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Inorganic Chemistry Y13 > 5.2.3 Redox and Electrode Potentials > Flashcards

5.2.3 Redox and Electrode Potentials Flashcards

1
Q

what is reduction

A

gain of electrons
- decrease in oxidation number

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2
Q

what is oxidation

A

loss of electrons
- increase in oxidation number

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3
Q

what occurs in redox reactions

A

oxidation and reduction

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4
Q

what is key when assigning oxidation numbers

A
  • group 1,2,3 are always just their charge
  • fluorine is always -1
  • H is always +1, except for metal hydrides
  • O is always -2, except with peroxides
  • for ions, all the oxidation numbers always equal the overall charge number
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5
Q

what is an oxidising agent

A

takes electrons from the species being oxidised
- is REDUCED itself

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6
Q

what is a reducing agent

A

adds electrons to the species being reduced
- is OXIDISED itself

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7
Q

how can you form redox equations from half equations

A

1) balance out the number of electrons
2) add the equations together, and cancel out the electrons
3) cancel out all other species present on both sides

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8
Q

what is a way to balance out redox equations that involve many O’s and H’s

A
  • via balancing changes in oxidation numbers
  • rather than just trial and error
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9
Q

how do you form redox equations using oxidation numbers

A

1) assign oxidation numbers to all species
2) identify the species that experience a change in oxidation number
3) balance out ONLY the species that have changed in oxidation number(e.g. if on increases by 6, and one decreases by 1, multiply the decreased one by 6 to match the overall change)
4) balance out all of the remaining atoms

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10
Q

what may be missing in redox equations

A

you may not know all of the species involved in the reaction, so may have to predict products/reactants

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11
Q

what is a common reactant/product of aqueous redox reactions

A
  • H2O is often formed
  • H+ and OH- are also likely
  • WHEN ADDINGS, ALWAYS DO A FINAL CHECK TO MAKE SURE BOTH SIDES OF THE EQUATION BALANCE BY CHARGE
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12
Q

how would you balance redox equations, where you are predicting products/reactants

A

1) assign oxidation numbers, and identify the species that change in oxidation numbers
2) balance out the electrons, by finding which side is more positive, so needs them added on
3) balance the remaining species, and add on your predicted products by using which elements are left out of overall equation
4) do a final balance of everything

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13
Q

what are the 2 common redox titrations

A
  • potassium manganate (VII) [KMnO4(aq)]
  • under acidic conditions
  • sodium thiosulphate [Na2S2O3]
  • through the determination of iodine I2
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14
Q

why do manganate (VII) titrations not need an indicator

A

manganate is self indicating

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15
Q

what is the basis of manganate (VII) titrations

A
  • involves the reduction of MnO4-(aq)
  • to Mn2+
  • under acidic conditions (through addition of H2SO4)
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16
Q

how do you set up a manganate (VII) titration

A
  • add a standard solution of KMnO4(aq) to a burette
  • use a pipette to add a measured volume of the solution being analysed to a conical flask
  • as well as an excess of dil. H2SO4 (provides the H+ ions needed for the reduction of MnO4-)
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17
Q

explain how you carry out the manganate titration and the colour changes

A
  • during titration, the manganate solution reacts and is decolourised as it is being added
  • end point is reached when the first permanent pink colour appears
  • indicating and excess of MnO4- ions
  • repeat the titration until you obtain 2 concordant titres
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18
Q

where do you read the meniscus from in a managantate (VII) titration

A
  • KMnO4 is a deep purple colour
  • so difficult to read the bottom of the meniscus
  • can just read it from the top
  • as we only need the difference for the titre, so subtracting 2 readings read from the top is the same as subtracting 2 from the bottom
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19
Q

why would you use a manganate VII titration

A

to analyse different reducing agents
- which are able to reduce MnO4- into Mn2+
- e.g. Fe2+ and Fe (II) (aq) ions
- e.g. ethanedioic acid ((COOH)2 (aq))

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20
Q

what are the equations involved in the manganate (VII) titration for Cu2+ ions

A

REDUCTION OF MnO4-:
MnO4- + 8H+ + 5e- →Mn2+ + 4H2O

OXIDATION OF Fe2+:
Fe2+ → Fe3+ + e-

ALL TOGETHER:
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O

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21
Q

how can you find the purity of a Cu(II) compound using a manganate (VII) titration

A
  • prepare a solution of your impure iron sample
  • add some of this sample into a conical flask, as well as an excess of H2SO4
  • use a burette and titre against KMnO4
  • can work out the moles of MnO4- used
  • so moles of the iron compound
  • compared to how many actually should have reacted (1:5 molar ratio)
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22
Q

what other things can you use in titrations instead of Manganate (VII)

