AN EXAMPLE OF A DANIEL CELL ELECTROCHEMICAL CELL Redox Equilibria Electrochemical cells An electrochemical cell has two halfcells The two half cells have to be connected with a salt bridge ID: 529282
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Slide1
REDOX EQUILIBRIASlide2
AN EXAMPLE OF A DANIEL CELL (ELECTROCHEMICAL CELL)Slide3
Redox Equilibria
Electrochemical cells
An electrochemical cell has two half–cells.
The two half cells have to be connected with a salt bridge.
Each half cell consists of a metal (acts an electrode) and a 1.00moldm
-3
solution of its own ions (
eg
Cu electrode in 1.00moldm
-3
Cu
2+
(
aq
)) .
These two half cells will produce a small voltage if connected into a circuit. (i.e. become a Battery or cell).
Standard conditions MUST apply i.e. 298K; 100kPa; 1.00moldm
-3
solutions
Electrons flow in the external circuit and
NOT
through the salt bridge
The voltmeter which measures the potential difference between the two half-cell must be of high resistanceSlide4
Why does a voltage form?
In the cell pictured above
When connected together the zinc half-cell has more of a tendency to oxidise to the Zn
2+
ion (as the Zn electrode is more reactive) and release electrons than the copper half-cell.
Zn Zn2+ + 2e- More electrons will therefore build up on the zinc electrode than the copper electrode.
Zn
2+
Zn
2+ Slide5
The copper half-cell has more of a tendency to reduce (as copper is less reactive) and
the Cu
2+
ions gain electrons from the copper electrode.
Cu
2+
(aq) + 2e- Cu(s)This would leave a positive charge on the copper electrode
This would leave the Zinc strip with a negative charge (negative electrode)And the copper strip with a positive charge (positive electrode)A potential difference is set up between the two half-cells (electrodes)
This potential difference is measured with a high resistant voltmeter, and is given the symbol E.
The E for the electrochemical cell above is E = +1.1V Slide6
IN GENERAL
W
hen drawing a diagram of a Daniel Cell the
MORE
reactive half-cell will be placed on the
LHS
.
OXIDATION always occurs in the LHS half-cellThe LESS reactive half-cell is always placed on the
RHS.REDUCTION always occurs in the
RHS half-cell
The LHS
electrode is known as the negative electrode The RHS
electrode is known as the positive electrodeElectrons always flow from in the external wires (circuit) from the
LHS
half-cell to the
RHS
half-cellSlide7
Metal
atoms lose electrons at the one electrode (oxidation), making it -
ve
, which travel through the wire to the other electrode, adding to ions to produce metal atoms (reduction).
Remember
: oxidation / anode / negative.
Remember also that usually the positive electrode is the right hand electrode. Slide8
LHS
OXIDATION OCCURS
This half-cell is MORE reactive
R
HS
REDUCTION OCCURS
This half-cell is LESS reactiveSlide9
Why use a High resistance voltmeter?
The
voltmeter needs to be of very high resistance to stop the current from flowing in the circuit
.
In this state it is possible to measure the maximum possible potential difference (E).
The
reactions will not be occurring because the very high resistance voltmeter stops the current from flowing.Slide10
Salt Bridge
The
salt bridge is used to connect up the circuit. The free moving ions conduct the
charge.
A
salt bridge is usually made from a piece of filter paper (or material) soaked in a salt solution, usually Potassium Nitrate.
The
salt should be unreactive with the electrodes and electrode solutions.. E.g. potassium chloride would not be suitable for copper systems as Chloride ions can form complexes with copper ions.Another example is that we would not use potassium chloride with silver systems as chloride ions react with silver ions to form a white precipitate of AgCl
.A wire is not used because the metal wire would set up its own electrode system with the solutions. Slide11
What happens if current is allowed to flow?
If
the voltmeter is removed and replaced with a bulb or if the circuit is short circuited, a current flows
.
The reactions will then occur separately at each electrode.
