Transcript Examples of Redox - wellswaysciences

Week 25
•
Explain the terms: redox, oxidation number, half-reaction, oxidising agent and
reducing agent for simple redox reactions.
•
Construct redox equations using relevant half equations or oxidation
numbers.
•
Interpret and make predictions for reactions involving electron transfer.
© Pearson Education Ltd 2009
This document may have been altered from the original
Redox
• Remember OIL
of electrons
•
RIG
• There are 3 definitions of oxidation (and so 3
for reduction)
• Oxidation is
• Loss of electrons
• Gain in oxidation number
• Loss of hydrogen
• What are the 3 definitions of reduction?
• Redox reactions are best thought of as
electron transfer reactions.
Oxidation numbers
• Using the rules from the table on page 182 assign
oxidation numbers in the following compounds:
• CO2
Mg(NO3)2
• Σ(C) + 2(O) = 0
Since Mg 2+ has +2 ox no
•  C + (2x-2) =0
Each NO3 = -1
•
C + (-4) = 0
Σ (N) + (3xO) = -1
•
C
= +4
N + (-6)
= -1
•
N -6 = -1
•
N = +5
•
For some transition metals
• Work out the oxidation
number of the metal in
each of the following:
• MnO4CrO3
• Cr2O72• MnO2
• FeCl4• V2O5
•
•
•
•
•
•
+7
+6
+6
+4
+3
+5
Redox Reactions
• Any competition or displacement reaction is a redox
reaction.
• E.g. Mg + CuO → MgO + Cu
• In this reaction the O2- remains unchanged and is
simply passed from the Mg to the Cu.
• It is a spectator ion.
Spectator ions are left out and IONIC EQUATIONS
are written.
• i.e. Mg + Cu2+ → Mg2+ + Cu
• Write HALF EQUATIONS
• Mg → Mg2+ + 2e- Loss of e- = oxidation
• Cu2+ + 2e- → Cu gain of e- = reduction
Oxidising and reducing agents
• In a competition reaction between metals the MORE
REACTIVE metal ends up as the compound. (See
previous slide)
• The more reactive metal is oxidised.
• The less reactive metal ion is reduced.
• The more reactive metal gives away its electrons to
the less reactive metal ion and thus is the reducing
agent.
• The less reactive metal ion takes the electrons and so
is the oxidising agent in the reaction.
• So oxidising agents are always reduced and lose
oxidation number.
• Reducing agents are always oxidised and increase
their oxidation number.
General Application
• These principles can be applied to any
reaction as long as oxidation numbers can be
assigned.
Examples of Redox
• PCl3 reacts with chlorine to form PCl5.
• PCl3 + Cl2 → PCl5
• i) What are the changes in oxidation
number for each element?
ii) Which substance is the oxidising
agent?
Examples of Redox
• PCl3 reacts with chlorine to form PCl5.
• PCl3 + Cl2 → PCl5
• i) What are the changes in oxidation
number for each element?
• PCl3 + Cl2 → PCl5
• +3 -1
0
+5 -1
ii) Which substance is the oxidising
agent?
Cl2 is the oxidising agent
• For each of the following reactions use
the change in oxidation number to
deduce the oxidising and reducing
agents.
• i) 2Cu2+ + 4I- → 2CuI + I2
• ii) MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
• iii) 2Cu + 4HCl + O2 → CuCl2 + 2H2O
• iv) 2NaOH + Cl2 → NaOCl + NaCl
•
•
•
•
•
i) oxidising agent Cu2+; reducing agent Iii)MnO2 oxidising agent; HCl red agent
iii)O2 oxidising agent; Cu reducing agent
iv)Cl2 oxidising agent;Cl2 reducing agent.
Above iv) is an example of
disproportionation- simultaneous redox.
The oxidation number goes up AND
down.
Constructing Equations From Half
Equations
• See lilac boxes on p. 182 and 183.
• Do the questions on p. 183
Simple Cells
•
•
•
•
•
•
•
•
Simple Cells.
e.g. Mg + 2HCl → MgCl2 + H2
This is a redox reaction.
