9.1 Oxidation & Reduction

Redox Reactions

• Reduction and Oxidation may be defined in 3 ways. Learn them!

Consider this experiment:

It is clear that the Lead Oxide was Reduced because it lost ________________.

The Hydrogen was oxidised because it gained _____________________.

So, any reaction involving the transfer of Oxygen is easily understood in Redox terms

But what about this reaction...? H₂S + Cl₂ --> 2HCl + S

• We could say that the Chlorine atoms have been ___________________ because ___________________________________

• We could say that the Sulphur atoms have been ____________________ because ___________________________________

And what about...? H₂ + Cl₂ --> 2HCl

• We could say that the Chlorine atoms have been ___________________ because ___________________________________

But we can’t really say that the Hydrogen atoms have been oxidised because they have lost Hydrogen.

So we have to split up the overall equation into half-equations to look at what is happening to the individual atoms as they turn into ions.

Split the following reactions into half-equations and use them to same what has been Oxidised/reduced in each case.

1. Ca + Br₂ --> CaBr₂

2. 2Al + 3S --> Al₂S₃

3. P₄ + 12Na --> 4Na₃P

4. N₂ + 3Sr --> Sr₃N₂

Looking at electrons is time-consuming and not always necessary.

Oxidation states are a system of keeping track of what is Oxidised/Reduced.

• Increasing Oxidation State is Oxidation

• Reducing Oxidation State is Reduction

• First, you need to learn the rules for assigning Oxidation States.

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1. What is the Oxidation State of Oxygen in:

a) O₂ b) Na₂O c) H₂O d) O₃ e) H₂O₂

2. What is the Oxidation state of Nitrogen:

a) N₂ b) N₂O c) NO₂ d)NO e)N₂O₄

3 Assign Oxidation States to Sulphur in

a. SO₂

b. SO₃

c. SO₄^2-

d. SO₃^2-

e. H₂S

f. HSO₄^-

g. H₂SO₄

3. What is the Oxidation State of Chromium in:

a. CrO₂

b. [CrO₄]^2-

c. CrF₆

d. [CrO₄]^3-

e. Cr₂O₇^2-

As you’ve seen, Chromium has many Oxidation states and so do most Transition Elements and lots of non metals.

· ClO^-, ClO₂^-, ClO₃^- and ClO₄^- - are all ions containing Chlorine and Oxygen. So, they are all Chlorate ions.

· NO₃^- and NO₂^- are both ions containing Nitrogen and Oxygen. So, they are all Nitrate ions.

· SO₃^2- and SO₄^2- are both ions containing Sulphur and Oxygen. So, they are all Sulphate ions.

If there was no way of naming them differently then confusion would result. Since the Oxygen atoms are all in the -2 Oxidation State we simply add the Oxidation State of the other atom (in Roman numerals) to the name of the ion.

SO₃^2- ---- Sulphate (IV) ion

SO₄^2- ---- Sulphate (VI) ion

Name the following ions:

a. ClO^-

b. ClO₂^-

c. ClO₃^-

d. ClO₄^-

e. NO₂^-

f. NO₃^-

A typical question might ask you to suggest a formula for a Sulphate (VI) ion

o Start by assuming S⁶⁺ even though this doesn’t really exist.

o Now add O^2- ions to it until the ion becomes negative.

o SO⁴⁺ --> SO₂²⁺ --> SO₃ --> SO₄^2-

Or, suggest a formula for a Phosphate (V) ion

o P⁵⁺ --> PO³⁺ --> PO₂⁺ --> PO₃^-

o Actually, the Phosphate (V) ion is really PO₄^3- but your first suggestion would still be a correct answer.

Suggest formulae for the following ions:

a. Bromate (VII)

b. Bromate (I)

c. Bromate (III)

d. Manganate (VII)

e. Ferrate (VI)

f. Vanadate (V)

If you know the reactants and products in a redox equation you can’t necessarily just balance the equation.

