IB syllabus > redox (hl) > 19.2 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

19.1 - Standard electrode potentials


19.1.1: Describe the standard hydrogen electrode.


The Standard Hydrogen Electrode (SHE)

This is an apparatus setup that is used to compare the electrode potentials of metal and other electrode systems. It comprises hydrogen gas at atmospheric pressure brought into contact with a platinum black electrode in 1molar (mol dm-3) acid solution. The hydrogen ions from the acid solution set up an equilibrium with the hydrogen gas.

2H+ + 2e H2

If the other half cell connected to the standard hydrogen electrode releases electrons then it can be said to be relatively negative with respect to the standard hydrogen electrode (which is assigned a value of 0 volts)

For example, metals that are more reactive than hydrogen gas preferentially release electrons (forcing the SHE to accept the electrons) and have standard electrode potentials with negative values.


19.1.2: Define the term standard electrode potential.


The standard electrode potential

This is the 'potential' of a redox system to lose or gain electrons when compared to the standard hydrogen electrode - assigned a value of 0 volts.

In any reduction-oxidation half equation the electrons are gained by the species on the left hand side:

Cu2+ + 2e Cu

This is an equilibrium and so if a more powerful reducing agent is allowed enters into electrical contact with the above system it can force the copper ions to accept electrons and push the equilibrium to the right hand side.

Conversely, if a weaker reducing agent is brought into contact with the above equilibrium then the copper can force it to accept electrons allowing its own equilibrium to move to the left hand side.

The electrode potential measures the tendency of electrons to flow away from or towards a redox equilibrium. They are always measured with respect to the standard hydrogen electrode (which is assigned a value of zero volts).

Equilibrium redox systems with the reduced side (usually a metal) more reactive than hydrogen have a negative electrode potential, i.e. they can lose electrons more easily than hydrogen.

Equilibrium redox systems with the reduced side less reactive than hydrogen have a positive electrode potential, i.e. they can lose electrons less easily than hydrogen.

Example: Zinc has a standard electrode potential of - 0.76 volts

Consequently the equilibrium...

Zn Zn2+ + 2e

has more of a tendency to move to the right hand side than the equilibrium...

H2 2H+ + 2e

Hence if the two equilibria are brought into electrical contact using an external wire and a salt bridge, the electrons will be pushed from the zinc equilibrium to the hydrogen equilbrium with a force of - 0.76V (the negative sign simply indicates the direction of flow - from zinc to hydrogen ions)

The two equations then may be summed together to give the reaction occuring in the whole cell.

Zn Zn2+ + 2e

2H+ + 2e H2

overall cell reaction

Zn + 2H+ Zn2+ + H2

Using the SHE to measure electrode potential

In the actual experimental setup the two half cells are connected together via an external circcuit wire and a salt bridge to make the whole cell.

In the above apparatus setup the copper | copper sulphate (aq) half cell is connected to the SHE via an external circuit containing a high resistance voltmeter (high resistance to prevent passage of current)

There is a salt bridge to complete the circuit - this allows the ions to flow from one side to another to equalise the movement of charge.

Once the apparatus is setup the reading on the high resistance voltmeter records the electrode potential of the Cu|Cu2+(aq) system

In this particular case the voltmeter will read +0.34 V indicating that copper is less reactive than hydrogen and that there is a force pushing electrons around the external circuit from the hydrogen half cell to the copper half cell.


Representing the cell

The whole cell can be represented by showing the half cells in order of phase (solid, | solution, |salt bridge | solution | solid)

The above cell diagram can be represented as:

H2(g),Pt | H+(aq) || Cu2+(aq) | Cu (s)


Electrochemical series

By comparing many redox systems with the SHE and other reference electrodes, a series can be drawn up showing the reductions in order of their standard electrode potential (usually from negative at the top to positive at the bottom)

click here for an example of the electrochemical series:

Using the electrochemical series

The species at the top on the right hand side are good reducing agents and the species at the bottom on the left hand side are good oxidising agents

Feasible reactions can be identified by any combination of an oxidising agent with a reducing agent, where the difference in electrode potentials is greater than 0.3 V. Basically something on the left will react with something higher up from the right hand side.

