Topic 14 - Redox 2 Flashcards
What is oxidation?
the loss of electrons, oxidation numbers increase
What is reduction?
The gain of electrons, oxidation numbers will decrease
How do s-block metals react in a redox reaction?
lose electrons to form positive ions with charges the same as their group number
How do p-block metals react in a redox reaction?
Can react by losing electrons, but non-metals react by gaining electrons to form negative ions with charges the same as their group number
How do d-block metals react in a redox reaction?
Form on with variable oxidation states , but tend to form positive ions with positive oxidation numbers
How is an electrochemical cell formed?
Two different metals dipped into salt solution of their own ions and connected by a wire
What is the redox process in an electrochemical cell?
Electrons flow from the anode to the cathode, where oxidation occurs at the anode and reduction occurs at the cathode
How does reactivity affect which metal is oxidised or reduced?
Reactive metals form ions more readily than unreactive metals, so more reactive metal is oxidised (anode) and the less reactive metal becomes the cathode
What is the cell potential (E cell)?
The voltage between the two half-cells
Why is platinum used as an electrode for solutions?
To provide an electrical connection to the external circuit, so needs to conduct electricity and be very inert (won’t react with anything)
Why is graphite used instead of platinum?
As platinum is very expensive
What does electrode potential measure?
How easily the substance in the half-cell is oxidised
How does a potential difference build up?
As the substances in the half-cells are either oxidised or reduced, due to the difference in charge between the electrode and the ions in solution
What is the standard electrode potential of a half-cell?
The voltage measured under standard conditions when the half-cell is connected to a standard hydrogen electrode
What happens to more negative E values?
- equilibrium lies further to the left than hydrogen
- metal atoms lose electrons more readily to form ions
What happens to more positive E values?
- Equilibrium lies further to the right than hydrogen
- metal atoms lose electrons less readily to form ions
What is the equation for E cell?
E cell = E reduced - E oxidation
What is the point of standard conditions?
Always get the same value for the electrode potential and can compare values for different cells
What are the rules of representing electrochemical cells in short hand?
- half-cell with more negative potential goes on the left
- oxidised forms go in the center
- double vertical lines represent the salt bridge
- Standard hydrogen half-cell should always go on the left
- if half-cells use inert electrodes show on outside of diagram
How does metal reactivity affect oxidation and electron potentials?
more reactive metal more easily loses electrons and have more negative standard electrode potentials
How can you determine thermodynamic feasibility from E values?
If the E cell is positive
When does disproportionation occur?
An element in a single species is simultaneously oxidised and reduced in the same reaction
What happens if you increase concentration of the more negative electrode potential?
- fewer electrons will be released
- electrode potential will be less negative
- E cell becomes less feasible
What happens if concentration of less negative electrode potential is increased?
- More electrons gained
- electrode potential will be more positive
- E cell becomes more feasible
What happens if temperature is increased in an electrochemical cell?
-shifts in the endothermic direction
- fewer electrons are released
- E cell becomes less feasible
What happens to the prediction when the reaction kinetics are not favorable?
- rate of reaction may be so slow that the reaction may not appear to happen
- has high activation energy, may stop reaction from happening
How is cell potential related o entropy and equilibrium constant?
the bigger the cell potential, the bigger the total entropy change taking place during the reaction (proportional)
How do fuel cells work?
- Use energy from a reaction of a fuel (H2) with oxygen to create a voltage
- Chemicals are stored separately outside the cell and fed in when electricity is required
How do oxygen-hydrogen fuel cells work in alkaline conditions?
- hydrogen and oxygen gas are fed into two separate platinum electrode
-electrodes are separated by an anion-exchange membrane, which the OH- pass through - electrolyte used is (kOH)
- H2 undergoes oxidation at anode (forms H+)
- O2 is reduced at the cathode (forms OH-)
- electrons flow from the positive electrode through an external circuit to the negative electrode
What is the ionic equation for the reaction at the positive electrode in alkaline conditions?
2H2(g) + 4OH-(aq) => 4H2O(l) + 4e-
What is the ionic equation for the reaction at the negative electrode in alkaline conditions?
O2(g) + 2H20(l) +4e- => 4OH-(aq)
How do hydrogen-oxygen fuel cells work in acidic conditions?
- H2 is split into H+ ions and e- by the platinum catalyst at the anode
- Polymer electrode membrane allows the H+ across, forcing e- to travel around he circuit
- nickel oxide catalyst splits O2 into O2-
- at the cathode O2 combines with H+ and e- to form water (waste product)
What is the ionic equation for the reaction that occurs at the anode in acidic conditions?