For instance Manganate (VII) ions oxidise Copper (I) to Copper (II) and become Mn²⁺ in the process .

But we can’t write Cu⁺ + MnO₄^- --> Cu²⁺ + Mn²⁺ because it doesn’t account for the Oxygen atoms and, anyway, they don’t react 1:1

So, we write half equations, then balance the electrons, then add them together.

1. Cu⁺ --> Cu²⁺ no atoms other than Cu so just balance charge with electrons

Cu⁺ --> Cu²⁺ + e^-

2. MnO₄^- --> Mn²⁺ to account for extra O atoms add H₂O

MnO₄^- --> Mn²⁺ + 4H₂O

To account for extra H atoms add H⁺ ions

MnO₄^- + 8H⁺--> Mn²⁺ + 4H₂O

Now that the atoms balance we can balance charges with electrons

MnO₄^- + 8H⁺ +5e^- --> Mn²⁺ + 4H₂O

3. Make sure both equations contain the same number of electrons

MnO₄^- + 8H⁺ +5e^- --> Mn²⁺ + 4H₂O

(Cu⁺ --> Cu²⁺ + e^- ) x5 ==> 5Cu⁺ --> 5Cu²⁺ + 5e^-

4. Add the two equations together and cancel

5Cu⁺ + MnO₄^- + 8H⁺ +5e^- --> Mn²⁺ + 4H₂O + 5Cu²⁺ + 5e^-

5. Answer is:

5Cu⁺ + MnO₄^- + 8H⁺ --> Mn²⁺ + 4H₂O + 5Cu²⁺

1. Iron(III) oxidises Aluminium to Al³⁺ and becomes Iron (II)

a. Complete: Al --> Al³⁺

b. Complete: Fe³⁺ --> Fe²⁺

c. Balance electrons

d. Add equations and cancel

2. Lead(II) oxidises Lithium to Li⁺ and becomes Iron

a. Complete: Pb²⁺--> Pb

b. Complete: Li --> Li⁺

c. Balance electrons

d. Add equations and cancel

3. Hydrogen Peroxide (H₂O₂) oxidises SO₂ to SO₄^2- and becomes water

a. Complete: H₂O₂ --> 2H₂O

b. Complete: SO₂ --> SO₄^2-

c. Balance electrons

d. Add equations and cancel

British students will remember this as the Reactivity Series from Year 9

Metals at the top react quickly, those at the bottom slowly (if at all)

This is because metals at the top lose electrons easily, and those at the bottom with more difficulty.

Which is why Gold & Platinum don’t tarnish, while Copper & Silver do slowly.

It also explains why reactive metals can displace less reactive metal ions from solution. The reactive metals lose their electron(s) easily and force the electron onto the less reactive metal ion.

1. Complete & balance those equations in which displacement happens

Ca_(s) + Cu²⁺_(aq) --> Cu_(s) + Ag⁺_(aq) -->

Mg_(s) + Ca²⁺_(aq) --> Sn_(s) + Pb²⁺_(aq) -->

Mg_(s) + Al³⁺_(aq) --> Al_(s) + Pb²⁺_(aq) -->

Like acid-base titrations these are all about using a known volume of a solution of known concentration to react with a known volume of some other reactant of unknown concentration.

They don’t generally need an indicator because the change of oxidation state will cause a change in colour to indicate the end-point.

Eg

14 cm³ Manganate (VII) solution of unknown concentration is required to completely oxidise 25cm³ of 0.02 mol/dm³ Iron (II) to Iron (III). Find the concentration of the Manganate.

