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More Enthalpy, more periodic table and the joy of redox potentials. Hydration. Bond enthalpies. Entropy. Free energy. Period 3. Redox potentials. Note: this PowerPoint may need the Royal Society of Chemistry font installed.

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more enthalpy more periodic table and the joy of redox potentials

More Enthalpy, more periodic table and the joy of redox potentials

Hydration

Bond

enthalpies

Entropy

Free

energy

Period 3

Redox

potentials

Note: this PowerPoint may need the Royal Society of Chemistry font installed.

It can be found at the RSC website: http://www.rsc.org/education/teachers/learnnet/RSCfont.htm

remember enthalpy
Remember Enthalpy?
  • Enthalpy is the amount of energy taken in or given out during any change (physical or chemical) under standard conditions
  • Symbol “ʅ” (“plimsoll”)
  • Assume standard conditions
    • 100kPa
    • 298K
    • concentration 1 mole dm-3
    • Standard amount – mole
h f formation
∆Hf Formation
  • The enthalpy change involved in the production of one mole of a compound from its elements under standard conditions, reactants and products being in their standard states.
  • E.g. ½ H2 + ½ Cl2 HCl
  • Note, this is NOT the normal balanced equation.
h i ionisation
∆Hi Ionisation
  • The molar enthalpy change involved in the removal of an electron from a species in the gas phase to form a positive ion and an electron, both also in the gas phase.
  • Keywords cation and anion
  • E.g. Na(g) Na+(g) + e-(g)
  • Can omit (g) for electron
h ea electron affinity
∆Hea Electron affinity
  • The standard molar enthalpy change when an electron is added to an isolated atom in the gas phase.
  • E.g. ½ Cl2(g) + e- Cl-(g)
  • ∆Hea = -364 kJ mol-1
  • Strongly electronegative elements attract electrons, they “like” being negative, and so ∆Hea is negative.
h diss bond dissociation enthalpy
∆Hdiss Bond dissociation enthalpy
  • The standard molar enthalpy change which accompanies the breaking of a covalent bond in a gaseous molecule to form two free radicals, also in the gas phase.
  • E.g. Cl2(g)  2Cl•(g) ∆H = +242 kJ mol-1
  • “homolytic fission”
  • The dot for the radicals is often omitted in thermodynamic equations, but you have to remember it should be there
h at enthalpy of atomisation
∆Hat Enthalpy of atomisation
  • The standard enthalpy change which accompanies the formation of one mole of gaseous atoms.
  • For an atomic solid, this is the same as the standard enthalpy of sublimation
  • E.g. Na(s) Na(g)
  • This is always endothermic
  • If the element is already a gas, this should be zero and ignored. (See over)
slide8
Note that for a diatomic molecule, you get 2 moles of atoms, so ∆Hat = ½∆Hdiss
  • E.g. in ½Cl2(g) Cl(g)
  • ∆Hat = +121 kJ mol-1
h l or h latt lattice enthalpy
∆HL (or ∆Hlatt) Lattice enthalpy
  • The enthalpy of lattice dissociation is the standard enthalpy change which accompanies the separation of one mole of a solid ionic lattice into its gaseous ions.
  • NaCl(s) Na+(g) + Cl-(g)∆HL = +771 kJ mol-1
  • If you write this as above, then it is dissociation and enthalpy is always positive.
slide10
If, however, you write it the other way round, then this value is the enthalpy of formation of the lattice:
  • Na+(g) + Cl-(g) NaCl(s)∆HL = -771 kJ mol-1
born haber diagrams
Born-Haber Diagrams…..
  • Show all of the enthalpy values involved in the creation of a compound, used to find the lattice enthalpy, or anything else you don’t know
  • It’s basically Hess’s law (remember it?)
    • (Not copy, unless you’re desperate for reassurance!)
  • It’s just a bunch of numbers again, but there’s lots of them. It’s a six-variable problem, one unknown and five known.
  • (occasionally eight-variable, but you would know seven)
what s in it
What’s in it?

