DM def: Electrochemistry; rusting Flashcards
- Describe how electrochemical cells work.
- Draw a diagram of their general arrangement.
- Two half-reactions (oxidation + reduction) occur in separate half-cells
- Electrons flow from one to the other through external wire
- Define electrical current
- Explain what an Ampere represents
- Rate of flow of charge
- One coulomb flowing per second
- What is meant by “potential difference”?
- How is it represented symbolically?
- A measure of the relative extents to which two electrodes release and accept electrons.
- Ecell
- What is the voltage of a cell?
- What equation links voltage, energy and charge?
- A measure of the energy transferred per coulomb flowing around a circuit
- E = QV
Why is a high-resistance voltmeter used when measuring a cell’s potential difference?
- Negligible current flows
- So concentrations of ions remain constant
- Maximum Ecell value is achieved
Drawing of current lowers potential difference
What does a half-cell consist of?
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
What is meant by “electrode potential”?
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.
Explain what determines the magnitude and relative sign of the potential of an electrode.
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
- When making an electrochemical cell, the two half-cells are first connected with a wire. Explain why another connection needs to be made.
- Explain why the solutions should not be mixed in order to make the second connection.
- Describe how the second connection is made.
- To complete the circuit
- Mixing would cause no potential difference (i.e. equal distribution of electrons), so no current would flow
- 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
Draw a diagram of an electrochemical cell whose half-cells contain copper and zinc.
298 K
What are the standard conditions under which electrode potentials are measured?
- 298 K
- 100 kPa (1 atm)
- 1 mol dm-3
- What is the name of the half-cell used as a reference against which all others are measured?
- What is its potential?
- What reaction occurs?
- Draw the set-up of the half-cell. Indicate how standard conditions are achieved.
- Standard hydrogen half-cell / electrode
- 0.0 V
- 2H+(aq) + 2e- ⇌ H2(g)
- Define “standard electrode potential”.
- How is it represented symbolically?
- The potential difference between a half-cell and the standard hydrogen half-cell.
- E⦵
Sign of potential depends on whether half-cell is more positive or negative than standard hydrogen half-cell
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.
- Most reactive metals have highest tendency to release e-
- Most negative E⦵ value
- Top of series
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.
Electrode is made of an unreactive solid, e.g. graphite / platinum (since half-reaction does not involve a metal).
Draw the standard half-cell for the following reaction:
Fe3+(aq) + e- → Fe2+(aq)
A
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 cell is set up using the half-reactions below. Calculate Ecell.
Ecell = 0.34 - (-0.76) = +1.10 V
Use a diagram to show how the value 1.10 for the copper/zinc cell could be predicted.
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
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.
Use data from the table below to explain:
- Why copper corrodes in water containing dissolved oxygen.
- How the iron prevents the copper from corroding.
- O2 can oxidise Cu to Cu2+, because Ecell = 0.4 - 0.34 = +0.06 V, which is positive so reaction is feasible.
- 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+.
- 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
Some standard electrode potentials are shown below.
Which statement(s) is/are correct?
- Mg(s) is the best reducing agent in the table
- V2+(aq) is the best oxidising agent in the table
- V3+(aq) will reduce Cr2+(aq)
A 1, 2 and 3
B Only 1 and 2
C Only 2 and 3
D Only 1
D
- Correct: Mg2+ is most -ve so is best oxidising agent; therefore Mg is best reducing agent
- Incorrect: V2+ acts as a reducing, not oxidising, agent. Best oxidising agent is V3+ since it is most +ve
- Ecell would be -0.41 - (-0.26) = -0.15 V, which is -ve so not feasible
- 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
Use a diagram of a water droplet on a steel surface to demonstrate and describe how rusting occurs.
[O2] is lower in centre so iron is oxidised (E⦵ becomes 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 (E⦵ becomes 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)
- Where in the figure below will the reduction of oxygen occur, and is this an anode or cathode site?
- Where will the oxidation of iron occur, and is this an anode or cathode site?
- On metal surface at shallow edges of droplet. Cathode site.
- On metal surface at centre of droplet. Anode site.
Explain which of the following could be used as a sacrificial metal to protect iron from rusting.
- 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+
Explain how this setup uses impressed current to protect the iron pipeline.
- Inert anode oxidises water in soil, and supplies electrons to pipeline
- Iron made into cathode site, accepting electrons
- Prevents oxidation of iron to Fe2+
