5.5 Redox & Electrochemistry

Cobalt, like many transition elements, can form multiple oxidation states. Some states are more stable than others.

The table below shows a range of half-equations possible for cobalt and its ions:

half-equation	Eθ / V
Co2+ (aq) + 2e- Co (s)	-0.28
[Co(NH3)6]3+ (aq) + e- [Co(NH3)6]2+ (aq)	+0.10
Co3+ (aq) + e- Co2+ (aq)	+1.80
O2 (g) + 4H+ (aq) + 4e- 2H2O (l)	+1.20
2H+ (aq) + 2e- H2 (g)	0.00

Use the values to explain why the Co³⁺ ion and cobalt metal are unstable in an acidic solution, whilst the Co²⁺ ion is stable.

Explain how the Co³⁺ ion can be stabilised, using information from part a).

Iron can also form two ions, Fe³⁺ and Fe²⁺ with the following half-equation: 

Fe³⁺ (aq) + e^- Fe²⁺ (aq)       E^θ = +0.77 V

[2]

Suggest why decreasing the pH increases the electrode potential of the reaction:

O₂ (g) + 4H⁺ (aq) + 4e^- 2H₂O (l)

Electrochemical redox reactions can occur whenever suitable ions are present in an electrolyte. This can be within a synthetic system or within a biological system.

Below are some half-reactions of metals:

half-reaction	Eθ / V
Cu2+ (aq) + 2e-  Cu (s)	+0.34
Ag+ (aq) + e- Ag (s)	+0.80
Fe2+ (aq) + 2e- Fe (s)	-0.41
Al3+ (aq) + 3e- Al (s)	-1.60
Hg2+ (aq) + 2e- Hg (s)	+0.79
Sn2+ (aq) + 2e- Sn (s)	-0.14

Amalgam is an alloy commonly containing mercury, silver, tin and copper and is commonly used for fillings in damaged teeth.  

[2]

[1]

When a variety of metal/ metal ion systems a present there are many different potential reactions. 

[1]

[1]

Two of the metal/ metal ion half-reactions from the table in part a) were set up in separate cells and were tested against an unknown metal ion/ metal cell, X²⁺ (aq)/ X(s).

 When the X²⁺ (aq)/ X(s) was paired with one of the metal ion/ metal cells from the table, it formed the negative electrode and an E^θ cell value of 0.62 V was recorded. 

When the X²⁺ (aq)/ X(s) was paired with the second of the metal ion/ metal cells from the table, it formed the negative electrode and an E^θ cell value of 1.10 V was recorded.

Suggest whether using mercury in amalgam fillings is dangerous, using data from the table.

Manganese is one of the metals that can be added to iron in small quantities to make different types of steel.

It is possible to calculate the % of manganese used in a particular steel by first oxidising the manganese to manganate(VII) and then performing a titration.

Nitric acid is added initially, but another chemical is needed to complete the oxidation. The following half-equations and standard cell potentials can help to identify how to complete the oxidation:

half-equation	Eθ / V
Mn2+ (aq) + 2e- Mn (s)	-1.17
MnO4- (aq) + 8H+ (aq) + 5e- Mn2+ (aq) + 4H2O (l)	+1.51
NO3- (aq) + 2H+ (aq) + e- NO2 (g) + H2O (l)	+0.80
IO3- (aq) + 6H+ (aq) + 5e- I2 (s) + 3H2O (l)	+2.00
IO3- (aq) + 3H2O (l) + 6e- I- + 6OH- (aq)	+0.26

[1]

[2]

 Phosphoric acid is used to complete the dissolving of the piece of steel and to prevent the precipitation of iron compounds. The solution containing MnO₄^- ions is then ready for titration, with a sample of the solution being placed in a conical flask.

[5]

A 0.800 g sample of high-manganese steel was dissolved in acid and the manganese oxidised to manganate(VII).

The acidified solution was made up to 250 cm³ and 25.0 cm³ portions were titrated against a 0.050 mol dm^-3 solution of FeSO₄.

An average titre of 17.50 cm³ was recorded.

Calculate the % mass of manganese in the steel.

Manganate(VII) ions cause a solution, even when very dilute, to take on a pink hue.

Identify a method, other than titration, that could be used to identify the concentration of a solution containing manganate(VII) ions.

Electrochemical cells can be non-rechargeable or chargeable. A common type of non-rechargeable cell is a small zinc-carbon cell used to power electronic devices.

An example is shown in Figure 1.

Add the missing label to Figure 1 and state its function.

At the positive electrode, manganese(IV) oxide reacts with the ammonium chloride in water to form manganese(III) oxide and one other product.

Write the balanced symbol equation for this reaction.

The zinc is oxidised by the chloride ion at the negative electrode according to the following equation, including spectator ions and assuming that no side reactions occur.

Zn + 2Cl^- ZnCl₂ + 2e^-

Using your answer from part b), write an equation for the overall reaction that occurs when the zinc-carbon cell discharges.

Lead-acid cells are used in cars and other vehicles. One way that the lead-acid cell can discharge is by the following overall equation:

PbO₂ (s) + 2H⁺ (aq) + 2HSO₄^- (aq) + Pb (s) 2PbSO₄ (s) + 2H₂O (l)     (E^θ cell = +2.15 V) 

The half-equation at the positive electrode is:

PbO₂ (s) + 3H⁺ (aq) + HSO₄^- (aq) + 2e^- PbSO₄ (s) + 2H₂O (l)     (E^θ cell = +1.69 V) 

Deduce the half equation at the negative electrode and calculate the standard electrode potential, E^θ. Write your answer according to the chemical convention for displaying half-equations.

