Unit 5- Topic 16 Flashcards
Describe the term oxidation in terms of electron transfer and oxidation number
Oxidation is the loss of electrons
An element in a species is oxidised when its oxidation number increases
Describe the term reduction in terms of electron transfer and oxidation number
Reduction is the gain of electrons
An element in a species is reduced when its oxidation number decreases
To what elements of the Periodic table does the definitions of oxidation and reduction apply to
They apply to all elements whether they are in the s-, p- or d- block of the Periodic Table
Definition of standard electrode potential
The measured potential difference of a cell when no current is flowing (emf) when a half-cell is connected to a standard hydrogen electrode under standard conditions
What are the standard conditions that the standard electrode potential are measured by
298K temperature
Gas pressure: 100kPa
Concentration of ions: 1 mol dm-3
These standard conditions apply to all equilibria
What reactions are involved in the measurement of electrode potential
All redox reactions and therefore equilibria
What is the dynamic equilibrium formed of a metal immersed in water
The rate at which ions are leaving the surface of the metal to go into solution is the same as the rate at which they are joining it from the solution so:
M x+ (aq) + xe- <——> M (s)
Why is the standard hydrogen electrode necessary
Because it is not possible to measure the absolute potential difference between a metal electrode and its solution since to establish a electrical connection with the solution a metal would be needed therefore you would be ensuring the potential difference between the two pieces of metal.
The conditions for only the SHE
Hydrogen gas at pressure: 100kPa
Platinum foil dipped in solution
Solution with concentration of [H+] = 1 mol dm-3
At 298K temperature
What is needed to connect a SHE to a metal half-cell
A high resistant voltmeter connected with platinum wire to metal and platinum foil of SHE
A salt bridge
Why is the salt bridge needed in electrochemical cells
Allows movement of ions
The ions in the salt ridge should not interfere with the ions in both half-cells
Why is the platinum foil covered in porous platinum
Porous platinum has a large surface area and allows an equilibrium between hydrogen ions in solution and hydrogen gas to be established quickly
Why is a high-resistance voltmeter used
To achieve no flow of electrons in the external circuit so that the voltmeter represents the difference in potential between the two half-cells when both reactions are in equilibrium
In electrochemical cells, inn which electrode does oxidation occur
Anode
Equation for emf (E cell)
E cell = E rhs - E lhs
How to measure E wit a systems involving gases
Gas is bubbled into a solution containing its ions. A piece of platinum covered in porous platinum is dipped into the solution to connect the gas molecules and its ions to the externa circuit through an electrical connection
How to calculate E cell of two non-metal elements and their ions
Each non- metal is compared to a SHE. A half-cell containing a solution of both ions (of one element) with each of concentration 1 mol dm-3 is connected to the SHE and E annotated. Same is done with the other non-metal, and redox reaction calculated.
Components needed to measure the E of ions of same element but different oxidation numbers
-platinum wire
-platinum foil covered in porous platinum
-solution with both ions at a concentration of 1 mol dm-3
What half cell goes to the left of the electrochemical cells
The negative electrode, the one with the most negative E value
What does the vertical solid line mean in a cell diagram
A phase boundary
What are the two rules to follow when drawing a cell diagram
1-the two reduced forms of the species are shown on the outside of the cell diagram
2-the positive electrode is shown on the right-hand side of the cell diagram
What does double vertical lines represent in cell diagram
Salt bridge
Draw the standard hydrogen electrode diagram
H+(aq) l 1/2 H2 (g) l Pt(s)
Draw the cell diagram of Fe 2+ and Fe 3+ half cell
Pt(s) l Fe 3+ (aq), Fe 2+ (aq)
Where does the S.H.E always go in cell diagram
Always written in the left-hand side
Write the cell diagram for the S.H.E and zinc metal
Pt(s) l 1/2H2 (g) l H+ (aq) ll Zn 2+ (aq) l Zn (s)
What may be the reasons for a thermodynamic feasible reaction to not take place
The reactants may be kinetically stable because the activation energy of the reaction is very large
The reaction may not be taking place under standard conditions since:
A reaction that is not thermodynamically feasible under standard conditions may become feasible when the conditions are altered
Why is the electrode potential altered by a change in the conditions of the reaction
Because the position of equilibrium of the half-cell reaction may change
How can you use standard electrode potentials to predict thermodynamic feasibility of a reaction
Change the sign of the reversed reaction and calculate the new Ecell for the whole reaction. If the Ecell is positive then th reaction is likely to be thermodynamically feasible if not then it is not thermodynamically feasible.
What is the relationship between total entropy (Stotal) and emf of a cell (Ecell)
T change in Stotal = nFEcell
n and F are both constants therefore at a given temperature, the total entropy change is proportional to the emf of the cell
How can entropy change and Ecell be used to predict direction of reaction
If Ecell is positive, change S total is also positive and therefore reaction will occur.