The
voltage will fall to zero as the reactants are used up.The most positive electrode will always undergo reduction. Cu2+ (aq) + 2e
Cu(s) (positive as electrons are used up)The most negative electrode will always undergo oxidation.
Zn(s
)
Zn2+(aq
) + 2e- (negative as electrons are given off)Slide12
Measuring
the standard
electrode potential of a
cell (E
O)
It
is not possible to measure the absolute potential of a half electrode on its own.
It is only possible to measure the potential difference between two electrodes.To measure it, it has to be connected to another half-cell of known potential, and the potential difference between the two half-cells measured.
by convention we can assign a relative potential to each electrode by linking it to a reference electrode (hydrogen electrode), which is given a potential of zero VoltsSlide13
The S
tandard
H
ydrogen
Electrode (SHE)
The standard electrode
potential
(E0) of all electrodes are measured by comparing their potential to that of the standard hydrogen electrode. The standard hydrogen electrode (SHE) is assigned the potential of 0 volts.NOTE:Because the equilibrium does not include a conducting metal surface a platinum wire is used which is coated in finely divided platinum. (The platinum black acts as a catalyst, because it is porous and can absorb the hydrogen gas
.)Slide14
The hydrogen electrode equilibrium is
:
H
2 (g
)
2H+ (aq) + 2e- In a cell diagram the hydrogen electrode is represented by:
Pt|H2(g)|H+
(aq
)Components of a standard hydrogen electrode.
To make the electrode a standard reference electrode some conditions apply
: 1.
Hydrogen gas at pressure of
1atm
2.
Solution containing the hydrogen ion at 1
moldm
-3
(solution
is usually
1moldm
-3
HCl
)
3.
Temperature
at 298KSlide15
Secondary standards
The Standard Hydrogen Electrode is difficult to use, so often a different standard is used which is easier to use.
These
other standards are themselves calibrated against the SHE.
This
is known as using a secondary standard - i.e. a standard electrode that has been calibrated against the primary standard.
The common ones are: silver / silver chloride E = +0.22V calomel electrode E = +0.27V Slide16
Standard Electrode
Potentials (E
O
)
When an electrode system is connected to the hydrogen electrode system, and standard conditions apply the potential difference measured is called the standard electrode
potential
(EO)REMEMBER - The standard conditions are : All
ion solutions at 1moldm-3temperature 298K
gases
at 100kPa pressure No current
flowingSlide17
When finding the potential of a half-cell under test, the standard electrode is always the left hand
electrode.
Measuring E
O
for the Cu
2+
/Cu half-cell against the SHE.This diagram shows how we can measure EO for the Cu2+/Cu half-cellE
O cell = E (Cu2+/Cu) = + 0.34 V
Note:
in the electrode system containing two solutions it is necessary to use a platinum electrode and both solutions must be of a 1M concentration.Slide18Slide19
A diagram showing how the E
O
value of Zn
2+
/Zn half-cell is worked out
E
O
cell = EO(Zn2+/Zn) = – 0.76V
Pt(s)|H2(g)|H+(aq
)||Zn2
+(aq)|Zn(s)Slide20Slide21
Standard electrode potentials are found in data books and are quoted
as:
Li
+
(aq
)|Li(s
)
EO= -3.03V more oxidised form on leftThey may also be quoted as half equations Li+(aq
) + e- Li(s) E
O=
-3.03V but again the more oxidised form is on the leftSlide22
BACK TO ELECTROCHEMICAL CELLS
RULES:
The half-cell placed on the
LHS
- is the more reactive system
- is the system which is always
OXIDISED - is always the negative electrode2. The half-cell placed on the RHS
- is the less reactive system - is the system which is always REDUCED
- is always the positive electrodeSlide23
HOW DO WE KNOW WHICH SYSTEM IS MORE REACTIVE?
