F325: Module 2: Redox Reactions and Electrode Potentials

Question 1

I) What is oxidation in terms of electron transfer and oxidation number?

Answer 1

The loss of electrons

An increase in oxidation number

Question 2

I) What is reduction in terms of electron transfer and oxidation number?

Answer 2

The gain of electrons

A decrease in oxidation number

Question 3

I) What is the main idea behind using oxidation numbers to balance redox equations?

Answer 3

The main idea is to remember that the total decrease in oxidation number for one element of the reactants must equal the total increase in oxidation number for another element in the reactants.

Question 4

I) What happens when a metal is in contact with a solution of its ions?

How does the reactivity of the metal affect this?

Answer 4

An equilibrium is established that results in electrons accumulating on the surface and the metal surface developing a negative charge.

The more reactive the metal, the greater the tendency to loose electrons, the more negative the charge on the metal surface and hence the more negative the electrode potential of the metal electrode.

Question 5

I) What is the name for the metal in contact with its ions in aqueous solution?

Answer 5

A half cell

Question 6

I) How do your form an electrochemical cell (in simple terms)?

Answer 6

By connecting two half cells together.

Question 7

I) What is used to connect two half cells together in an electrochemical cell?

Answer 7

A salt bridge

This is usually filter paper soaked in saturated potassium nitrate solution. This completes the circuit and maintains the ionic balance within each half cell.

Question 8

I) In which direction do the electrons flow in an electrochemical cell?

Answer 8

From the more negative electrode to the more positive electrode.

Question 9

I) Describe the half cell arrangement if you are comparing a redox system with aqueous ions present?

Answer 9

You would insert a platinum electrode into a solution containing the aqueous ions.

Question 10

I) Describe the half cell arrangement if you are comparing a redox system with gaseous elements and their ions?

Answer 10

You would insert a platinum electrode into a solution containing the aqueous ions of the gas.

A sample of the gas would be passed into the solution of ions.

Question 11

I) What is the standard hydrogen electrode, what are its properties and why is it useful?

Answer 11

It is a half cell composed of:

• An inert electrode of platinum.
• A solution of 1 mol dm^3 HCl.
• Hydrogen gas is bubbled into the solution under a pressure of 1 atm (100 kPa).
• 298 K

The equation for the half-cell is:
2H+(aq) + 2e- H2(g)

It is assigned a standard electrode potential of 0V and is used as a reference electrode.

Question 12

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

Answer 12

The emf of the half-cell with a standard hydrogen electrode used as a reference electrode.

Question 13

I) What are standard conditions?

Answer 13

298K
100 kPa
1 mol dm^3 solutions

Question 14

I) Does the standard electrode potential of a half-cell depend of the number of electrons involved?

Answer 14

No

Question 15

I) How would you calculate the emf of a cell?

Answer 15

Positive electrode - Negative electrode
+Ep - (-En)
The emf is Always positive

Question 16

I) What is the effect on the emf of a cell if you dilute a solution of M^n+ ions?
The equation for the equilibrium is:
M^n+(aq) + ne- M(s)

Answer 16

This decreases the concentration of M^n+ ions which shifts the equilibrium to the left to compensate, forming for electrons and therefore making the half-cell electrode potential more negative.

Question 17

I) What is an oxidising agent?

Answer 17

An electron acceptor

It is reduced and in the process, oxidises another chemical.

Question 18

I) What is a reducing agent?

Answer 18

An electron donor

It is oxidised and in the process, reduces another chemical.

Question 19

I) If a metal electrode is being oxidised, is it the positive or negative terminal in a cell?

Answer 19

Negative terminal (Cathode)

Question 20

I) If a metal electrode is being reduced, is it the positive or negative terminal in a cell?

Answer 20

Positive terminal (Anode)

Question 21

I) Why is a high resistance voltmeter used in a electrochemical cell?

Answer 21

The voltmeter needs to be of very high resistance to stop the current from flowing in the circuit. In this state it is possible to measure the maximum possible potential difference (E).
The reactions will not be occurring because the very high resistance voltmeter stops the current from flowing.

Question 22

I) Why is a salt bridge used in an electrochemical cell instead of a wire?

Answer 22

A wire is not used because the metal wire would set up its own electrode system with the solutions.

Question 23

I) What happens if current is allowed to flow and the voltmeter is replaced with a bulb?

Answer 23

If the voltmeter is removed and replaced with a bulb or if the circuit is short circuited, a current flows.
The reactions will then occur separately at each electrode. The voltage will fall to zero as the reactants are used up.

Question 24

I) Why is a platinum electrode used in a standard hydrogen electrode?

