OCR

Chemistry A H432 Redox and Electrode Potentials

Redox Reactions

Reducing and oxidizing agents
In any redox reaction there will be an oxidizing agent and a reducing agent.
• The oxidizing agent contains the species being reduced and therefore decreasing in oxidation
number (it is the whole molecule/ion containing that species)
• The oxidizing agent takes electrons from the species being oxidized

• The reducing agent contains the species being oxidized and therefore increasing in oxidation
number (it is the whole molecule/ion containing that species)
• The reducing agent donates electrons to the species being reduced

Constructing equations for redox reactions
Method 1: combining half-equations (which show numbers of electrons transferred)

Example:
Acidified manganate(VII) ions are a powerful oxidizing agent. They can oxidize Fe2+ ions to Fe3+ ions. At
the same time the manganate(VII) ions are reduced to Mn2+ ions.

We can represent what happens to the iron ions with a half-equation. The number of electrons to
include comes from the change in oxidation number of the species being oxidized or reduced:

Oxidation is loss of electrons: Fe2+(aq) ! Fe3+(aq) + e-
oxidation number of Fe: +2 +3

We can construct a half equation for the reduction of acidified manganate(VII) ions
Reduction is gain of electrons: MnO4-(aq) + 5e- ! Mn2+(aq)
oxidation number of Mn +7 +2

This is not balanced, however. The charges are different on each side, and we need to deal with the
four oxygens in the MnO4- ion. The clue here is the reagent is acidified managanate(VII). If we include
H+ ions on the left hand side, we can produce water with the oxygens as a product. Four oxygens will
require eight H+ ions:

Reduction is gain of electrons: MnO4-(aq) + 8H+ + 5e- ! Mn2+(aq) + 4H2O(l)
This is balanced for both charges and atoms present.

To combine the two half equations into an overall equation, we now need to have the same number of
electrons lost in the oxidation as are gained in the reduction, so we need to multiply the first half-
equation by 5 throughout:

5 Fe2+(aq) ! 5 Fe3+(aq) + 5 e-

Then we can simply add the half equations together, and cancel the five electrons on each side:

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OCR Chemistry A H432 Redox and Electrode Potentials

MnO4-(aq) + 8H+(aq) + 5e- ! Mn2+(aq) + 4H2O(l)

2+
5 Fe (aq) ! 5 Fe (aq) + 5 e-
3+

- + 2+ 3+ 2+
So MnO4 (aq) + 8H (aq) + 5Fe (aq) ! 5 Fe (aq) + Mn (aq) + 4H2O(l)

Example:
We are also familiar with acidified potassium dichromate as a powerful oxidizing agent. The
dichromate ions act in a similar way to the manganate(VII) ions. For example, they can oxidize Sn2+
ions to Sn4+ ions.

You should recall that dichromate ions are an intense orange colour, and that when they act as an
oxidizing agent, their colour changes to dark green. This is because they have been reduced to Cr3+
ions.

Again we can represent what happens with half-equations

Sn2+(aq) ! Sn4+(aq) + 2e- Oxidation

Cr2O72-(aq) + 6e- ! 2Cr3+(aq) Reduction
+6 +3 (Number of electrons to include from
+6 +3 change in oxidation numbers)

Cr2O72-(aq) + 14H+(aq) + 6e- ! 2Cr3+(aq) + 7H2O(l) Balancing by adding H+ and H2O

And again we can combine these into a single redox equation by getting the same number of electrons
in each half equation – i.e. multiplying the oxidation equation by 3 throughout, and adding the two half
equations so the electrons cancel on each side.
3Sn2+(aq) + Cr2O72-(aq) + 14H+(aq) + 6e- ! 3Sn4+(aq) + 6e- + 2Cr3+(aq) + 7H2O(l)

N.B. we can also be asked to work out the oxidation or reduction half equation if we are given the
overall equation and the other half-equation:

Using the same example as above, we might be told that the overall equation is:
3Sn2+(aq) + Cr2O72-(aq) + 14H+(aq) ! 3Sn4+(aq) + 2Cr3+(aq) + 7H2O(l)

and that the oxidation going on is Sn2+ ! Sn4+. We are required to work out the reduction half-
equation:

Firstly multiply the given half-equation up to get the same stochiometry as in the overall equation:
3Sn2+ + 6e- ! 3Sn4+

Now position this above the overall equation and work out what is missing on each side:
Oxidation 3Sn2+ + 6e- ! 3Sn4+
Reduction Cr2O72- + 14H+ ! 6e- + 2Cr3+ + 7H2O
Overall 3Sn (aq) + Cr2O72-(aq) + 14H+(aq) + 6e- ! 3Sn4+(aq) + 6e- + 2Cr3+(aq) + 7H2O(l)
2+

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OCR Chemistry A H432 Redox and Electrode Potentials

Method 2: Constructing redox equations using oxidation numbers

We can balance a redox equation using oxidation numbers, without having to construct separate half-
equations if we choose.

This works because the total increase in oxidation numbers in the balanced equation must be equal to
the total decrease in oxidation numbers – a consequence of OIL RIG !

Example:
Hydrogen iodide (HI) is oxidized to iodine by concentrated sulphuric acid. The sulphuric acid is reduced
to hydrogen sulphide. Construct a balanced equation for this.

Step 1: Write an equation using the formulae for reactants and products you know about
HI + H2SO4 ! I2 + H2S - don't worry about balancing or missing atoms yet !

