemicals to be oxidized (to lose electrons). Therefore the oxidizing agent itself is gaining electrons (reduction). An oxidizing agent is itself reduced. Where the oxidation arrow begins, that’s where the reducing agent is.

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15
Q

Reducing agents

A

Reducing agent is itself oxidized. Where the reduction arrow begins, that’s where the oxidizing agent is.

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16
Q

How to turn a formula into a name

A

In a compound, the second element’s subscript is multiplied to its oxidation number divided by the first element’s subscript.

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17
Q

How to turn a name into a formula

A

In a compound, the oxidation number of the elements are their own subscripts using the crossover method.
**Empirical formula is used so double check!!!

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18
Q

Deducing whether an element undergoes oxidation or reduction

A
  1. Writing out the oxidation number of elements
  2. Looking at which elements have changed the most
  3. Deduce whether the change is oxidation or reduction
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19
Q

Metal activity series (in the data booklet)

A

Top:
- most reactive
- “easy” to be oxidized
- best reducing agent
- low i.e.

Bottom:
- least reactive

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20
Q

Activity series tell you which metals react with which ions

A

“Most reactive ends up as an ion”

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21
Q

Zinc + Copper sulfate

A

Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq)
Zinc metal(s) + copper sulfate(aq) = copper + zinc sulfate

Solid zinc metal dipped in aqueous copper sulfate solution makes copper on the zinc metal and zinc sulfate in the copper sulfate solution.

Zinc becomes an ion. Reactivity: zinc > copper

22
Q

Silver nitrate + copper

A

2AgNO3(aq) + Cu(s) -> Ag(s) + Cu(NO3)2(aq)
Silver nitrate(aq) + copper(s) = silver + copper nitrate

Solid copper wire dipped into silver nitrate solution = silver nitrate turning into silver coating around the copper and the liquid solution making copper nitrate

Copper becomes an ion. Reactivity: copper > silver

23
Q

Zinc nitrate + copper

A

Zn(NO3)2(aq) + Cu(s) = no reaction because zinc > copper in the reactivity series

24
Q

Zinc nitrate + silver

A

Zn(NO3)2(aq) + Ag(s) = no reaction because zinc > silver in the reactivity series

25
Q

Reactivity series for metals

A
  • Top is most reactive metal
  • Metals hate e- (electrophobic)
  • Wants to lose electrons
  • Wants to oxidize
  • Is a good reducing agent
26
Q

Ways of depleting oxygen dissolved in water

A
  1. Raw sewage when mixed with water become oxidized, thus removing oxygen from the water itself
  2. Nitrate fertilizers will encourage the growth of algae which will reduce oxygen levels
  3. Phosphates that are used in many detergents
27
Q

Winkler experiment for dissolved oxygen

A
  1. Pouring water from a lake into a volumetric flask, fill it up completely and measure the amount of liquid contained after. Slow pouring, funnel touching glass to avoid aeration.
  2. Take the temperature of the liquid.
  3. Add manganous sulfate monohydrate (white powder) (not specified how much) into the volumetric flask. Supplies the manganese 2 ions (Mn2+)
  4. Add alkaline iodide azide pillow (small white powder packet) into the volumetric flask. Supplies iodide and increases the pH cause we need hydroxide ions.
  5. Cork up the volumetric flask. Make sure there are no air bubbles and shake.

**If it turns brown there is dissolved oxygen.
**The brown is the manganese 2 turning to manganese 4

  1. Set the volumetric flask down and let the precipitate settle down halfway.
  2. Once it is halfway, shake it up again and set the volumetric flask down and let the precipitate settle down halfway.
  3. Carefully remove the stopper.
  4. Add 2 ml of concentrated sulfuric acid (H2SO4). The new brown colour is the iodine molecules being formed. Manganese 4 turns into manganese 2.
  5. Carefully transfer solution into a beaker. DO NOT AERATE.
  6. Do a titration with sodium thiosulfate (Na2S2O3). Sodium thiosulfate is gonna turn the brown I2 iodine molecules into colorless iodide molecules.
  7. Getting closer to the end of the reaction (when the solution turns lighter in colour), add starch to sharpen the end point (two pinches approx.).

**Make sure to write down the titration readings

  1. Use a graduated cylinder (or another accurate glassware) to measure the liquid inside the volumetric flask.
28
Q

Winkler method using equations

A

2Mn^2+(aq) + 4OH-(aq) + O2(aq) -> 2MnO2(s) + 2H2O(l)

MnO2(s) + 2I-(aq) + 4H+(aq) -> Mn^2+(aq) + I2(aq) + 2H2O(l)

I2(aq) + 2S2O3^2-(aq) -> S4O6^2-(aq) + 1I-(aq)

  1. If you know the volume and concentration of sodium thiosulfate (Na2S2O3 / 2S2O3 ^2-)(aq) you can get the moles. Then you can get the moles for I2(aq)
  2. Knowing the moles of iodine, you can get the moles for manganese oxide (MnO2)(s)
  3. Once you know that, you can work out the dissolved oxygen content (mol) and knowing the volume of water used, you can get the concentration
29
Q

3 measurements of concentration

A
  1. conc (mol dm^-3) = mol / dm^3
  2. conc (g dm^-3) = mass of solute / volume
  3. conc (ppm) = (mass of constituent / mass of sample) x 10^6
30
Q

Biological oxygen demand

A
  • Get sample of water (which probably has pollution and bacteria)
  • Saturate it with oxygen until it can’t take any more oxygen
  • Close the lid

The organisms are going to digest their food and use the oxygen up. The pollution is food for the organisms so they use up more oxygen.

