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Electrode potentials are very useful in that they allow us to work out whether a reaction is feasible. Feasible means that we know the natural direction of the reaction concerned. However, we are not predicting anything about the rate of reaction. It may be an extremely slow reaction, so whilst it may be feasible, it may be very slow. It is also important to realize that the predictions we make have been made using standard electrode potentials, so the predictions are only valid under standard conditions. We may predict that a reaction is not feasible, but by changing the temperature (from 298 K) and/or changing the concentrations (from 1 M) the reaction may become feasible.
In order to predict reaction feasibility you must write down a cell diagram which matches up with the reaction you are looking at. Work out Ecell and then:
|
Ecell is positive for a feasible reaction |
A few examples should make this clear:
Will manganese displace nickel(II) from its salts?
This is the reaction:
|
Mn(s) + Ni2+(aq) ⇒ Mn2+(aq) + Ni(s) |
The cell diagram for this is:
|
Mn(s)½Mn2+(aq)::Ni2+(aq)½Ni(s) |
We have the manganese as the left hand electrode as this had to be turned around so that, reading from left to right, we have manganese turning into manganese(II) as shown in the equation. The nickel is the right hand electrode as, reading from left to right, we have nickel(II) turning into nickel as shown in the equation. All the standard electrode potentials in the data book are shown as right hand electrodes. This means when we assemble a cell diagram one of the cells is the right hand electrode just as shown in the book. The other must be the left hand electrode, and so needs to be turned around from the data book version. Make sure you do this carefully - simply read the version from the book backwards, do not swap anything around.
Ecell = - 0.25 - - 1.19 = + 0.94 V
so this reaction is feasible.
Will bromide reduce copper(II) to copper(I)?
This is the reaction:
|
Cu2+(aq) + Br-(aq) ⇒ Cu+(aq) + ½Br2(aq) |
The cell diagram for this is:
|
Pt½Br-(aq), ½Br2(aq)::Cu2+(aq), Cu+(aq)½Pt |
Again, note that the copper half cell forms the right hand electrode as it already reads copper(II) to copper(I) as required by the equation. However, the bromine half cell had to be turned around to form the left hand electrode, as it originally read bromine to bromide which was not in the equation.
Ecell = + 0.15 - + 1.09 = - 0.94 V
so this reaction is not feasible.
Will acidified manganate(VII) oxidize chloride to chlorine?
This is the reaction:
|
MnO4- (aq) + 8H+(aq) + 5Cl-(aq) ⇒ Mn2+(aq) + 4H2O(l) + 2½Cl2(aq) |
The cell diagram is:
|
Pt½Cl-(aq), ½Cl2(aq)::[MnO4-(aq) + 8H+(aq)], [Mn2+(aq) + 4H2O(l)]½Pt |
Ecell = + 1.51 - + 1.36 = + 0.15V
so this reaction is feasible.
Will acidified manganese(IV) oxide oxidize chloride to chlorine?
This is the reaction:
|
MnO2(s) + 4H+(aq) + 2Cl-(aq) ⇒ Mn2+(aq) + 2H2O(l) + Cl2(aq) |
The cell diagram is:
|
Pt½Cl-(aq), ½Cl2(aq)::[MnO2(s) + 4H+(aq)], [Mn2+(aq) + 2H2O(l)]½Pt |
Ecell = + 1.23 - + 1.36 = - 0.13V
so this reaction is not feasible. However, this is a reaction which is used to generate chlorine gas. Remember, this prediction only applies to standard conditions. If the Ecell is only slightly negative it may be possible to make it positive by changing the conditions. This reaction does take place when concentrated hydrochloric acid (greater than 1 mol dm-3) is heated (greater than 298K) with manganese(IV) oxide.
You may be asked about spontaneous reactions. There is a little confusion here depending on whether you are American or British. The Americans use the terms spontaneous and feasible interchangeably. However, in Britain a spontaneous reaction is a feasible reaction which takes place at a reasonable rate of reaction. So a spontaneous reaction is always feasible, but a feasible reaction is not necessarily spontaneous. The term spontaneous has been used on the exam before with the American meaning. Either approach to the answer should gain you credit. If the reaction is not feasible then neither is it spontaneous, so there is no confusion. If the reaction is feasible, then I would prefer the response which states that whilst the reaction is feasible no prediction can be made about the rate, and therefore the spontaneity, of the reaction.