Step 2: Use the oxidation numbers to see what is oxidized and what is reduced
HI + H2SO4 ! I2 + H2S
+1 -1 +1 +6 -2-2 0 +1 -2
+1 -2-2 0 +1

Keep only the oxidation numbers that have changed during the reaction:
HI + H2SO4 ! I2 + H2S
-1 +6 0 -2
0
Step 3: Balance the equation FOR ONLY THOSE ATOMS BEING OXIDISED OR REDUCED. Don't worry
about anything else in the equation being balanced yet.
2HI + H2SO4 ! I2 + H2S
-1 +6 0 -2
-1 0

Step 4: Calculate the total increase and decrease in oxidation numbers:
Reduction (S) is +6 to -2 = -8
Oxidation (I) is -1,-1 to 0 = +2

Step 5: Multiply up the species in the oxidation and/or reduction to get the same increase and
decrease in oxidation numbers – in this case we need to multiply the 2HI ! I2 by 4
8HI + H2SO4 ! 4 I2 + H2S

Step 6: Complete the balancing. If we are missing oxygen atoms, add water to that side. If we are
missing hydrogen atoms, add H+ to that side (look for 'acidic conditions' in the question as another clue
as to when to include H+ as a reactant). Under alkaline conditions we might add OH-.

8HI + H2SO4 ! 4 I2 + H2S + 4 H2O

Finally: Check it is balanced for all elements
Check it is balanced for charge – total charge on each side the same.

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OCR Chemistry A H432 Redox and Electrode Potentials

Practice:
Hydrogen sulphide, H2S is oxidized to sulphur, S, by nitric acid, HNO3, which is itself reduced to nitrogen
monoxide, NO.
Step 1 H2S + HNO3 ! S + NO
Step 2 -2 +5 0 +2

Step 3,4 no further balancing needed oxidation (S) -2 to 0 = +2
reduction (N) +5 to +2 = -3

Step 5 need 3x (H2S!S) and 2x (HNO3! NO) to balance
3H2S + 2HNO3 ! 3S + 2NO

Step 6 missing 4 x O on RHS so add 4x H2O (then no need to add H+)
3H2S + 2HNO3 ! 3S + 2NO + 4H2O

In acidic conditions, silver metal, Ag, is oxidized to silver(I) ions, Ag+ by NO3- ions, which are reduced to
nitrogen monoxide (NO).
Step 1 Ag + NO3- + H+ ! Ag+ + NO
Step 2 0 +5 +1 +2

Step 3,4 no further balancing needed oxidation (Ag) 0 to +1 = +1
reduction (N) +5 to +2 = -3

Step 5 Need x3 (Ag+ ! Ag)
3Ag + NO3- + H+ ! 3 Ag+ + NO

Step 6 Need 2 more O on RHS, so add 2H2O. Balancing then needs 4H+ on LHS.
3Ag + NO3- + 4H+ ! 3Ag+ + NO + 2H2O

Now prove to yourself that the half-equation method used for manganate(VII) ions and iron(II) can be
done this way as well, with the same result. Acidified managanate(VII) ions are capable of oxidizing
iron(II) ions to iron(III) ions. The manganite(VII) is reduced to Mn2+ ions:
Step 1 MnO4- + Fe2+ +H+ ! Mn2+ + Fe3+
Step 2 +7 +2 +2 +3

Step 3,4 no further balancing oxidation +2 to +3 = +1 reduction +7 to +2 = -5

Step 5 Need 5x (Fe2+ ! Fe3+)
MnO4- + 5 Fe2+ +H+ ! Mn2+ + 5 Fe3+

Step 6 Need four O on RHS, so add 4H2O. Balancing then requires 8H+ on LHS
MnO4- + 5 Fe2+ +8 H+ ! Mn2+ + 5 Fe3+ + 4H2O


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OCR Chemistry A H432 Redox and Electrode Potentials

Redox titrations
In a redox titration we are titrating one solution in which something is going to be oxidized, against
another in which something is going to be reduced. We don't use a pH indicator in a redox titration.
The changes in colour of the substances when they are oxidized and reduced tell us when we have
reached the endpoint. If we know the concentration of one substance, we can work out the
concentration of the other from a balanced equation for the redox reaction.

There are two examples we should understand fully, although we may meet other examples of redox
titrations where information about the titration is provided in the question.

Titration of iron(II) ions using manganate(VII) ions under acidic conditions
Purpose: To determine the concentration of Fe2+(aq) in a solution.
e.g. determining the mass of iron in iron tablets
determining the %purity of iron in an alloy containing iron
determining the concentration of iron(II) ions in contaminated water
determining the molar mass and formula of an iron(II) salt

How to do it:
If the sample to be investigated isn't already a solution of iron(II) ions, the sample may need to be
crushed and dissolved in water and sulphuric acid. We'll need to measure the mass of the sample
before dissolving. The resulting solution will be pale green, almost colourless if the concentration of
iron(II) is fairly low.

We titrate a known volume of the sample solution in a conical flask against potassium managate(VII)
solution of known concentration in the burette. As the deep purple manganate(VII) ions react with the
iron(II) ions they are decolourised. This continues until all the iron(II) ions have reacted. The next drop
of potassium manganate(VII) stays purple, so the endpoint is when the first hint of pink/purple persists
in the solution.
N.B. if the iron(II) solution is in the burette, the endpoint will be when the purple colour of the
manganate(VII) in the flask disappears.

The chemistry:
The managanate(VII) ions are reduced to managese(II), while the iron(II) is oxidized to iron(III) ions.
We have already constructed an equation for this:

MnO4-(aq) + 8H+(aq) + 5e- ! Mn2+(aq) + 4H2O(l)
2+
5 Fe (aq) ! 5 Fe (aq) + 5 e-
3+

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ! 5 Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Mole ratio: 1 : 5
One mole of MnO4- ions reacts with five moles of Fe2+ so we have the mole ratio of 1:5 when we do the
titration calculation.




