Topic 14 Redox II

Question 1

How to set up an electrochemical cell?

Answer 1

Step 1 - Obtain the metals under investigation and clean them with sandpaper to ensure surface impurities are removed
Step 2 - Some metals have grease on the surface. Wash the surface with propanone. Wear gloves to prevent contamination moving forward
Step 3 - Place each metal into a solution containing the ion of the same metal
Step 4 - Make the salt bridge from filter paper soaked in saturated KNO3 or Kcl. This should link the 2 beakers and each end submerged in each beaker
Step 5 - Connect the electrodes with wires, crocodile clips and a voltmeter. A reading will appear if it is set up correctly.
If you are using an oxidising agent containing oxygen you need to add acid too

Question 2

What is a half cell?

Answer 2

A half cell is one-half of an electrochemical cell. They can be constructed of a metal dipped in its ions, or a platinum electrode with 2 aqueous ions

If we had an iron electrode dipped into a solution of Fe2+ then a reaction will take place when connected to another half-cell. The reaction is in equilibrium

Fe2+ (aq) + 2e- <——–> Fe(s)

If we have a half cell with 2 aqueous ions we must use an inert but electrically conductive electrode. Platinum is commonly used. The equation to represent this half cell is - Fe3+(aq) + e- <——> Fe2+(aq)

Question 3

What are electrochemical cells made of?

Answer 3

They are made of 2 half-cells joined by a wire, a voltmeter and a salt bridge

A voltmeter is used to measure the voltage between the 2 half cells. This is called the EMF or E cell
Electrons flow from a more reactive metal to a less reactive one

Question 4

What reaction do you get when connecting 2 half cells?

Answer 4

One side undergoes a reduction process, and the other undergoes an oxidation process. This is a REDOX reaction

The less reactive (zinc) half cell shows a loss of electrons as it loses electrons easier compared to the more reactive other half cell (copper). Oxidation has occurred.
ZN<—–>Zn2+ +2e-
Observation - The zinc electrode will become thinner as more zn2+ is produced to make the electrons

The less reactive half cell (copper) accepts the electrons produced by the more reactive half cell (zinc). Reduction has occurred.
Cu2+ +2e- <—–>Cu
Observation - The copper electrode will get thicker as cu2+ ions receive the electrons and turn into copper

Question 5

What is the salt bridge made of?

Answer 5

The salt bridge is just filter paper with saturated KNO3 solution. Ions flow through, which balances the charges.

Question 6

What is electrode potential?

Answer 6

Each half cell has an electrode potential value that is measured in Volts. It tells us how easily the half cell gives up electrons (oxidised)

Question 7

How do we write the eqations in electrochemical cells?

Answer 7

You always write them in the reduced form. This means we always show reduction in the forward direction (electrons on the left side)

Zn2+ + 2e- <—–> Zn
Cu2+ +2e- <——> Cu

Question 8

What is the way to tell if something is being oxidised or reduced from the data booklet value?

Answer 8

Remember NO PR
the more negative half cell will undergo oxidation
The more positive half cell with undergo reduction

The more negative one( the one being oxidised) is losing electrons so you need to flip the equation.
Zn <——> Zn2+ + 2e-

Then you combine the 2 equations
Zn + Cu2+ <——> Zn2+ + Cu

Question 9

What is the Standard Hydrogen electrode? (SHE)

Answer 9

It is used as a reference to measure standard electrode potentials.

Electrode potentials of half cells can’t be measured on their own. We can measure them against a reference half cell called a SHE. The SHE has E theta value of 0.00Volts

Question 10

What does the standard stand for in SHE?

Answer 10

In order to compare E theta values we have to have the following condition

Temp at 298K
Pressure at 100kPa
Conc of ions at 1 moldm^-3

Make sure u always have 1 mole of hydrogen ions -
1 moldm-3 of HCL or 0.5 moldm-3 of H2So4

Question 11

What is the trend in the electrochemical series?

Answer 11

As you go up the series - They have an increasing tendency to gain electrons (more powerful oxidising agents)
The most powerful oxidising agent is Cl2 and the weakest oxidising agent is mg2+

As you go down the series - They have an increasing tendency to lose electrons (more powerful reducing agents)
The most powerful reducing agent is Mg and the weakest reducing agent is Cl-

Question 12

How to calc standard cell potential?

Answer 12

standard cell potential = Standard electrode potential of the reduced - standard electrode potential of the oxidised

Question 13

What affects standard cell potential?

