14 Redox 2Equations and electron transfer Half reactions (equations) Use of redox wrt stoichiometry Titration
except in compounds with O or F (see below)usually -1Chlorine
 always -1Fluorine
except in metal hydrides where it is -1 (see below)usually +1Hydrogen
except in peroxides and F2O (see below)usually -2Oxygen
 always +2Group 2 metals
 always +1Group 1 metals
Write 1/2 eqns. They are good receivers of electrons
Reducing agents: Reactive metals eg….
They are good donors of electrons
However, we use a scale to measure how good an aqueous species is at reducing or oxidising. This is called the
ELECTRODE POTENTIAL SCALE
This is a classic displacement reaction
It also involves REDOX. Let’s write the half equations for the above reaction.
Zn (s) Zn2+ + 2e-
Reduced form Oxidised form Can reduce other species Can oxidise other species
Cu2+ (aq) + 2e- Cu (s)
Oxidised form Reduced form Can oxidise other species Can reduce other species
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Notice that we now write these as reversible reactions, and that in each case there is both a reducing and oxidising agent.
Write half equations for the redox reactions involving chlorine and sodium:
Note: The situation is similar to that of a Lowry-Bronsted acid/base conjugate pair:
Weak acid Strong base + H+ (aq)
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Reducing agent
Reducing agent
Oxidising agent
Identify the strongest and weakest oxidising and reducing agents on the table
Draw arrows to represent the trends
Discuss - the AAC “Allen Anticlockwise rule)
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Best reducing agents
Best oxidising agents
Worst reducing agents
Worst oxidising agents
Notes:
1. The scale represents the tendency for the oxidised state to accept electrons and be reduced.
Try to remember the species at each end of the scale and the general trends, eg.
K+ (aq) + e- K (s) -2.92 V
Cl2 (aq) + 2e- 2Cl- (aq) +1.36 V
Strong reducing agent
Strong oxidising agent
2. The greater the positive value, the greater this tendency.
K+ (aq) + e- K (s) -2.87 V
Cl2 (aq) + 2e- 2Cl- (aq) +1.36 V
“Standard electrode potentials”
Spontaneous tendency
Spontaneous tendency
The standard electrode potentials are measured by placing the system in question in a circuit with the standard hydrogen electrode
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The standard hydrogen electrode consists of a platinum plate in a 1M solution of hydrogen ions with hydrogen gas at 1 atm. being bubbled through it:
This is the left hand component of the cell.
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To do this, we set up a cell with the system whose standard electrode potential we want to measure and the standard hydrogen electrode.
The right hand electrode (called a 1/2 cell) corresponds to the electrode potential to be measured.
We measure the voltage, V which gives us the SEP if the conditions are standard.
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This represents the standard electrode potential, Eº, if the following conditions are met:
• The standard hydrogen electrode forms the left-hand half cell
• Any concentrations are 1M
• Temperature 298K
Note: Changes in concentration and temperature will affect the rate of the reactions, but not the likely direction of reaction. Effects of these changes will be small but voltages of the cells can be slightly changed.
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CELLS
In general,
The half cell with the more negative standard electrode potential produces electrons and is the negative pole of the cell.
The half cell with the more positive standard electrode potential accepts electrons and is the positive pole.
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The Daniell cell…
These produce an electric current due to chemical reactions and are formed when 2 different half cells are connected in a circuit.
The Daniell cell…
The more negative electrode potential shifts to the left
Eqn.
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Eqn.
Zn (s) Zn2+ + 2e-
By convention:
• The more positive half cell is always on the right (there is an exception to this rule).
• The vertical dotted lines represent the salt bridge.
• The oxidised forms are written next to these dotted lines.
• The e.m.f. of the cell is calculated using the following equation:
Ecell = ERHelectrode - ELHelectrode
pp84-91 13
We can represent this
Ecell = ERHelectrode - ELHelectrode
Some possible systems:
1. If we arrange the table of standard electrode potentials with the most negative at the top, then the reactions will go in an ANTI- CLOCKWISE direction.
Eg: Fe2+ and MnO4-
Redox and Standard electrode potentials
We can use standard electrode potentials to work out whether REDOX reactions will occur and which directions they are most likely to go.
You need to develop a system for yourself as to how to remember to do this.
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2. The equilibrium with the more negative value for ºE reveals the reducing agent as this half equation is more likely to release electrons.
pp90-91 15
3. Try to remember some classic examples like K and F2, Zn and Cu or S2O32- and MnO4-
Conversely, the equilibrium with the more positive electrode potential is more likely to accept electrons
Feasibility of reactions. As we have seen above, the half cell with the more negative electrode potential will release electrons and reduce the oxidised form of the system with the more positive electrode potential.
