ISSR CLASSES
Checkpoint Science
iGCSE Chemistry
iGCSE Physics
iGCSE Coordinated Science
A-Level Chemistry
PRACTICALS
Practicals Home
Practicals A-Z
Study Plan for Practical Work
In each of the following reactions we shall consider either the oxidation of iron(II) to iron(III) or the reduction of iron(III) to iron(II). We, therefore, need the following standard electrode potential:
|
Fe3+(aq), Fe2+(aq)|Pt |
E = + 0.77 V |
The value of Ecell also enables us to predict the value of the equilibrium constant. I have used :: to represent the salt bridge in each case.
The following links will take you to the appropriate analysis:
a Iron(II) sulfate and bromine water
c Iron(II) sulfate and silver nitrate
d Iron(III) chloride and sodium chloride
e Iron(III) chloride and sulfur dioxide solution
f Iron(II) sulfate and acidified potassium manganate(VII)
g Iron(III) chloride and potassium iodide
h Iron(II) sulfate and chlorine water
The possible half-equations are:
|
Fe2+(aq) ⇒ Fe3+(aq) + e- |
|
Br2(aq) + 2e- ⇒ 2Br-(aq) |
Giving the balanced equation:
|
Br2(aq) + 2Fe2+(aq) ⇒ 2Fe3+(aq) + 2Br-(aq) |
The cell diagram representing this reaction is:
|
Pt|Fe2+(aq), Fe3+(aq)::Br2(aq), 2Br-(aq)|Pt |
Ecell = + 1.09 - + 0.77 = + 0.32V
As this is a positive value, this reaction is feasible. This reaction will not go to completion, but if it takes place at a reasonable rate the orange bromine colour should be lost and replaced by the pale yellow of iron(III). The video shows that this is the case. However, it is fairly slow and takes a few seconds.
The possible half-equations are:
|
Fe3+(aq) + e- ⇒ Fe2+(aq) |
|
Zn(s) ⇒ Zn2+(aq) + 2e- |
Giving the balanced equation:
|
Zn(s) + 2Fe3+(aq) ⇒ 2Fe2+(aq) + Zn2+(aq) |
The cell diagram representing this reaction is:
|
Zn(s)|Zn2+(aq)::Fe3+(aq), Fe2+(aq)|Pt |
Ecell = + 0.77 - - 0.76 = + 1.53V
As this is a positive value, this reaction is feasible. This reaction will go to completion. We would expect the zinc to dissolve and the solution to change colour from yellow to green. However, the colour is often quite pale, and so any colour change can be difficult to observe. The video shows that this reaction does take place, although it takes a few minutes. The presence of iron(II) at the end of the reaction is confirmed when the added sodium hydroxide solution produces a green precipitate of iron(II) hydroxide. Effervescence is also seen. This is the reaction of the zinc with the acidic iron(III) solution which produces hydrogen gas.
The possible half-equations are:
|
Fe2+(aq) ⇒ Fe3+(aq) + e- |
|
Ag+(aq) + e- ⇒ Ag(s) |
Giving the balanced equation:
|
Ag+(aq) + Fe2+(aq) ⇒ Fe3+(aq) + Ag(s) |
The cell diagram representing this reaction is:
|
Pt|Fe2+(aq), Fe3+(aq)::Ag+(aq)|Ag(s) |
Ecell = + 0.80 - + 0.77 = + 0.03 V
As this is a positive value, this reaction is feasible. This value is rather small, so the reaction will be close to equilibrium, with slightly more products than reactants. The video shows that this reaction does take place, as a precipitate of silver is seen. Again, the reaction is fairly slow, taking a few seconds for the precipitate to first appear.
The possible half-equations are:
|
Fe3+(aq) + e- ⇒ Fe2+(aq) |
|
2Cl-(aq) ⇒ Cl2(aq) + 2e- |
Giving the balanced equation:
|
2Cl-(aq) + 2Fe3+(aq) ⇒ 2Fe2+(aq) + Cl2(aq) |
The cell diagram representing this reaction is:
|
Pt|2Cl-(aq), Cl2(aq)::Fe3+(aq), Fe2+(aq)|Pt |
Ecell = + 0.77 - + 1.36 = - 0.59 V
As this is a negative value, this reaction is not feasible. This is fairly obvious if you think about it. If iron(III) ions reacted with chloride ions, it would not be possible to have the two ions together in iron(III) chloride!
The possible half-equations are:
|
Fe3+(aq) + e- ⇒ Fe2+(aq) |
|
SO2(aq) + 2H2O(l) ⇒ SO42-(aq) + 4H+(aq) + 2e- |
Giving the balanced equation:
|
SO2(aq) + 2H2O(l) + 2Fe3+(aq) ⇒ 2Fe2+(aq) + SO42-(aq) + 4H+(aq) |
The cell diagram representing this reaction is:
|
Pt|[SO2(aq) + 2H2O(l)], [SO42-(aq) + 4H+(aq)]::Fe3+(aq), Fe2+(aq)|Pt |
Ecell = + 0.77 - + 0.17 = + 0.60 V
As this is a positive value, this reaction is feasible. This reaction will go to completion.
The possible half-equations are:
|
Fe2+(aq) ⇒Fe3+(aq) + e- |
|
MnO4-(aq) + 8H+(aq) + 5e- ⇒Mn2+(aq) + 4H2O(l) |
Giving the balanced equation:
|
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ⇒5Fe3+(aq) + Mn2+(aq) + 4H2O(l) |
The cell diagram representing this reaction is:
|
Pt|Fe2+(aq), Fe3+(aq)::[MnO4-(aq) + 8H+(aq)], [Mn2+(aq) + 4H2O(l)]|Pt |
Ecell = + 1.51 - + 0.77 = + 0.74 V
As this is a positive value, this reaction is feasible. With a value greater than + 0.60 V this reaction will go to completion. The video shows that this is the case, with the purple manganate(VII) being rapidly decolourised.
The possible half-equations are:
|
Fe3+(aq) + e- ⇒Fe2+(aq) |
|
2I-(aq) ⇒I2(aq) + 2e- |
Giving the balanced equation:
|
2I-(aq) + 2Fe3+(aq) ⇒2Fe2+(aq) + I2(aq) |
The cell diagram representing this reaction is:
|
Pt|2I-(aq), I2(aq)::Fe3+(aq), Fe2+(aq)|Pt |
Ecell = + 0.77 - + 0.54 = + 0.23 V
As this is a positive value, this reaction is feasible. This reaction will not go to completion, but there will be more products than reactants. It should be easy to check in practice, as brown iodine solution will be produced. The production of iodine is seen to be rapid in the video.
The possible half-equations are:
|
Fe2+(aq) ⇒Fe3+(aq) + e- |
|
Cl2(aq) + 2e- ⇒2Cl-(aq) |
Giving the balanced equation:
|
Cl2(aq) + 2Fe2+(aq) ⇒2Fe3+(aq) + 2Cl-(aq) |
The cell diagram representing this reaction is:
|
Pt|Fe2+(aq), Fe3+(aq)::Cl2(aq), 2Cl-(aq)|Pt |
Ecell = + 1.36 - + 0.77 = + 0.59 V
As this is a positive value, this reaction is feasible. This reaction will go virtually to completion. It will be difficult to see any colour change. The smell of the chlorine should disappear, but this is rather unreliable as a test. We shall check that iron(III) has been formed by adding sodium hydroxide solution.