Give the formulae of the following complex ions:
Dichromate (VI)
Manganate (VII)
Sulphate (IV)
Thiosulphate
Cr2O72MnO4SO32S2O32-
Ethanedioate
Iodate (V)
Sulphate (VI)
Nitrate (V)
C2O42IO3SO42NO3-
Oxidation Number
The oxidation state of an ion is the number of electrons that must either be added or
removed from that ion in order to give a neutral atom. It is like the valency but written the
other way round. For non-ionic compounds, we imagine the more electronegative atoms
‘takes’ the electrons. There are a number of rules that can be used to determine the
oxidation state of a particular element. These are:







The sum of all the oxidation states in a compound is ____0_____.
The sum of all the oxidation states in an ion is equal to the __charge__on that ion.
The oxidation states of all the uncombined elements are ____0______
The oxidation state of oxygen, in a compound, is ___-2___ (except in peroxides)
The oxidation states of hydrogen in compounds is __+1___ ( except in hydrides)
When in compounds, group one elements have an oxidation state of __+1___,
group two elements have an oxidation state of _+2__ and group three _+3__.
Unless asked to find the oxidation state of a group seven element you can normally
assume that it will be ___-1____.
Student Activity 1
Find the oxidation state of the underlined elements in the following compounds.
K2O
Oxygen
= -II
(this is Roman numeral 2 not 11)
This is a compound and so all of the
oxidation states must add up to zero.
Oxygen has an oxidation state of –II.
The potassium must be worth +2 but as there are two of them each potassium is worth +I.
MnO4-.
This is an ion and so all of the
oxidation states must add up
to the charge on the ion.
Oxygen has an oxidation
state of –II.
Oxygen 4  -2 = -8
Manganese
= ?
Ion ( charge ) = -1
The manganese must be worth +7 so that the sum adds up!
By applying oxidation numbers to a chemical reaction it is possible to identify whether a
species is being oxidised or reduced.
When an element is oxidised in an equation, its oxidation state __increases__.
When an element is reduced in an equation, its oxidation state ___decreases__.
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Student Activity 2
Write an equation for the reaction between sodium and chlorine and use
oxidation states to show the redox.
2Na +
Oxidation states
0
Cl2 →
0
2NaCl
+1 -1
1. What is the change in the oxidation state of the sodium?
0 → +1
2. What is the change in the oxidation state of the chlorine?
0 → -1
Constructing redox equations from half-equations
An overall redox equation for a reaction can be constructed by combining the half equations showing the
transfer of electrons in the reaction.
Example1 : The reaction of silver ions with zinc metal.
The half equations are:
Ag+(aq) +
Zn(s) →
→
e-
Ag(s)
Zn2+(aq) + 2e-
To combine the two equations the number of electrons must be the same in each half equation and they must
be on opposite sides of the equations. To do this we multiply the silver half equation by 2.
2Ag+(aq) +
Zn(s)
2e→
→
2Ag(s)
Zn2+(aq) + 2e-
We can then add the two equations together:
2Ag+(aq) + Zn(s)
→
2Ag(s) + Zn2+(aq)
This is a fully balanced redox equation
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Student Activity 3
Write half equations for the following reactions:
(i)
Oxidation of ethanedioate ions to carbon dioxide;
C2O42- → 2CO2 + 2eReduction of dichromate (VI) in the presence of H+ solution to
chromium (III);
(ii)
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
(iii)
Reduction of iodate (V) ions to iodide ions in acid solution
IO3- + 6e- + 6H+ → I- + 3H2O
(iv)
Oxidation of iodide ions;
2I- → I2 + 2eReduction of sulphate (VI) ions in the presence of H+ solution to
sulphur dioxide and water;
(v)
SO42- + 4H+ + 2e- → SO2 + 2H2O
(vi)
Reduction of tetrathionate (S4O6 2-) to thiosulphate (S2O32-)
S4O62- + 2e- → 2S2O32(in tetrathionate, formally two sulfurs have ON = +6, two have ON = -1)
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Student activity 4
Combine the following half equations to make the full ionic equation.
a) The oxidation of iron (II) ions by manganate.