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OCR Chemistry A H432 Redox and Electrode Potentials

The calculations:
i) 25.0cm3 of a solution of iron(II)sulphate required 23.0cm3 of 0.0020 mol dm-3 potassium
manganate(VII) to completely oxidize it in acidic solution. What was the concentration of the
iron(II)sulphate solution ?

Step 1: Calculate moles of the permanganate
moles = conc. x vol = 0.00200 x (23/1000) = 4.60 x 10-5 moles

Step 2: Use the mole ratio from the balanced equation to get moles of iron(II) ions
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ! 5 Fe3+(aq) + Mn2+(aq) + 4H2O(l)
1 : 5
4.60x10-5 2.30 x 10-4 moles of Fe(II)

Step 3: Find concentration of iron(II) solution
Conc of Fe2+ = moles / vol (in dm3) = 2.3x10-4 / 0.0025 = 0.00920 mol dm-3

ii) The solution was made by dissolving a tablet containing hydrated iron(II) sulphate and sucrose
(which does not react with permanganate ions) in 250cm3 of water. Calculate the mass of FeSO4 in the
tablet.

Moles of FeSO4 = conc x vol in dm3 = 0.00920 x 0.25 = 0.0023 mol (or 10 x the moles in 25cm3)

Mass of FeSO4= 0.0023 x Mr = 0.0023 x (55.8 + 32.1 + (16.0 x 4)) = 0.0023 x 151.9
= 0.350g

iii) The formula for the hydrated iron sulphate used in the tablet is FeSO4.7H2O. If the mass of the
tablet was 8.5g, calculate the % by mass of hydrated iron(II) sulphate in the tablet.

Mr of FeSO4.7H2O = 151.9 + (7 x 18) = 277.9
Mass in tablet = moles x Mr = 0.023 x 277.9 = 0.6392g

% by mass = (0.6392 / 8.5) x 100 = 7.5 %

Practical: Nuffield expt 19.1c p.458

Titration of iodine in solution using thiosuphate ions
Purpose: To determine the concentration of an oxidizing agent in a solution e.g. Cu2+(aq) ions.
e.g. determining the % of copper in an alloys such as brass or bronze
determining concentration of copper ions in a solution
determining concentration of dichromate ions in a solution
determining the concentration of chlorate(I) ions ClO- in bleach
determining the concentration of hydrogen peroxide solution

How to do it:
The analysis is done in two stages.

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OCR Chemistry A H432 Redox and Electrode Potentials

In the first stage, the oxidizing agent to be determined is reacted with excess iodide ions. This results in
iodine being formed in the solution, the amount of iodine being directly related to the amount of the
oxidizing agent present.
e.g. 2Cu2+(aq) + 4I-(aq) ! 2CuI(s) + I2(aq)
so in this case 2 moles of Cu2+ ions produce 1 mole of iodine.

In the second stage, the concentration of iodine in the solution is determined by titrating it against
sodium thiosulphate of known concentration. The red-brown colour of the iodine in solution fades to a
pale straw colour, and to colourless when all the iodine has been reduced to iodide ions. This makes
the endpoint difficult to see, so close to the endpoint an amount of starch is added as an indicator.
Starch is blue-black when iodine is present, but at the endpoint the starch becomes colourless.

The chemistry:
I2(aq) + 2e- ! 2I-(aq) reduction

At the same time the thiosulphate ions are oxidized to tetrathionate ions:
2 S2O32-(aq) ! S4O62-(aq) + 2e- oxidation

The overall equation is therefore:
I2(aq) + 2 S2O32-(aq) ! 2I-(aq) + S4O62-(aq)
Mole Ratio: 1 : 2
This means that 2 moles of thiosulphate ions react with 1 mole of iodine.

The calculation
The calculation is also done in two stages. In the first stage, the titre for the thioulphate and its
concentration are used along with the mole ratio above to work out the moles of iodine in the portion
of solution that was titrated. This is then scaled up to get moles of iodine in the whole solution.

In the second stage, the moles of iodine in the whole solution are used along with the mole ratio of
iodine to oxidizing agent to work out the concentration of oxidizing agent in the whole solution. This
can then be used further if needed to work out the mass of the oxidizing agent given a formula etc.

e.g. 30.00cm3 of bleach was reacted completely with iodide ions in acidic solution. The iodine formed
was titrated against 0.2000 mol dm-3 sodium thiosulphate, requiring 29.45cm3 of the thiosulphate
solution to reach the endpoint.

The chlorate(I) ions in the bleach react with iodide ions according to the equation:
ClO-(aq) + 2I-(aq) + 2H+(aq) ! Cl-(aq) + I2(aq) + H2O(l)
Calculate the concentration of the chlorate ions in the bleach.

Step 1: Calculate moles of the thiosulphate ions in titration
moles = conc x vol (in dm3) = 0.2000 x (29.45/1000) = 5.890 x 10-3 moles

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OCR Chemistry A H432 Redox and Electrode Potentials

Step 2: Use the equation for the titration to get moles of iodine in the titre
I2(aq) + 2 S2O32-(aq) ! 2I-(aq) + S4O62-(aq)
1 : 2
2.945 x 10-3 : 5.89x10-3

The whole bleach solution was titrated, so moles of I2 released by the ClO- ions in the bleach
= 2.945 x 10-3 mol.