The equations for the reactions can be constructed by first balancing the number of electrons in each half equation and then adding the half equations together.

Example: Construct the equation for the reaction between dichromate ions and tin 2+ ions

The dichromate half equation is:

Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l) (Eº = + 1.33 V)

And the tin 2+ half equation is:

Sn4+(aq) + 2e- Sn2+(aq) (Eº = 0.55 V)

By inspection we can see that the dichromate equation needs six electrons on the left hand side whereas the tin 2+ equation has only two electrons on the right hand side. We must, then, multiply the tin 2+ equation by three before adding them.

Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l)   1
Sn4+(aq) + 2e- Sn2+(aq) multiply by three 2
3Sn4+(aq) + 6e- 3Sn2+(aq)   3
  add 1 and 3
Cr2O72-(aq) + 14H+(aq) + 3Sn4+(aq) 2Cr3+(aq) + 7H2O(l) + 3Sn2+(aq)  
   

Whether or not the reaction will be spontaneous is decided by applying E = E (red) - E (ox)

In this case the Cr2O72- is reduced and the Sn2+ oxidised therefore

E = 1.33 - 0.55 = 0.78 V

This is a positive value greater than 0.3V therefore the reaction is spontaneous.

 


Important points to note:

1. The value of the electrode potential is a relative value for the equilibrium compared with the standard hydrogen electrode. Its sign never changes regardless of which way round the equilibrium is written. By convention the above equilibria are written as reductions (left to right) but they could just as easily be written as oxidations without changing the sign of the electrode potential.

2. These standard electrode potential values refer to standard conditions i.e. 1 Molar concentrations at 25ºC and atmospheric pressure. If these conditions change then so does the electrode potential. In other words, for calculating cell potential or spontaneity of reaction it is important to understand that variations occur.

For example, according to standard electrode potentials, manganese IV oxide will not react spontaneously with Hydrochloric acid, however this is the standard preparation of chlorine in the laboratory.

MnO2(s) + 4H+(aq) + 2e- Mn2+ + 2H2O(l) Eº = 1.23
Cl2(g) + 2e- 2Cl-(aq) Eº = 1.36
Predicting spontaneity, E = E(red) - E(ox) = 1.23 - 1.36 E = - 0.13
Negative value therefore no reaction!!

In the lab preparation the manganes(IV) oxide is heated with the concentrated HCl - these are not standard conditions, the temperature is much greater than 25ºC and the concentration of the acid much greater than 1 mol dm-3. Under these new conditions the reaction becomes spontaneous and proceeds at a comfortable rate to collect the chlorine gas produced.

MnO2(s) + 4H+(aq) + 2Cl-(aq) Mn2+ + 2H2O(l) + Cl2(g)


19.1.3: Calculate cell potentials using standard electrode potentials


Cell potential

When two half cells are combined using an external circuit and salt bridge, a potential difference arises due to the different electrode potentials of the two half cells.

The cell with the more negative potential will attempt to push electrons around the external circuit towards the less negative half cell. The measured potential (using a high resistance voltmeter) is called the cell potential - it is equal to the DIFFERENCE between the electrode potentials of the two half cells.

The direction of (attempted) flow is from the cell with the most negative potential to the cell with the least negative (most positive) potential.

Example

In the following cell made up of two half cells

The standard electrode potential of zinc = -0.76V and that of copper is +0.34V

The cell potential is the difference between the standard electrode potential values

= -0.76 - +0.34 = -1.10V

The zinc half cell has the most negative potential and so the direction of electron flow would be from the zinc half cell to the copper half cell.

Reactions occuring in the half cells

As the zinc half cell releases electrons then..

