Summary Edexcel IAL Chemistry Unit 5 Notes – Color-Coded Mind Maps, Topic Breakdowns & Exam-Ready Guide

IAL Chemistry Unit 5

by QueenCramsC

, Non-Metal-Metal in half cell Non metal-Nonmetal in half cell

high resistance voltmeter
Standard Hydrogen Electrode high resistance voltmeter
+O 80V
.



(KNO3 <ags)
Salt bridge salt
bridge
Absolute potential difference : a Silver electrode
+ 1 09V .




)
N a Brecgs 100
Kpa
potential difference
↑

the blu Dt (s) 3 Ptiss wire
↑
= mils wire -




a metul and solution of its ions
DAs ire -
3
-


Platinum
7 platinum
a is (s)
Platinum
Platinum is

EH2(y) =
"
# (ag) + e = 0 00 H 2504(ag) [H - =
1 00 moldms AgNOs lug) [Ay] =
1 00 moldm'3 [Br] = 1 00 moldn's
7
. .
.
.




AgingA set
H 2504(ag) [H - 1 00 moldms
Chap
=

2 = Hey)
Standard Electrode Potential
.


+
:
= CHTay)
Brelay C
0 00
.
+ 2 = H2(y) i
& 0 00 .




eme measured when a half cell * As Ag"/Ag half cell has the more the value * As Bra/r half cell bas the more the Evalue it is the the electrode (cathode


is connected to a standard it is the the electrode (cathode) So
Ag
"
cons are more So Bry molecules are more
likely to get reduced and He cys is more
likely t
get oxidised



electrode under standard conditions
hydrogen likely to get reduced and He cys is more
likely t
get oxidised = cell = 1 .

09 -
0 = + 1 .

09 V




Excell = 0 .
80 -0 = + 0 .
801



Fuel cells Role of Salt
bridge
7 Fon -Fon half cell
high resistance voltmeter
Basic Medium -
allow ins to move blw cells/solutions salt
bridge
+ 1 09V




·
.



J




-

maintains neutrality Ha,a
↑ Ptiss wire

-
kNO3/K(1 used baz NO3 o Cl Dt (s) 3
wire
usually soluble
-
are
#
platinum (s)

He0 (e)
y Platinum is -




Electrochemistry
H 2504(ag) [H - =
1 00. moldms [MnOi cag] & [Mn2" (ag] 7 00 mol di = .




CHTay) + 2 = H2(y) MnO4ag) 8HT(ay) 5 = Mn2(y) 4120()
+ + +



&H
electron flow Acidic Medium = 0 00 .

required = 1 52V
+
ions are
= + .




Relationship AS Relationship
J
between Eocell and total between equilibrium constant ,
K a Excell * MnO4r2 Ecell is the the pole * 12/1+ ve pole



M
Stotal =
Blak
TDS total nF Ech Ecl 1 52V



stee
* =
+
2 52 -0
~


= = .
.




(4) (t) n =
number of moles of electrons




what


D
==
Faraday Constant
-

* both are constant ,
therefore ,
at a
given temperature * n ,
F & R are
RT
constants. Therefore, do values tell us ?
metal loses e
Hacy) * Sototal & E cell Ink a cell * - ve Evalve >
-

eg lies more to the left
>
-




Coxidation b in)
Greducing agent


proton exchange membrane Disproportionation Reactions Reaction between
Manganese (IV) Oxide and tC * thevalue >
-

eg lies more t the right ->Metal gains oxidising agea




CellDiagr
-



Preparation of Chlorine :




Hab (e) is
electrolyte cutcags
220m
acidic
zc
"
-

will + Cu(s) Equation : MnO2(s + 4HD (ap
>
- Mm + 2( + 21620 + D2
(g)
%


G2
+
+ i = Cut + 0 .
1SV
-value :
equilibriat MnOz (ss +
4Hlap +
2e & Mrtcap +

Zin + 25 = zn(s) = 0 76
.




2- =
2C- caps
(4) Cu
+
+ =
m 52)
equilibriu 2
Clacg) +

Curing + ze - ( h ( s )= + 0 34V
14 0
.


+ .




& salt bridge
E cell value for the Full Enuation is 13V y
2 - O2(y 21
0
Cutcaps Intag)
Heggs "says
.



=

2H20(1) Zinc
-




>
- 21 + + + 2 - -
> a + G+ e 0 .
1
copper
In(s) + > + Cu(s)
feasible
thermodynamically
-




so reaction is NOT under standard conditions ·




ai Cu + 0 52
To
+
.