Using the moles of iodine to find moles of chlorate ions:
ClO-(aq) + 2I-(aq) + 2H+(aq) ! Cl-(aq) + I2(aq) + H2O(l)
1 : 1
2.945 x 10-3
2.945 x 10-3

Therefore the 30cm3 of bleach contained 2.945 x 10-3 moles of chlorate(I) ions.
Step 3: Calculate the concentration of the chlorate(I) ions in the bleach
conc of chlorate(I) = moles / vol in dm3 = 2.945 x 10-3 / 0.030 = 0.0982 mol dm-3

Practical: Nuffield expt. 6.8c (potassium iodate(V) with 10% potassium iodide by mass)

Electrode Potentials
Electron transfers in redox reactions
If we dip a zinc strip into copper II sulphate, we observe a layer of copper deposited on the zinc. We
might also observe that the blue colour of the copper sulphate solution fades towards colourless as
zinc sulphate is formed in solution. CuSO4(aq) + Zn(s) ! Cu(s) + ZnSO4(aq)
+2 0 0 +2
The oxidation of zinc is producing electrons, while the reduction of copper ions is consuming them. If
we could separate these two reactions, we could have the electrons flow though an electrical circuit to
get from where they were produced to where they are consumed. This is what happens in the cells of a
battery.

Cells and half cells
An electrical cell comprises two half-cells, one in which oxidation is taking place to supply electrons,
and one in which reduction is taking place to consume them.

Each half cell therefore contains an element which changes oxidation state.

electrical cell

half-cell in which half-cell in which
reduction consumes oxidation releases flow of
electrons electrons electrons





external circuit

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OCR Chemistry A H432 Redox and Electrode Potentials

The simplest type of half-cell comprises a metal placed in an aqueous solution of its ions. An
equilibrium exists at the surface of the metal between the oxidation state of the metal and the
oxidation state in its ion.

By convention, the equilibrium is always written with the electrons on the left side:
e.g. for copper and copper II ions: Cu2+(aq) + 2e- ⇌ Cu(s)

and for zinc and zinc ions: Zn2+(aq) + 2e- ⇌ Zn(s)

This convention means that the forward reaction is always REDUCTION (gaining electrons) and the
reverse reaction is always OXIDATION (losing electrons).

We have written these as reversible reactions because either oxidation or reduction could take place in
any half cell, it is only when it is connected to another half cell that one goes in the oxidation direction
and one goes in the reduction direction. To decide which way each reaction will go, we need to know
the electrode potential of each half cell.

Electrode Potential
The tendency of a half cell to gain electrons is measured using a value called standard electrode
potential. The larger (more positive) the electrode potential, the greater the tendency of that half-cell
to gain electrons. It is measured in Volts, V.

We compare the electrode potentials of the two half cells:
In the half-cell with the most positive electrode potential,
• the reaction will go in the forward direction
• reduction will take place
• electrons will be gained
This will be the positive terminal of the cell.

In the half-cell with the smallest (most negative or least positive) electrode potential
• the reaction will go in the reverse direction
• oxidation will take place
• electrons will be released to flow around the circuit
This will be the negative terminal of the cell.

The electrode potentials for the two half cells we have considered so far are:
Cu2+(aq) + 2e- ⇌ Cu(s) + 0.34 V
Zn2+(aq) + 2e- ⇌ Zn(s) - 0.76 V

Here we can see that the Cu2+/Cu half-cell has the more positive electrode potential:
- so this will be the positive terminal of the cell
- the reaction will be Cu2+(aq) + 2e- ! Cu(s)
- copper ions will be reduced
- the electrons to do this will flow IN from the external circuit
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OCR Chemistry A H432 Redox and Electrode Potentials

The Zn2+/Zn half-cell has the least positive electrode potential
- so this will be the negative terminal of the cell
- the reaction will be Zn2+(aq) + 2e- " Zn(s)
- zinc will be oxidized to zinc ions
- electrons are given up to flow OUT into the external circuit

Metal/Metal ion half cells
The experimental setup for the half cells described above would require two half cells in which each
metal is in contact with a solution containing ions of the same metal:

It is the difference between the electrode
potentials for the two half-cells which drives
one half-cell to produce electrons and the
other to consume them – we can measure this
with a voltmeter. It is referred to as the
potential difference (same is in physics).

As the reactions proceed, both half-cell ends up
with an imbalance between positive and
negative ions. The Zn2+/Zn half-cell ends up
with extra positive ions, and the Cu2+/Cu half-
cell ends up with fewer positive ions. The salt
bridge allows ions to move from one half-cell to
the other in order to correct this imbalance,
otherwise the electrons would stop flowing in the external circuit. The salt bridge could be as simple
as a strip of filter paper soaked in an aqueous solution of an ionic compound, however the ionic
compound chosen mustn't react with the solutions in either of the half cells. Often aqueous KNO3 or
NH4NO3 is used.

Non-metal/non-metal ion half cells
A half-cell simply has to support an equilibrium where an element is
in two different oxidation states. We could therefore have a non-
metal element in contact with a solution containing ions of that
element.

e.g. a hydrogen half-cell comprises hydrogen gas, H2, in contact
with aqueous hydrogen ions:

2H+(aq) + 2e- ⇌ H2(g)


An inert platinum electrode is used to transfer the electrons into or
out of the half-cell and make the connection to the rest of the
electrical circuit. The platinum does not react at all. The platinum

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OCR Chemistry A H432 Redox and Electrode Potentials

electrode is immersed in a solution containing the H+ ions (i.e. an acid), and hydrogen gas is bubbled
over the electrode surface.

Non-metal/non-metal ion half-cells are not limited to hydrogen and positively charged ions. We could
just as well have e.g. chlorine gas in contact with aqueous chloride ions:
Cl2(g) + 2e- ⇌ 2Cl-(aq)

A metal ion/metal ion half cell
This type of half-cell contains an aqueous solution with ions of the same
element in two different oxidation states, e.g. Iron(II) and iron(III).
Fe3+(aq) + e- ! Fe2+(aq)


As before, an inert platinum electrode is needed to transfer the electrons
into or out of the half-cell.