What is the relationship between the equilibrium constant and Ecell
Since change Stotal = RlnK and change in Stotal= nFEcell / T
nFEcell lnK=\_\_\_\_\_\_\_\_\_\_\_\_ RT
What other name can standard electrode potentials be referred as
Standard reduction potentials
How is the electrochemical series ordered
The most negative E values are placed at the top and most positive at the bottom
Where are reducing agents located in the electrochemical series
The species on the right-hand side of the half-equations are all capable of behaving as reducing agents as they can lose electrons
Where are oxidising agents located in the electrochemical series
The species on the left-hand side of the half-cell equations are all capable of acting as oxidising agents as they can gain electrons
What does a very negative E value mean in the electrochemical series
The more negative the E value, the more the equilibrium lies towards the left- and the more readily the species on the right loses electrons therefore the more powerful the reducing agent
What does a very positive E value mean in the electrochemical series
The more positive the E value, the more the equilibrium lies towards the right - and the more readily the species on the left loses electrons therefore the more powerful the oxidising agent
What is a disproportionation reaction
A reaction in which an element in a species is both oxidised and reduced in the same reaction at the same time
If the E value is the most negative of the two reactions proposed, to which side will the equilibrium fall
Equilibrium will move to the left, releasing electrons
If the E value is the most positive of the two reactions proposed, to which side will the equilibrium fall
Equilibrium will move to the right, accepting electrons
How to calculate percentage uncertainty
uncertainty
Percentage uncertainty =_______________ x 100
measured value
How to reduce percentage uncertainty in brunette readings
Dilute the chemical in the burette
This means that more will be required for the titration
Therefore, the overall uncertainty in the burette readings is reduced
Is potassium manganate (VII) a powerful reducing or oxidising agent?
A powerful oxidising agent
Give the half equation for the reduction of manganate (VII) ions in acidic solution
MnO4 - (aq) + 8H+ (aq) + 5e- -> Mn2+ (aq) + 4H2O (l)
Why is potassium manganate(VII) not used in redox titrations in alkaline solutions
Because it yields manganese (IV) oxide as a brown precipitate which interferes with te end point colour of the titration. So it is used in acidic solutions to prevent the formation of manganese (IV) oxide
Describe the behaviour (colour) of potassium manganate (VII) in a redox titration
As the titration proceeds, manganese (II) ions, Mn2+, accumulate but, at the dilution used, they give a colourless solution. As soon as the potassium manganate (VII) is in excess, the solution becomes pink. It therefore acts as its own indicator- the end point is the first permanent pink colour
Give the ionic half-equations of the titration of potassium manganate (VII) and iron (II) ions
MnO4- (aq) + 8H+ (aq) + 5e- -> Mn2+ (aq) + 4H2O (l)
Fe2+ (aq) -> Fe3+ + e-
Give the overall equation for the titration of potassium manganate (VII) and iron (II) ions
MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) -> Mn2+ (aq) +4H2O (l) +5Fe3+ (aq)
Method for performing the titration of potassium manganate (VII) and iron (II) ions
-Pipette an accurately measured volume, usually 25.0cm3, of the iron (II) solution into a conical flask and then add a small volume of dilute sulfuric acid
-Slowly add potassium manganate(VII) solution of accurately known concentration from a burette and swirl the mixture
-The potassium manganate(VII) solution will turn colourless until all of the iron (II) ions have been oxidised
-The addition of one more drop of potassium manganate(VII) solution will turn the mixture pale pink
What occurs to ethanedioic acid in the titration of this with potassium manganate (VII), give the ionic half-equation
The ethanedioic acid is reduced to carbon dioxide
H2C2O4 (aq) -> 2H+ (aq) + 2CO2 (g) +2e-
Give the overall equation for the titration of ethanedioic acid and potassium manganate (VII)
2MnO4- (aq) + 16H+(aq) + 5C2O4 2+ (aq) -> 2Mn2+ (aq) + 8H2O (l) + 10CO2 (g)
Method for the titration of ethanedioic acid and potassium manganate (VII)
-Aqueous ethanedioic acid heated to 60C is pipetted into the conical flask. (The acid is highly poisonous so a pipette filler must be used) and is the acidified with dilute sulfuric acid.
-Aqueous potassium manganate (VII) is added from the burette.
-Since the reaction is slow, the pink does not immediately disappear.
-After the initial sample is added, the Mn2+ ions produced act as a catalyst and the reaction speeds up, allowing you to titrate normally and obtain an accurate end point.
