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A level- Chemistry > Redox > Flashcards

Redox Flashcards

1
Q

Define oxidation

A

Loss of electrons or increase in oxidation number

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2
Q

Define reduction

A

Gain of electrons or decrease in oxidation number

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3
Q

What is an oxidising agent

A
  1. A reagent that oxidises (takes electrons from) another species.
  2. It contains the species that is reduced.
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4
Q

What is a reducing agent

A
  1. A reagent which reduces (adds electrons to) another species
  2. Contains the species that is oxidised
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5
Q

Write the steps for writing a redox equations from half-equations

A
  1. Write out the half equations
  2. Balance the electrons
  3. Add the half equations and cancel the electrons
  4. Cancel any species that are on both sides of the eqaution
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6
Q

Some ideas of how to predict products of redox reactions

A
  1. In aqueous redox reactions H2O is often formed.
  2. H+ and OH- ions are also likely products
  3. Check that both sides are balanced by charge
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7
Q

Write the half-equations for the following reactions and state whether oxidation or reduction:

  1. Manganate (VII) ions, MNO4- reacting in acidic conditions to form manganese(II) ions
  2. VO3- reacting in acidic conditions to produce Vanadium (III) ions
  3. Chromium (III) ions reacting in alkaline solution to produce chromate (VI) ions
A
  1. MnO4- + 8H+ + 5e- → Mn2+ + 4H2O = Reduction
  2. VO3- + 6H+ + 2e- →V3+ +3H2O = Reduction
  3. Cr3+ + 8OH- →CrO4 2- + 3e- + 4H2O = Oxidation
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8
Q

Describe the procedure of a manganate (VII) titration

A
  1. A standard solution of potassium manganate is added to the burette
  2. Using a pipette, add a measured volume of the solution being analysed to the conical flask.
  3. An excess of dilute sulfuric acid is also added to provide the H+ ions required for the reduction of MnO4- ions.- no indicator needed
  4. During the titration the manganate solution reacts and is decolourised and the endpoint is judged by the first permanent pink colour, indicating when there is an excess of MnO4- ions present.
  5. Repeat until you obtain concordant titres. 6. The manganate titrations can be used for the analysis of many different reducing agents e.g Fe2+ and COOH2
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9
Q

Describe how you would analyse the purity of an iron (II) compound

A
  1. Do a manganate titration with impure iron compound e.g. FeSO4.7H2O -this will be in a conical flask and have the KMnO4 added
  2. Calculate the meant titre- volume of KMnO4- and from this calculate the mols of MnO4- reacted
  3. Then determine the mols of Fe2+ that reacted using the equation and molar ratios.
  4. Scale up to find the amount of Fe2+ in the original solution
  5. Find mass of FeSO4.7H2O in the impure sample
  6. Find percentage purity= mass of FeSO4.7H2O/mass of impure sample* 100
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10
Q

What is a half-cell

A
  1. A half-cell contains the chemical species present in a redox half-equation.
  2. A voltaic cell can be made by connecting together two different half-cells, which then allow electrons to flow.
  3. A voltaic cell is a type of electrochemical cell which converts chemical energy into electrical energy - takes place in modern cells and batteries that power devices such as mobile phones. 4. In the cell the chemicals in the two half-cells must be kept apart- if allowed to mix, the electrons would flow in an uncontrolled way and heat energy would be released rather than electrical energy.
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11
Q

Describe what a metal/metal ion half-cell looks like

A
  1. A metal rod dipped into a solution of its aqueous metal ion. A vertical line is used for the phase boundary between the aqueous solution and the metal e.g Zn2+(aq)|Zn(s)
  2. At the phase boundary an equilibrium will be set up
  3. Convention is that the equilibrium is written so that the forward reactions shows reduction and the reverse shows oxidation.
  4. In an isolated half-cell there is no net transfer of electrons either into or out of the metal.
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12
Q

Describe what an ion/ion half-cell looks like

A
  1. Contains ions of the same element in different oxidaiton states e.g Fe2+ and Fe3+ Fe3+ (aq) + e- ↔ Fe2+ (aq)
  2. There is no metal to transport electrons in this half-cell so and inert metal electrode made out of platinum is used.
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13
Q

How do you know which electrode has a greater tendency to gain or lose electrons in a cell with two metal/metal ion half-cells.