Positive (metal) and negative (non-metal)gaseous ions

  • Jobs to do:
  • Atomise the metalΔHat or ΔHsub
  • Ionise the metalΔHie1
  • atomise the gasΔHat
  • Add electron to non-metalΔHea
  • Join ions to make solid compoundΔHL

Elements join to make solid compound ΔHf

Solid ionic compound

a full cycle
A full cycle

“Highest enthalpy level”

Na+(g) + Cl(g) + e-

Electron affinity of non-metal

∆Hea

Ionisation of metal

∆Hi

Na(g) + Cl(g)

Na+(g) + Cl-(g)

Atomisation of non-metal

½∆Hdiss(same as ∆Hat)

Na(g) + ½Cl2(g)

Lattice formation of crystal

Atomisation or sublimation of the metal

∆Hsub

∆HL

Na(s) + ½Cl2(g)

∆Hf

Formation

NaCl(s)

“Lowest enthalpy level”

slide14
Note:
  • Lattice enthalpy determined experimentally is not the same as the theoretical value.
  • The Lattice enthalpy calculation assumes that the bonding is 100% ionic.
  • It isn’t.
  • Don’t forget to ionise G2 metals (Mg etc.) twice, Al three times and oxygen has two electron affinities.
  • Think about the ion sizes and charges – they affect lattice enthalpy
look at the diagram notice
Look at the diagram – notice:
  • Endothermic (as drawn) – ionisation, atomisation, sublimation
  • Exothermic (as drawn) – Formation, electron affinity, lattice formation
  • And all of the relationships follow Hess’s law, any two paths will have the same enthalpy.
  • As one equation, LHS = RHS:
  • ΔHsub + ΔHi + ΔHat – ΔHf = – ΔHL – ΔHea
  • But remember you may have more than one ΔHat, ΔHi, ΔHdiss or ΔHea
trends
Trends
  • The LE is larger for:
    • Smaller anions like F-
    • Closer ions
    • More attraction (e.g. 2+ cation)
so try this na s cl 2 g nacl s
So, try this: Na(s) + ½Cl2(g) NaCl(s)
  • Given:
    • 1st i.e.of Na = +496
    • Dissociation enthalpy of Cl2 = +242
    • Atomisation enthalpy of Na = +107
    • Lattice enthalpy of NaCl = -786
    • Electron affinity of Cl = -349
  • Find the value for enthalpy of formation for NaCl
h hyd hydration enthalpy
∆Hhyd Hydration enthalpy
  • The enthalpy of hydration is the standard molar enthalpy change for the process:
  • X±(g) X±(aq)∆H = ∆Hhyd
  • In which 1 mole of gaseous ions is completely hydrated in water to infinite dilution, under standard conditions
  • X± represents any cation (X+) or anion (X-)
  • Usually exothermic
h sol solution enthalpy
∆Hsol Solution enthalpy
  • The enthalpy of solution is the standard molar enthalpy change for the process in which 1 mole of an ionic solid dissolves in enough water to ensure that the ions are separated and do not interact
  • NaCl(s) Na+(aq) + Cl-(aq) ∆Hsol=+2 kJ mol-1
  • So, hydration enthalpy = lattice formation enthalpy + solution enthalpy.
  • Usually endothermic
typical problems involve

Na+(g) and Cl-(g)

∆Hhyd for Na+

Na+(aq) + Cl-(g)

∆HL for NaCl

∆Hhyd for Cl-

Na+(aq) + Cl-(aq)

∆Hsol for NaCl

NaCl(s)

Typical problems involve....
  • You will be expected to find one of these, given the others, as usual. But you’ll have to remember how.
  • Remember:

hydration enthalpy = solution enthalpy + lattice formation enthalpy.

slide21

Na+(g) and Cl-(g)

∆Hhyd Na+

Na+(aq) + Cl-(g)

∆HL for NaCl

∆Hhyd Cl-

Na+(aq) + Cl-(aq)

∆Hsol NaCl

NaCl(s)

E.g.
  • Given:
  • ∆HL for NaCl = +771 kJmol-1
  • And ∆Hhyd for Na+ = -405 kJmol-1
  • And ∆Hhyd for Cl- = -364 kJmol-1
  • Calculate the enthalpy change of solution for NaCl
  • Draw a cycle:
  • Calculate ∆Hsol= +771 +(-405) + (-364)
  • = +2 kJmol-1
you do
You do:
  • Given:
  • ∆HL for KCl = +701 kJmol-1
  • And ∆Hhyd for K+ = -322 kJmol-1
  • And ∆Hhyd for Cl- = -364 kJmol-1
  • Calculate the enthalpy change of solution for KCl
bonds
Bonds
  • Energy is needed to break a bond between two atoms (endothermic)
  • Making new bonds releases energy (exothermic)
  • Total reaction energy = energy of bonds broken – energy of bonds made
slide25
Or
  • Etotal = Ereactants – Eproducts
  • Unit : kilojoules per mole
  • kJ per mole
  • kJ mol-1
slide26