Standard electrode potentials
We need to measure and compare electrode potentials under controlled and reproducible conditions.
We therefore choose to compare the electrode potentials of different half-cells to the electrode
potential of a hydrogen half-cell (our reference standard), which we define as having an electrode
potential of zero volts when under standard conditions of 298K (25°C) and with a H2 gas pressure of
101kPa (1 atmosphere) and an H+ solution concentration of 1 mol dm-3.

Standard electrode potentials of other half-cells also need to be measured with the same standard
conditions: 298K, 101kPa gas pressure, and 1 mol dm-3 solution concentrations, or with equal
concentrations of each ion in a metal ion/metal ion half cell.

To measure the standard electrode potential of other half-cells, we combine them with a standard
hydrogen electrode to form a complete cell, then measure the overall cell potential, (the potential
difference across the cell) with a high resistance voltmeter. This gives a direct reading of the standard
electrode potential of the half-cell being measured.


Definition: The standard electrode potential of
a half-cell, Eө, is the electrode potential of a
half-cell compared with a standard hydrogen
half cell, measured at 298K with solution
concentrations of 1 mol dm-3 and a gas
pressure of 101kPa.

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OCR Chemistry A H432 Redox and Electrode Potentials

Cell potential
Electrode potentials can be used to predict the cell potential (potential difference) produced by any
combination of half-cells.

For example, if we used our Zn2+/Zn and Cu2+/Cu half cells together to make a cell:
Cu2+(aq) + 2e- ⇌ Cu(s) Eө = + 0.34 V most positive electrode potential => reduction
Zn2+(aq) + 2e- ⇌ Zn(s) Eө = - 0.76 V least positive electrode potential => oxidation

We can then calculate the standard cell potential, Eөcell.
Eөcell = Eө (reduction reaction) - Eө (oxidation reaction)

So Eөcell = 0.34 - (-0.76) = +1.10 V SHOW THE + SIGN!

N.B. Cell potentials should be positive – if it comes out negative, you've got the wrong half-cells doing
oxidation and reduction !

Worked example:
A silver/copper cell is made by connecting together an Ag+/Ag half cell and a Cu2+/Cu half cell.
The half-cell equilibria and standard electrode potentials are:
Ag+(aq) + e- ⇌ Ag(s) Eө = +0.80V reduction reaction
2+ - ө
Cu (aq) + 2e ⇌ Cu(s) E = +0.34V oxidation reaction
Eөcell = 0.80 - 0.34 = +0.46V

Practice:
For each of the half-cell combinations below, give the value of Eөcell

Ca2+/Ca | 2H+/H2 Given Eө values: Ca2+/Ca -2.87V
Ans: +2.87V 2H+/H2 0.00V

Mn2+/Mn | Cl2/2Cl-
Mn2+/Mn -1.19V
Ans: +2.55V Cl2/2Cl- +1.36V

Cr3+/Cr2+ | Mn2+/Mn Cr3+/Cr2+ -0.41V
Ans: +0.78V

Cell reactions
We may also be asked to derive the overall equation for the redox reaction taking place in the two
half-cells. This is just the same as combining a reduction and an oxidation half-equation to get the
overall equation for a redox reaction, which we've done before.

Step 1: Write the two half equations
Ag+(aq) + e- ⇌ Ag(s)
Cu2+(aq) + 2e- ⇌ Cu(s)

Step 2: Change the ⇌ to ! to show which way each is going, based on the electrode potentals

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OCR Chemistry A H432 Redox and Electrode Potentials

Ag+(aq) + e- ! Ag(s)
Cu (aq) + 2e " Cu(s) better written as Cu(s) ! Cu2+(aq) + 2e-
2+ -

Step 3: Get the same number of electrons in each half-equation
2Ag+(aq) + 2e- ! 2Ag(s)
2+ -
Cu(s) ! Cu (aq) + 2e

Step 4: Add the two equations and cancel the electrons on each side
Cu(s) + 2Ag+(aq) + 2e- ! Cu2+(aq) + 2e- + 2Ag(s)

Practice:
For each of the half-cell combinations below, give the overall cell reaction:

Ca2+/Ca | 2H+/H2 Given Eө values: Ca2+/Ca -2.87V
+ 2+ +
Ans: Ca(s) + 2H (aq) ! Ca (aq) + H2(g) 2H /H2 0.00V

Mn2+/Mn | Cl2/2Cl-
Mn2+/Mn -1.19V
2+ - -
Ans: Cl2(g) + Mn(s) ! Mn (aq) + 2Cl (aq) Cl2/2Cl +1.36V

Cr3+/Cr2+ | Mn2+/Mn Cr3+/Cr2+ -0.41V
3+ 2+ 2+
Ans: 2Cr (aq) + Mn(s) ! 2Cr (aq) + Mn (aq)

We might also be given the overall equation and one of the half equations, and asked to work out what
is going on in the other half-cell:

e.g. In a cell, the following overall reaction occurs:
Cu(s) + 2Ag+(aq) ! Cu2+(aq) + 2Ag(s)

The half-equation for the Ag/Ag+ electrode is Ag+ + e- ! Ag
Work out the half-equation for what is taking place in the other half-cell:

Step 1: Calculate the changes in oxidation number taking place in the overall equation, and multiply up
the known half equation to the corresponding number of electrons.