A
  1. The more reactive metal releases electrons more readily and is oxidised
  2. The electrode with the more reactive metal loses electrons and is oxidised- negative electrode
  3. The electrode with the less reactive metal gains electrons and is reduced- positive electrode.
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14
Q

Define standard electrode potential

A

The e.m.f of a half-cell compared with a standard hydrogen half-cell measured at 298 K with solution concentrations of 1 mol dm-3 and a gas pressure of 100kPa

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15
Q

Describe the standard hydrogen half-cell

A
  1. Beaker containing 1 mol dm-3 of H+(aq)
  2. Glass tube with holes to allow bubbles of H2 (g) to escape
  3. H2(g) Going into the glass tube at the top
  4. Platinum electrode.
  5. The standard electrode potential of a standard hydrogen electrode is exactly 0V
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16
Q

Describe how to measure a standard electrode potential

A
  1. Connect the half-cell to a standard hydrogen electrode
  2. The two electrodes are connected by a wire to allow a controlled flow of electrons
  3. The two solutions are connected with a salt bridge which allows ions to flow. The salt bridge typically contains a concentrated solution of an electrolyte that does not react with either solution e.g. a strip of filter paper soaked in aqueous potassium nitrate
  4. A voltmeter is connected in the middle of the wires.
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17
Q

Why is potassium nitrate used as a salt bridge over potassium iodide and chloride and sulfate for an Ag+ cell

A
  1. All the other chemicals will react with Ag+ to produce precipitates.
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18
Q

Suggest why, when measuring a standard electrode potential, it is necessary to arrange things such that no current is allowed to flow

A
  1. Once current starts to flow, solution concentrations will have changed, so conditions will no longer be standard.
19
Q

Describe what the standard electrode potential values mean

A
  1. The more negative the E value- the greater the tendency to lose electrons and undergo oxidation and the less the tendency to gain electrons and undergo reduction.
  2. The more positive the E value- the greater the tendency to gain electrons and undergo reduction and the less the tendency to lose electrons and undergo oxidation.
  3. Metals tend to have negative E values and lose electrons
  4. Non-metals tend to have positive E values and gain electrons.
20
Q

Describe how to set up cells to measure standard cell potentials Ecells

A
  1. Prepare two standard half-cells using standard conditions
  2. Connect the metal electrodes of the half-cells to a voltmeter using wires.
  3. Prepare a salt bridge by soaking a strip of filter paper in KNO3
  4. Connect the two solutions of half-cells with a salt bridge
  5. Record the standard potential from the voltmeter
  6. The cell potential is the potentials difference between the two half-cells
21
Q

Write the equation to find the standard cell potential from the standard electrode potentials

A
  1. Ecell= E(reduction)-E(oxidation)
  2. Measured in Volts- remember to have capital letter
22
Q

Suggest two reasons why, even when half-cells are correctly connected together, the predicted result may not take place.

A
  1. The activation energy may be very high, so the rate of reaction is too slow to be observed
  2. The actual conditions (often concentration) used may not be standard, so the predictions are no longer valid (the half reactions concerned are reversible equilibria so their position and hence the electrode potential will change.)
  3. Need to be aqueous equilibria but many reactions take place that are not aqueous.
23
Q

How should you answer questions about reactions feasibility

A
  1. Talk about the E values and talk about them as being more positive or more negative than each other.
  2. Use this to say whether species will oxidise or reduce each other.
  3. If the Ecell value is negative then the reaction is not feasible
  4. The more positive E value species will oxidise the more negative species.
24
Q

Explain whether Fe2+ can oxidise magnesium metal and write the equation. Fe2+ 2e-↔Fe E= -0.44 Mg2+ + 2e- ↔ Mg E= -2.37

A
  1. Fe2+ has a more positive E value than Mg2+/Mg
  2. Therefore Fe2+ can oxidise Mg 3. Fe2+ + Mg → Mg2+ + Fe
25
Q

Manganate(VII) ions in acidic solution are often used as an oxidising agent. Sulfuric acid is frequently used. Explain why it is not a good idea to use hydrochloric acid in such solutions.

