Module 5 Section 2: Energy Flashcards
Combine the two half equations
Why are transition metals good oxidising and reducing agents
They are good at changing oxidation state
So they’ll give out or receive electrons
Also change colour so it’s easy to spot when the reaction is finished
How potassium manganate is used as an oxidising agent
Contains manganate (VII) ions (MnO4 -) in which manganese has an oxidation state of +7
Can be reduced to Mn2+ ions during a redox reaction
MnO4 - is purple and [Mn(H2O)6]2+ is pale pink, but looks colourless so the colour change in the reaction is purple to colourless
How is potassium dichromate used as an oxidising agent
Contains dichromate(VI) ions (Cr2O7 2-) in which chromium has an oxidation state of +6
They can be reduced to Cr+3 ions during a redox reaction
Cr2O7 2- is orange and [Cr(H2O)6]3+ looks green so colour change is from orange to green
How to work out the concentration of a reducing agent
Titrate a known volume of it against an oxidising agent of known concentration
Can work the other way
How to carry out redox titration with Fe2+ as a reducing agent and MnO4 - as a oxidising agent
Measure out a quantity of the Fe2+ reducing agent
Add dilute sulfuric acid to the flask (this is an excess to make sure there are enough H+ to allow oxidising agent to be reduced)
Gradually add the aqueous MnO4 - to the reducing agent using a burette, swirling the conical flask as you do so
Stop when the mixture in the flask just becomes tainted with the colour of the MnO4 - (end point) and record volume of oxidising agent added
Run more titrations and calculate mean volume of MnO4 -
How can iodine - sodium thiosulfate titrations be used
They are a way of finding the conc of an oxidising agent
The more concentrated an oxidising agent is, the more ions will be oxidised by a certain volume of it
How to prepare for a titration of iodine and sodium thiosulfate
Use a sample of oxidising agent to oxidise as much iodide as possible by measuring out 25cm3 of potassium iodate solution (KIO3 and oxidising agent)
Add this to excess acidified potassium iodide solution (KI) where the iodate(V) ions oxidise some of the iodide ions to iodine
Then find out how many moles of iodine have been produced by titrations the solution with a known conc of sodium thiosulfate (Na2S2O3)
Method of titration of iodine with sodium thiosulfate
Take flask containing solution of oxidised iodide ions (iodine and water)
Add sodium thiosulfate from a burette to the flask drop by drop
Colour fades to pale yellow (hard to see at end point) so add 2cm3 of starch solution to detect presence of iodine so solution goes dark blue showing there still iodine there
Add sodium thiosulfate one drop at a time until blue colour disappears
When this happens, all the iodine has just been reacted
Then calculate number of moles of iodine in the solution
How to do titrate calculation to find number of moles of iodine produced
How to calculate the concentration of oxidising agent using the original equation (6.6x10-4 moles of iodine and 25cm3 oxidising agent)
Definition of lattice enthalpy
Lattice enthalpy is the enthalpy chance when 1 mole of an ionic lattice is formed from its gaseous ions under standard conditions of 298K and 100kPa
What is lattice enthalpy
Ionic compounds can form regular structures called giant ionic lattice
The positive and negative ions are held together by electrostatic attractions
Lattice enthalpy is when gaseous ions combine to make a solid lattice
What does lattice enthalpy measure
This measures the ionic bond strength
The more negative the lattice enthalpy the stronger the bonding
What makes the lattice enthalpy more exothermic
The higher the charge on the ions the more energy is released when an ionic lattice forms
Due to stronger electrostatic forces between the ions
This means that the lattice enthalpy will be more negative
The smaller the ionic radii of the ions involved, the more exothermic the lattice enthalpy
Smaller ions have a higher charge density and the smaller ionic radii means the ions can sit closer together in the lattice
What is the born Haber cycle used for
Lattice enthalpy can’t be measured directly so this cycle is used to calculate the enthalpy change if you took another less direct route
Meaning of the born Haber cycle diagram
Route 1:
Formation of gaseous atoms
• changing the elements in their standard states into gaseous atoms (endothermic process)
Formation of gaseous ions
• changing gaseous atoms into positive and negative gaseous ions (endothermic)
Lattice formation
• changing gaseous ions into the solid ionic lattice (this is lattice enthalpy and is exothermic)
Route 2
Converts elements in their standard states directly to the ionic lattice
Includes one enthalpy change which is enthalpy or formation and is exothermic
Standard enthalpy change of atomisation
The enthalpy change that takes place for the formation of one mole of gaseous atoms from the element in it’s standard state under standard conditions
What sign is in front of the enthalpy of atomisation
Standard enthalpy of atomisation is always an endothermic process so + value
Bonds are broken to form gaseous atoms
What is electron affinity
This is the opposite of ionisation energy
Electron affinity measures the energy to gain electrons
Always exothermic as the electron being added is attracted to nucleus
Definition of the first electron affinity
The enthalpy change that takes place when one electron is added to each atom in one mole of gaseous atoms to form one mole of gaseous 1- ions
Diagram example of successive electron affinities of oxygen
Why are second electron affinities endothermic
Second electron affinities are endothermic
This is because an electron is being gained by a negative ion which repels the electron away
Means that energy must be put in to force the negative electron onto the negative ion
How can electrochemical cells be made
Electrochemical cells can be made from two different metals dipped in salt solutions of their own ions and connected by a wire (the external circuit)
What are the two reactions taking place in an electrochemical cell
Oxidation and reduction
What happens in a zinc/copper electrochemical cell
Zinc loses electrons more easily than copper. So in the half-cell on the left, zinc (from the zinc electrode) is oxidised to form Zn2+ ions.