H-H = 436,

Br-Br = 193,

H-Br = 368

e.g.
  • H2 + Br2 2HBr
  • broken: made:
  • H-H H-Br x 2
  • Br-Br

Total = (436 + 193) – (368 + 368)

= -107  exothermic

steps
Steps:
  • Obtain balanced equation
  • Draw molecules for reactant and product
  • Count each bond
  • Total each bond
  • Final calculation
slide28

H-H = 436

O=O = 500

H-O = 464

e.g.
  • 2H2 + O2 2H2O
  • broken: made:
  • H-H (twice) H-O x 4
  • O=O

Total = (436 *2 + 500) – (464 x 4)

= 872 + 500 – 1856

= - 484  exothermic

( For one water it is -242 kJ mol-1)

try this

H-H = 436

I-I = 151

H-I = 297

Try this:
  • H2 + I2 2HI
  • Think – what bonds are in the reactant molecules, what bonds are in the product molecules
  • Broken: Made:
  • H-H, I-I H-I x 2
  • Total = (436 + 151) – (297 *2)
  • = -7 kJ mol-1 for the reaction as shown
  • (-3.5 kJ mol-1 for each mole of HI)
try this1

C-H = 414

O=O = 500

H-O = 464

C=O = 805

Try this:
  • CH4 + 2O2 CO2 +2H2O
  • Think – what bonds are broken,what bonds are made.
  • Broken: Made:
  • C-H x4, O=O x2 C=O x2, H-O x 4
  • Total = (1656 +1000) – (1610 + 1856)
  • = 2656 – 3466
  • = - 810 kJ mol-1
try this2

C-H = 414

O=O = 500

H-O = 464

C=O = 805

C=C = 613

Try this:
  • C2H4 + 3O2 2CO2 +2H2O
  • Think – what bonds are broken,what bonds are made.
  • Broken: Made:
  • C-H x4, O=O x3, C=C C=O x4, H-O x 4
  • Total = (1656 +1500 + 613) – (3220 + 1856)
  • = 3769 – 5076
  • = - 1307 kJ mol-1
exothermic endothermic
Exothermic Endothermic

atoms

atoms

Products

Reactants

ΔHreact

ΔHreact

Reactants

Products

a new quantity entropy
A new quantity - Entropy
  • Degree of “disorder” in a system.
  • Zero entropy at 0K
  • Absolute scale – not just increments like ∆H
  • Depends on temperature and solution.
  • Gases most, then liquids, then solids
  • Simple substances – low entropy values
  • Not zero for elements – depends on temperature
  • Different entropy values for same chemical
    • e.g. CO2 as solid or gas,
  • Water as solid, liquid or gas
discuss these entropy values
Discuss these entropy values:

Consider: which is the most/least organised; how big each molecule is; what type of intermolecular forces there are; how strong they are, etc.