A
  1. Cl2/Cl- has a more negative E value than MnO4- and H+/ Mn2+
  2. Therefore MnO4- ions in acid solution can oxidise the Cl- ions to Cl2 gas, which is toxic
  3. Also, MnO4- will have been used to oxidise the Cl- ions rather than what it is supposed to be oxidising
26
Q

Explain why an excess of sulfruic acid should be added to either the manganate solution or the test solution before titrating

A

To provide the H+ ions needed for MnO4- to be reduced

27
Q

What happens to the ‘.xH2O’ when a sample of (COOH)2.xH2O is dissolved in water?

A

It becomes part of the solvent

28
Q

Experiments involving iodine/thiosulfate titrations

A
  1. Thiosulfate ions S2O3 2- (aq) are oxidised: 2S2O3 2- →S4O6 2- + 2e-
  2. Iodine is reduce: I2 + 2e- → 2I-
  3. The concentration of aqueous iodine can be determined by titration with a standard solution of sodium thiosulfate.
  4. These titrations can be used to determine: - ClO- concentration in bleach - Cu2+ content in copper (II) compounds - The Cu content in copper alloys
29
Q

Describe the procedure of a iodine/thiosulfate titration

A
  1. Add standard solution of Na2S2O3 to the burette
  2. Prepare a solution of the oxidising agent to be analysed, and using a pipette add this solution to a conical flask
  3. Then add an excess of Potassium iodide. The oxidising agent reacts with iodide ions to produce iodine - solution turns yellow-brown colour
  4. Titrate this solution with Na2S2O3. During the titration the iodine is reduced back to I- ions and the brown colour fades quite gradually- difficult to decide end-point
  5. Use starch indicator- when end-point is being approached the iodine colour has faded enough to become a pale straw colour.
  6. A deep blue-black colour forms to assist with the identification of the end point.
  7. At the end point, all the iodine will have just reacted and the blue-black colour disappears.
30
Q

How do you use iodine-thiosulfate titrations to analyse oxidising agents

A

e. g ClO- concentration in bleach
1. React with I- and H+ to form I2 ClO- + 2I- + 2H+ →Cl- + I2 + H2O
2. In the titration I2 reacts with the S2O3 2- ions 2S2O3- + I2 → 2I- + S4O6 2-
3. 1 mol ClO- produces 1 mol I2 which reacts with 2 mol S2O32-. So 1 mol ClO- is equivalent to 2 mol S2O3 2-
4. So you find out the mols of S2O3 2- reacted from the titre volume
5. Then Determine the amount of I2 and ClO- that reacted using molar ratio
6. Then determine the amount of ClO- in the 250cm3 solution. Multiply by 10
7. Determine concentration of ClO- in bleach. See how much bleach was used to prepare the standard solution. Multiply to get how many mols 1dm3 contains and that is equal to the concentration

31
Q

Why is starch not added to the titration before the titration begins in an iodine-thiosulfate titration

A

Once added, it is impossible to judge how close the endpoint of the titration is. Once the iodine is straw coloured and the endpoint is approaching, the starch is added and from this point the thiosulfate can be added dropwise.

