F325 Redox Equations and Titrations

7/31/2019 F325 Redox Equations and Titrations

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F325 Redox Equations and Titrations

Redox Equations and Titrations

TerminologyRedox reactions are ones in which both reduction and oxidation takes place simultaneously.

We recognize this because one reacting species will have an increase in oxidation number(i.e. it will be oxidized, by losing electrons) while another reacting species will show a decreasein oxidation number (i.e. it will be reduced, by gaining electrons).

The species which causes the other to be oxidized is called the oxidizing agent (it will itselfget reduced during the reaction) and the species which causes the other to be reduced is thereducing agent (and it will itself get oxidized).

Consider the rather spectacular reaction between magnesium and copper oxide. When heatedtogether, a violently exothermic displacement takes place:

Mg + CuO MgO + CuWe can see what is going on by writing two half equations one for the oxidation and one forthe reduction. Note that the oxide ion is a spectator ion, unchanged by the reaction. It thereforedoes not show up in either half-equation:

Oxidation: Mg Mg2+ + 2e-

Reduction: Cu2+ + 2e- CuSo the magnesium is acting as the reducing agent, and the copper oxide is acting as the

oxidizing agent.

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Redox

explain, for simple redox reactions, the terms redox, oxidation number, half-reaction, oxidising agentand reducing

agent

construct redox equations using relevant half equationsor oxidation numbers

interpret and make predictions for reactionsinvolving electron transfer.

Redox reactions and titrations

describe, using suitable examples, redox behaviour in transition elements

carry out redox titrations, and carry out structured calculations, involving MnO4 and I2/S2O3

2

perform non-structured titration calculations, based on experimental results.


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Constructing redox equations method using half-equations

Example: oxidation of Fe2+ and reduction of MnO4-

Acidified manganate(VII) ions are a powerful oxidizing agent. They can oxidize Fe2+ ions to

Fe3+

ions. At the same time the manganate(VII) ions are reduced to Mn2+

ions.

We can represent what happens to the iron ions with a half-equation

The number of electrons to include comes from the change in oxidation number of the speciesbeing oxidized or reduced:

Oxidation is loss of electrons: Fe2+(aq) Fe3+(aq) + e-pale green yellow

oxidation number of Fe: +2 +3

There is a corresponding half equation for the reduction of acidified manganate(VII) ions:

Reduction is gain of electrons: MnO4-(aq) + 8H

+(aq) + 5e

- Mn2+(aq) + 4H2O(l)purple very pale pink

oxidation number of Mn: +7 +2

To write an overall equation for what happens, we need to have the same number of electronslost in the oxidation as are gained in the reduction, so we need to multiply the first half-equation by 5 throughout:

5 Fe2+(aq) 5 Fe3+(aq) + 5 e-

Then we can simply add the half equations together, and cancel the five electrons on eachside:

MnO4-(aq) + 8H

+(aq) + 5e

- Mn2+(aq) + 4H2O(l)5 Fe2+(aq) 5 Fe3+(aq) + 5 e-

So MnO4-(aq) + 8H

+(aq) + 5Fe

2+(aq) 5 Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Example: reduction of dichromate ions

We are also familiar with acidified potassium dichromate as a powerful oxidizing agent. Thedichromate ions act in a similar way to the manganate(VII) ions. As an example, they canoxidize Sn2+ ions to Sn4+ ions.

You should recall that dichromate ions are an intense orange colour, and that when they act asan oxidizing agent, their colour changes to dark green. This is because they have beenreduced to Cr3+ ions.

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Again we can represent what happens with half-equations:

Sn2+(aq) Sn4+(aq) + 2e- OxidationCr2O7

2-(aq) + 14H

+(aq) + 6e

- 2Cr3+(aq) + 7H2O(l) Reduction

And again we can combine these into a single redox equation by getting the same number ofelectrons in each half equation i.e. multiplying the oxidation equation by 3 throughout.