continued
Continued....
  • Symbol S – units J K-1 mol-1
  • Yes, Joules, not kJ
  • So units will be a problem (and catch you out!)
  • Sɵ - usualconditions – can be looked-up
  • E.g. CO2(g), S=214
  • For NaCl(s), S=72
  • For C(diamond), S=2.4
  • ∆S = Sɵ (products) - Sɵ (reactants)
  • Note direction of subtraction
a sample question from yahoo answers
A sample question from Yahoo!answers
  • State whether ΔS is positive, negative, or zero for each of the following processes:(a) 46 g of liquid water goes from 60°C to 50°C at 1 atm pressure.(b) 7 g of N2(g) goes from 50°C and 2.0 atm pressure to 50°C and 1.0 atm pressure.Please give an explanation and the general relationship. Thank you.
a simple question
A simple question
  • Each answer may be used once, more than once, or not at all.
    • a) CO(g) + H2O(g) = CO2(g) + H2(g)
    • b) KCl(s) = K+(aq) + Cl-(aq)
    • c) Na2CO3(s) = Na2O(s) + CO2(g)
    • d) N2(g) + 3H2(g) = 2NH3(g)
    • e) H2O(s) = H2O(l)
  • In which reaction would the entropy change be closest to zero?
  • In which reaction(s) would the entropy change have a positive value?
  • Which would be most positive?
  • In which reaction(s) would the entropy change have a negative value?
calculate
Calculate:
  • The entropy change in:
  • CaCO3 CaO + CO2
  • Given: Sʅ (CaCO3) = 92.9 J K-1 mol-1
  • Sʅ (CaO) = 39.7 JK-1 mol-1
  • Sʅ (CO2) = 213.6 JK-1 mol-1
  • ∆Sʅsystem = 39.7+213.6-92.9
  • = 160.4 JK-1 mol-1
calculate1
Calculate:
  • The entropy change in:
  • 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l)
  • Given: Sʅ (NaHCO3(s)) = 101.7 J K-1 mol-1
  • Sʅ (Na2CO3(s)) = 135.0 JK-1 mol-1
  • Sʅ (CO2(g)) = 213.6 JK-1 mol-1
  • Sʅ (H2O(l)) = 69.9 JK-1 mol-1 .
  • ∆Sʅsystem = 135.0+213.6+69.9 – 203.4
  • = 215.1 JK-1 mol-1 .
calculate2
Calculate:
  • The entropy change in:
  • N2(g) + 3H2(g) 2NH3(g)
  • Given: Sʅ (N2(g)) = 191.6 J K-1 mol-1
  • Sʅ (H2(g)) = 130.6 J K-1 mol-1
  • Sʅ (NH3(g)) = 192.3 J K-1 mol-1
  • ∆Sʅsystem = 384.6 – 583.4
  • = -198.8 JK-1 mol-1
  • This system becomes less disorganised.
why do reactions happen
Why do reactions happen?
  • E.g.
  • HCl + NaOH Exothermic neutralisation
  • 4Fe + 3O2 Exothermic rusting
  • 2Mg + O2 Exothermic combustion
  • CaCO3 CaO + CO2 Endothermic
  • but
  • NaCl + H2O  Na+(aq) + Cl-(aq) Endothermic
  • 2HCl + NaHCO3 Endothermic

So why do these two happen spontaneously?

so how can we know which reactions happen
So how can we know which reactions happen?
  • A “balance” – more entropy could “beat” a positive enthalpy value
  • New quantity – Gibbs free energy change:
  • ∆G = ∆H - T∆S
  • A negative Gibbs value means the reaction should be spontaneous (feasible)
  • It has nothing to do with kinetics, a feasible reaction still might not noticeably happen, high Ea being a likely obstacle.
examples
Examples:
  • C(s) + O2(g) CO2(g)
  • ΔH = -394 kJ mol-1
  • ΔS = +3.3 J mol-1
  • What is the free energy change?
  • ΔG = ΔH - TΔS
  • = -394 x 103 – 298 x 3.3 J mol-1
  • = -395 kJ mol-1
  • Negligible TΔS term in spite of 298

Could do in kJ instead

examples1
Examples:
  • 2Fe(s) + 1½O2(g) Fe2O3(s)
  • ΔH = -825 kJ mol-1
  • ΔS = -272 J mol-1
  • What is the free energy change?
  • ΔG = ΔH - TΔS
  • = -825 x 103 – 298 x -272 J mol-1
  • = -744 kJ mol-1
  • Very feasible. Negative entropy term as gas goes to solid. Large enthalpy term
examples2
Examples:
  • 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)
  • ΔH = +130 kJ mol-1
  • ΔS = +335 J mol-1
  • What is the free energy change for 1 mole of NaHCO3?
  • ΔG = ΔH - TΔS
  • = 130 x 103 – 298 x 335 J mol-1
  • = +30 kJ mol-1 for 2 moles, +15 kJ mol-1