ANSWER
From the E
O
data of each half-cell
The more negative the E
O
value the more reactive the system this system will be placed on the LHS and is OXIDISED
Slide24
From the data book:
E
O
Zn
2+
/Zn = -0.76V
E
O Cu2+/Cu = +0.34V The Zn2+/Zn half-cell is more reactive
Zinc Half Cell (Negative electrode)Oxidation occurs
Zn(s)
Zn2+(
aq) + 2e-
Copper Half Cell (Positive electrode) Reduction occurs
Cu
2+
(
aq
) + 2e-
Cu(s)
OVERALL EQUATION
Zn
Zn
2+
+ 2e-
Cu
2+
+ 2e- Cu
Zn(s) + Cu
2+
(
aq
)
Zn
2+
(
aq
) + Cu(s)Slide25
Cell notation (cell diagrams)
Rather than drawing complicated diagrams of electrochemical
cells each time we can write
a shorthand
form instead in the form of a cell diagram
In a cell diagram the symbols
where | represents a phase boundary (i.e. between species in different states)
|| represents a salt bridge
Also we can use the word ROOR
(Red/Ox/Ox/Red)
To know the order of the symbolsSlide26
REMEMBER –
there are 3 types of electrodes
Metal electrodes
-
These are the type met above, which consist of a metal surrounded by a solution of its ions,
e.g
. Zn(s) | Zn
2+(aq). Gas electrodes This is for a gas and a solution of its ions. Here an inert metal (usually platinum) is the actual electrode, e.g. Pt(s) | H2
(g) | H+ (aq)
Redox electrodes
This is for two different ions of the same element (e.g. Fe2+ and Fe
3+), where the two types of ions are present in solution with an inert electrode (usually Pt).
e.g. Pt(s) | Fe2+
(
aq
), Fe
3+
(
aq
)Slide27
So for the Zn/Zn
2+
and Cu/Cu
2+
electrochemical cell:
The cell
digram
is Zn(s)/Zn2+(aq)//Cu2+ (aq)/Cu(s)
R 0 O RSlide28
Calculating the EMF of a cell
In order to calculate the
Ecell
, we must use ‘standard electrode potentials’ for the half cells.
So for this electrochemical cell:
EMF = E
RHS
– ELHS = 0.34 – (-0.76) = 1.10VA positive value indicates that a reaction is feasible.Slide29
EXAMPLE 1 – Given the following E
0
values:
K+
(aq) + e-
K(s) EO = - 2.93V Ag+ (aq) + e- Ag(s) EO
=+ 0.80V
Draw an electrochemical cell to show how these half-cells are connected
Write appropriate half equations for each half cellGive the overall equationWhich way does the current flow in this electrochemical cell
How is the salt bridge madeWhat is the purpose of having a salt bridge
Write a cell diagram for this electrochemical cellGive the value of the EMFIn reality why would we NOT carry out this electrochemical processSlide30
High resistant voltmeter
Silver
1 moldm
-3
Ag
+
ions
Potassium
1moldm
-3
K
+
ions
Salt bridge
298K
100kPa
K(s)
K
+
(
aq
) + 1e-
Ag
+
(
aq
) + 1e-
Ag(s)
2.
3.
Overall equation K(s) + Ag+(
aq
)
K(
aq
) + Ag(s)
4.
From left to right in the external wires
5
.