We know that the more negative equilibrium will be oxidised itself
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Cl2 (aq) + 2e- 2Cl- (aq) +1.36 V
I2 (aq) + 2e- 2I- (aq) +0.54 V
So, we can use this information to devise a general method for deciding whether a reaction is feasible.
Eg. Will Chlorine gas oxidise iodide ions in solution?
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Consider the reverse process: Will I2 oxidise Cl- ions?
We could also postulate that the greater the difference in the SEPs, the more likely the reaction is to happen. We will come back to this point.
Feasible reactions may not happen for various reasons:
1. Eº and ΔH are related to the energetics of a reaction. The kinetics of a reaction can be an important factor (Eact). Energetics determine whether it is feasible, whether it should happen, but kinetics determine the rate (which may be insignificant).
Consider the following equilibria:
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We have known that copper metal does not react with aqueous hydrogen ions (acids) for a long time and now we know why! The reaction isn’t feasible.
However, the reaction between Copper(II) ions and hydrogen gas appears feasible, but it doesn’t happen under standard conditions. Why?
The strong single covalent bond in the diatomic molecules would have to be broken.
Ea is too high and the reaction does not take place.
Cu2+(aq) + 2e- Cu(s) +0.34 V
2H+(aq) + 2e- H2(g) 0 V
2. Conditions may not be standard, affecting the position of equilibrium
Changes in concentration, temperature, pressure…will affect the reactions taking place, and a reaction in which the Eº are not favourable may actually take place. An excellent example is the reaction used to produce Chlorine gas in the lab…
The spontaneous tendency of the above set of equilibria is shown.
However, we know that concentrated hydrochloric acid produces chlorine gas on reaction with Manganese(IV)Oxide.
Why?
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Conditions are not standard.
The increase in chloride ion concentration shifts the bottom equilibrium to the left
The increase in hydrogen ion concentration shifts the top equilibrium to the right
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In fact, these changes in concentration affect the electrode potentials.
Increasing the hydrogen ion concentration in the top equilibrium makes it more likely to accept electrons and therefore would make the electrode potential more positive (> +0.23V)
Increasing the chloride ion concentration in the bottom equilibrium makes it more likely to release electrons and therefore would make the electrode potential more negative (< +1.36V)
NB: - In addition, if a gas escapes from the system (as is the case above), a reaction may proceed in an unexpected direction.
- Temperature will also affect the position of equilibrium.
Disproportionation reactions Copper (I) ions and compounds are not stable under standard conditions…
Try to generate the same equations to show whether the Iron(II) oxidation state is stable under standard conditions
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Entropy and Cells
Intuition will tell us that Eºcell is directly proportional to the ΔStot
Try and show this relationship using the above equation and the equation for the total entropy change. Remember, to get from total entropy change to Gibbs free energy, we multiplied by (-T), so if we divide the above expression by (-T) we get…
The equation to use is…
ΔStotal = nFE So total entropy change is proportional to the Eºcell
T
As we saw in the last topic ΔG º = -RTlnK
Derive the relationship between lnK and Eºcell using these equations and make lnK the subject…
So clearly, lnK is directly proportional to Ecell and we can use this equation to calculate K for a reversible redox reaction
pp92-95 23
RT
Titrations Acidified potassium manganate (VII) - a self indicator - is a strong oxidising agent used in titrations (burette). The end point is given by the first permanent pink colour.
Half reaction?…
Solutions must be standardised for use, as KMnO4 is a strong oxidising agent
Sodium ethanedioate Na2C2O4 is used for this purpose. At first the reaction is slow so warm ethanedioic is used but then the Mn2+ ions act as a catalyst for the reaction (this is “autocatalysis” - the product of a reaction catalyses the reaction)
Half equation for the acid and overall equation? …
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What is formed if insufficient acid is used in the titration, or in alkaline conditions? What effect would this have on the titration?
Potassium Manganate(VII) and Fe2+ ions Write the 2 half equations and the overall equation…
NB1: The acid used in this reaction is boiled to make sure that dissolved oxygen does not oxidise the Fe2+
NB2: Mn2+ ions in solution give a pale pink colour but this is not noticeable in this titration due to the low concentrations involved
Look at the investigation on p99
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Sodium Thiosulphate
This titration is usually used to calculate how much iodine has been produced in another reaction. Starch solution is used to indicate the end point…
Why can the iodine itself not be used and a self indicator?
Why is the starch not added until near the end point?
Look at the investigation on p100
pp98-101 26
Core Practical 11: Redox titration
Used for titrations with iodine. Try and write the half equations and the overall equation…
http://www.youtube.com/watch?v=esuAlB4NVi0
In ACIDIC conditions the equations for the reactions are:
In ALKALINE conditions the equations for the reactions are:
NB1: Mistake in book p97 (check the cathode equation in alkaline conditions) NB2: Read through the advantages, disadvantages and issues surrounding the use of fuel cells
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Equations not required… pp96-97 the Exam Style Qs
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