# DM def: Electrochemistry; rusting Flashcards

1
Q
1. Describe how electrochemical cells work.
2. Draw a diagram of their general arrangement.
A
• Two half-reactions (oxidation + reduction) occur in separate half-cells
• Electrons flow from one to the other through external wire
2
Q
1. Define electrical current
2. Explain what an Ampere represents
A
1. Rate of flow of charge
2. One coulomb flowing per second
3
Q
1. What is meant by “potential difference”?
2. How is it represented symbolically?
A
1. A measure of the relative extents to which two electrodes release and accept electrons.
2. Ecell
4
Q
1. What is the voltage of a cell?
2. What equation links voltage, energy and charge?
A
1. A measure of the energy transferred per coulomb flowing around a circuit
2. E = QV
5
Q

Why is a high-resistance voltmeter used when measuring a cell’s potential difference?

A
• Negligible current flows
• So concentrations of ions remain constant
• Maximum Ecell value is achieved

Drawing of current lowers potential difference

6
Q

What does a half-cell consist of?

A

An electrode in contact with a solution:

• Where half-reaction involves an elemental metal, electrode is made of this metal + solution contains one of its cations
• Where half-reactions are between ions or between molecules and ions, electrode is made of inert substance + solution contains these molecules and/or ions
7
Q

What is meant by “electrode potential”?

A

The potential difference in a half-cell between the electrode and solution of ions / of ions + molecules.

Measures the tendency of a half-reaction to accept or release electrons.

8
Q

Explain what determines the magnitude and relative sign of the potential of an electrode.

A

The position of equilibrium of a half-reaction, e.g.:

Zn2+(aq) + 2e- ⇌ Zn(s)

• Further to right → greater tendency to accept e- → more positive
• Further to left → greater tendency to release e- → more negative
9
Q
1. When making an electrochemical cell, the two half-cells are first connected with a wire. Explain why another connection needs to be made.
2. Explain why the solutions should not be mixed in order to make the second connection.
3. Describe how the second connection is made.
A
1. To complete the circuit
2. Mixing would cause no potential difference (i.e. equal distribution of electrons), so no current would flow
3. Salt bridge: strip of filter paper soaked in an ionic solution, e.g. KNO3 (carries current, but avoids mixing)

May also be called “ion” bridge because circuit is completed by movement of ions, not electrons

10
Q

Draw a diagram of an electrochemical cell whose half-cells contain copper and zinc.

A

298 K

11
Q

What are the standard conditions under which electrode potentials are measured?

A
• 298 K
• 100 kPa (1 atm)
• 1 mol dm-3
12
Q
1. What is the name of the half-cell used as a reference against which all others are measured?
2. What is its potential?
3. What reaction occurs?
4. Draw the set-up of the half-cell. Indicate how standard conditions are achieved.
A
1. Standard hydrogen half-cell / electrode
2. 0.0 V
3. 2H+(aq) + 2e- ⇌ H2(g)
13
Q
1. Define “standard electrode potential”.
2. How is it represented symbolically?
A
1. The potential difference between a half-cell and the standard hydrogen half-cell.
2. E

Sign of potential depends on whether half-cell is more positive or negative than standard hydrogen half-cell

14
Q

In the electrochemical series, negative and positive E values are positioned at the top and bottom respectively, with the standard hydrogen half-cell in the middle.

Deduce where the most reactive metals are positioned.

A
• Most reactive metals have highest tendency to release e-
• Most negative Evalue
• Top of series
15
Q

The reaction between a metal and its ions is only one example of redox. Other viable half-reactions are between ions or between molecules and ions, some of which are shown below.

Describe how the half-cells for these reactions are set up.

A

Electrode is made of an unreactive solid, e.g. graphite / platinum (since half-reaction does not involve a metal).

16
Q

Draw the standard half-cell for the following reaction:

Fe3+(aq) + e- → Fe2+(aq)

A
17
Q
A

A

18
Q

A student uses a Cu2+(aq)/Cu(s) half-cell to confirm the Eo of a Cl2(aq)/Cl-(aq) half-cell.

Draw a diagram of the apparatus the student would set up, showing state symbols. Indicate how standard conditions are achieved.

A
19
Q

How is Ecell worked out from the standard electrode potential of two half-cells?

A

Always positive: Eoreduced - Eooxidised

If feasible, this means subtracting more negative value from more positive value.

20
Q

A cell is set up using the half-reactions below. Calculate Ecell.