The reactions are:
Mg →Mg2+ + 2e- OXIDATION
2H+ + 2e- → H2
REDUCTION.
These are half equations.
If the 2 half equations take place in separate
containers electrons will flow in a wire
connecting the 2 half reactions.
• This arrangement is called a SIMPLE CELL
• In the example above electrons flow FROM
the Mg to the H+ in the wire.
• (The salt bridge makes the electrical
connection between the 2 half cells, allowing
ions to transfer between them.
• It is usually a piece of filter paper soaked in
potassium nitrate solution. (All nitrates are
soluble so no precipitation reactions can occur
by accident.).)
• Current flows. If a bulb is placed in the
circuit this would light up.
What is a half cell?
• It is made up of an element in 2 oxidation
states – generally a metal in contact with an
aqueous solution of its salt.
• An equilibrium is set up on the surface of the
metal
• E.g. Cu2+(aq) + 2e- D Cu(s)
• The electrode potential is the tendency of
this half cell to lose or gain electrons and
gives an idea of the position of the
equilibrium.
Week 25
Standard hydrogen half cell
© Pearson Education Ltd 2009
This document may have been altered from the original
Standard Electrode Potentials
• Half cell potentials cannot be measured in
isolation.
• Standard electrode potentials are defined by
definition as the VOLTAGE (or EMF)
MEASURED UNDER STANDARD
CONDITIONS WHEN THE HALF CELL IS
PART OF AN ELECTROCHEMICAL CELL
WITH THE OTHER HALF CELL BEING A
STANDARD HYDROGEN ELECTRODE.
• Or as written on p. 186
Standard conditions:
•
•
•
•
1 atmosphere pressure
1.0M solution
25oC (298K)
A standard hydrogen electrode is a reference
electrode. By definition its potential is taken as 0
volts.
• When a standard half cell is connected to a hydrogen
half cell through a high resistance voltmeter (to
prevent current flow) the measured potential is taken
to be that of the half cell in question.
• This is the emf of the cell –the voltage produced by a
cell when no current flows.
Standard Hydrogen Half Cell
• See diagram
• The platinum wire is merely to allow for
electrons to pass from one half cell to the
other in the simple cell.
• The half cell equilibrium is:
• 2H+ + 2e- D H2(g)
• This sets up on the platinum foil which is
covered in porous platinum black in which the
electron transfer takes place.
Measuring Standard Cell Potentials
• The standard cell potential of a cell is
the e.m.f. between 2 half cells making
up the cell under standard conditions.
• The standard cell potential of any cell is
the difference between the standard
electrode potentials for each half cell.
• The cell reaction is the overall chemical
reaction taking place in the cell.
• This is the sum of the oxidation and
reduction half reactions taking place.
Week 25
Measuring the standard electrode potential of a Zn/Zn2+ half cell
© Pearson Education Ltd 2009
This document may have been altered from the original
Standard Zinc Half Cell
• See previous slide.
• The measured potential for a standard zinc half cell
(against the hydrogen half cell) is-0.76v.
• What does this mean?
• For the equation:
• Zn2+(aq) + 2e- D Zn(s) EӨ = - 0.76v
• The negative sign indicates that this equilibrium lies
LEFT.
• The zinc terminal is the negative terminal i.e. it loses
electrons which are used to oxidise the H+ ions in the
hydrogen half cell.
• Zinc is reduced in this standard cell.
• Since zinc is a reactive metal which reacts with acid
to form hydrogen gas and a zinc salt this is as
expected.
Calculating Cell Potentials
• Once standard electrode potentials have been found
from simple cells using the standard hydrogen
electrode the potentials from combinations of half
cells can be worked out.
• Using the worked examples on p. 187 in the text book
work out the standard cell potential and thus the cell
reaction for the silver-copper cell.
• Standard electrode potentials are always written as
REDUCTION POTENTIALS.
• i.e. oxidised species + electrons → reduced species.
• It is a measure of the tendency for reduction to
occur.
• Values range from about -3 to +3 volts.
