Transcript Redox - Plusnet

Redox
Difficult but necessary
Obviously:
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Oxidation is adding oxygen
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Reduction is removing oxygen
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2H2 + O2  2H2O
2FeO + C  2Fe + CO2
But also
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oxidation is removal of hydrogen
And reduction is adding hydrogen
And Oilrig
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Oxidation is loss of electrons:
Cu – 2eˉ  Cu2+
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Reduction is gain:
Cu2+ + 2eˉ  Cu
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Notice the charge increases
Notice the charge has reduced
Both often happen in one reaction, so it
is a redox reaction
More…
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A redox reaction:
2Al2O3  4Al + 3O2
aluminium reduced +3 to 0,
oxygen oxidised, -2 to 0
Not a redox reaction:
PbCl2 + 2NaI PbI2 + 2NaCl
Pb stays at +2, Cl stays at -1,
Na stays at +1, I stays at -1
e.g.
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Mg + CuO  MgO + Cu
As ions:
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Note the oxygen ions are spectator ions, they
aren’t actually involved, so we get:
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Mg + Cu2+ + O2ˉ  Mg2+ + Cu + O2ˉ
Mg + Cu2+  Mg2+ + Cu
So the magnesium has reduced the copper
And the copper has oxidised the magnesium
Oxidation numbers / states
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Represent charges where there aren’t
any
They are an “accounting trick” to keep
track of how atoms have control over
electrons
Apply to ions and covalently bonded
atoms
The oxidation numbers of elements are
zero e.g.. Fe(s), and even O2
Working them out
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Rules for assigning: (these rarely change)
F is always -1
O is -2, except in OF2
Group 7 are -1, except with O or F
Group 1 metals are +1
Group 2 metals are +2
H is +1, except in hydrides, e.g. NaH
Al is +3
The total for an ion is its charge (e.g. -1 for CN-)
More electronegative atoms get negative numbers
The total for a compound is 0, even in O2, Cl2 etc.
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Note: the allocation of a high oxidation
number does not necessarily mean that
electrons have been lent and borrowed.
E.g. in CrO42ˉ the oxidation number of
chromium is +6, yet it is covalently bonded to
the oxygens and the energy required to
remove 6 electrons would be prohibitive.
All the +6 tells us is that the electrons
probably spend more time near the oxygens.
Working out
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Overall charge on a compound ion is the
sum of the oxidation states:
E.g. for MnO4ˉ in KMnO4
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Oxidation state of Mn is +7 because overall
charge is -1 and oxygens are -8 (-2 x 4)
So -1 = +7 – 8
Write MnO4ˉ as manganese(VII) oxide or
manganate(VII)
Manganate(VII) compounds are common
oxidising agents
e.g. oxidation states in CaSO4
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Ca is +2
O is -2
X 4 =-8
Uncharged compound so total oxidation
number is 0
So sulphur is +6 (0 = +2 -8 +6)
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Ca O4 S
Call it calcium sulphate(VI)
e.g. the thiosulphate anion S2O32ˉ
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This is a common reducing agent, it
donates electrons to reduce other
chemicals
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Overall = oxygens + sulphurs
-2 = (3 x -2) + (2 x sulphur)
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-2 = (-6) + 4
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2 x sulphur = +4
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sulphur = +2
This is the sulphur(II) oxide (or
thiosulphate) anion
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Practice:
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What is the oxidation state of:
Chromium in CrO42Hydrogen and magnesium in MgH2
Both elements in water H2O
Chlorine in HClO
Sodium and chlorine in Na2ClO3
Carbon in carbonate CO32Iron in Fe3O4
Naming:
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If there is any doubt about the oxidation
state, usually transition metals, it must be
given:
CuCl2 Copper(II) chloride
CuCl3 Copper(III) chloride
NaNO3 Sodium nitrate(V)
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in NO3ˉ we count nitrogen using -6 for 3
oxygens,
to make -1 for the negative charge, so N is +5
From
-1=+5-6
(remember, overall charge is the total of
oxidation states)
Redox or not?
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Cl2 + 2KBr  2KCl + Br2
Cl: 0 to -1, Br: -1 to 0
Cl reduced, Br oxidised
MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O
Mn from +4 to +2, some Cl from -1 to 0
Mn reduced, Cl oxidised
2CrO42ˉ +2H+  Cr2O72ˉ + H2O
Cr is +6 before and after, nothing else
changes either – not redox
Balancing
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Just when you thought you had got it....
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Consider this redox change:
MnO4-(aq)  Mn2+(aq)
Continued
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Continued....
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MnO4-(aq)  Mn2+(aq)
In water
Add oxygen in H2O to balance....
Giving MnO4-(aq)  Mn2+(aq) + 4H2O(l)
Assume an acidic solution to balance H....
Giving MnO4-(aq) + 8H+  Mn2+(aq) + 4H2O(l)
Sort-out electrons for charge and redox....
MnO4-(aq) + 8H+ + 5e-  Mn2+(aq) + 4H2O(l)
+7
+2
In fact we’ve always done this, but it was easy examples...
Try:
VO43-(aq)  V2+(aq)
VO43-(aq) + 8H+(aq) + 3e-  V2+(aq) + 4H2O
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MnO4-(aq)  MnO2(s)
MnO4-(aq) + 4H+(aq) + 3e-  MnO2(s) + 2H2O
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CrO42-(aq)  Cr2+(aq)
CrO42-(aq) + 8H+(aq) + 4e-  Cr2+(aq) + 4H2O
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SO42-(aq)  S8(s)
8SO42-(aq) + 64H+(aq) + 48e-  S8(s) + 32H2O
SO42-(aq) + 8H+(aq) + 6e-  S(s) + 4H2O
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Some specific half-equations of
oxidising agents:
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Oxygen plus metal:
O2 + 4e-  2O2chlorine plus metal:
Cl2 + 2e-  2ClSulphur plus metal:
S + 2e-  S2In hydrogen peroxide, oxygen is in a -1 state.
Is this likely to be a stable compound?
H2O2 + 2H+ +2e-  2H2O
More....
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Concentrated sulphuric acid:
2H2SO4 + Cu  CuSO4 + 2H2O + SO2
½ equation: SO42- + 2e-+4H+  SO2 +2H2O
Conc. nitric acid:
Cu + 4HNO3  Cu(NO3)2 + 2H2O + 2NO2
Some specific half-equations of
reducing agents:
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Oxygen plus metal:
O2 + 4e-  2O2chlorine plus metal:
Cl2 + 2e-  2ClSulphur plus metal:
S + 2e-  S2In hydrogen peroxide, oxygen is in a -1 state.
Is this likely to be a stable compound?
H2O2 + 2H+ +2e-  2H2O