Could do in kJ instead

slide46
Using the same information, prove that this reaction is feasible above around 400K
  • Hint:
  • using ΔG = ΔH – TΔS you need a point where you change from not feasible to feasible, so ΔG = 0
melting fusion
Melting (fusion)
  • At equilibrium, e.g. ice at 0°C, there should be no change in G, or ΔG =0
  • Note le Chatelier’s principle here – more heat energy would make more water from ice, so temperature does not change.
  • This gives ΔG = ΔH - TΔS = 0
  • So ΔH = TΔS or ΔS(fus) = ΔH(fus) / T(fus)
  • Same for boiling
boiling anything
Boiling anything
  • Same as for melting, ΔG = ΔH - TΔS = 0
  • So ΔH = TΔS or ΔS(vap) = ΔH(vap) / T(vap)
  • Given that the enthalpy change of fusion of ice is 6.0 kJ mol-1, calculate the entropy change involved and explain why it is small.
  • 6 = 273 ΔS
  • Therefore ΔS = 6/273 = 0.22
try this3
Try this:
  • Hydrogen gas is a non-polluting fuel. Hydrogen gas may be prepared by electrolysis of water. 2H2O(l) 2H2(g) + O2(g)
  • Predict the signs of ΔH, ΔG and ΔS for the production of hydrogen gas by electrolysis of water.
periodicity
Periodicity
  • Chemical reactions of period 3 elements
  • (Because they ring all of the bells)
  • Na, Mg, Al, Si, P, S, Cl, Ar
  • Some obvious facts:
  • Ar – no reactions
  • Na, Mg, Al metals (Al is a metalloid really)
with water
With water
  • Only Na with cold water
  • 2Na(s) + 2H2O(l) H2(g)+ 2Na+(aq) + 2OH-(aq)
  • Al, Si, P, S do not react with water
  • Mg with steam only:
  • Mg(s) + H2O(g) H2(g) + MgO(s)
  • Cl as discussed in unit 2
  • Ok, if I must.....
  • Cl2(g) + H2O(l)Ý HCl(aq)+ HClO(aq)
  • (sunlight) 2Cl2 + 2H2O  4H+ + 4Cl- + O2
with oxygen
With oxygen
  • Na, Mg, Al, Si:
  • 2Na(s) + ½O2(g) Na2O(s) (White oxide, yellow flame)
  • Mg(s) + ½O2(g) MgO(s) (white oxide)
  • 2Al(s) + 1½O2(g) Al2O3(s) (white oxide)
  • Si(s) + O2(g) SiO2(s) (white oxide)
  • P4(s) + 5O2(g) P4O10(s) (white oxide)
  • All very exothermic
  • S(s) + O2(g) SO2(g) (pale blue)(+SO3)
nature of the oxides
Nature of the oxides`
  • Physical properties:
slide54
The oxides can be acidic, basic or amphoteric
  • Na2O, MgO basic, make hydroxides (14,9)
  • Al2O3 SiO2, are insoluble – no pH value (7) but we use limestone to remove SiO2 as an acid impurity in steelmaking Al2O3 is amphoteric – can neutralise acids and alkalis
  • P4O10, SO2 (&SO3) acidic (0,3,(0))
  • Because: SO2 moderately solubleweak, SO3 very solublestrong
  • AlkaliAcid across period.
need to know equations
Need to know equations:
  • AlCl3(s) Al3++ 3Cl-(aq) followed by
  • [Al(H2O)6]3+(aq) [Al(H2O)5(OH)]2+(aq) + H+(aq)

(this is a partial hydrolysis of the aluminium ion)

  • Al2O3 is also amphoteric:
  • Al2O3(s) + 6H+(aq) 2Al3+(aq) + 3H2O(l)
  • Or
  • Al2O3 + 3H2O + 2OH-(aq) 2[Al(OH)4- (aq)]
simpler
Simpler:

SiCl4(s) + 4H2O(l) Si(OH)4(s) + 4H+(aq) + 4Cl-(aq)

PCl5(s) + 4H2O(l) H3PO4(aq) + 5H+(aq) + 5Cl-(aq)

Total hydrolysis of the molecule in both cases.