A

Ecell = 0.34 - (-0.76) = +1.10 V

21
Q

A cell is set up using the half-reactions below:

Fe3+(aq) + e- ⇌ Fe2+(aq) E= +0.77 V

MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l) E= +1.51 V

1. Write the equation for the overall reaction.
2. Calculate Ecell.
A

Electrons flow towards half-cell with more positive E, MnO4-/Mn2+

Iron oxidised: 5Fe2+(aq) → 5Fe3+(aq) + 5e-

Manganese reduced: MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)

Overall: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Ecell = 1.51 - 0.77 = +0.74 V

22
Q

Use a diagram to show how the value 1.10 for the copper/zinc cell could be predicted.

A
23
Q

A redox reaction is deduced to be feasible from its standard electrode potentials.

1. Why may it not be spontaneous?
2. How could you make it spontaneous?
A
1. High activation enthalpy - would mean negligable rate with no observable change
2. Use a catalyst (to lower Ea)

“Entropy has not been taken into account” is not a valid answer to 1 since the use of potentials accounts for entropy

24
Q

A redox reaction is deduced to be unfeasible from its standard electrode potentials.

How could you make it feasible?

A

Change the conditions: (standard electrode potentials are measured at standard conditions)

• Temperature
• Pressure
• Concentration
25
Q
A

Ce4+ reduced to Ce3+ so subtract 1.36 from each value:

HCl: Ecell = 1.28 - 1.36 = -0.08 V

H2SO4: Ecell = 1.44 - 1.36 = +0.08 V

H2SO4 since Ecell is positive, so the reaction is feasible

26
Q
A

V2+ ions in VCl2 are oxidised to green V3+ ions by O2 in air.

Ecell = 0.4 - (-0.26) = +0.66 V, which is positive so reaction is feasible.

V3+ are further oxidised by H2O in air to blue VO2+ ions.

Ecell = 0.34 - (-0.26) = +0.60 V, which is positive so reaction is feasible.

VO2+ are not further oxidised to VO2+ ions, since:

Ecell = 0.4 - 1 = -0.6 V, which is negative so reaction is not feasible.

27
Q

Use data from the table below to explain:

1. Why copper corrodes in water containing dissolved oxygen.
2. How the iron prevents the copper from corroding.
A
1. O2 can oxidise Cu to Cu2+, because Ecell = 0.4 - 0.34 = +0.06 V, which is positive so reaction is feasible.
2. O2 oxidises Fe in preference to Cu, because Ecell for O2 and Fe/Fe2+ = 0.4 - (-0.44) = +0.84 V, which is more positive than for O2 and Cu/Cu2+.
28
Q
A
• High [H+] gives reduction of ClO3- a more positive electrode potential
• High [Cl-] gives oxidation of Cl- a more negative electrode potential
• Such that Ecell becomes positive; reaction becomes feasible
29
Q

Some standard electrode potentials are shown below.

Which statement(s) is/are correct?

1. Mg(s) is the best reducing agent in the table
2. V2+(aq) is the best oxidising agent in the table
3. V3+(aq) will reduce Cr2+(aq)

A 1, 2 and 3

B Only 1 and 2

C Only 2 and 3

D Only 1

A

D

1. Correct: Mg2+ is most -ve so is best oxidising agent; therefore Mg is best reducing agent
2. Incorrect: V2+ acts as a reducing, not oxidising, agent. Best oxidising agent is V3+ since it is most +ve
3. Ecell would be -0.41 - (-0.26) = -0.15 V, which is -ve so not feasible
30
Q
A
• On addition of ammonia, [Co(H2O)6]2+ is converted by ligand exchange to [Co(NH3)6]2+
• Oxygen can oxidise [Co(NH3)6]2+ to [Co(NH3)6]3+ since O2/OH- has more positive Eo
• Oxygen cannot oxidise [Co(H2O)6]2+ since O2/OH- has more negative Eo
31
Q
1. What is corrosion?
2. What is rusting?
A
1. The slow, chemical conversion of a refined metal into a stabler form.
2. The corrosion of iron.
32
Q

Rusting is an electrochemical process.

1. Give the 2 half-equations involved.
2. Indicate the direction of each reaction and the relative values of E.
A

Fe2+(aq) + 2e- ⇌ Fe(s)

Backward reaction; E is more -ve

1/2O2(g) + H2O(l) + 2e- ⇌ 2OH-(aq)

Forward reaction; E is more +ve

33
Q

Use a diagram of a water droplet on a steel surface to demonstrate and describe how rusting occurs.