Why is the aqueous ethanedioic acid heated to around 60 C before its titration with potassium manganate(VII)
Because the reaction is very slow, help overcome the activation energy.
How is the sulfuric acid treated when it is going to be used in a titration with iron (II)
The aqueous sulfuric acid is boiled to remove dissolved oxygen that might otherwise oxidise some of the iron(II) ions to iron (III)
State the ionic half-equations involved in the titration of iodine and sodium thiosulfate
Thiosulfate ions reduce iodine to iodide ions.
2S2O3 2- (aq) -> S4O6 2- (aq) + 2e-
I2 (aq) + 2e- -> 2I- (aq)
State the ionic half-equations involved in the titration of iodine and sodium thiosulfate
Thiosulfate ions reduce iodine to iodide ions.
2S2O3 2- (aq) -> S4O6 2- (aq) + 2e-
I2 (aq) + 2e- -> 2I- (aq)
Overall equation for the redox titration of iodine and sodium thiosulfate
2S2O3 2- (aq) + I2 (aq) -> S4O6 2- (aq) + 2I- (aq)
What is the indicator used non the redox titration of iodine and sodium thiosulfate
The indicator is starch solution. With free iodine it produces a deep blue-black colour. The blue-black colour disappears as soon as sufficient sodium thiosulfate has been added to react with all of the iodine
What can the redox titration of sodium thiosulfate and iodine be used for
-the direct estimation of iodine
-the estimation of a substance that can take part in a reaction that produces iodine
Method of the redox titration of sodium thiosulfate and iodine
-Add sodium thiosulfate solution from the burette to the iodine solution until the original brown colour of the iodine changes to pale yellow
-Add a few drops of starch solution to produce a blue colouration
-Add the sodium thiosulfate solution drop by drop until the blue-black solution turn colourless
Why is starch added to the redox titration of sodium thiosulfate and iodine
Because if not, it would be very difficult to detect the end point. The colour of the iodine solution becomes very faint towards the end of the reaction, and this makes it difficult t accurately assess when the end point has been reached
Why do you need to start the redox titration of sodium thiosulfate and iodine without the starch solution and add it later
Because if starch is added too early, it adsorbs some of the iodine and reduces the accuracy of the titration.
What effect does increasing the concentration have in the emf
Makes it more positive as fewer electrons are made in the reaction
What effect does increasing the pressure have in the emf
More negative as more electrons are produced
Which electrode is the cathode in a fuel cell
The positive electrode
Which electrode is the anode in a fuel cell
The negative electrode
What is a fuel cell
A fuel cell produces a voltage from the chemical reaction of a fuel with oxygen. In the hydrogen-oxygen fuel cell, hydrogen is supplied externally as a gas, and the cell can operate as long as the fuel supply is maintained.
What fuels are used in fuel cells
Hydrogen, methanol/ethanol, hydrogen-rich methanol
Why are both metal electrodes coated with platinum in the hydrogen-oxygen fuel cell
Because it catalyses the reactions that take place at the electrodes.
Reaction in negative electrode [anode] in acidic conditions in fuel cells
H2 (g) -> 2H+ (aq) + 2e-
Reaction in positive electrode [cathode] in acidic conditions in fuel cells
1/2 O2 (g) + 2H+ (aq) + 2e- -> H2O (l)
Overall reaction for fuel cells in both conditions
H2 (g) + 1/2 O2 (g) -> H2O
Overall reaction for fuel cells in both conditions
H2 (g) + 1/2 O2 (g) -> H2O
What happens in a hydrogen-oxygen fuel cell
The hydrogen ion (protons) pass through the proton exchange membrane, which allows them to enter the compartment containing the positive electrode, where they can react with oxygen.
Reaction in negative electrode [anode] in alkaline conditions in fuel cells
H2 (g) + 2OH- (aq) -> H2O (l) + 2e-
Reaction in positive electrode [cathode] in alkaline conditions in fuel cells
1/2 O2 (g) + H2O (l) + 2e- -> 2OH- (aq)
Advantages of hydrogen-oxygen fuel cells
-offer alternative to fossil fuels
-the only waste product is water, so no pollutants like CO or CO2 are produced
-they are lighter and more efficient than engines that use fossil fuels
Disadvantages of hydrogen-oxygen fuel cells
-hydrogen explodes very easily so great care is needed when transporting it. You need to either: compress the gas, adsorb it onto the surface of a suitable solid material, absorb it into a suitable material
-compressing the gas is dangerous as transporting hydrogen under pressure is hazardous
-there is no firm conclusion as to what metal should be used to adsorb hydrogen
-metal hydride absorb hydrogen but high temperatures are required to release the hydrogen.
-supply of hydrogen is a problem since we are now extracting it from methane which is a limited source. Extracting it from renewable sources is possible but expensive.