You need to wait 5 days until you can do the Winkler test / BOD

31
Q

BOD

A

Is the measurement of nitrates, phosphates and sewage allowing organisms to remove oxygen as they grow and die. A high BOD is unhealthy for the ecosystem.

32
Q

Salt bridge

A
  • Completes the circuit
  • Balances the charge in the half cells
33
Q

Electrolysis / electrolytic cell

A

Put electricity IN (add electricity). Electrical energy to chemical energy.

34
Q

Voltaic / electrochemical / galvanic cell

A

Get electricity OUT (make electricity). Chemical energy to electrical energy.

35
Q

Wires

A

Conductor (allows the passage of electricity and is unchanged by it

36
Q

Electrode

A

Normally graphite or platinum. Solids used in the electrolytic cell.

37
Q

Electrolyte

A

Substance that conducts electricity as a liquid / aqueous and is chemically decomposed in the process. As a solid, it is an insulator.

38
Q

Positive electrode

A

Anode

39
Q

Negative electrode

A

Cathode

40
Q

Diatomic elements

A

Hydrogen (H), nitrogen (N), oxygen (O), fluorine (F), chlorine (Cl), bromine (Br), iodine (I)

41
Q

Convention on writing the cell (cell notation)

A

LHS || RHS

LHS = oxidation half-cell. Loss of electrons
RHS = reduction half-cell. Gain of electrons
|| = salt bridge

42
Q

Standard electrode potential

A
  1. Make a standard half-cell
  2. Make a 1 mol solution of an ion you’re interested in
  3. Attach an electrode with a metal you’re interested in / platinum electrode
    At 100 kPa (standard pressure)
    At 298 K (standard temperature)
  4. Attach a voltmeter on the half-cell
  5. Attach a standard hydrogen electrode on the voltmeter

The voltage that the voltmeter reads is the standard electrode potential = Eθ

43
Q

Standard hydrogen electrode

A

Only HCl and HNO3 (any other acids won’t work)

1 mol dm^-3 H+(aq)
100 kPa hydrogen gas
Platinum electrode
25 degrees C or 298K

44
Q

If the battery works

A

= spontaneous battery
= + voltage

45
Q

If the battery will not work

A

= non-spontaneous battery
= - voltage

46
Q

Electrolysis experiments

A
  • Diluted NaCl(aq)
  • Concentrated NaCl(aq)
  • CuSO4(aq) with inert electrodes (such as graphite or platinum)
  • CuSO4(aq) with copper electrodes
47
Q

Dilute NaCl(aq)

A
  1. Label the positive electrode and negative electrodes
  2. Write down the ions that are present (Na+, Cl-, H+, OH-)
  3. Looking at the positive ions because they are attracted to the negative electrodes. **lowest on the list (of standard electrode potential) which is H+
  4. Looking at the negative ions because they’re attracted to the positive electrodes. **preferentially discharged which is OH-

You can observe colourless bubbles from each electrodes

48
Q

Concentrated NaCl(aq)

A
  1. Label the positive electrode and negative electrodes
  2. Write down the ions that are present (Na+, Cl-, H+, OH-)
  3. Looking at the positive ions because they are attracted to the negative electrodes. **lowest on the list (of standard electrode potential) which is H+
  4. In a concentrated reaction, the more concentrated one is discharged (forms an element), Cl-

You can observe green bubbles

49
Q

CuSO4(aq) with inert electrodes (such as graphite or platinum)

A
  1. Label the positive electrode and negative electrodes
  2. Write down the ions that are present (Cu2+, SO4^2-, H+, OH-)
  3. Looking at the positive ions because they are attracted to the negative electrodes. **lowest on the list (of standard electrode potential) which is Cu2+
  4. Looking at the negative ions because they’re attracted to the positive electrodes. **preferentially discharged which is OH-

Copper 2+ is gonna stick on the negative electrode (plated). Oxygen gas is gonna be released at the positive electrode

50
Q

CuSO4(aq) with copper electrodes

A
  1. Label the positive electrode and negative electrodes
  2. Cu2+ will go to the negative electrode making the electrode bigger (reduction)
  3. Positive electrode will dissolve into Cu2+ (oxidation)

The colour of the electrolyte stays the same (blue)

Chemistry HL (20 decks)

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