This releases electrons into the external circuit.
In the other half-cell, the same number of electrons are taken from the external circuit, reducing the Cu2+ ions to copper atoms
Which way do electrons flow in electrochemical cells
Flow from more reactive metal to least reactive metal
What would adding a volt meter into the circuit allow you to find
Shows the voltage between the two half-cells.
This is the cell potential or e.m.f.
Add a high resistance voltmeter
Other types of electrochemical cells
You can also have half-cells involving solutions of two aqueous ions of the same element, such as Fe2+/Fe3+
The conversion from Fe2+ to Fe3+, or vice versa, happens on the surface of a platinum electrode.
Electrochemical cells can also be made from non-metals.
For systems involving a gas (e.g. chlorine), the gas can be bubbled over a platinum electrode sitting in a solution of its aqueous ions (e.g. Cl-)
Platinum used as its inert and conducts electricity
What determines what direction the reaction goes in at each electrode
The reversible arrows show that both reactions can go in either direction.
Which direction each reaction goes in depends on how easily each metal loses electrons (how easily it’s oxidised).
What are electrode potentials
This measures how easily a metal is oxidised
A metal that’s easily oxidised (losing electrons) has a very negative electrode potential
One that’s harder to oxidise has a less negative or a positive electrode potential.
Example of how electrode potentials can determine the direction of the reaction in the cell
The zinc half-cell has a more negative electrode potential, so zinc is oxidised (the reaction goes backwards)
Copper is reduced (the reaction goes forwards)
What does the Θ symbol mean
The Θ symbol next to the E means it’s under standard conditions - 298K and 100 kPa.
Shorthand for writing electrochemical cell
Use example of Zn/Cu cell
The half-cell with the more negative potential goes on the left.
The oxidised forms go in the centre of the cell diagram
How to calculate overall cell potential
What voltage will the cell potential always be
The cell potential will always be a positive voltage
Because the more negative E° value is being subtracted from the more positive E° value.
E.g. the cell potential for the Zn/Cu cell - +0.34 - (-0.76) = +1.1 v
How do you measure the electrode potential
You measure the electrode potential of a half-cell against a standard hydrogen electrode
Definition of the standard electrode potential
The standard electrode potential, E°, of a half-cell is the voltage measured under standard conditions when the half-cell is connected to a standard hydrogen electrode.
What are the standard conditions for electrochemical cells
Any solutions must have a concentration of 1.00 mol dm3 or be equimolar (i.e. contain the same number of moles of ions).
The temperature must be 298 K (25 °C).
The pressure must be 100 kPa
What side is the standard hydrogen electrode shown
The standard hydrogen electrode is always shown on the left
Doesn’t matter whether or not the other half-cell has a more positive value.
The standard hydrogen electrode half-cell has a value of 0.00 V.
How to work out whole cell potential while using standard hydrogen electrode
Whole cell potential = E°RHS -E°LHS
E°LHS is 0V so voltage reading will be equal to E°RHS
Reading could be positive or negative depending on the way the electrons flow
How can conditions effect the value of the electrode potential
Like regular reversible reactions the equilibrium position in a half-cell is affected by changes in temperature, pressure and concentration.
Changing the equilibrium position changes the cell potential.
Standard conditions are used to measure electrode potentials
Using these conditions means you always get the same value for the electrode potential and you can compare values for different cells
Shifting a position of equilibrium can increase or decrease the cell potential depending on whether the position shifts to produce more or less electrons
Overall equation for oxidation of iodide to iodine using iodate ions
Ionic equation for iodine reacting with sodium thiosulfate