A
  • other oxidising agents
  • like Cr2O7 2-/H+
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23
Q

what are the ionic equations behind the iodine/thiosulfate titration

A

THIOSULFATE IS OXIDISED:
2S2O3 2- → S4O62- + 2e-

IODINE IS REDUCED:
I2 + 2e- → 2I-

OVERALL:
2S2O3 2- + I2→ S4O62- + 2I-

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24
Q

what can iodine/thiosulfate titrations be used form

A
  • determining the ClO- concentration in household bleach
  • the Cu2+ content in copper (II) compounds
  • the Cu content of copper alloys
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25
how do you set up for an iodine/thiosulfate titration
- add a standard solution of Na2S2O3 to a burette - add a solution of the oxidising agent you are analysing using a pipette into a conical flask - add an excess of potassium iodide into the flask as well - the oxidising agent will react with the iodide ions and produce iodine: forming a yellow-brown solution - titrate the solution with the Na2S2O3
26
how is the iodine/thiosulfate titration carried out, and which colour changes accompany this
- as the Na2S2O3 is added - the iodine is reduced back to I- ions - and the brown colour fades gradually
27
how do you determine the iodine/ thiosulfate titration end point
- difficult to determine the exact endpoint of iodine disappearing - so add starch indicator once the iodine colour has faded to a pale straw colour - forms a blue-black colour due to the iodine left - when the end-point is finally reached (through adding more thiosulfate) - all the iodine will have reacted - so the blue-black colour disappears
28
what do you use iodine/thiosulfate titrations to test
- to test any oxidising agent - as long as capable of oxidising I- into I2 - e.g. chlorate (I) ions - e.g. copper (II) ions
29
what are the equations for using iodine/thiosulfate titrations to test household bleach
- the active ingredient in bleach in chlorate (I) ions ClO- (also known as hypochlorite) - ClO- + 2I- + 2H+ → Cl- + I2 + H2O - 2S2O32- + I2 → 2I- + S4O62- - giving the chlorate:thiosulfate ratio of 1:2
30
how do you carry out the iodine/thiosulfate titration to determine the bleach concentration
- add some bleach into a volumetric flask, and use water to make up a 250cm3 SS - add 25cm3 into a conical flask using a pipette - add 10cm3 of KI - and 1 mol dm-3 HCl (to acidify the solution and provide the H+ ions for reaction) - use burette and titrate against SS of Na2S2O3 - repeat until you have achieved concordant titres - analyse the results for the conc. of chlorate ions
31
how do you make up standard solutions for analysing the copper content of a compound
- for Cu2+ salts, can just dissolve compound in water to produce the Cu2+ ions - for insoluble compounds, need to react with acids - e.g. copper alloys (brass/bronze) - alloy is reacted and dissolved in conc. HNO3, and then neutralised to form the Cu2+ ions - for Cu(s) to Cu2+
32
what is the 2 step process of finding the copper content in iodine/thiosulfate titrations
FIRST: - the Cu2+ reacts with the iodide I- ions to form a solution of I2(aq) and a white precipitate of CuI (brown colour mixture) - 2Cu2+ + 4I- → 2CuI + I2 (copper is reduced and iodine is oxidised) SECOND: - the iodine is titrated with the Na2S2O3 standard solution to form nitrate ions - 2S2O32- + I2 → 2I- + S4O62-
33
what is the ratio for the iodine/thiosulfate titration with copper and chlorate
COPPER:THIOSULFATE = a 1:1 ratio CHLORATE:THIOSULFATE = 1:2
34
what is a voltaic cell
a type of electrochemical cell which converts chemical energy into electrical energy - takes place in modern cells and batteries
35
how do electrochemical cells generate electricity, and what does this mean for the type of reaction
- the electrical energy results from movement of electrons - so requires a chemical reaction which transfers electrons from one species to another - = redox reactions
36
what do half cells contain
the chemical species present in the redox half equations
37
how is a voltaic cell made
- by connecting 2 half cells
38
what is important about the 2 half cells in voltaic cells
- the chemicals in each one must be kept apart - if they mixed, the electrons would flow in an uncontrolled way - so heat energy would be released - rather than electrical
39
how would you draw the Zn2+(aq)/Zn(s) half cell
- beaker containing solution of Zn2+ - a rod in the middle of Zn(s)
40
what are the two types of half cells
metal/metal ion metal ion/metal ion
41
how is a metal/metal ion half cell set up
a metal rod dipped in a solution of its aqueous ion - SIMPLEST, where both same element
42
what is the phase boundary of a metal/metal ion half cell
where the metal is in contact with its ion - an equilibrium is set up here
43