Overall Cu + 2Ag+ ! Cu2+ + 2Ag so 2 electrons are being transferred
0 +1 2+ 0
+1


2Ag+ + 2e- ! 2Ag multiplying up the known half-equation

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OCR Chemistry A H432 Redox and Electrode Potentials

Step 2: Subtract the known half-equation from the overall equation to reveal the other half-equation:
Cu + 2Ag+ ! Cu2+ + 2Ag
+ -
minus 2Ag + 2e ! 2Ag

= Cu - 2e- ! Cu2+
= Cu ! Cu + 2e- because by convention we use + not – in chemical equations.
2+


Practice:
The example above was rather trivial, and they can be quite a lot harder to figure out, but the method
is as described above e.g.

In a methanol fuel cell, the overall equation is CH3OH + 1½ O2 ! CO2 + 2 H2O
At one electrode, the reaction taking place is 4H+ + 4e- + O2 ! 2H2O
Work out what happens at the other electrode:

Oxidation number changes:
C H3 O H + 1½ O2 ! C O2 + 2 H2O
-2 +1 -2 +1 0 +4 -2 +1 -2
+1 -2 +1
+1

-2 to +4 0 to -2 (3x)
= 6 electrons = 6 electrons

Now multiply up known half-equation to get 6 electrons transferred:
6H+ + 6e- + 1½O2 ! 3H2O

Subtract this half equation from the overall equation (don't worry about – signs yet)
CH3OH + 1½ O2 ! CO2 + 2 H2O
+ -
6H + 6e + 1½O2 ! 3H2O

CH3OH – 6H+ - 6e- ! CO2 - H2O

Finally rearrange to get rid of subtractions (and check if the half-equation will simplify)

CH3OH + H2O ! CO2 + 6H+ + 6e-

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OCR Chemistry A H432 Redox and Electrode Potentials

Storage Cells (batteries)

Half-cells are combined to produce two types of commercial electrical storage cell, or battery.

i) non-rechargeable cells e.g. alkaline cells
These provide electrical energy until the oxidation and reduction has been carried out to such an
extent that the potential difference can no longer be maintained. The cell is then "flat", and must be
disposed of.

Alkaline cell construction: graphite
cathode (+)


zinc anode (-)
Details not required


MnO2 and
NaOH paste




The zinc anode supplied electrons, as the zinc is oxidized by the hydroxide ions:

ZnO(s) + H2O(l) + 2e- ⇌ Zn(s) + 2OH-(aq)

At the graphite cathode, electrons are supplied to the manganese IV oxide which is reduced. Hydroxide
ions are produced, replacing those which were consumed at the anode

2MnO2(s) + H2O(l) + 2e- ⇌ Mn2O3(s) + 2OH-

Eөcell is + 1.5V (typical of AA cells)


ii) Rechargeable cells e.g. NiCd, NiMH, Lithium ion batteries (in laptops), Lithium ion polymer batteries
(iPod etc.)
The reduction and oxidation in the two half-cells which supplies the electrical current can be reversed
during recharging. This reverses the two redox equilibria, recreating the reactants so that the cell can
be used again and again.

Cadmium, along with other substances commonly used in batteries, is toxic. There are environmental
problems associated with its disposal, hence its use only in rechargeable and therefore reusable cells.

Lithium is highly reactive, and reacts exothermically, so there is a risk of fire if the contents of a lithium-
based battery are exposed by breaking open the battery.
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OCR Chemistry A H432 Redox and Electrode Potentials

In NiCd batteries the positive and negative electrodes, isolated from each other by the separator, are
rolled in a spiral shape inside the case, and surrounded with potassium hydroxide as the electrolyte.
This allows a NiCd cell to deliver a much higher maximum current than an equivalent size alkaline cell.
The internal resistance for an equivalent sized alkaline cell is higher which limits the maximum current
that can be delivered.

The oxidation half-cell reaction at the negative terminal in a NiCd battery during discharge is:


and the reduction half-cell reaction at the positive terminal in a NiCd battery is:

Details not required
This gives a cell reaction of:


When the storage cell is recharged, these two reactions go in the opposite direction.

Fuel Cells
A fuel cell releases the energy from the reaction of a fuel with oxygen in electrical form, producing a
voltage rather than heat. The most common fuel cells use hydrogen, although other hydrogen-rich
fuels such as methanol and natural gas can be used.

Pure hydrogen fuel emits only water as the product of the reaction, whilst hydrogen-rich fuels produce
only small amounts of CO2 and other gaseous pollutants, so vehicles fitted with fuel cells have the
potential to be cause less damage to the environment than those burning petrol or diesel.

How a fuel cell works:
• The reactants (hydrogen and oxygen) flow in and the
products (water) flow out, while the electrolyte remains in
the cell
• Fuel cells do not have to be recharged and can operate
virtually continuously as long as the fuel and oxygen
continue to flow into the cell
• The electrodes are made of a catalyst material such as a
titanium sponge coated in platinum
• The electrolyte is an acid or alkaline membrane that allows
ions to move from one compartment of the cell to the other
(like a salt bridge)

The equilibria involved in an alkaline fuel cell are:

2H2O (l) + 2e- ⇌ H2 (g) + 2OH-(aq) Eө = -0.83 V (negative electrode)
- -
½O2 (g) + H2O (l) + 2e ⇌ 2OH (aq) Eө = +0.40 V (positive electrode)

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OCR Chemistry A H432 Redox and Electrode Potentials

The equilibria involved in an acidic fuel cell are:

2H+(aq) + 2e- ⇌ H2 (g) Eө = +0.00 V (negative electrode)
½O2 (g) + 2H+ (aq) + 2e-⇌ H2O (l) Eө = +1.23 V (positive electrode)


The overall reaction and cell potential in both cases is the same:
H2 (g) + ½O2 (g) → H2O (l) Eөcell = 1.23 V

Feasibility of reactions
The benefits of using electrode potentials are not limited to calculating the cell potential of a cell. More
generally, we can use electrode potentials to predict whether a redox reaction will take place.