Fe3+ +
e→ Fe2+
MnO4- +
8H+ +
5e→ Mn2+
+
4H2O
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
b) The oxidation of iodide ions by manganate.
I2
+ 2e→ 2IMnO4- + 8H+
+ 5e→ Mn2+ + 4H2O
10I- + 2MnO4- + 16H+ → 2Mn2+ + 8H2O + 5I2
c) The oxidation of tin (II) ions by dichromate.
Sn4+ + 2e→ Sn2+
Cr2O72+ 14H+ + 6e- → 2Cr3+ + 7H2O
3Sn2+ + Cr2O72- + 14H+ → 3Sn4+ + 2Cr3+ + 7H2O
d) The redox reaction of copper (II) ions with iodide. Note – you
need some iodide to react with the Cu (I) formed to make a solid ppt
of CuI
I2
+ 2e→ 2ICu2+ + e→ Cu+
2Cu2+ + 2I- → 2Cu+ + 2I2
Working out half equations from an overall equation
Example: 2Mg(s) + O2(g)→ 2MgO(s)
Mg has been oxidised (from 0 to +2) and oxygen has been reduced (0 to -2).
- Write out the starting and end charge of the species.
- Work out whether electrons have been lost or gained
- Add in the correct number of electrons to balance the charge
- Remember oxidation: electrons on RHS; reduction electrons on LHS
Mg → Mg2+ + 2eO2 + 4e- → 2O24
Student Activity 5
Deduce the half equations from the overall equations given.
a)Zn(s) + Cu2+(aq)→ Cu(s) + Zn2+(aq)
Zn → Zn2+ + 2eCu2+ + 2e- → Cu
b)2FeCl2(aq) + Cl2(g)→2FeCl3(aq)
Fe2+ → Fe3+ + eCl2 + 2e- → 2Clc)Mg(s) + 2HCl(aq)→ MgCl2(aq) + H2(g)
Mg → Mg2+ + 2e
2H+ + 2e- → H2
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Constructing redox equations using oxidation numbers
This is an alternative method to constructing half-equations and then adding two half
equations together
Example: oxidation of copper to copper (II) by concentrated nitric acid which is reduced to
nitrogen monoxide.
Step 1: Write a ‘draft’ equation based on the information given in the question. (don’t worry
if there are atoms missing or it’s not balanced at this stage)
Cu +
HNO3
→
Cu2+
+
NO
Step 2: Identify the changing oxidation numbers and balance any atoms that are changing
oxidation number.
-3
Cu +
0
HNO3
+5
→
Cu2+
+2
+
NO
+2
+2
Step 3: Balance the equation so that changes in oxidation number are balanced, i.e. total
increase in oxidation number = total decrease in oxidation number.
Since copper increases by 2 and nitrogen decreases by 3, we must multiply Cu/Cu 2+ by 3
and HNO3/NO by 2.
-3 x 2 = -6
3Cu +
0
2HNO3
+5
→
3Cu2+
+2
+
2NO
+2
+2 x 3 = +6
Step 4: Balance the remaining atoms by adding H+ or H2O as needed to either side. In this
case, (a) we add 4 x H2O to the right hand side to balance the oxygens and then (b) we add
6 x H+ to the left hand side to balance the hydrogens.
→
3Cu2+
a) 3Cu +
2HNO3
b) 3Cu +
2HNO3 + 6H+
→
+
2NO
3Cu2+
+
+
4H2O
2NO
+ 4H2O
This means that the final equation is:
3Cu +
2HNO3 + 6H+
→
3Cu2+
+
2NO
+ 4H2O
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Student activity 6
Use oxidation numbers to construct a redox equation for the reaction where hydrogen
iodide is oxidised to elemental iodine, and sulphuric acid is reduced to sulphur dioxide.
HI +
H2SO4 →
O.N. -1
2HI +
+6
H2SO4 →
1/2 I2 +
SO2
0
+4
I2 +
SO2
I2 +
SO2 + 2H2O
Then add water
2HI +
H2SO4 →
Using redox equations in titration calculations
If you can construct a redox equation, you can use this to calculate quantities relating to
titration experiments.