Normally made by soaking filter paper in aqueous
KCl
BUT in this case we cant use
KCl
as it precipitates with Ag
+
ions
KNO
3
is used instead
Slide31
6. The salt bride allows IONS to be conducted between the two half cells
allowing electrical connection to be made between the two half-cells
K(s)/K
+
(
aq
)//Ag
+(aq)/Ag(s)EMF = ERHS –ELHS = 0.8 – (-2.93) = +3.73V
9. Since K is highly explosive in water and would explode in a 1.00moldm-3
solution of its own ionsSlide32
Example 2 – Given
MnO
4
-
(aq) +
8H
+ (aq) + 5 e- Mn2+(aq) + 4H2O(l) EO = +1.51 Fe
3+(aq) + e-
Fe2+
(aq) EO = +0.77
Draw an electrochemical cell to show how these half-cells are connected
Write appropriate half equations for each half cell
Give the overall equation
Write a cell diagram for this electrochemical cell
Give the value of the EMFSlide33
High resistance voltmeter
Platinum electrode
Platinum electrode
1 moldm
-3
Fe
2+(
aq
)
1 moldm
-3
Fe
3+(
aq
)
1 moldm
-3
MnO
4
-
(
aq
)
1moldm
-3
Mn
2+
(
aq
)
1moldm
-3
H
+
(
aq
)
Salt bridge
Fe
2+
(
aq
)
Fe
3+
(
aq
) + 1e-
MnO
4
-
(
aq
) + 8H
+
(
aq
) + 5e-
Mn
2+
(
aq
) + 4H
2
O(l)
OVERALL : 5Fe
2+
(
aq
) + MnO
4
-
(
aq
) + 8H
+
(
aq
)
5Fe
3+
(
aq
) + Mn
2+
(
aq
) 4H
2
O(l)
Pt(s)/Fe
2+
(
aq
);Fe
3+(
aq
)//MnO
4
-
(
aq
);Mn
2+(
aq
)/Pt(s)
R
O
O
RSlide34
EMF = E
RHS
– E
LHS
= 1.51 – 0.77 = 0.74VSlide35
OXIDATION AND REDUCING AGENTS
The most powerful reducing agents will be found at the most negative end of the series on the
right, since it will itself be OXIDISED in the process
(
ie
the one with the lower oxidation number)
The
most powerful oxidising agents will be found at the most positive end of the series on the left since it will itself be REDUCED in the process (ie the one with the higher oxidation number)Slide36
So,
In general
the more positive the E
O
value the more powerful the oxidising agent
The more negative the E
O
value the more powerful the reducing agentSlide37
Example 1
Use electrode data to explain why fluorine reacts with water. Write an equation for the reaction that occurs.
ANSWER
First apply idea that more positive
Eo
will reduce (go forward) and more negative
Eo
will oxidise (go backwardsExplanation to write As EO
F2/F- >
EO
O2/H2
O, and Ecell is a positive value of +1.64V,
F2
will oxidise H
2
O to O
2
work out
Ecell
and quote it as part of your answer
Ecell
=
Ered
-
Eox
= 2.87-1.23
= +1.64V
Remember
to cancel out electrons in full equation
Equation
2F
2
(g) + 2H
2
O(I
) → 4F
–
(
aq
) + O
2
(g) + 4H
+
(
aq
) Slide38
Example 2
Use data from the table to explain why chlorine should undergo a redox reaction with water. Write an equation for this reaction
First select relevant half equations by considering the
E
O
values and applying the idea that more positive
EO will reduce (go forward) and more negative EO will oxidise (go backwards)
Explanation to write As EO Cl2
/Cl-
> EO
O2/H2O, and
Ecell is a positive value of +0.13V, Cl2 will oxidise H
2
O to O
2
Equation
2Cl
2
(g) + 2H
2
O(I
) → 4Cl
–
(
aq
) + O
2
(g) + 4H
+
(
aq
) Slide39
Example 3
Suggest what reactions occur, if any, when hydrogen gas is bubbled into a solution containing a mixture of iron(II) and iron(III) ions. Explain your answer.
Fe
3+
(
aq
) + e
– → Fe2+ (aq) EO = +0.77V2H+(aq) + 2e– → H2(g) EO = 0.00VFe2+ (aq) + 2e– → Fe(s) EO = –0.44VFirst select relevant half equations by considering the
EO values and applying the idea that more positive EO will reduce (go forward) and more negative EO will oxidise (go backwards)
Explanation to write
Fe3+
will be reduced to Fe2+ by H2 oxidising to H+ because EO
Fe3+ /Fe2+ > EO
H
+
/H
2
and
Ecell
is a positive value of +0.77V
Equation
2Fe
3
+
(
aq
) + H
2
(g)
→ 2Fe
2
+
(
aq
) + 2H
+
(
aq
) Slide40
Example 4
Use the half-equations to explain in terms of oxidation states what happens to hydrogen peroxide when it is reduced.