For example, "Will a solution containing iron(III) ions oxidise nickel to nickel(II) ions, and be reduced to
iron(II)?”

We need to look at the electrode potentials for the oxidation and reductions involved.
# We use the same convention of writing each half equation with the electrons on the left
# We change the equilibrium signs to arrows to show which way the reaction would have to go in
each half cell for the reaction to go as described: in this case we want Ni ! Ni2+ and Fe3+ ! Fe2+

Ni2+(aq) + 2e- ⇌ Ni(s) Eө = -0.25V
Fe3+(aq) + e- ⇌ Fe2+(aq) Eө = +0.77V

• Now check that one arrow is going in each direction. If not, you don't have a redox reaction, you
have two species that both want to be oxidized, or both reduced so no reaction will occur.

• Work out the value Eөcell would have if the reaction did happen.
Eөcell = Eө reduction reaction (forward arrow) - Eөoxidation reaction (backward arrow)

• If the calculated Eөcell has a positive value, the reaction is feasible. If the value is negative, the
reaction is not feasible. In our example Eөcell = 0.77 – (-0.25) = +1.02V

So we can say that iron(III) ions CAN oxidize nickel to nickel(II) ions, but we need to be able to explain
what the calculation has shown:

General explanation if the reaction is feasible:
The electrode potential for the <REDUCTION HALF CELL> is more positive than the electrode potential
for the <OXIDATION HALF CELL> so <SPECIES BEING REDUCED> is a sufficiently powerful oxidizing agent
to oxidise <SPECIES BEING OXIDISED> to <SPECIES AFTER OXIDATION>, releasing the electrons
needed to reduce the <SPECIES BEING REDUCED> to <SPECIES AFTER REDUCTION>.

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OCR Chemistry A H432 Redox and Electrode Potentials

"Why does Fe3+ react with Ni ?"
The electrode potential for the Fe3+/Fe2+ half cell is more positive than the electrode potential for
Ni2+/Ni, so Fe3+ is a sufficiently powerful oxidizing agent to oxidise Ni to Ni2+, releasing the electrons
needed to reduce the Fe3+ to Fe2+.

Now let's look at the opposite effect. Another possible redox reaction might be "Will nickel(II) ions
oxidize iron(II) ions to iron(III), themselves being reduced to nickel?"
We approach this the same way: We want Fe2+ ! Fe3+ and Ni2+ ! Ni

Fe3+(aq) + e- ⇌ Fe2+(aq) Eө = +0.77V
2+ -
Ni (aq) + 2e ⇌ Ni(s) Eө = -0.25V
Eөcell = -0.25 – 0.77 = -1.02V

Check: Yes we've got one arrow going each way
Since Eөcell is negative we can say that nickel(II) ions CAN'T oxidize iron(II) to iron(III). Again we need to
be able to explain what the calculation has shown:

General explanation if reaction not feasible:
The electrode potential for the <OXIDATION HALF CELL> is more positive than the electrode potential
for <REDUCTION HALF CELL>, so the <SPECIES BEFORE REDUCTION> is not a powerful enough oxidizing
agent to oxidise <SPECIES BEFORE OXIDATION> to <SPECIES AFTER OXIDATION> and the necessary
electrons will not be available to reduce <SPECIES BEFORE REDUCTION> to <SPECIES AFTER
REDUCTION>.

“Why do nickel(II) ions not oxidise iron(II) to iron(III)?”
The electrode potential for the Fe3+/Fe2+, half cell is more positive than the electrode potential for
Ni2+/Ni, so the Ni2+ is not a powerful enough oxidizing agent to oxidize Fe2+ to Fe3+ and the necessary
electrons will not be available to reduce Ni2+ to Ni.

So long as we know the standard electrode potentials, we can deal with quite complicated redox
systems.
most likely to lose
Zn2+(aq) + 2e- ⇌ Zn(s) Eө = -0.77V electrons and be
2+
Fe (aq) + 2e -
⇌ Fe(s) E = -0.44V oxidised
ө
agents
i.e. reducing

2H+(aq) + 2e- ⇌ H2(g) Eө = 0.00V

Cu2+(aq) + 2e- ⇌ Cu(s) Eө = +0.34V
3+ - 2+
Fe (aq) + e ⇌ Fe (aq) Eө = +0.77V
Ag+(aq) + e- ⇌ Ag(s) Eө = +0.80V
NO3-(aq) + 2H+(aq) + e- ⇌ NO2(g) + H2O(l) Eө = +0.80V most likely to gain
+ -
O2(g) + 4H (aq) + 4e ⇌ 2H2O(l) Eө = +1.23V electrons and be
Cl2(g) + 2e- ⇌ 2Cl-(aq) Eө = +1.36V reduced i.e. oxidising
agents

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OCR Chemistry A H432 Redox and Electrode Potentials

Example: Does copper react with acids ?

If we consider the usual reaction of a metal with an acid, i.e. metal + acid ! metal salt + hydrogen then
we want H+ ! H2 and Cu ! Cu2+

2H+(aq) + 2e- ⇌ H2(g) Eө = 0.00V Reduction
2+ - ө
Cu (aq) + 2e ⇌ Cu(s) E = +0.34V Oxidation

Eөcell for this reaction = 0.00 – 0.34 = -0.34V so copper won't react with the H+ ions.

Explanation:
The electrode potential for the Cu2+/Cu half cell is more positive than the electrode potential for
2H+/H2 half cell, so the H+ ion is not a powerful enough oxidizing agent to oxidise Cu to Cu2+ and the
necessary electrons will not be available to reduce H+ to H2.