· Write half-equations MnO₄^- + 8H⁺ +5e^- -- Mn²⁺ + 4H₂O

Fe²⁺ -- Fe³⁺ + e^-

· Write full equation MnO₄^- + 8H⁺ +5Fe²⁺ --> Mn²⁺ + 4H₂O + 5Fe³⁺

· Moles of Iron (II) = vol x conc =25/1000 x 0.02 = 0.0005

· Moles of Manganate = 0.0005/5 = 0.0001 (5:1 ratio)

· Conc. of Manganate = mols/vol = 0.0001/(14/1000) = 0.007 mol/dm³

Given the following two half–reactions

(a) Given (i) S₄O₆^2–_(aq) + 2e^– --> 2S₂O₃^2–_(aq)

and (ii) I_2(aq) + 2e^– --> 2I^–_(aq)

Construct the full ionic redox equation for the reaction of the thiosulphate ion S₂O₃^2– and iodine I₂.

(b) What mass of iodine reacts with 23.5 cm³ of 0.0120 mol dm^–3 sodium thiosulphate solution.

(c) 25.0 cm³ of a solution of iodine in potassium iodide solution required 26.5 cm³ of 0.0950 mol dm^–3 sodium thiosulphate solution to titrate the iodine.

What is the molarity of the iodine solution?

Question 2: Given the half–reaction C₂O₄^2–_(aq) --> 2CO_2(g) + 2e-

or H₂C₂O_4(aq) --> 2CO_2(g) + 2H⁺_(aq) + 2e-

1. write out the balanced redox equation for manganate(VII) ions oxidising the ethanedioate ion (or ethane–dioic acid).

(b) 1.520 g of ethanedioic acid crystals, H₂C₂O₄.2H₂O, was made up to 250.0 cm³ of aqueous solution and 25.00 cm³ of this solution needed 24.55 cm³ of a potassium manganate(VII) solution for oxidation.

Calculate the molarity of the manganate(VII) solution and its concentration in g dm^–3.

Question 3:

2.83 g of a sample of Haematite iron ore [iron (III) oxide, Fe₂O₃] were dissolved in concentrated hydrochloric acid and the solution diluted to 250 cm³.

25.0 cm³ of this solution was reduced with Tin(II) chloride (which is oxidised to Sn⁴⁺ in the process) to form a solution of iron(II) ions.

This solution of iron(II) ions needs 26.4 cm³ of 0.0200 mol dm^-3 potassium dichromate(VI) solution for complete oxidation back to iron(III) ions.

(a) given the half–cell reactions

(i) Sn⁴⁺_(aq) + 2e^– --> Sn²⁺_(aq)

and (ii) Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6e^– --> 2Cr³⁺_(aq) + 7H₂O_(l)

deduce the fully balanced redox equations for the reactions

(i) the reduction of iron(III) ions by tin(II) ions

(ii) the oxidation of iron(II) ions by the dichromate(VI) ion

(b) Calculate the percentage of iron(III) oxide in the ore.

Aquatic life needs dissolved Oxygen to survive even though its solubility is low. The higher the Oxygen concentration in water the less polluted it’s likely to be.

• Why should Oxygen dissolve poorly in water?

The BOD (Biological Oxygen Demand) is the additional Oxygen you have to add to water to completely oxidise all Organic matter in 5days, measured in parts per million (ppm).

The Winkler method saturates an alkaline water sample with Oxygen and leaves it for 5 days . It measures the Oxygen content before and after. This allows them to see how much Oxygen was needed to oxidise any organic matter.

The reactions.

1. Mn²⁺ oxidised to MnO₂ by Oxygen 2Mn²⁺ + O₂ + 4OH^- --> 2 MnO₂ + 2H₂O

2. MnO₂ oxidises I^- to I₂ 2 MnO₂ + 4I^- + 4H⁺ --> 4I₂ + 2Mn²⁺ + 4H₂O

3. I₂ reacts with starch and goes black

4. Thiosulphate removes black colour 4 I₂ + 4S₂O₃^2- --> 4I^- + 2 S₄O₆^2-

Overall, 1 mole of Oxygen required 4 moles of Thiosulphate to be used in the final reaction