32
Q

What are the 3 main types of cell

A
  1. Primary
  2. Secondary
  3. Fuel cells
33
Q

Describe the features of a primary cell

A
  1. Non-rechargeable so are designed to be used once only
  2. Electrical energy is produced by oxidation and reduction at the electrodes
  3. However the reactions cannot be reversed so eventually the chemicals will be used up, voltage will fall, the battery will go flat and the cell will be discarded or recycled.
  4. Primary cells are used for low-current, long-storage devices e.g wall clock and smoke detectors
34
Q

Give a common example of a primary cell

A
  1. Most are alkaline based on zinc and manganese dioxide and a potassium hydroxide alkaline electrolyte
  2. Zn (s) + 2MnO2(s) → ZnO (s) + MnO3 (s)
35
Q

Describe the features of secondary cells

A
  1. Rechargeable
  2. The cell reaction producing chemical energy can be reversed during recharging. The chemicals in the cell are then regenerated and the cell can be used again
36
Q

Give 3 common examples of secondary cells

A
  1. Lead-acid batteries used in car batteries
  2. Nickel-cadium (NiCd) cells and nickel-metal hydride (NiMH)- the cylindrical batteries used in radios and torches etc.
  3. Lithium-ion and lithium-ion polymer cells used in our modern appliances- laptops, tablets, cameras, mobile phones
37
Q

Describe the lithium-ion and lithium-ion polymer cells

A
  1. Lithium is light metal which translates into a very high energy density
  2. Cells can be regular shape- but because the solid polymer is flexible, flexible batteries can be easily formed into various shapes and sizes. ideal for fitting around other components in laptops and phones.
  3. When a lithium-ion cell charges and discharges, electrons move through the connecting wires to power the appliances, whilst Li2+ ions move between the electrodes within the cell.
  4. The negative electrode is made of graphite covered in lithium metal - Li →Li+ + e-
  5. The positive electrode is made out of a metal oxide e.g CoO2 - Li+ + CoO2 + e-→ LiCoO2
  6. They can become unstable at high temperatures and on rare occasions have ignited laptops and phones.
38
Q

Describe the features of fuel cells

A
  1. Use energy from the reaction of a fuel with oxygen to create a voltage
  2. The fuel and oxygen flow into the fuel cell and the products flow out. The electrolyte remains in the cell
  3. Fuel cells can operate continuously provided that the fuel and oxygen are supplied into the cell
  4. Fuel cells do not have to be recharged
39
Q

What is the main type of fuel cell

A
  1. Hydrogen as they produce no CO2 as water is the only combustion product
  2. They can have either an alkali or acidic electrolyte`
40
Q

Show the equations for an alkali fuel cell

A
  1. OH- electrolyte
  2. Redox systems: - 2H2O + 2e- ↔ H2 + 2OH- - 1/2O2 + H2O +2e- ↔ 2OH-
  3. Oxidation- H2 + 2OH- → 2H2O + 2e-
  4. Reduction- 1/2O2 + H2O + 2e- → 2OH-
  5. Overall- H2 + 1/2O2 → H2O
41
Q

Show the equations for an acidic fuel cell

A
  1. H+ electrolyte
  2. Oxidation is the same as a standard hydrogen half cell
  3. Redox system - 2H+ + 2e- ↔ H2 - 1/2O2 + 2H+ + 2e- ↔ H2O
  4. Oxidation- H2→ 2H+ + 2e-
  5. Reduction- 1/2O2 + 2H+ + 2e- → H2O
  6. Overall- H2 + 1/2 O2 → H2O
42
Q

What are the advantages of secondary over primary cells

A
  1. They can be recharged and reused, reducing environmental impact (waste & raw materials) and overall long term costs.
43
Q

What are the advantages of the hydrogen fuel cell

A
  1. Only waste is water- not contribute to climate change
  2. More efficient
  3. Oxygen used is extracted from the air and the hydrogen can be extracted from methanol (renewable resource)
44
Q

What are the disadvantages of the hydrogen fuel cell

A
  1. Petroleum is easy to store liquid. But Hydrogen is a gas with low boiling point needs to be stored in cooled and pressurised containers.
  2. Cell has limited lifespan and would need to be replaced
  3. But it is not more dangerous to transport hydrogen than petroleum.

A level- Chemistry (31 decks)

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