Overall + 02(y) 1620
make the reaction feasible increase the concentration of
hydrochloric acid

equation 12 >
(2) Cell
Diagram
,


(y)
-
: *
za-
:

C + Cus + 0 .
37V
moldi. [CI]
about 20 So the CHT] and ins
both increase


*
so position of equilibrium - shifts to the
right >
- Ex value increases
zis(cug) : Incap 1 cuis
*
Cr/cut
&

As E cut/cu > E and Ecel of Shifts to the E2 value decreases 2981

moldmsT
and
position equilibrium 2 left >
- =




Advantages of Fuel cells have positive of [Cu ]
"
value + o 37
Ea cell cap) moldn
+
a
value becomes
sitive (2n 1 Rules
.



so =
> cap
= :


.
so it is
thermodynamically feasible Cut
-




(i) less pollution and less CO2. (Pure hydrogen emits only water while eam shift to
RAS
hydrogen-rich fuels produce only small amounts of air pollutants and I f cell the then from LS-RAS is feasible
-Thermodynamic
feasibility 1 reduced species outside oxidised species inside (ROOR)
Mo
is rxn




MnO
, ,
!
.




CO2


Zens
run occur
>
S
(ii) lighter with greater efficiency than engines using fossil fuel Eal
o


1 101
zag,Cing(cue
=
+ .




Zncs)
electrode is the RHS of cell
2 .
the on
diagram
= =
+ 0 .

34V

If Excell then from RHS > IS is feasible
Disadvantages of Fuel cells
-

is -
ve , rxn


S
ran occur so chloride ins can now release electronsb MnOz <
forming Clig) so reaction is now In(s) + Cu*"cap -
> Cuisi +
Inig Reaction in feasible Exceptions :




* Explosive "Cap(Zness E
+

PE /H2(g) /H cagsi
*
In =
-0 .
76V


*Storing and transporting hydrogen at high pressure to be in liquid state is hazardous "lay
*




* limited lifetime (regular replacement and disposal) and high production costs/ Expensive
thermodynamically feasible under non-standard conditions. Cu + 2 = Cu(
+ when
measuring standard electrode potential ,
standard
hydrogen

*Use of toxic chemicals in their production / nanotubes of carbons are needed to adsorb SUMMARY
6. 2 (ap + 20- H2(g)E = 0 .
000



Cu(s) Cn2 2 [0 left
always
>
- + 0 34 the
(aq) electrode is
.




hydrogen • The thermodynamic feasibility of a chemical reaction can be predicted using standard electrode potentials.
-
=
on




• Although the standard electrode potentials indicate that a reaction is thermodynamically feasible, it may not take place for two reasons: =
* Hydrogen is readily available by the electrolysis of water, but this is expensive. To be a el = 0 .
34 -
0 00. = 0 .
34V


green fuel the electricity necded would need to be produced from renewable resources 1. the reactants may be kinetically stable because the activation energy for the reaction is very large /Hays /Hap E cell
+
2H +
(ag) + Cu(s) - > Cu2 +
He(y) Reaction is not feasible Pt is : Znay /Ins =
-0 .
760

2. the reaction may not be taking place under standard conditions. (4) (7

• A reaction that is not thermodynamically feasible under standard conditions may become feasible when the conditions are altered.
Excell
• Changing the conditions may alter the electrode potential, E, of a half-cell because the position of equilibrium of the half-cell reaction may change. Pt( la Hap Cu (cue : + 0 34
.

=

, Redox Titrations
♡Note ♡

this topic mostly comes as systematic and repetitive calculations, you can find worked examples at pearson’s ial
chemistry student book 2. This document will only cover important questions and their answers about the topic ♡!


Potassium manganate(VII) titrations
Most common Redox Reactions
A known concentration of potassium manganate(VIl) can be used to determine the quantity of a reducing agent
Redox titrations with potassium manganate (VI) KMnO4 present in a sample.
• In titrations involving potassium manganate(VII) the half reaction is: With hydrogen peroxide
2"
Mn84 +
-




Mr
Mnags
8H + 5 - + 4 He
+ +


Mn04-cap) + 8 + 5 - + 4 e0 cel Ox :
H202 >
-


02 + 24
+
+ 22

This may react with ethanedioic ions following the half reaction: MnO4 cap Mn2+
+ -




Red + S Cay) + Se > +
4H2
ag)
:




Pink Colourless Overall 6H +
502 2Mn2(ay) 8Hz)
2 Mn04-cap) 5
-




C20-2002
: + + > + +

+ 25




To combine these half equations we must x Equation 1 by 2 and x Equation 2 by 5 and add. With ethanedioate
The full reaction is:
(Mn04 Mnz" 20)x2
?] Why is potassium manganate always used in acidic solutions? 4+
+
+ 5 + 8H - +