Example: Iron (II) salts reacting with potassium manganate(VII) solution
Potassium Manganate (VII) solution is a powerful oxidising agent.
Reduction half-equation: The manganate (VII) ion is reduced to the
manganese (II) ion
MnO4- + 5e- + 8H+ → Mn2+ + 4H2O
Oxidation half –equation: The iron(II) ion is oxidised to the iron (III) ion
Fe2+ → Fe3+ + eThe overall equation does not contain any electrons. The iron equation has to
be multiplied by five so that the electrons can be cancelled out when the two
equations are added together.
5Fe2+ → 5Fe3+ + 5eThe overall equation is found by combining the oxidation and reduction half
equations once the electrons are the same on either side of the equation.
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
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Student Activity 7
1. 25cm3of a solution of an iron(II) salt required 20.0cm 3 of 0.02moldm-3 potassium manganate (VII) for
complete oxidation in acid solution.
Calculate the concentration of the iron(II) salt solution.
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
moles MnO4- used = 0.02 x 20/1000 = 0.0004 moles
moles Fe2+ reacted = 0.0004 x 5 = 0.002 moles
conc. Fe2+ in original solution = 0.002 x 1000/25 = 0.08 moldm-3
2. 2.225g of wire was dissolved in sulphuric acid and the solution made up to 250cm 3. All the iron was
converted into Fe2+(aq) ions. 25.0cm 3 of the solution was titrated against 0.02moldm -3 potassium manganate
(VII) solution. 24.80cm 3 of potassium manganate VII) solution was required. Calculate the percentage by
mass of iron in the wire.
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
moles MnO4- used = 0.02 x 24.80/1000 = 0.000496 moles
moles Fe2+ reacted = 0.000496 x 5 = 0.00248 moles
moles of Fe2+ in original solution = 0.00248 x 10 = 0.0248 moles
when wire dissolved, Fe → Fe2+ + 2emoles of Fe in wire = 0.0248 moles
mass of Fe in wire = 0.0248 x 55.8 = 1.384 g (4 s.f.)
%mass of Fe in wire = 1.384/2.225 x 100 = 62.20% (4 s.f.)
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3. 0.312g of powdered iron tablets was dissolved in water and a little dilute sulphuric acid
added. 20.10cm3 of 0.0100 moldm-3 potassium manganate (VII) solution was titrated until a
permanent pale pink colour was observed. Calculate the percentage mass of iron in the
tablets.
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
moles MnO4- used = 0.01 x 20.10/1000 = 0.000201 moles
moles Fe2+ reacted = 0.000201 x 5 = 0.001005 moles
mass of Fe in tablet = 0.001005 x 55.8 = 0.0561 g (3 s.f.)
%mass of Fe in tablet = 0.0561/0.312 x 100 = 18.0% (3 s.f.)
2MnO4- +16H+ + 5C2O42- → 2Mn2+ + 8H2O + 10CO2
[2]
moles of C2O42- = 0.04 x 25/1000 = 0.001 moles
moles MnO4- needed to react = 0.001 x 2/5 = 0.0004 moles
volume MnO4- = 0.0004 x 1000/0.02 = 20 cm3
[3]
10
5.
colourless
pink
moles Fe2+ used = 0.05 x 25/1000 = 0.00125 moles
moles MnO4- reacted = 0.00025 moles
conc. MnO4- = 0.00025 x 1000/12.3 = 0.0203 moldm3 (3 s.f.)
6.