H2(g) => 2H+ + 2e-
What is the ionic equation for the reaction that occurs at the cathode in acidic conditions?
1/2O2(g) + 2H+ +2e- => H2O
Where does hydrogen come from?
- CH4 (finite , expensive)
- from renewable resources (expensive)
- Wind to hydrogen via electrolysis of water
What is the new generation of fuel cells that uses alcohols?
- Alcohol is oxidised at the anode in the presence of water =>
CH3OH + H2O => CO2 + 6e- +6H+ - H+ ions pass through the electrolyte and oxidised to water =>
6H+ +6e- + 3/2O2 => 3H2O
What is the first type of redox titrations and what is it used for?
- The use of the powerful oxidising agent potassium manganate
- used for the quantitative extimation of reducing agents
- Iron(ii)
- Ethanedioic acid
What happens to potassium manganate in redox titrations in acidic conditions?
MnO4- + 8H+ +5e- => Mn^2+ + 4H2O
- MnO4 = purple
- Mn^2+ = colourless /pale pink
- no indicator needed
What happens to potassium manganate in redox titrations in alkali conditions?
Yields manganese oxide (iv) as a brown precipitate which interferes with end point
What is the method for the potassium manganate titration?
- using a pipette, transfer known volume of iron(ii) solution (reducing agent) into a conical flask
- Add some dilute sulfuric acid (in excess)
- Slowly add MnO4- (in burette) to flask, swirling mixture as you do
- Stop when the mixture just becomes tainted with a purple colour (MnO4-), recording volume (of oxidising agent)
- Repeat until get concordant results
Why is the sulfuric acid added to the reducing agent?
To make sure there are sufficient number of H+ ions for the oxidising agent (MnO4-) to be reduced
What is the half equation of the reduction of potassium manganate?
MnO4^-(aq) + 8H+(aq) + 5e- => Mn^2+(aq) + 4H2O(l)
What is the half equation of the oxidation of Iron (ii)?
Fe^2+(aq) => Fe^3+(aq) + e-
What happens in the redox titration between potassium manganate and ethanedioic acid?
- Aqueous ethanedioic acid (extremely poisonous) is added to the conical flask
- Acidified with sulfuric acid
- Potassium manganate added to burette
- Reaction slow, heat in 60’C water bath
- Once sufficient Mn2+ is produced, reaction proceeds as normal
What is an autocatalyst?
When one of the products of the reaction acts as a catalyst for the reaction
e.g. Mn2+ for titration potassium manganate and ethanedioic acid
What is the half - equation of ethanedioic acid being oxidised?
H2C2O4 => 2H+ + 2CO2 +2e-
What is the iodine-sodium thiosulfate titration used for?
- Finding the concentration of an oxidising agent
What is the method for the iodin-sodium thiosulfate reaction?
- Using pipette, transfer volume of iodine solution into conical flask (will be brown)
- Slowly add the thiosulfate (S2O32-), swirling the mixture as you do
- iodine will turn pale yellow (reduced from iodide)
- Add a few drops of starch solution (goes deep blue) showing presence of iodine
- End point will be determined when the deep blue colour goes colourless
- Repeat for concordant results
What happens if the starch solution is added too early?
Absorbs some of the iodine - reducing accuracy
What happens when the starch solution isn’t added?
- Difficult to detect end point
- Colour of iodine solution becomes very faint - towards end of reaction
What happens in the process of making the iodine?
- Measure out certain volume of potassium iodate(v) solution - oxidising agent
- add to excess of acidified potassium iodide solution
- Iodate(v) ions oxidise the iodide ions to iodine
What is the equation for the formation of iodine?
IO3- + 5I- +6H+ => 3I2 + 3H2O
How are titrations used to find the percentage of copper in an alloy?
- Copper(ii) ions will oxidise iodide ions to iodine
- Used to find percentage
2Cu^2+ +4I- => CuI + I2
What is the method to find the percentage of copper in an alloy?
- Dissolve known mass of alloy in concentrated nitric acid (oxidises copper into ions)
- Make up to 250cm3 using distilled water in a volumetric flask
- Pipette 25cm3 of solution into conical flask
- Add sodium carbonate solution to neutralise any remaining nitric acid (until precipitate forms - removed by ethanoic acid)
- Add excess of potassium iodide solution (reacts with copper ions)
- White precipitate of copper ions form
- Titrate against sodium thiosulfate (find no. moles of iodine)
What are the sources of error in the titrations?
- Adding starch indicator at the right point
- Starch solution needs to be freshly made
- Precipitate of copper iodide makes seeing colour change difficult
- Iodine produced can evaporate
- Titration reading false
- Final figure of copper too low
- helps if the solution is cool