• So:
• Cu2+ + 2e- → Cu E = +0.34v
• Zn2+ + 2e- → Zn E = -0.76v.
• The more positive the potential the more likely it is
that reduction will occur i.e. go right.
• Reactions which go RIGHT more readily than the
reduction of hydrogen ions are always given the
POSITIVE electrode potential.
Another Way of Calculating Cell EMF
• Whenever 2 standard half cells are connected
together an overall potential is set up and
electrons will flow in the wire to the
CATHODE (the positive terminal) where
reduction takes place.
• We can write a short hand for the cell in
question.
• e.g. a zinc/copper cell
• The two half equations are
• Zn2+ + 2e- → Zn E = -0.76v
• Cu2+ + 2e- → Cu E = +0.34v
• Zn(s)│Zn2+(aq)││Cu2+(aq)│Cu(s)
• Apply these rules.
• The half cell with the more POSITIVE potential is placed on the
RIGHT.
• This is where reduction occurs.
• EMF of the cell is given by:
• Ecell = ER –EL
• Where ER is the potential of the right hand cell
• And
EL is the potential of the left hand cell.
• When 2 electrodes combine to form a cell the value for Ecell
must be at least +0.4v if the reaction is to occur spontaneously
• So Ecell = +0.34- -0.76
•
= +1.1v
• This value tells us that the above redox reaction will take place
as the cell is written.
• Zn → Zn2+ + 2e• Cu2+ + 2e- → Cu
• i.e. Zn + Cu2+ → Zn2+ + Cu
• The more positive potential tells us that that reaction
proceeds left to right but that the less positive
reaction goes the other way i.e. right to left.
Using Standard Electrode Equations to
Predict Whether a Reaction Will Occur
•
•
•
•
•
•
•
•
•
•
•
•
Is aqueous iodine able to oxidise bromide ions to aqueous bromine?
1. Write down the reaction you are testing.
I2(aq) + 2Br- → 2I-(aq) + Br2(aq)
2. Split it into 2 half equations (taking account of the DIRECTION of
each reaction) and show the electrode potentials.
I2(aq) + 2e- → 2I-(aq) E = + 0.54v
2Br-(aq) → Br2(aq) + 2e- E = -1.07v*
*This is an oxidation potential. The reaction is going in the opposite
direction from the standard form and so the sign on the potential is
changed.
3. Add the half reactions and the electrode potentials together
I2(aq) + 2Br-(aq) → 2I-(aq) + Br2(aq)
E = 0.54 + (-1.07)
= -0.53v
The negative value for the cell emf indicates that the reaction will NOT
go spontaneously as written and will go back the other way with a
positive emf +0.53v.
Predict whether the following cell should
proceed spontaneously.
•
•
•
•
•
•
Cr3+ + Fe2+ → Cr2+ + Fe3+
Cr3+ + e- → Cr2+ E = -0.41v
Fe2+ → Fe3+ + e- E =-0.77v
Cell potential = -0.41+(-0.77)
= -1.18v.
The reaction will NOT go.
The Effect of Changing Conditions
• Each redox potential is actually an equilibrium
constant with the sign and size of the
• potential indicating the position of the
equilibrium under standard conditions.
• When conditions are not standard the
position of the equilibrium and therefore the
size of the potential can also change.
• This is important in reactions where the
overall cell potential is low and close to +0.3v.
It is possible to make reactions proceed by
changing conditions which would NOT proceed
under standard conditions.
Limitations
• Consider Le Chatelier when deciding the
effect of non standard conditions on the size
of a cell potential.
• E.g. from p.189
• Cu2+(aq) + 2e- D Cu(s)
• Increasing the concentration of the aqueous
ions causes the equilibrium to shift RIGHT to
remove them.
• This in turn causes the electrode potential to
become more positive and this may cause
predicted reactions to be wrong.
Reaction Rate
• Redox potentials are about equilibrium NOT
rate.
• Some reactions with a positive Ecell greater
than 0.3v may go so slowly as to be useless.
• The larger the difference between the EӨ
values the more likely a reaction will take
place.