Answer 13

Electrode potentials can change if the conditions deviate away from the standard conditions

As electrode potentials involve reversible reactions, position of equilibrium can change depending on reaction conditions.

So if we change the equilibrium position, the cell potential value changes too

Question 14

What is cell notation?

Answer 14

They simplify how we draw cells.

Reduced form |Oxidised from ||Oxidised form | Reduced form

The most negative half cell potential goes to the left of the double line

Single solid lines show a physical state change
Double solid lines shows a salt bridge

Zn(s) | Zn2+ (aq) ||Cu2+(aq) | Cu(s)

Question 15

What if you have 2 aqueous ions?

Answer 15

Don’t use straight lines, use commas

Mg(s)|Mg2+(aq)||Fe3+(aq) , Fe2+(aq)|Pt(s)

Question 16

How to predict reaction feasibility

Answer 16

Standard electrode potentials can be used to predict if a stated reaction is likely to proceed under standard conditions

1 - identify which is being oxidised, the most negative value
2 - Take the oxidised equation and flip it. Make sure number of electrons are equal. Write the 2 equations next to each other
3 - Combine the 2 equations to obtain the feasible reaction
4 - Compare this equation to the reaction stated in the question. If they match, then they will react
5 - confirm this by calculating the standard cell potential. All feasible reactions will have a + value

Question 17

Just because we calc Electrode potential to state a feasible reaction doesn’t mean it will actually work. WHY?

Answer 17

Because of non-standard conditions. If we change conc or temp this can cause the electrode potential to change
Even if you get a + cell electrode potential value, it doesn’t guarantee success as let’s say you increase concentration or temp and equilibrium shifts to a not favourable side. This could make your cell potential more negative and less likely to be feasible

Question 18

How would kinetics affect the feasability of a reaction?

Answer 18

The rate of reaction may be too slow so it appears like nothing is happening
If the reaction has a high activation energy it may stop the reaction happening all together unless this is breached

Question 19

What is the formula for the link?

Answer 19

Electrode potential is directly proportional to total entropy change

Electrode potential is directly proportional to the natural log of the equilibrium constant

Question 20

Energy storage cells (batteries)

Answer 20

find the half equations
Li+ +e- <——> Li
Li+ + CoO2 + e- <—–> Li+[CoO2]-

The top one has the more negative value so oxidation occurs there so you flip the equation
Negative electrode - Li<—–> Li+ +e-
Positive electrode - Li+ + CoO2 + e- <—–> Li+[CoO2]-

Overall equation on discharge - Li + CoO2 <—-> Li+ [CoO2]

E cell = Reduced - oxidised

Question 21

Fuel cells

Answer 21

Electricity is generated by a continuous external supply of chemicals rather than a “ready store” like in batteries

Question 22

What is an example of a fuel cell and how does it work?

Answer 22

An alkaline hydrogen-oxygen fuel cell is one example of fuel cell

1. Hydrogen is fed in here. It reacts with OH- ions in the solution in the middle
   2H2 (g) + 4OH-(aq) —–> 4H2O (l) + 4e-
2. Flow of electrons - Electrons produced in reaction 1 travel through a platinum electrode.
3. Component - The flow of electrons is used to power something
4. Oxygen feed - Oxygen is fed here. It reacts with water and the 4 electrons made from step 1 to make OH- ions. O2(g) + 2H2O(l) +4e- ——> 4OH-(aq)
5. Negative electrode (cathode) - electrons flow to the negative electrode which is made from platinum
6. Electrolyte - The electrolyte is made from KOH solution. It carries the OH- ions from the cathode to the anode
7. Positive electrode (Anode) - Electrons flow from the + electrofe which is made from platinum
8. Water emitted - The product of the reaction in step 1 is now released into the surroundings.
9. Movement of OH- ions - The ions produced from traction 4 are carried towards the anode via the electrolyte
10. Ion exchange membrane - These line the platinum electrodes and these allow OH- ions to pass through but not hydrogen and oxygen gas

Equation you get is 2H2 + O2 —-> 2H2O

Question 23

Other types of fuel cells

Answer 23

The car industry developed alcohol fuel cells which create hydrogen gas using a reformer. The main alcohols used are methanol and ethanol
Some newer fuel cells now use the fuel directly without the need to convert to hydrogen first. This is how it works…..
The alcohol is oxidised at the anode with water present CH3OH + H20 —-> CO2 + 6e- + 6H+
The H+ ions pass through the electrolyte and are oxidised to water. 6H+ + 6e- + 1.5O2—> 3H2O

Question 24

What are redox titrations used for?