oh no not redox again
Oh, no, not redox again....
  • OILRIG – oxidation is loss, reduction is gain
  • Elements – oxidation state is always 0
  • Compounds always 0, BUT
  • Compound ions and any other ions add up to their formal charge.
  • Remember rules:
    • F -1 (always)
    • O -2 (except with F)
    • Halogens -1 (except with F or O)
    • Group1 +1
    • H +1 (but -1 in metal hydrides)
    • Group2 +2
  • Remember this is a hierarchy, top rules rule and if you need to know ones that aren’t here, it’s based on electronegativity.
slide59
e.g.
  • In SO42-, O is -2, so that’s -8, so the sulphur must be +6, to make the total -2
  • In Cr2O72-, O is -2 so that’s -14, so the two chromiums add up to +12, that’s +6 each
and some more
And some more...
  • D-block elements in compounds and ions
  • Calculate using same rules, don’t be surprised by drastic oxidation numbers, they do that
  • E.g. in dichromate ion Cr2O72-
  • -14 for oxygen plus the metal ions = -2
  • Therefore for Cr = ½ of 14-2 = 6 each.
slide61
e.g.
  • What is the oxidation state of Cr in [CrCl2(H2O)4]+ ?
  • Easy – we know:
  • State of Cr + twice Cl- + 4x (H2O) = +1
  • Water is neutral (zero)
  • “Cl-” is -1
  • So “Cr” -2 = +1
  • Therefore “Cr” = +3 .
balancing again
Balancing (again)
  • Look for the changes in oxidation state – there should be only one element changing in a half-equation, two or more in a full equation. O and H probably don’t change.
  • Balance any oxygens by adding water
  • Assuming acid conditions, balance any hydrogens by adding H+
  • Balance oxidation numbers with electrons
  • E.g. MnO4- MnO2 becomes

MnO4- + 4H+ + 3e-  MnO2 + 2H2O

let s practice balance these
Let’s practice; balance these:
  • VO42- V2+
  • Cr2O72-  2Cr3+
  • SO2  SO42- +2e-
  • VO42- + 8H+ + 4e- V2+ + 4H2O
  • Cr2O72- + 14H+ + 6e-  2Cr3+ + 7H2O
  • SO2 + 2H2O  SO42- +2e- + 4H+
adding half equations to get full equations
Adding half-equations to get full equations
  • The trick here is to get the electrons right
  • Perhaps get them all on one side so they cancel
  • E.g. displacement of copper from one of its compounds by magnesium – add the ½ equations
  • : Mg(s) – 2e- Mg2+(aq)
  • : Cu2+(aq) + 2e- + Cu(s)
  •  + :Mg(s) – 2e- + Cu2+(aq) + 2e- Mg2+(aq) + Cu(s)
  • Or Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)
adding half equations part 2
Adding half-equations part 2
  • The alternative is to get equal numbers of electrons on each side
  • E.g. displacement of copper from one of its compounds by magnesium
  • ½ equation : Mg(s) – 2e- Mg2+(aq)
  • ½ equation : Cu(s) – 2e-  Cu2+(aq)
  •  + (in this case) the reverse of :Mg(s) – 2e- + Cu2+(aq) Mg2+(aq) + Cu(s) – 2e-
  • Or Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)
  • Looks harder but that depends on what they give you in the question...
sometimes you have to multiply
Sometimes you have to multiply:
  • E.g. given:
  • Cl2(g) + 2e- 2Cl-(g) 
  • And Al(s) – 3e-  Al3+(g) 
  • Note that these are already balanced for charge and change of oxidation states
  • We have to multiply  by 3 before adding to  doubled :
  • 2Al(s) - 6e- + 3Cl2(g) + 6e-  2AlCl3(g)
  • Or 2Al(s) + 3Cl2(g)  2AlCl3(g)
  • Again, it depends on the question ......
slide67
Try:
  • “What are the equations for the oxidation of sulphur dioxide to sulphate by the dichromate anion, in which chromate is reduced to a chromium(III) ion?”
  • First know your compounds and ions and balance them:
  • SO2 (aq) SO42- (aq) + 2e- And
  • Cr2O72-(aq) +6e- 2Cr3+(aq)
balance multiply and add
Balance, multiply and add
  • These balance to:
  • SO2 +2H2O  SO42- + 2e- + 4H+ And
  • Cr2O72- + 6e- +14H+ 2Cr3+ + 7H2O
  • Then you need to get the electrons equal:
  • 3SO2 + 6H2O  3SO42- + 6e- + 12H+
  • And then add, letting the electrons cancel:
  • 3SO2 + 6H2O + Cr2O72- +14H+ 3SO42- + 12H+ + 2Cr3+ + 7H2O
  • And we can cancel the water and protons:
  • 3SO2 + Cr2O72- + 2H+ 3SO42- + 2Cr3+ + H2O
  • And I left out the state symbols on purpose to make it easier ....
try this show all working
Try this (show all working):
  • In a titration, potassium permanganate will oxidise iron from iron(II) to iron(III) and is reduced to Mn2+(aq).
  • Write out both half-equations
  • Balance both, adding oxygens and H+ as needed
  • Work out the oxidation state changes
  • Get electrons equal in both (multiply as necessary)
  • Add
  • Cancel as necessary.
metals and their solutions
Metals and their solutions
  • For a metal in a solution of one of its own salts, an equilibrium exists:
  • Metal(s) Ý Metal+ ion + electron(s)
  • E.g. Mg(s)Ý Mg2+(aq) + 2e-(aq)