Explanation to write
As E
O
H
2O2/H2O > EO O
2/H2O2 and
Ecell is a positive value of +1.09V , H
2O2 disproportionates
from -1 oxidation state to 0 in O2 and -2 in H2
OEquation
2H
2
O
2
(
aq
)
→ O
2
+ 2H
2
O(I) Slide41
Will
a reaction take place when zinc is added to silver nitrate solution?
Relevant
half equations are
Zn
2
+
+ 2e- Zn Eθ = -0.76V Ag+ + e-
Ag Eθ
= +0.80V
Example 5
If a reaction takes place the zinc will become zinc ions.
This involves losing electrons so it corresponds to the left hand cell. (This is the reverse of the usually written half equation, so this will be the left hand cell.)
So
the Ag is the
RHS
and Zn is the
LHS
E
θ
CELL =
E
θ
RHC
-
E
θ
LHC
E
θ
CELL = +0.8 - -0.76 = +1.56
The
positive value indicates that the reaction is feasible:
2Ag
+
(
aq
) + Zn(s) → Zn
2+
(
aq
) + 2Ag(s)Slide42
Example 6
Will
a reaction take place when acidified hydrogen peroxide is added to bromide ions? Relevant half equations are:
Half equation 1: Br2 + 2e- 2Br
-
Eθ = +1.07Half equation 2: O2 + 2H+ + 2e- H2O2 Eθ = +0.68V Half equation 3 H2O2 + 2H
+ + 2e- 2H2O Eθ = +1.77V
The oxidation of bromide ions is 2Br-
→ Br2 + 2e- Since this process involves losing electrons, it is the left hand cell There are two half equation for hydrogen peroxide
. Looking at half equation 2 first; EθCELL =
EθRHS - Eθ LHS
E
θ
CELL
= +0.68 - +1.07 = -
0.39
The negative value tells us that this reaction is not feasible.
Inspection
of the half equation would also tell us this of course since in a redox equation one half equation always has to be the reverse of the usual form. Slide43
Looking now at the other hydrogen peroxide half equation
E
θ
CELL = E
θ
RHS
- EθLHS EθCELL = +1.77 - +1.07 = +0.70 The positive value indicates that the reaction is feasible:
H2O2(aq) + 2H
+ (
aq) + 2Br- (
aq) → 2H2O(l) + Br2
(aq)Slide44
Effect of conditions on Cell voltage
E
cell
- Limitations
of Eθ
values
The
electrode potential values have limitations because; • they refer to standard conditions • they indicate the energetic feasibility of a reaction, not the kinetics. If the conditions are different from the standard, the emf can change
. For example, the measurement of Eθ
is made at a concentration of 1moldm-3
. If the concentration of one of the solutions is changed, this will change the
emf of the reaction.
For example 2Ag
+
(
aq
) + Cu(s) → Cu
2+(
aq
) + 2Ag(s)
For
the Cu and Ag cell, the standard value for the reaction is
:
E
θ
CELL
=
E
θ
RHS
-
Eθ
LHS
E
θ
CELL
=
+0.8 - +0.34 = +0.46VSlide45
Values of
emf
for different silver ion concentrations are shown in the table below
At low silver ion concentration, the reaction will tend to go in the opposite direction
2Ag
+
(
aq) + Cu(s) → Cu2+(aq) + 2Ag(s) Slide46
The effects of changing conditions on E cell can be made by applying le
Chatelier’s
principle
If current is allowed to flow, the cell reaction will occur and the
Ecell
will fall to zero as the reaction proceeds and the reactant concentrations drop.
Ecell
is a measure of how far from equilibrium the cell reaction lies. The more positive the Ecell the more likely the reaction is to occur. Effect of concentration on EcellLooking at cell reactions is a straight forward application of le
Chatelier. So increasing concentration of ‘reactants’ would increase Ecell and decreasing them would cause Ecell to decrease.