2I- → I2 + 2e-
2I- → I2 + 2eFeO42- + 4e- + 8H+ → Fe2+ + 4H2O
Overall: FeO42- + 8H+ + 4I- → Fe2+ + 4H2O + 2I2
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7.
moles of MnO4- = 0.02 x 38/1000 = 0.00076 moles
moles of H2O2 in portion used = 0.00076 x 5/2 = 0.0019 moles
moles of H2O2 in original solution = 0.0019 x 10 = 0.019 moles
conc. of original solution = 0.019 x 1000/25 = 0.76 moldm-3
Determination of copper ion concentration using the thiosulphate titration:
combining two molar ratios
Copper (II) salts react with iodide ions to produce iodine which is then titrated against
sodium thiosulfate solution. Copper (II) salts are reduced by iodide ions
Cu2+ + e- → Cu+ and 2I- → I2 + 2eBut, copper (I) salts are unstable in solution (they disproportionate to form Cu 2+ and Cu)
and will only form as the solid – precipitated out as the iodide salt. So the overall equation
is:
2Cu2+(aq) + 4I- (aq) → 2CuI(s) + I2 (aq)
Note: You need 4 iodide ions
All copper (I) salts are white solids but this in solution with iodine will appear as a brownish
solid due to the iodine colour. The iodine liberated is titrated against sodium thiosulfate
I2(aq) + 2S2O32- (aq) → S4O62- (aq) + 2I- (aq)
The end point will be when the iodine colour has disappeared, but this is difficult to see so
we add starch solution when the colour becomes a pale straw colour and the iodine makes
the starch go blue-black; when all the iodine has been used up by the thiosulfate the blueblack colour will disappear. However, the solution is not colourless because there is still the
white solid of copper (I) iodide present.
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Student Activity 8 (Jun 2011)
13
Using quantitative data to find the reacting ratios
Iodate (V) ions can be estimated by reacting them with iodide ions, producing iodine
solution and then titrating the iodine with thiosulfate as above.
1. 20.0 cm3 of KIO3 solution containing 3.00 g dm-3 KIO3 was reacted with excess potassium
iodide. The liberated iodine needed 16.80 cm3 of 0.100 moldm-3 sodium thiosulfate solution
for complete reaction.
a) find the number of moles of thiosulfate used in the titration
moles thiosulfate = 0.1 x 16.8/1000 = 0.00168 moles
b) Find the number of moles of iodine liberated
I2 + 2S2O32- → 2I- + S4O62moles iodine = 0.00168/2 = 0.00084 moles
c) Find the number of moles of potassium iodate in the original
solution
in 1 dm3, 3 g of KIO3, moles of KIO3 in 1 dm3 = 3/214 = 0.01402 (4
s.f.)
in 20 cm3, moles of KIO3 = 0.01402 x 20/1000 = 0.0002804 moles
d) Find the ratio of iodine:iodate as a whole number
iodine : iodate
0.00084 : 0.00028 → 3 : 1
e) Suggest an ionic equation for the reaction between iodate and
iodide ions.
5I- + IO3- + 6H+ → 3I2 + 3H2O
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Using quantitative data to predict oxidation state change.
You can use redox titration data and half equations to calculate the number of moles of electrons being added
to or being taken from a species
Example: 0.18 moles of dichromate ions will react with 0.54 moles of nickel forming chromium (III) ions. What
is the new oxidation number of nickel?
Step 1: Write down or construct a half equation for the known change:
Cr2O72-
+
14H+
+
6e-
→
2Cr3+
+
7H2O
Step 2: Calculate the number of moles of electrons being transferred to the dichromate ions.
Moles of electrons = 0.18 x 6 = 1.08 moles
Step 3: Calculate how many moles of electrons are being lost from each mole of nickel
1.08/0.54 = 2
Each mole of nickel is losing 2 moles of electrons, i.e. each atom of nickel is losing 2 electrons, and therefore
the oxidation number will have increased by 2.
The new oxidation number of nickel will be +2.
Student Activity 9
If 3 moles of manganate react with 5 moles of X+, what will be the new oxidation state of X?
………… MnO4- + 5e- + 8H+ → Mn2+ + 4H2O ……………………..............................
…………moles of electrons transferred to MnO4- = 3 x 5 = 15 moles……………………
…………moles of electrons lost from X+ per mole of X+ = 15/5 = 3 moles…………...….
…………each X+ loses 3 electrons, so new oxidation state = +4....………………………
Or a more visual approach, if each of the manganates below is taking in 5 electrons, how
many electrons are being given out per X+.
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