The diagram shown below represents the standard hydrogen electrode.

A student set up an electrochemical cell consisting of copper and zinc. 

A student set up another electrochemical cell consisting of the Cu²⁺ / Cu (E^θ = + 0.34 V) and Ag⁺ / Ag (E^θ = + 0.80 V) half cells.

An electrochemical cell is shown below:

Lithium ion cells are a type of rechargeable cell. In the cell lithium ions flow, enter and exit the solid electrodes.

The half-equations for the reaction at the electrodes can be represented as follows.

   Positive electrode:   Li⁺ + CoO₂ + e^– → Li⁺[CoO₂]^–   E^θ = +0.36 V

   Negative electrode:   Li⁺ + e^– → Li    E^θ = -3.04 V

Calculate the standard cell potential of the cell. 

During discharge the lithium ions move from the negative electrode to the positive electrode; whilst recharging a lithium ion cell, the lithium ions move from the positive electrode to the negative electrode.

Give the half equations for the electrodes during recharging.

The aircraft Pathfinder is an example of a solar rechargeable aircraft (SRA). Solar cells are powered during the day by capturing energy from sunlight and converting it into electricity. 

Explain why rechargeable cells are often attached to the solar cells.

A rechargeable nickel–cadmium cell is an alternative to a lithium ion cell. The half-equations for this cell are given below:

Fuel cells are used to generate an electric current and do not need to be electrically recharged. The Gemini and Apollo moon probes use hydrogen-oxygen fuel cells. The product of this reaction can be used to supplement the drinking water for astronauts.

Deduce the half equations for the reactions at each electrode in a hydrogen oxygen fuel cell, and then the overall equation for the hydrogen-oxygen fuel cell, to show the product used to supplement drinking water.

A fuel cell is an electrochemical device which converts chemical energy into electrical energy. A continuous supply of fuel is supplied to one electrode and an oxidant to the other. 

The electrodes used in hydrogen fuel cells are often made of a porous mixture of carbon-supported platinum or a porous ceramic material coated in platinum.

State why the electrodes must be porous.

The General Motors Electrovan was built in 1966. It was the first vehicle powered by a hydrogen fuel cell and could travel at up to 70 mph for 30 seconds.

A 500 cm³ standard solution was made using 13.0 g of hydrated iron(II) sulfate, FeSO₄.xH₂O in acidic solution. 

A 25.0 cm³ sample of this solution was titrated against 0.0200 mol dm^-3 potassium manganate(VII) solution. 

26.75 cm³ of the potassium manganate(VII) solution was required.

The percentage of iron in a steel wire can be determined from a redox titration using acidified potassium manganate(VII) solution.

The steel wool is placed in an excess of sulfuric acid to dissolved the iron.

Explain why the iron wire is dissolved in an excess of sulfuric acid and not ethanoic acid.

1.60 g of steel wire was dissolved in an excess of dilute sulfuric acid and the solution made up to 250 cm³.

A 25.0 cm³ portion of this solution required 26.0 cm³ of 0.0200 mol dm^–3 potassium manganate(VII) solution for complete reaction. 

Calculate the percentage of iron in the steel wire.

In acidic conditions, potassium manganate(VII) solution can be used to analyse sodium sulfite, Na₂SO₃, samples according to the following procedure:

Procedure

1. Dissolve approximately 2.00 g of sodium sulfite in 250.0 cm³ of deionised water
2. Acidify 25.0 cm³ portions of the sodium sulfite solution
3. Titrate the sodium sulfite solution with 0.0150 mol dm^-3 potassium mangante(VII) solution

The equation for the oxidation of sodium sulfite is given below.

2MnO₄^– (aq) + 6H⁺ (aq) + 5SO₃^2– (aq) → 2Mn²⁺ (aq) + 5SO₄^2– (aq) + 3H₂O (l)

A student used 1.97 g of impure sodium sulfite to make their 250.0 cm³ solution. The student recorded an average titre of 27.80 cm³ for their titration experiments.

Determine the percentage of sodium sulfite, Na₂SO₃, in the impure sample. Show all your working.

Hydrated copper(II) sulfate can have different amounts of water of crystallisation, x. Compound A has the formula CuSO_4•xH₂O.

An iodine–thiosulfate titration is performed to find the value of x and the formula of Compound A, using the following procedure.

Procedure

   Step 1 A 5.60 g sample of A is dissolved in water to make a 250.0 cm³ standard    solution.

   Step 2 A 25.00 cm³ sample of this solution is pipetted into a conical flask.

   Step 3 An an excess of KI (aq) is added.

2Cu²⁺ (aq) + 4I^– (aq) → 2CuI (s) + I₂ (aq)

   Step 4 The resulting mixture is titrated with 0.105 mol dm^–3 Na₂S₂O₃ (aq).

2S₂O₃^2– (aq) + I₂ (aq) → S₄O₆^2– (aq) + 2I^– (aq)

Titration readings

Complete the table and calculate the mean titre. 

In Step 3, suggest why an excess of KI (aq) is added.

Near the end point of the titration is approached, a solution that can accurately detect the end point is added.

State the solution and explain the colour change observed at the end point.

Determine the formula of compound A. Show your working.