Answer 24

Because the equilibrium does not include a conducting metal surface a platinum wire is used which is coated in finely divided platinum. (The platinum black acts as a catalyst, because it is porous and can absorb the hydrogen gas.)

Question 25

I) What does it mean if the electrode potential of a half cell is positive or negative (in terms or redox)?

Answer 25

As more +ve increasing tendency for species on left to be reduced, and act as oxidising agents. As more -ve increasing tendency for species on right to oxidise, and act as reducing agents.

Question 26

(1 mark) A chlorine half cell has a standard electrode potential of +1.36V and a standard hydrogen electrode has an electrode potential of 0.00V. In which direction would electrons flow if the two half cells were connected together?

Answer 26

-Electrons would flow to the half cell with the more positive electrode potential, in this case, the chlorine.

Question 27

(1 mark) | What are standard conditions when measuring a standard electrode potential?

Answer 27

- 298K - 100kPa - 1 moldm^-3 solutions

Question 28

How do you measure the electrode potential when the redox involves two ions rather than a metal in a solution of its ions?

Answer 28

- Use a platinum electrode | - In a solution of both ions both 1 moldm^-3

Question 29

(3 marks) | Define the term standard electrode potential.

Answer 29

- emf of a cell - comprising of a half cell combined with a standard hydrogen electrode - conc 1 moldm^-3, 298K, 100kPa

Question 30

(2 marks) An Fe 3+/Fe2+ half cell was combined with a Cr2O7^2- + 14H+/Cr3+ half cell - Cr2O7^2- + 14H+/Cr3+ = +1.33V - Fe 3+/Fe2+ = +0.77V Which species are oxidised and which are reduced and why?

Answer 30

- Cr2O7^2- + 14H+/Cr3+ has a more positive electrode potential - Therefore Cr2O7^2- is the stronger oxidising agent - Which oxidises Fe2+ to Fe3+

Question 31

(2 marks) Cl2/Cl- = +1.36V I2/I- =+0.54V Fe2+/Fe3+ = +0.77 Chlorine will oxidise Fe2+ to Fe3+ but iodine will not. Explain why.

Answer 31

- Electrode potential for chlorine is more positive than the electrode potential of Fe3+ therefore is a better oxidising agent than Fe3+ - Electrode potential for iodine is less positive than Fe3+ therefore is a poorer oxidising agent than Fe3+

Question 32

(2 marks) Give two examples why there may be no visible reaction when two half cells are connected despite being theoretically feasible.

Answer 32

- Activation energy too high | - Reaction too slow

Question 33

(2 marks) | Explain the function of a salt bridge.

Answer 33

- Maintain/balance charge - To complete the circuit - (but does not chemically react)

Question 34

(1 mark) Explain why sulphate ions should not be able to oxidise bromide ions. -Sulphate half cells =+0.17V -Bromine half cells = +1.07 V

Answer 34

-Electrode potential for sulphate is less than bromine

Question 35

I) How would increasing the concentration of reactants affect the Ecell?

Answer 35

Using Le Chatelier's principle, increasing the concentration of reactants would shift the equilibrium to right causing Ecell to increase.

Question 36

I) How would an increase in temperature affect the Ecell?

Answer 36

Most cells are exothermic in the spontaneous direction. Therefore, applying Le Chatelier's principle, an increase in temperature would result in the position of equilibrium shifting to the left causing a decrease in in Ecell.

Question 37

I) What does a positive Ecell tell you about the likelihood of a reaction taking place?

Answer 37

If the Ecell is positive it means that a reaction might occur.

Question 38

I) If Ecell is positive why might the reaction not occur?

Answer 38

It might occur very slowly (so slowly that ineffectively doesn't happen. The reaction might have a very high activation energy.

Question 39

(2 marks) In a copper half cell (copper electrode in copper ions) why would the electrode potential decrease if the concentration of copper ions is less than 1M? Explain in terms of equilibrium.

Answer 39

- Because concentration of Cu2+ is less than 1 mol dm–3 / less than standard - equilibrium moves to left (reducing +ve value of E)

Question 40

(3 marks) Cu2+ + 2e- Cu +0.34V Ag+ + e- Ag +0.80V A standard Cu half cell is connected to a standard Ag half cell and the potential of the cell is measured. Water is then added to the Cu2+/Cu half cell. This changes the position of equilibrium of the half cell: explain why, in terms of equilibrium, why the cell potential increases.

Answer 40

- Conc of Cu2+ decreases (as solution is diluted) causing the equilibrium to shift to the left - So the cell has a bigger difference in E (electrode potential becomes less positive) - More electrons released by Cu (to form more Cu2+)