3Sn2+(aq) + Cr2O72-

(aq) + 14H+

(aq) + 6e- 3Sn4+(aq) + 6e- + 2Cr3+(aq) + 7H2O(l)

N.B. we can also be asked to work out the oxidation or reduction half equation if we aregiven the overall equation and the other half-equation:

Using the same example as above, we might be told that the overall equation is:

3Sn2+

(aq) + Cr2O72-

(aq) + 14H+

(aq) 3Sn4+(aq) + 2Cr3+(aq) + 7H2O(l)and that the oxidation part is Sn2+ Sn4+ + 2e-. We are required to work out the reductionhalf-equation:

Firstly multiply the given half-equation up to get the same stochiometry (numbers in front of the

formulae) as in the overall equation: 3Sn2+ 3Sn4+ + 6e-Now compare this to the overall equation and work out what is missing on each side thismust be what is in the reduction half-equation:

Oxidation 3Sn2+ 3Sn4+ + 6e-Reduction Cr2O7

2- + 14H+ + 6e- 2Cr3+ + 7H2OOverall 3Sn2+(aq) + Cr2O7

2-(aq) + 14H

+(aq) 3Sn

4+(aq) + 2Cr

3+(aq) + 7H2O(l)

Constructing redox equations Method using oxidation numbersWe can balance a redox equation by making sure that the increase in oxidation numbers dueto the oxidation taking place is balanced by the decrease in oxidation numbers due to thereduction taking place, without having to construct separate half-equations.

The key is that the total increase in oxidation numbers in the balanced equation must be equalto the total decrease in oxidation numbers.

e.g. hydrogen iodide (HI) is oxidized to iodine by concentrated sulphuric acid. The sulphuricacid is reduced to hydrogen sulphide. Construct a balanced equation for this.

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Step 1: Write an equation using the formulae for reactants and products you know about

HI + H2SO4 I2 + H2S - don't worry about balancing or missing atoms yet !

Step 2: Add the oxidation numbers to see what is oxidized and what is reduced

HI + H2SO4 I2 + H2S+1 -1 +1 +6 -2-2 0 +1 -2

+1 -2-2 +1

You may want to get rid of those oxidation numbers which don't change, or don't have a'partner' on both sides of the equation, so you can see the reduction and oxidation clearly

HI + H2SO4 I2 + H2S-1 +6 0 -2

Step 3: Balance the equation FOR ONLY THOSE ATOMS BEING OXIDISED OR REDUCED.Don't worry about O or H atoms which are still not balanced, yet.

2HI + H2SO4 I2 + H2S

Step 4: Add the oxidation numbers for each species being oxidized or reduced, too see whatthe total increase and decrease in oxidation numbers is:

2 HI + H2SO4 I2 + H2S-1 +6 0 -2-1 0

-2 0

So total reduction (S) is +6 to -2 = -8Total oxidation (I) is -2 to 0 = +2

Step 4: Balance the increase and decrease in oxidation numbers in this case we need the

2HI I2 to happen four times to balance one H2SO4 H2S

8 HI + H2SO4 4 I2 + H2S

Step 5: Check if anything else needs to be balanced. If we are missing oxygen atoms, add

water to that side. Then if we are missing hydrogen atoms, add H + to that side (look for 'acidicconditions' in the question as another clue as to when to include H + as a reactant !)

8HI + H2SO4 4 I2 + H2S + 4 H2OFinally: Check it is balanced for all elements

Check it is balanced for charge total charge on each side the same.

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Practice:1) Use oxidation numbers to write balanced equations for the following redox reactions:

a) Hydrogen sulphide, H2S is oxidized to sulphur, S, by nitric acid, HNO3, which is itselfreduced to nitrogen monoxide, NO.

b) In acidic conditions, silver metal, Ag, is oxidized to silver(I) ions, Ag+ by NO3- ions, which are

reduced to nitrogen monoxide (NO).

2) Prove that oxidation numbers method produces the same result as we previously obtainedusing the half-equation method for the reaction between manganate(VII) ions and iron(II).

Acidified managanate(VII) ions are capable of oxidizing iron(II) ions to iron(III) ions. Themanganate is reduced to Mn2+ ions:

Redox titrations

In a redox titration we are titrating one solution in which something is going to be oxidized,against another in which something is going to be reduced. If we know the concentration ofone substance, we can work out the concentration of the other from a balanced equation forthe redox reaction.

We don't use a pH indicator in a redox titration. Changes in colour of the substances beingoxidized and reduced tell us when we have reached the equivalence point.

There are two examples we should understand fully, although we may meet other examples ofredox titrations where information about the titration is provided in the question.

Titration of iron(II) ions using manganate(VII) ions under acidic conditionsPurpose: To determine the concentration of Fe2+(aq) in a solution.

e.g. determining the mass of iron in iron tabletsdetermining the %purity of iron in an alloy containing irondetermining the concentration of iron(II) ions in contaminated waterdetermining the molar mass and formula of an iron(II) salt

How to do it:If the sample to be investigated isn't already a solution containing iron(II) ions, it may need tobe crushed and dissolved in water and sulphuric acid. We'll need to measure the mass of thesample before dissolving. The resulting solution will be pale green (almost colourless if the

concentration of iron(II) is fairly low).