 This is a half-equation that describes the equilibrium.

slide71
For a reactive metal, this equilibrium lies mostly to the right, the metal is oxidised.
  • These electrons “want” to move away from the metal, but must be replaced by electrons from the solution.
  • The weird thing is that the electrons still end-up stuck in the solid metal electrode, which ends up being negatively charged
  • So electrons can’t get into the solution from this metal.
slide72
Another metal-ion equilibrium can be connected such that it provides a way for electrons to flow.
  • For an unreactive metal, the equilibrium lies mostly to the left:
  • E.g. Cu(s)Ý Cu2+(aq) + 2e-(aq)
  • So this equilibrium “takes” electrons from the equilibrium of the more-reactive metal.
  • Each is a half-cell.
  • Two half-cells make a cell.
electrode potentials
Electrode potentials
  • Conventions:
    • Half-equations given to you are written as reductions – always adding electrons
    • Electrodes written as reductions, ion|metal, eg: Fe2+|Fe
    • This is writing a half-cell
    • The bar represents a phase change
    • A full cell will be two of these, but with oxidation on the left:Zn|Zn2+||Cu2+|Cu
definition
Definition:
  • We can’t find the voltage (p.d.) of half a cell, need a reference.
  • Standard electrode potential of an element Eʅ is the potential difference between a standard hydrogen half-cell and a half-cell of the element under standard conditions.
  • Standard conditions for Eʅ :
    • 298K
    • 1 molar concentration solutions
    • 1 atm. for any gases involved
    • No current allowed (or only little ones!)
standard hydrogen electrode
Standard hydrogen electrode
  • Platinum surrounded by gaseous hydogen in a solution of H+
  • Gas electrode written as:Pt|H2(g)|H+(aq) if it’s left electrode, (usual) H+(aq)|H2(g)|Pt if it’s right electrode
  • Total cell example:
  • Pt|H2(g)|H+(aq)|| Ag+(aq) | Ag
  • Problems with the hydrogen cell – inconvenient, potentially dangerous, relies on standard conditions, including pressure.
slide76
Left-hand cell is the more negative of the two in most diagrams, that is, it’s the one with the most negative voltage when written as a reduction.
  • That way, you have a working cell, with the electrons going from the metal in the left half-cell to the metal in the right half-cell.
  • So the left side “wants” to be oxidised, the right side “wants” to be reduced.
  • This reversible reaction only gives a voltage when it is unbalanced, i.e. not at equilibrium, if equilibrium is attained, the cell voltage will be zero.
  • Concentration changes that push away from equilibrium will increase the voltage.
slide77
Potential of a cell is opposite polarity from electrolysis
  • Similar: reduction occurs at the cathode and oxidation at the anode in both cells and electrolysis.
  • Different: the anode will be negative.
  • Think about rechargeable cells, one polarity “charges” and one “discharges”
for a tin hydrogen cell which of the following actions would change the measured cell potential
For a tin/hydrogen cell, which of the following actions would change the measured cell potential?
  • a.) Increasing the pH in the cathode compartment
  • b.) Reducing the pH in the cathode compartment
  • c.) Increasing [Sn2+] in the anode compartment
  • d.) Increasing the pressure of hydrogen gas in the cathode compartment
  • e.) All of the above
calculating cell potentials
Calculating cell potentials

Given a list of reduction potentials, you can use the standard voltages to predict the result of redox reactions

These give you the voltage of the cell that could be formed by the redox couples specified.

If the cell has a positive voltage, it can work spontaneously (ΔG is negative)

how many roads
How many roads….

There are lots of ways to do these calculations. Don’t lose sight of the predictive nature of the calculation.

One (of many) ways:

Check the two redox half equations. The question should say (or make you work out) which is being reduced and which is being oxidised.

Reverse the equation for the oxidised one and reverse its sign, too.

Put the oxidation on the left and the reduction on the right.

Then add the potentials

If the result is positive, you have a cell.

If it’s negative, there can be no reaction (unless it’s an electrolysis).