Redox (II) Redox titration calculations

Edexcel A level Chemistry (2017)

Topic 14: Redox (II) Redox titration calculations

Here are the two learning objectives:

14/18. To be able to carry out both structured and non-structured titration calculations including Fe2+/MnO4−, and I2/S2O32−

14/19. To be able to understand the methods used in redox titrations

Redox Titrations

Two examples of redox titrations are required for the Edexcel course.  You can see from the learning objectives the two redox reactions required are those between manganate(VII) and iron(II) and between iodine and thiosulphate. 

How to carry out these two redox titrations: methodology

Redox titrations rely on following changes in the redox reaction chemistry to determine the titration end point.

1. The Manganate(VII)/iron(II) titration

In a manganate(VII)—iron(II) titration, the purple manganate(VII) solution is added to the almost colourless iron(II) solution so that at the end point the iron(II) solution is slightly pink from the manganate(VII).

It is easier to follow a change from colourless to pink than it is a colour change from pink to colourless which would happen if the iron(II) solution was added to a solution of pink manganate(VII) ions.

No indicator is needed because the purple manganate(VII) is self indicating.

2. The Iodine/thiosulphate titration

In the case of the iodine—thiosulphate titration, an indicator is required since the end point is not easy to see.  In this titration, you usually add sodium thiosulphate solution from the burette to an aqueous solution of iodine in the conical flask.  Aqueous solutions of iodine are brown but fade to yellow then colourless at the end point so to follow the colour change and serve a sharp end point is impossible.

Instead starch solution is added to the iodine solution once it becomes golden coloured (see the video below).  This results in a deep blue–black starch–iodine complex.  Adding thiosulphate solution brings an endpoint with a sudden loss of the blue–black colour: a very sharp endpoint.   

You can watch a very brief iodine thiosulphate titration here.

Calculations using redox titration results.

I’m going to talk you through a non-structured calculation on the reasonable assumption that if you can carry out these kinds of calculation then the structured calculations you will find fairly straightforward. 

Here is a typical non–structured calculation question:

Iron(II)ethanedioate dihydrate can be analysed by titration using potassium manganate(VII) in acidic solution.

In this reaction, manganate(VII) ions oxidise iron(II) ions and ethanedioate ions.

A 1.381g sample of impure FeC₂O₄.2H₂O was dissolved in an excess of dilute sulphuric acid and made up to 250ml of solution.

25ml of this solution decolourised 22.35ml of a 0.0193 mol.dm^–3 solution of potassium manganate(VII).

Use the half-equations given below to calculate the reacting ratio of moles of manganate(VII) ions to moles of iron(II) ethanedioate.

MnO₄^—   +   8H⁺   +  5e^—   ⟶  Mn²⁺   +   4H₂O

Fe²⁺    ⟶     Fe³⁺   +   e^—

C₂O₄^2—   ⟶   2CO₂   +  2e^—

Then calculate the percentage by mass of FeC₂O₄.2H₂O in the original impure sample.

First, calculate the reacting molar ratio:

Each iron(II) ion provides one electron on oxidation and each ethandioate ion provides two electrons: a total of three moles of electrons per iron(II)ethandioate.

5 moles iron(II)ethandioate provide 15 electrons and these reduce three moles manganate(VII) ions.

So the overall equation has to be:

5FeC₂O₄  + 3MnO₄^— + 24H⁺ ⟶ 3Mn²⁺ + 10CO2 + 5Fe³⁺ + 12H₂O

Now you have the reacting ratio of manganate(VII) to iron(II)ethandioate as 3 moles to 5 moles.

Second, you calculate the moles of manganate(VII) actually used in the titration and work back from this value.

Use n=cV

Moles MnO₄^—  =  0.0193  × 22.35/1000   =   0.00043 moles

Third calculate the number of moles of the iron(II) salt in the 25ml of its solution:

Moles FeC₂O₄  =   0.00043 ×  5/3   =   0.00072 moles

Fourth, note that 25ml sample was a tenth of the total sample dissolved in the 250ml volumetric flask.  Therefore:

Total moles FeC₂O₄    =  0.0072moles  

Fifth, calculate the molar mass of the iron(II) salt:

Fe—                55.8

C—12 × 2  =   24

O—16 × 4  =   64  +

H2O—18 × 2= 36

Giving a total Mr (FeC₂O₄ ) of 179.8 g.mol^—1

Sixth, calculate the mass of pure iron(II) salt using mass =  molar mass ×  moles

Mass of FeC₂O₄   =   179.8  ×  0.0072  = 1.295g    

Lastly, calculate the percentage by mass of the pure iron(II) salt  since

% purity   =    pure mass   ×   100  /total mass

% purity   =    1.295  ×   100/1.391   =   93.2%

And there you have it.  Step by step you can work through the calculation and see that the stages are common to most types of calculation like this one. 

Redox(II): Fuel Cells

Edexcel A level Chemistry (2017)

Topic 14: Redox(II): Fuel Cells

Here are the learning objectives relating to fuel cells:

14/16. To be able to understand that the energy released on the reaction of a fuel with oxygen is utilized in a fuel cell to generate a voltage.

Knowledge that methanol and other hydrogen-rich fuels are used in fuel cells is expected.

14/17. To know the electrode reactions that occur in a hydrogen-oxygen fuel cell.

Knowledge of hydrogen-oxygen fuel cells with both acidic and alkaline electrolytes is expected.

The Proton Exchange Fuel Cell

Proton exchange membrane fuel cells are a type of fuel cell being developed to replace conventional alkaline fuel cells of the type used in the Space Shuttle.

Their distinguishing features include lower temperature ranges (50 to 100 °C) and a special polymer electrolyte membrane.