what is the conventional way of writing an equilibrium
- FORWARD REACTION: shows reduction - BACKWARD REACTION: shows oxidation - e.g. Zn2+ + 2e- ⇌ Zn
44
what happens in an isolated half cell
there is no net transfer of ions either into or out of the metal
45
what is the direction of electron transfer dependent on when 2 half cells are connected
e- flow depending on the relative tendency at each electrode to release electrons
46
how are ion/ion half cells set up
- contain ions of the same element in different oxidation states as the solution - there is no metal to transfer electrons into or out of the half cell, so an inert metal electrode is used, e.g. Pt - shown via beaker with solution, e.g. Fe2+/Fe3+ and the electrode labelled with Pt
47
how do you find out which electrode has the greatest tendency to lose or gain electrons
- electrode with MORE REACTIVE metal: loses electrons and is oxidised, forming the NEGATIVE electrode - electrode with LESS REACTIVE metal: gains electrons and is reduced, forming the POSITIVE electrode
48
what is the sign for standard electrode potential
E°
49
what is E°
the tendency to gain electrons, and be reduced
50
what is the standard chosen for E° values
- hydrogen gas - and a solution of H+ ions - with an inert platinum electrode - the standard electrode potential of this half cell is 0V
51
what are the standard conditions needed for E°
- all solutions at 1 moldm-3 (e.g. the acidic H+ solution) - 298K (25C) for gases - pressure of 100kPa
52
how would you draw the standard hydrogen half cell set up
- beaker containing you H+ solution - the platinum electrode - a glass tube to allow H2 bubbles to escape - the H2 gas entering from this glass tube at the side - REMEMBER TO LABEL ON STANDARD CONDITIONS
53
what is standard electrode potential
the emf (voltage) of a half cell connected to a standard hydrogen half-cell under standard conditions
54
how do you measure the standard electrode potential
connect the half cell to a standard electrode (hydrogen)
55
how would you set up apparatus to find the E° of a half cell
- set up both of your electrodes - connect them with a wire attached to the electrodes (to allow a controlled flow of electrons) - have a voltmeter in the middle of this wire - ALWAYS SHOW THE FLOW OF ELECTRONS IN THIS WIRE USING AN ARROW - connect the solutions with a salt bridge (to allow the ions to move and flow)
56
what are typical examples of salt bridges
- usually contain a concentrated solution of an electrolyte that does not react with either solution on filter paper - e.g. filter paper soaked in KNO3(aq)
57
where can you find the E° values of many redox systems
many have already been measured and listed in data reference tables - REMEMBER: in data tables, they always show the equilibrium so that the forward reaction is reduction (ALWAYS GAINING ELECTRONS)
58
what does a more negative E° value show
- a greater tendency to lose electrons - and undergo oxidation - (the less the tendency to gain electrons and undergo reduction)
59
what does a more positive E° value show
- the greater the tendency to gain electrons - and undergo reduction
60
what E° values do metals tend to have
- more negative E° - so more likely to lose electrons - the more negative this value is for a metal, the greater reactivity it has, in losing electrons
61
what E° values do non-metals tend to have
- more positive E° values - so more likely to gain electrons - the more positive this value is for non-metals, the greater their reactivity in gaining electrons
62
what is a cell potential
- the e.m.f value measured when connecting 2 half cells - via symbol Ecell - so just he potential difference between 2 half cells
63
how do you set up half cells to measure cell potentials
1) prepare 2 half cells at standard conditions 2) connect the metal electrodes of the half cells using a wire to a voltmeter 3) prepare a salt bridge (soak filter paper in saturated KNO3(aq) 4) connect the 2 solutions using a salt bridge 5) record the standard cell potential using a voltmeter
64
what is the flow of electrons in a cell potential Ecell
from the MORE negative half cell to the LESS negative half cell, so from the negative to positive electrode - less negative E° is more likely to gain electrons
65
how do you write an overall equation from electrode potentials
- combine the relevant E° values: 1) REDUCTION equation stays the same, as if gaining electrons 2) OXIDATION half equation is reversed, to show the electrons are being lost 3) overall, combine the 2 half equations, and get rid of the electrons
66
how do you calculate standard cell potential
- using the standard E° values, and the difference between E°cell = E°(positive electrode) - E°(negative electrode)

Inorganic Chemistry Y13 (8 decks)

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