Answer 24

They can be used to work out the conc of a reducing or oxidising agent.

Question 25

Example of a redox titration

Answer 25

Look at finding the conc of MnO4- by titrating against a reducing agent like Fe2+

Have your reducing agent Fe2+ solution with an unknown conc but known volume in a conical flask. Add dilute sulfuric acid into this too. This ensures you have sufficient H+ ions to allow the reduction of the oxidising agent

Have an oxidising agent in the burette with a known conc - MnO4- ions

Add the MnO4- ions in the burette to the conical flask until you see a feint colour of MnO4- appear. This is known as the endpoint. Add drop-by-drop near here.

Be careful as you get a sharp colour change. The MnO4- ions from KMnO4 are purple. They’ll immediately react with the reducing agent until all the agent is used up

Read how much oxidising agent was added. Read the bottom of the meniscus. Always read at eye level

Always record results to 2 dp and keep going until you have 2 concordant results

Question 26

How to calc unknown conc in a redox titrations?

Answer 26

Write out an equation and balance it
Calc the number of moles MnO4-
Use the equation to find out the molar ratio in order to work out moles of Fe2+
Now calc conc = moles/volume

Question 27

What else can redox titrations be used to calculate?

Answer 27

They can be used to work out the % of iron in iron tablets.

Write out an equation and balance it
Calc the number of moles of MnO4-
Use the equation to find out the molar ratio in order to work out moles of Fe2+
Now calc the number of moles of Fe2+ in the whole iron table by working out how many moles there are in 250cm3
No calc the mass of Fe in the tablet
Now calc the % of iron in the iron tablet

Question 28

What is the iodine sodium thiosulfate titration used for?

Answer 28

It is useful for finding out the conc of an oxidising agent

Question 29

How to complete the iodine sodium thiosulfate titration?

Answer 29

1. Measure out a volume of KIO3 (oxidising agent). This will produce the IO3- ion needed. Usually, you would use 25cm3 but it could be any volume. Add excess acidified potassium iodide solution (KI) to the KIO3 solution.
   IO3- (aq) + 5i- (aq) + 6H+ (aq) —-> 3H2O (l) + 3I2 (aq)
   The I- ions are oxidised to I2.
2. Add the solution from step 1 into the conical flask. Then add sodium thiosulfate into the conical flask and look out for a pale yellow colour. As the colour change is difficult to see - add 2cm3 of starch. This turns a deep blue colour if iodine is still present in the flask. Keep adding until the blue colour disappears. At this point all of the iodine has reacted and we can use the volume of sodium thiosulfate added to work out the number of moles of iodine. Calc the moles of iodine
3. Use the moles of iodine in step 2 to work out the concentration of IO3-. Use original equation

Question 30

What is the copper % titration and how do we do it?

Answer 30

This uses iodide ions as seen before. Copper ions oxidise iodide ions to iodine. This is particularly useful as it allows us to find the % of copper in an alloy such as brass.

1. Use an oxidising agent to oxidise as much of the iodide as possible. Dissolve a known mass of the alloy in some concentrated nitric acid and pour it into a 250cm3 conical flask. Make up to 250cm3 with water. Pipette 25cm3 of the solution into a flask. Add sodium carbonate solution to neutralise residual nitric acid. Keep adding until you just start to see a precipitate form. Then add several drops of ethanoic acid to remove this precipitate. Add excess acidified potassium iodide solution which will react with the copper ions
   2Cu2+ + 4I- —-> 2CuI + I2
   The I- ions are oxidised to I2. The Cu2+ ions are reduced to Cu+ and a white precipitate of copper iodide is formed
2. Work out the number of moles of iodine that have been produced
3. Calculate the concentration of the oxidising agent

Question 31

What are some of the erros in these titrations?

Answer 31

Ensure that the starch indicator is added at the correct point when most of the iodine has reacted. If this didn’t happen then the blue colour would take time to disappear
Make the starch solution only when you’re ready to use it
The copper iodide precipitate makes it difficult to see the colour of the solution
Keep the solution as cool as possible. Iodine produced in the reaction can evaporate readily at room temp which can lead to a false titre. If this has happened to the final, copper % would be too low