How the proton exchange membrane fuel cell works

Fuel cells essentially transform the chemical energy released in the reaction between hydrogen and oxygen into electrical energy.

A stream of hydrogen is delivered to the cathode side of the cell. At the cathode side it is catalytically split into protons and electrons. An oxidation half—cell reaction takes place reaction (loss of electrons) at the cathode:

H₂ ⟶  2H⁺   +   2e^—

The newly formed protons permeate through the polymer electrolyte membrane to the anode side. The electrons travel along an external circuit to the anode side of the cell.

Meanwhile, a stream of oxygen is delivered to the anode side of the cell.

At the anode side, oxygen molecules react with the protons permeating through the polymer membrane and the electrons arriving by the external circuit and combine to form water molecules.

A reduction half-cell reaction or oxygen reduction takes place at the anode:

½O₂  + 2H⁺   +  2e^—  ⟶  H₂O

The overall effect is for hydrogen and oxygen to combine to form water:

H₂  +   ½O₂    ⟶  H₂O

Properties of the polymer membrane

To function, the membrane must conduct hydrogen ions (protons) but not electrons as this would in effect "short circuit” the fuel cell.

The membrane must also not allow either gas to pass to the other side of the cell, a problem known as gas crossover.

Finally, the membrane must be resistant to the reducing environment at the cathode as well as the harsh oxidative environment at the anode.

The Direct Methanol Fuel Cell

Direct-methanol fuel cells or DMFCs are similar to proton exchange fuel cells but methanol is used as the fuel. Their main advantage is the ease of transport of methanol (over hydrogen gas) because methanol is an energy-dense yet reasonably stable liquid in most environmental conditions.

Methanol is a liquid from -97.0 °C to 64.7 °C at atmospheric pressure. The energy density of methanol is an order of magnitude greater than even highly compressed hydrogen, and 15 times higher than lithium ion batteries.

The chemistry of how the DMFC works

The DMFC relies upon the oxidation of methanol on a Platinum/Ruthenium catalyst layer to form carbon dioxide.  

Water is consumed at the anode and is produced at the cathode. Protons (H⁺) are transported across the proton exchange membrane—often made from polymer—to the cathode where they react with oxygen to produce water. 

Electrons flow through the external circuit from cathode (—) to anode (+), providing power to connected devices.

Cathode (—) reaction:  CH₃OH   +   H₂O  ⟶  6H⁺   + 6e^—  +  CO₂  Oxidation

Anode (+) reaction:  1½O₂   +   6H⁺   +   6e^—  ⟶   3H₂O

Reduction

Overall reaction:  CH₃OH    +   1½O₂   ⟶   2H₂O   +   CO₂

You can see that the overall reaction is just the combustion of methanol in pure oxygen.

Methanol and water are adsorbed on a catalyst usually made of platinum and ruthenium particles, and lose protons until carbon dioxide is formed. 

As water is consumed at the cathode in the reaction, pure methanol cannot be used without provision of water via either passive transport such as osmosis or pumping. The need for water limits the energy density of the fuel.

Cell efficiency is quite low, so they are targeted at portable applications, where energy and power density are more important than efficiency.

Redox (II): Storage cells: the Lithium-ion cell.

Edexcel A level Chemistry (2017)

Topic 14: Redox (II): Storage cells: the Lithium-ion cell.

14/15 To be able to understand the application of electrode potentials to storage cells.

The Lithium–ion cell.

Lithium ion (Li-ion) cells are the most popular type of rechargeable cell for most applications.

There is a global market of at least £8bn and predicted to grow to £50bn by 2020.

Li-ion battery offers many advantages over other secondary (or rechargeable) cells:

•  It is lighter than other rechargeable batteries for a given capacity

•  Li-ion chemistry delivers a high voltage

•  Low self-discharge rate (about 1.5% per month)

•  Do not suffer from battery memory effect

•  Environmental benefits: rechargeable and reduced toxic landfill

However Li-ion batteries have also struggled with issues such as:

•  Poor cycle life, particularly in high current applications

•  Rising internal resistance with cycling and age

•  Safety concerns over overheating or overcharging

•  Applications demanding more from Li-ion battery capacity

How a Lithium–ion cell works

In Li-ion batteries, lithium ions move from the anode to the cathode during discharge, and from cathode to anode when charging.

The materials used for the anode and cathode can dramatically affect a number of aspects of the battery’s performance, including capacity.

New higher capacity materials are urgently required in order to address the need for greater energy density, cycle life and charge lifespan, among the other issues faced by Li-ion batteries.

Graphite has traditionally been the anode of choice for commercial use, with typical first generation Li-ion chemistry working as follows:

At the cathode:

LiCoO₂ – Li⁺ – e^– ↔ Li_0.5CoO₂

At the anode:

6C + Li⁺ + e^– ↔ LiC₆

Overall reaction on a Li-ion cell:

                             C + LiCoO₂ ↔ LiC₆ + Li_0.5CoO₂

Materials other than graphite have been investigated, with silicon offering the highest gravimetric capacity.

The volumetric capacity of silicon, i.e. the capacity of silicon taking into account volume increases resulting from lithium insertion, is still significantly higher than that associated with carbon anode materials.

The potential contained within silicon holds great promise for the future of Li-ion batteries, if it can be used without compromising the battery cycle life.

When charging a lithium ion battery, lithium is inserted into the silicon, causing a dramatic increase in volume (up to 400%).

On discharge, lithium is extracted from the silicon which returns to a smaller size.

Repeated expansion and contraction places great strain on the silicon, causing silicon material to fracture or pulverise.

This, in turn, leads to the electrical isolation of silicon fragments from nearest neighbours and a loss of conductivity in the anode of the battery.

For this reason, charge-discharge cycle life for conventional silicon-based anodes is typically short.

This post comes with thanks to nexeon.co.uk