Zn Zn2+ + 2e

As the copper half cell acceptx electrons then..

Cu2+ + 2e Cu

Adding these two reactions gives the overall cell reaction as:

Zn + Cu2+ Zn2+ + Cu

cell potential = 1.10 V

 


19.1.4: Predict whether a reaction will be spontaneous using standard electrode potential (Eº) values.


Spontaneity

This term, in chemistry, is understood to mean whether or not a reaction is thermodynamically (energetically) feasible. Reactions may be spontaneous but do not proceed because of kinetic factors.

Prediction of spontaneity

Electrode potentials for redox systems can be used to predict whether or not the reaction could proceed. For a redox reaction there must be a reducing agent (to lose electrons) and an oxidising agent (to gain electrons). Both reagents may be considered as an equilibrium between the reduced state and the oxidised state, for example:

Zn2+ + 2e Zn

In this equation the reduced state is zinc and the oxidised state is the zinc ion. By convention such equilibria are always written as reductions going from left to right.

Considering the copper | copper ions equilibrium:

Cu2+ + 2e Cu

The copper is the reduced state and the copper ions the oxidised state.

Note: If the terminology is a little confusing consider the oxidation state of the species. Cu2+ has an oxidation state of 2+ whereas the element Cu has an oxidation state of zero - clearly the ion is the oxidised state.

To consider the spontaneity of reaction between the copper equilibrium and the zinc equilibrium whe must select one species from the left hand side (reduced stae) and one from the right hand side (oxidised state), for example

Spontaneity of the reaction between copper ions and zinc metal

Cu2+ + Zn is there a reaction?

If there were to be a reaction then the copper ions would get reduced and the zinc metal would get oxidised.

To find spontaneity we apply the equation:

Eº (species that gets reduced) - Eº (species that gets oxidised) = E (cell potential) and examine the result:

As:

Cu2+ + 2e Cu Eº = +0.34V
and
Zn2+ + 2e Zn Eº = -0.76V
then
E = E (red) - E (ox) E = +0.34 - (-0.76)V
E = + 1.10V


The answer is positive and greater than 0.3 therefore the reaction is spontaneous - zinc reacts with copper ions - and the equation can be constructed by adding together the two half reactions (being careful to place the equilibria the correct way round)

Cu2+ + Zn Zn2+ + Cu


Relationship between Eº and Gibb's free energy

Form the energetics unit it can be seen that for any reaction to proceed then Gibb's Free Energy ΔGº (a measure of the entropy of the universe) must be equal to a negative value. It follows that as the reqirement for spontaneity of a reaction according to electrode potentials is for the value of Eº for a redox reaction to be positive then this must be related to a negative Gibbs Free Energy change.

ΔGº = -nEº

where n is a proportionality constant.

In fact it turns out that ΔGº depends on the total electrical charge transferred during reaction. This may be given by the number of moles of electrons z and the total charge on one mole of electrons transferred during the redox process.

ΔGº = -zFEº

Where z is the moles of electrons transferred and F is the charge on one mole of electrons (1 Faraday = 96,500 Coulombs)

This equation allows calcualtion of the Gibb's Free Energy from a consideration of the electrode potentails of the redox components.

Example: Find the Gibb's Free Energy of the reaction between a solution containing Copper ions and Zinc metal

Cu2+ + 2e Cu Eº = +0.34V
and
Zn2+ + 2e Zn Eº = -0.76V
then
E = E (red) - E (ox) E = +0.34 - (-0.76)V
E = + 1.10V

 

ΔGº = -zFEº
and
z = 2 moles of electrons  
F = 96,500 Coulombs
Gibbs Free Energy = - 212.300 kJ

It must be remembered that this value only applies to standard conditions and that any change in conditions will cause a consequent change in the value of both E and ΔG

 


Resource

Printable list of standard electrode potentials (word)

Electrochemical series