A

[O2] is lower in centre so iron is oxidised (Ebecomes more -ve):

Fe(s) → Fe2+(aq) + 2e-

Electrons flow along steel surface to edges of droplet.

[O2] is higher at edges so oxygen is reduced (Ebecomes more +ve):

1/2O2(g) + H2O(l) + 2e- → 2OH-(aq)

Products diffuse away from surface + react to form rust in secondary processes:

Fe2+ + 2OH- → Fe(OH)2(s)

Fe(OH)2(s) + O2(aq) → Fe2O3 • xH2O(s)

34
Q
1. Where in the figure below will the reduction of oxygen occur, and is this an anode or cathode site?
2. Where will the oxidation of iron occur, and is this an anode or cathode site?
A
1. On metal surface at shallow edges of droplet. Cathode site.
2. On metal surface at centre of droplet. Anode site.
35
Q

Name two methods of protecting iron against rust.

A
• Barriers (oil, grease, paint, polymer, sacrificial metals)
• Impressed current (i.e. applied emf)
36
Q

Explain which of the following could be used as a sacrificial metal to protect iron from rusting.

A
• Metal is oxidised so E is always subtracted from value for reduction of O2. So metals with more negative E than iron - Mg / Zn / Cr - can be used since they give more positive Ecell value
• Equilibrium positions for these metals’ half-reactions are towards the left, producing electrons. This shifts Fe2+/Fe equilibrium right, preventing formation of Fe2+
37
Q

Car bodies are often made of galvanised steel, which contains zinc and is itself coated by a layer of zinc oxide.

Explain how the iron would be protected:

1. If the surface remained undamaged
2. If the coating were scratched
A
1. Zinc oxide coating prevents corrosion of zinc metal beneath.
2. Zinc is a sacrififial metal: it corrodes in preference to iron.
38
Q

Explain how impressed currect (applied emf) prevents the corrosion of iron.

A
• Ordinarily, iron is oxidised (rusts) at an anode site; reverse reaction occurs: Fe2+(aq) + 2e- ⇌ Fe(s)
• Water is oxidised at an inert, corrosion-resistant anode, providing electrons: H2O(l) → ½O2(g) + 2H+(aq) + 2e-
• External current is supplied from anode to make iron a cathode site. Prevents oxidation by shifting Fe2+(aq)/Fe(s) equilibrium to right
39
Q

Explain how this setup uses impressed current to protect the iron pipeline.

A
• Inert anode oxidises water in soil, and supplies electrons to pipeline
• Iron made into cathode site, accepting electrons
• Prevents oxidation of iron to Fe2+
40
Q

A steel bolt used to secure a wooden section of a marine boat was removed for inspection.

Although the head of the bolt was in good order, the internal section was severely corroded.

1. Suggest why the internal section corroded.
2. Suggest why the head was not significantly corroded.
A
1. Not in contact with air or water. Low [O2] means this is an anode site, where Fe is oxidised. Electrons flow to cathode sites with higher [O2], where O2 is reduced.
2. In contact with air + water. High [O2] + [H2O]. Unlikely to be anode site where corrosion would occur.
41
Q

Explain why corrosion occurs at a greater rate in salt water than in fresh water.

A
• Corrosion is electrochemical
• Ions act as a salt bridge between anode + cathode sites
• Higher [ions] in salt water
42
Q

A mercury thermometer is dropped and broken in an aluminium dinghy. The spillage takes a while to clean up.

Soon afterwards, small holes develop in the hull. Suggest why.

A
• Electrochemical cell is set up between mercury + aluminium in presence of sea water
• Al has a greater tendency to be oxidised (more negative E), so becomes anode site + corrodes
43
Q

A student is investigating the electrode potentials of iron. They set up two Fe2+(aq) + 2e- ⇌ Fe(s) half-cells and record the voltage as 0.

1. They then blow air through a straw into the left-hand beaker. The electrode becomes slightly more positive than the other one.

They reason that the oxygen concentration in the left-hand beaker is increased, causing this reaction to occur: O2 + 2H2O + 4e- → 4OH-

Discuss the student’s statement. (2 marks)

1. Iron rusts at the bottom of pits in steel where the oxygen concentration is lower than at the surface.

Explain how this relates to the observations in 1. (2 marks)

A

1.

• Student is correct
• Higher [O2] in left-hand beaker
• Uses electrons, so electrode is positive

2.

• In right-hand beaker, this reaction occurs: Fe → Fe2+ + 2e-
• This is where [O2] is lower