We titrate a known volume of the sample against potassium managate(VII) solution (potassiumpermanganate) of known concentration.

As the deep purple manganate(VII) ions mix with the iron(II) ions they are decolourised. Thiscontinues until all the iron(II) ions have reacted. The next drop of potassium manganate(VII)stays purple, so the endpoint is when the first hint of pink/purple persists in the solution.

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The chemistry:The managanate(VII) ions (MnO4

-) are reduced to managese(II), while the iron(II) is oxidized toiron(III) ions. We have already constructed an equation for this:

MnO4-(aq) + 8H

+(aq) + 5e

- Mn2+(aq) + 4H2O(l)5 Fe2+(aq) 5 Fe3+(aq) + 5 e-

So MnO4-(aq) + 8H

+(aq) + 5Fe

2+(aq) 5 Fe3+(aq) + Mn2+(aq) + 4H2O(l)

The important consequence of this equation is that one mole of MnO 4- ions reacts with five

moles of Fe2+ so we have the mole ratio of 1:5 when we do the titration calculation.

Example calculations:25.0cm3 of a solution of iron(II)sulphate required 23.0cm3 of 0.0020 mol dm-3 potassium

permanganate to completely oxidize it in acidic solution. What was the concentration of theiron(II)sulphate solution ?

Step 1: Calculate moles of the permanganate

moles = conc. x vol in dm3

= 0.00200 x (23/1000) = 4.60 x 10-5

moles

Step 2: Use the mole ratio from the balanced equation to get moles of iron(II) ions

MnO4-(aq) + 8H

+(aq) + 5Fe

2+(aq) 5 Fe

3+(aq) + Mn

2+(aq) + 4H2O(l)

1 : 54.60x10-5 2.30 x 10-4 moles of Fe(II)

Step 3: Calculate concentration of iron(II) sulphate solutionConc of Fe(II) = moles / vol (in dm3) = 2.3x10-4 / 0.0025 = 0.00920 mol dm-3

The solution above was made by dissolving a tablet containing hydrated iron(II) sulphate and

sucrose (which does not react with permanganate ions) in 250cm3

of water. Calculate themass of FeSO4 in the tablet.

Step 1: Convert concentration of iron(II) solution into moles0.00920 mol dm-3 of iron sulphate = 0.00920 * (250 / 1000) = 0.0023 moles FeSO4 weredissolved in the 250cm3 of water, so 0.0023 moles of iron(II) sulphate were in the tablet.

Step 2: Calculate mass of FeSO4 dissolved:Mass of FeSO4= 0.0023 x RFM = 0.0023 x (55.8 + 32.1 + (16.0 x 4)) = 0.0023 x 151.9

= 0.350g

The formula for the hydrated iron sulphate used in the tablet is FeSO 4.7H2O. If the mass of thetablet was 8.5g, calculate the % by mass of hydrated iron II sulphate in the tablet.

RFM of FeSO4.7H2O = 151.9 + (7 x 18) = 277.9Mass in tablet = moles x RFM = 0.023 x 277.9 = 0.6392g

% by mass = (0.6392 / 8.5) x 100 = 7.5 %

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Titration of iodine in solution using thiosuphate ionsPurpose: To determine the concentration of an oxidizing agent in a solution e.g. Cu 2+(aq) ions.

e.g. determining the % of copper in an alloys such as brass or bronzedetermining concentration of copper ions in a solutiondetermining concentration of dichromate ions in a solution

determining the concentration of chlorate(I) ions ClO-

in bleachdetermining the concentration of a hydrogen peroxide solution

How to do it:The analysis is done in two stages.In the first stage, the oxidizing agent to be determined is reacted with excess iodide ions. Thisresults in iodine being formed in the solution, the amount of iodine being directly related to theamount of the oxidizing agent present.e.g. 2Cu2+(aq) + 4I-(aq) 2CuI(s) + I2(aq)

so in this case 2 moles of Cu2+ ions produce 1 mole of iodine.

In the second stage, the concentration of iodine in the solution is determined by titrating itagainst sodium thiosulphate of known concentration. The red-brown colour of the iodine insolution fades to a pale straw colour, and to colourless when all the iodine has been reducedto iodide ions. This makes the endpoint difficult to see, so close to the endpoint an amount ofstarch is added as an indicator. Starch is blue-black when iodine is present, but at the endpointthe starch becomes colourless.