So why do we observe that copper does react with nitric acid ?
If the H+ ions aren’t the oxidizing agent, it must be something to do with the other ions in the solution:
the nitrate ions:


Cu2+(aq) + 2e- ⇌ Cu(s) Eө = +0.34V Oxidation
- + -
NO3 (aq) + 2H (aq) + e ⇌ NO2(g) + H2O(l) Eө = +0.80V Reduction

Eөcell = 0.80 – 0.34 = +0.46V

So copper will react when H+ and NO3- ions are both present.

Explanation:
The electrode potential for the NO3-/NO2 half cell is more positive than the electrode potential for the
Cu2+/Cu half cell so NO3- in the presence of H+ is a sufficiently powerful oxidizing agent to oxidise Cu to
Cu2+, releasing the electrons needed to reduce the NO3- to NO2.

Comparing Oxidising Agents and Reducing Agents
We might be asked to compare the species in a redox reaction and say which is the better oxidizing
agent, or reducing agent.

e.g. Which species is the best oxidizing agent?

Cu2+(aq) + 2e- ⇌ Cu(s) Eө = +0.34V
NO3-(aq) + 2H+(aq) + e- ⇌ NO2(g) + H2O(l) Eө = +0.80V

An oxidizing agent causes something else to be oxidized, so it is itself reduced. We look for the half
equation that is going in the reduction direction (the one with the most positive electrode potential, so
the half cell in which the reaction goes in the forward direction), and the better oxidizing agent will be
the reactant(s) for that reaction: in this case, acidified nitrate ions.

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OCR Chemistry A H432 Redox and Electrode Potentials

If we were asked for the best reducing agent, we look for the substance that is itself oxidized, in this
example the Cu, because in this half-equation the electrode potential is least positive, so the reaction is
going in the reverse (oxidation) direction.

Overall equations
Once we have determined that a redox reaction will work, and we know in which direction the reaction
in each half-cell is going we can multiply up to balance the electrons, and combine the half equations
to get an overall equation for the redox reaction EXACTLY as we have done before with half equations.

e.g. when copper reacts with concentrated nitric acid:

Cu2+(aq) + 2e- ⇌ Cu(s)
NO3-(aq) + 2H+(aq) + e- ⇌ NO2(g) + H2O(l)


Step 1: Cu(s) ! Cu2+(aq) + 2e- turning round equation
- + -
NO3 (aq) + 2H (aq) + e ! NO2(g) + H2O(l)

Step 2: Cu(s) ! Cu2+(aq) + 2e-

- + -
2NO3 (aq) + 4H (aq) + 2e ! 2NO2(g) + 2H2O(l) multiplying up electrons

Step 3: Cu(s) + 2NO3-(aq) + 4H+(aq) ! 2NO2(g) + 2H2O(l) + Cu2+(aq)


Limitations of predictions using standard electrode potentials
1) Non-standard conditions alter the values of the electrode potentials, so if the redox reaction is to
happen under any conditions other than 298K, 101kPa and concentrations of 1 mol dm-3 (or equal
concentrations in a metal ion/metal ion half cell) then the electrode potentials won't be correct.

e.g. Cu2+(aq) + 2e- ⇌ Cu(s) Eө = +0.34V

If we increase the concentration of Cu2+ ions, Le Chatelier tells us that the position of equilibrium
moves in the forward direction to oppose the change and use up the Cu2+ ions. The electrode potential
measures the tendency of the reaction to go in the forward direction (i.e. to gain electrons by
reduction), so the electrode potential increases compared to the standard Eө value.

2) Standard electrode potentials apply to aqueous equilibria. Many redox reactions don't happen
under these conditions, so the electrode potentials can't reflect those conditions.
e.g. 2Mg(s) + O2(g) ! 2MgO(s)

3) Even if the electrode potentials indicate that a redox reaction is feasible, it says nothing about the
RATE of the reaction, which may be so unobservably slow due to a very high activation energy.

As a general rule, the larger the difference between the electrode potentials of the two redox
equilibria, the more likely it is that a reaction will take place in practice.

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OCR Chemistry A H432 Redox and Electrode Potentials

Consider again the reaction of copper with nitric acid, using these rules.
Cu2+(aq) + 2e- " Cu(s) Eө = +0.34V
- + -
NO3 (aq) + 2H (aq) + e ! NO2(g) + H2O(l) Eө = +0.80V

Standard conditions means the concentration of nitrate ions must be 1 mol dm-3. So for copper
reacting with 1M nitric acid, the difference in electrode potentials is 0.80 – 0.34 = +0.46 V. This is a
fairly small difference, so the reaction might take place.

What if we use concentrated acid?
If we increase [H+] then the position of equilibrium for the NO3-/NO2 redox equilibrium shifts to the
right to use up the H+. This means the electrode potential of the reduction half-cell gets more positive.

Remembering that Eөcell = Eө reduction - Eөoxidation, if Eө reduction increases, then Eөcell increases and the
reaction is more likely to be feasible, and we can anticipate it is more likely take place under these
conditions (i.e. go faster).

Would the reaction go faster if we added Cu2+ ions to the solution?
If we increase [Cu2+] then the position of equilibrium for the Cu2+/Cu redox equilibrium shifts to the
right (forward direction) to use up the Cu2+, in accordance with LeChatelier’s principle. This means the
electrode potential of the oxidation half-cell will have become more positive. Since Eөcell = Eө reduction -
Eөoxidation, and increase in Eөoxidation will reduce Eөcell and the reaction is less likely to happen.

p. 21