2 OX : C204 >
-
2CO2 + 2e
Dxs
-




(20p 2002
-

2
>
-


MnO 4 lay
+
Mn2(ag)
-




+ Red S Cay) + Se > +
4H2

@] because if used in alkaline solution, a brown precipitate of
: +



Overall : 2MnO4 cap) +
16 tag) +
56204cay >
10002(g) +
2Mncy +
SH2U
+ 16H
2MnO + > 2Mn
*
+
@H20
MnO2 will form, interfering with the reaction’s end point colour.
-




*The reaction between MnOI
and C O is slow → the reaction is between two negative ions To
500 > +
-




e- -
2

do as a titration the conical flask can be heated to 60° C to speed up the initial reaction.
* Once Mn is produced the reaction becomes faster => it acts as an auto -catalyst
2t



5C2042-
*
2 MnO4
*
+ + 16H >
-

2Mn +
56044 + 820 + 10CO2
?] Explain Why no indicator is added to reactions
The reaction requires excess dilute sulphuric acid and a temperature of about 60°C
involving KMnO4. Manganate(VII) is a deep purple colour in solution, but the manganese(Il) ion, to which it is
Thiosulphate titrations
reduced, is almost colouriess.
@] KMnO4 is a self indicator. As MnO4- ions are As the manganate(VII) is added, from a burette, it reacts turning colourless.
reduced to Mn2+ ions, the colour changes from pink to When all the reducing agent has reacted, the manganate(VII) no longer reacts and its colour
Thiosulphate and lodine titrations are used to determine the
remains in the flask.
concentration of oxidising agents.
colourless.

-
• KMnO4 is a powerful oxidizing agent First of all the oxidising agent is added to a solution containing excess
• In alkaline solution MnO4- >
MnOe brown precipitate iodide ions. This oxidises the iodide ions to lodine giving a brown colour.
• Only use dilute sulfuric acid for manganate titration, Insufficient volumes of sulfuric acid
?] Explain why acidified potassium dichromate (VI) will produce MnOz instead of Mn2 2 - In + 2
MnOp-cag)
+
MnOz(s) CH20
-




+ 4H cays + 3 > +

cannot be used as an indicator in redox titrations even The brown MnO2 will mask the colour change, therefore inaccurate volume of manganate colourless pale pink
• Using a weak acid like ethanoic acid cannot supply the needed ( 81 )
+ Thiosulphate is then added from a burette; this reacts with the iodine to
though it produces a colour change. • It cannot be conc HCl as the Cl ions would be oxidised to Cl by MnO4- as the
-




form colourless product.
232835 12 5406" 25-
@] The colour change from green to orange has no clear & MnOc/Mnc
*
>
E Ck/p- +
-



> +


tetratheonate
MnOkcag) Mag
4HeOce
+
+ 81 cap
+ be > +



end point. Clacap + 2e- < 2C-
say) During the titration, the colour intensity decreases, eventually reaching a pale yellow colour Al this
This would lead to a greater volume of manganate being used and poisonous Cl 2 being
point, a few drops of starch solution are added to give the deep blue complex showing the last traces of
produced
lodine. Thiosulphate is then added dropwise, until the mixture becomes colourtess
?] Why must ethandioic acid be heated to about 60°C • It cannot be nitric acid as it is an oxidising agent. It oxidises Fe to Fe as
2+ 3t


2920scap > S406"cap 20
before the titration?
+
-

=
E NO3 / HNO2 > E fe3 /fezt
+




2[
-




↓2
-




H20e On
2e >
NO3 cap 3H (ay (g)
2 +
+ + >
HNOz cays +
cop
@] The reaction is very slow due to many reactant particles Fe3"cap) + e- > Felt lag)
252032 +
Encaps
>
-




S400 cap + 2
Flag
This would lead to a smaller volume of manganate being used
which have a low likelihood of successful collisions.
• The indicator used is starch that produces deep blue-black coloration. This coloration
disappears when sufficient sodium thiosulfate has been added
?] During the titration of KMnO4 with Iron (II) ions, why is
• If the starch is added too early: it adsorbs some iodine and reduces the accuracy of
the sulfuric acid boiled before the titration? titration.

@] To remove the oxygen dissolved in sulfuric acid, which • The colour of the iodine solution becomes very faint towards the end of the reaction so
difficult to accurately determine the end point.

would oxidise iron (II) ions to iron (III) ions.