Energy

Question 1

What are oxidising and reducing agents?

Answer 1

Oxidising agent: accepts electrons and get reduced.

Reducing agent: donates electrons and gets oxidised.

Question 2

Why are transition metals useful in redox titrations?

Answer 2

• They can be used as oxidising and reducing agents
• As they readily change oxidation state by donating or recieving electrons
• And change colour when they change oxidation state so endpoint is easy to spot.

Question 3

What is the half equation for the reduction of the MnO₄^- ion?

Answer 3

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

Question 4

What is the half equation for the reduction of the Cr₂O₇^2- ion?

Answer 4

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

Question 5

What are the steps of a redox titration?

Answer 5

1. Measure out a known volume of reducing agent using a pipette and place in conical flask.
2. Add dilute sulphuric acid (acidic conditions to facilitate redox)
3. Gradually add oxidising agent from burette, swirling conical flask constantly.
4. Stop when mixture in flask just turns colour of reducing agent (for MnO₄^- this is pink) and record volume of reducing agent added.
5. Repeat until you get concordant results and calculate mean titre.

Question 6

What are the steps of an iodine-thiosulphate titration?

Answer 6

1. Add a known volume of oxidising agent (thing you are trying to find conc. of) to excess acidified potassium iodide solution. Some proportion of the iodide ions will be oxidised to iodine (I_{2 (aq)}).
2. Titrate resulting solution with sodium thiosulphate solution (Na₂S₂O₃) of a known concentration. When you are close to the end point, the brown iodine colour will fade to pale yellow (as it is reduced to I^-). Add 2cm³ of starch solution; the solution will go dark blue initially as there will be some iodine remaining. Add thiosulphate one drop at a time until the blue colour disappears.
3. From volume of thiosulphate added you can calculate moles of iodine produced in step 1: I₂ + 2S₂O₃^2- → 2I^- + S₄O₆^2- . From the stoichiometry of the first reaction between the oxidising agent and potassium iodide, you can find the moles and therefore conc. of the oxidising agent.

Question 7

What is the standard electrode potential of a half-cell?

Answer 7

The emf of a half-cell, compared with a standard hydrogen half-cell measured at 298K with solution concentrations of 1 moldm^-3 and a gas pressure of 100 kPa.

Question 8

What is the convention for drawing cells?

Answer 8

[electrode] ｜reduced form (1) ｜ oxidised form (1) ‖ oxidised form (2) ｜ reduced form (2) ｜ [electrode]

where:

(1) is the substance which is oxidised in the overall cell reaction
(2) is the substance which is reduced in the overall cell reaction

Question 9

How is a standard cell potential calculated?

Answer 9

E^⦵ = E^⦵(+ve electrode) - E^⦵(-ve electrode)

+ve electrode is where reduction takes place

-ve electrode is where oxidation takes place

Question 10

How is reactivity linked to the electrochemical series?

Answer 10

• For metals: the more reactive, the more it wants to lose electrons (get oxidised) therefore the more negative its electrode potential (therefore the higher it is on the electrochemical series).
• For non-metals: the more reactive, the more it wants to gain electrons (get reduced) therefore the more positive its electrode potential is (therefore the lower it is on the electrochemical series).

Question 11

How do you calculate cell potentials when one of the half-cells contains an acid?

Answer 11

Use the hydrogen half-equation

Question 12

How can you determine the feasibility of a reaction?

Answer 12

1. Check what exactly the reaction that you are being asked about is (e.g. ions of one metal reacting with a different solid metal)
2. Write a half-equation for each reactant, with reduction as the forwards direction.
3. The one with the more positive electrode potential will go forward and the other will go backwards. Combine the half-equations bearing this in mind.
4. The resulting overall cell reaction is the feasible direction so if it matches the reaction you are being asked about, the reaction is feasible.

Question 13

What are the limitations of dermining feasiblity using the electrochemical series?

Answer 13

• Prediction using E^⦵ only states if a reaction is feasible under standard conditions:
  
  • If conc. of one of the reactants changes, overall cell reaction equilibrium will shift to counteract this change, increasing or reducing ease of electron loss and therefore changing cell potential.
• The kinetics of the reaction may not make it feasible:
  
  • rate of reaction too slow
  • high activation energy

Question 14

What is a fuel cell?

Write the equations for the process that take place at each of the electrodes in a conventional fuel cell.

Answer 14

A fuel cell produces an emf by reacting a fuel, usually hydrogen, with an oxidant, usually oxygen. [At the anode, the Pt catalyst splits the H₂ into protons and electrons. The polymer electrolyte membrane (PEM) only allows H⁺ to cross and this forces the e^- to travel around the external circuit to get to the cathode, creating a current. At the cathode, O₂ combines with the H⁺ and the e^- to form water.]

Anode: H₂ → 2H⁺ + 2e^-

Cathode: 1/2 O₂ + 2H⁺ + 2e^- → H₂O

Question 15

What are the advantages and disadvantages of fuel cells over conventional combustion engines?

Answer 15

• Advantages:
  
  • More efficient as energy isn’t wasted as heat.
  • Less pollution (CO_{2 (g)}); only waste product is water.
• Production of cells uses toxic chemicals, which need to be disposed of once the cell reaches the end of its life span.
• Chemicals used to make cells e.g. lithium are v. flammable & reactive.

Question 16

What is the standard lattice enthalpy?

Answer 16

The enthalpy change when 1 mole of an ionic lattice is formed from its gaseous ions under standard conditions. The more -ve the lattice enthalpy, the stronger the ionic bonding.

Question 17

How does ionic radius affect lattice enthalpy?

Answer 17

• Ionic radius increases
• Attraction between ions decreases
• Less energy required to make lattice
• LE become less exothermic

Question 18

How does ionic charge affect lattice enthalpy?

Answer 18

• Ionic charge increases
• Attraction between ions increases
• LE becomes more exothermic

Question 19

What is the enthalpy change of solution?

Answer 19

Enthalpy change when 1 mole of a compound is completely dissolved in water under standard conditions.

Question 20

What is the enthalpy change of hydration?

Answer 20

Enthalpy change when 1 mol of gaseous ions is completely dissolved in water to form 1 mol of aqueous ions.

Question 21

What is the standard enthalpy change of formation?

Answer 21

The enthalpy change when 1 mol of a compound is formed from its constituent elements in their standard states, under standard conditions.

Question 22

What is the relationship between enthalpies of solution, hydration and lattice enthalpy?

Answer 22

ΔH_soln = ΔH_hyd - ΔH_LE

Question 23

Explain how:

1. ionic radius
2. ionic charge

affect the enthalpy of hydration.

Answer 23

1. Ionic radius:
   
   • Ionic radius increases
   • Attration between ion and water molecules decreases
   • Hydration enthalpy less exothermic
2. Ionic charge:
   
   • Ionic charge increases
   • Attraction to water molecules increases
   • Hydration enthalpy becomes more exothermic

Question 24

What is entropy?

Answer 24

A measure of the dispersal of energy in a system, which is greater the more disordered a system is.

Question 25

What factors affect the entropy of a system?

Answer 25

• Physical states: solids are more order than liquids which are more ordered than gases.
• Number of particles: more particles means more disorder so higher entropy.

Question 26

When is a reaction:

• always thermodynamically feasible?
• never thermodynamically feasible?

Answer 26

• when ΔH is -ve and ΔS is +ve
• when ΔH is +ve and ΔS is -ve

Question 27

How can you calulate the threshold temperature at which a reaction becomes feasible?

Answer 27

Set ΔG equal to zero and rearrange to get:

T = ΔH/ΔS