The chemistry:

I2(aq) + 2e- 2I-(aq) reduction

At the same time the thiosulphate ions are oxidized to tetrathionate ions:

2 S2O32-

(aq) S4O62-(aq) + 2e- oxidationThe overall equation is therefore:

I2(aq) + 2 S2O32-

(aq) 2I-(aq) + S4O62-(aq)This means that 2 moles of thiosulphate ions react with 1 mole of iodine.

Example calculations

The calculation is also done in two stages. In the first stage, the titre for the thioulphate is usedalong with its concentration and the mole ratio above to work out the number of moles of iodinein the portion of solution which was titrated.

In the second stage, the moles of iodine in the whole solution are used along with the moleratio of iodine to oxidizing agent to work out the concentration of oxidizing agent addedoriginally present in the whole solution. This can then be used further if needed to work out themass of the oxidizing agent given a formula etc.

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e.g. 30cm3 of bleach was reacted completely with iodide ions in acidic solution. The iodineformed was titrated against 0.200 mol dm-3 sodium thiosulphate, requiring 29.45cm3 of thethiosulphate solution to reach the endpoint.

The chlorate(I) ions in the bleach react with iodide ions according to the equation:ClO-(aq) + 2I-(aq) + 2H

+(aq) Cl

-(aq) + I2(aq) + H2O(l)

Calculate the concentration of the chlorate ions in the bleach.

Firstly use the titration data to get the moles of iodine which was released when the bleachreacted with the excess iodide ions.

Step 1: Calculate moles of thiosulphate ions usedmoles = conc x vol (in dm3) = 0.200 x (29.45/1000) = 5.89 x 10-3 moles

Step 2: Use the balanced equation for the titration to get moles of iodineI2(aq) + 2 S2O3

2-(aq) 2I

-(aq) + S4O6

2-(aq)

1 : 22.945 x 10-3 : 5.89x10-3

Secondly use the moles of iodine to find moles of chlorate ions, using the first equation

ClO-(aq) + 2I-(aq) + 2H

+(aq) Cl

-(aq) + I2(aq) + H2O(l)

1 : 2 : 12.945 x 10-3 2.945 x 10-3

Step 3: Calculate concentration of chlorate ions:

conc of chlorate(I) = moles / vol (in dm3

) = 2.945 x 10-3

/ 0.030 = 0.0982 mol dm-3

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Answers to Practice Questions1a) Hydrogen sulphide, H2S is oxidized to sulphur, S, by nitric acid, HNO3, which is itselfreduced to nitrogen monoxide, NO.

Step 1 H2S + HNO3 S + NO

Step 2 -2 +5 0 +2

Step 3 no further balancing needed oxidation (S) -2 to 0 = +2reduction (N) +5 to +2 = -3

Step 4 need 3x (H2SS) and 2x (HNO3 NO) to balance

3H2S + 2HNO3 3S + 2NOStep 5 missing 4 x O on RHS so add 4x H2O (then no need to add H

+)

3H2S + 2HNO3 3S + 2NO + 4H2O

1b) In acidic conditions, silver metal, Ag, is oxidized to silver(I) ions, Ag+ by NO3- ions, which

are reduced to nitrogen monoxide (NO).

Step 1 Ag + NO3- + H+ Ag+ + NO

Step 2 0 +5 +1 +2

Step 3 no further balancing needed oxidation (Ag) 0 to +1 = +1reduction (N) +5 to +2 = -3

Step 4 Need x3 (Ag+ Ag)

3Ag + NO3- + H+ 3 Ag+ + NO

Step 5 Need 2 more O on RHS, so add 2H2O. Balancing then needs 4H+ on LHS.

3Ag + NO3- + 4H+ 3Ag+ + NO + 2H2O

2) Acidified managanate(VII) ions are capable of oxidizing iron(II) ions to iron(III) ions. Themanganate is reduced to Mn2+ ions:

Step 1 MnO4- + Fe2+ +H+ Mn2+ + Fe3+

Step 2 +7 +2 +2 +3

Step 3 no further balancing oxidation +2 to +3 = +1 reduction +7 to +2 = -5

Step 4 Need 5x (Fe2+ Fe3+)

MnO4- + 5 Fe2+ +H+ Mn2+ + 5 Fe3+

Need four O on RHS, so add 4H2O. Balancing then requires 8H+ on LHS

Step 5 MnO4- + 5 Fe2+ +8 H+ Mn2+ + 5 Fe3+ + 4H2O

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F325 Redox Equations and Titrations