Zn
2+(aq) + 2e- Zn(s)
EO= - 0.76VFe2+(aq) +
2e- Fe(s) EO= -
0.44V
Overall:
Zn
+
Fe
2+
Fe
+ Zn
2+
E
O
=
+
0.32
Increasing the concentration of Fe
2+
and decreasing the concentration of Zn
2+
would cause
Ecell
to increase
.Slide47
Effect of temperature on
Ecell
Most cells are exothermic in the spontaneous direction so applying Le
Chatelier
to a temperature rise to these would result in a decrease in
Ecell
because the equilibrium reactions would shift backwards.
If the Ecell positive it indicates a reaction might occur. There is still a possibility, however, that the reaction will not occur or will occur so slowly that effectively it does not happen. If the reaction has a high activation energy the reaction will not occur. Slide48
Ecell
is directly proportional to the total entropy change and to
lnK
(where K is equilibrium constant)for a reaction
A positive
Ecell
will lead to a positive total entropy changeSlide49
Breathalysers
The earliest breathalysers used the colour change that occurs when dichromate(VI) ions react with ethanol to measure the amount of alcohol. They could measure the extent to which dichromate turns green
Fuel cells measure the current from an ethanol fuel cell. More alcohol means larger current measured. Fuel cell breathalysers are portable
Infrared breath analysers can determine the amounts of alcohol from an infrared spectrum. More alcohol means greater absorbance. The infrared breathalysers do not use the OH absorption to detect the amount of alcohol on the breath because Water in the breath also has an OH bond.
Many countries will use evidence from both fuel cells and infrared breath analysers for evidence for prosecution, because additional evidence is more reliableSlide50
Fuel
cell
A fuel cell uses the energy from the reaction of a fuel with oxygen to create a voltage
Hydrogen Fuel cell (potassium hydroxide electrolyte)
4e- + 4H
2
O 2H2 +4OH- E=-0.83V 4e- + 2H2O +O2 4OH-
E=+0.4V Overall reaction: 2H2 + O2 2H2O E=1.23V
In acidic conditions these are the electrode potentials.
The Ecell
is the same as alkaline conditions as the overall equation is the same
2e- + 2H+ H2 E=0V
4e- + 4H
+
+
O
2
2H
2
O
E=1.23V
Overall 2H
2
+
O
2
2H
2
O
E=1.23V
Alkaline conditions
Using standard conditions: The rate is too slow to produce an appreciable current.
Higher temperatures are therefore used to increase rate but the reaction is exothermic so by applying le
chatelier
would mean the
emf
falls. A higher pressure can help counteract thisSlide51
Advantages of Fuel cells over conventional petrol or diesel-powered vehicles
less
pollution and less CO2. (Pure hydrogen emits only water whilst hydrogen-rich fuels produce only small amounts of air pollutants and CO2).
(
ii) greater efficiency;Limitations of hydrogen fuel cells
expensive
(
ii) storing and transporting hydrogen, in terms of safety, feasibility of a pressurised liquid and a limited life cycle of a solid ‘adsorber’ or ‘absorber’ limited lifetime (requiring regular replacement and disposal) and high production costs, use of toxic chemicals in their productionHydrogen is readily available by the electrolysis of water, but this is expensive. To be a green fuel the electricity needed would need to be produced from renewable resourcesHydrogen can be stored in fuel cells (i) as a liquid under pressure, (ii) adsorbed on the surface of a solid material, (iii) absorbed within a solid material;Slide52
Ethanol fuel cells
Ethanol fuel cells have also been developed. Compared to hydrogen fuel cells they have certain advantages including. Ethanol can be made from renewable sources in a carbon neutral way Raw materials to produce ethanol by fermentation are abundant Ethanol is less explosive and easier to store than hydrogen. New petrol stations would not be required as ethanol is a liquid fuel
Equation that occurs at oxygen
electrode
4e-
+ 4H+ +
O2
2H2O E=1.23V Equation that occurs at ethanol electrodeC2H5OH + 3H2O → 2CO2 + 12H+ + 12e Overall EquationC2H5OH + 3O2 → 2CO2 + 3H2O