Doc Brown's Advanced Level Chemistry  Quantitative redox reaction analysis

GCE Advanced Level REDOX Volumetric Analysis Titration Revision Questions

ORIGINAL QUESTIONS * acid–base & other non–redox titration Questions * Qualitative Analysis

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Most of the answers have been rounded up or rounded down to three significant figures (3sf)

Question 1: (a) MnO₄^–_(aq) + 8H⁺_(aq)  + 5Fe²⁺_(aq) ==> Mn²⁺_(aq) + 5Fe³⁺_(aq) + 4H₂O_(l) 

(b) (i) mol MnO₄^– = 0.0200 x 24.3 / 1000 = 0.000486

therefore mol Fe²⁺ = 5 x 0.000486 (1 : 5 in equation) = 0.00243 in 20 cm³

so scaling up to 1 dm³, the molarity of  Fe²⁺ = 0.00243 x 1000 / 20 = 0.122 mol dm^–3.

(ii) The end point is the first faint permanent pink due to a trace excess of KMnO₄.

(c) mol MnO₄^– = 0.0200 x 25.45 / 1000 = 0.000509

mol Fe = 5 x 0.000509 = 0.002545,

mass Fe = 0.002545 x 55.9 = 0.1423 g

total Fe in wire = 0.1423 x 10 = 1.423 g (only ¹/₁₀th of the made up solution used in titration)

so % Fe = 1.423 x 100 / 1.51 = 94.2 %

(d) The choice of acid is fully discussed in Ex 6.4 of Advanced Redox Chemistry Part 2 but basically you should know that ...

The titrations are done with (i) dilute sulfuric acid present to prevent side reactions e.g. MnO₂ formation (brown colouration or black precipitate).

(ii) Nitric acid is an oxidising agent and would oxidise iron(II) ions to iron(III) ions,

and (iii) chloride ions are oxidised to chlorine by manganate(VII) ions,

therefore, using either acid (ii) or (iii) would give a false titration result.

(e) (i) 23.86 cm³, i.e. the average titration value, which is statistically more accurate than an individual titration result.

(ii) molarity = mol / volume in dm³

BUT, remember, 1 dm³ = 1000 cm³,

burettes and pipettes work in cm³ and molarity works in dm³,

a bit of pain I know, but live with it and go with the flow!

therefore mol = molarity x vol in cm³ / 1000

mol MnO₄^– = 0.0200 x 25/1000 = 4.772 x 10^–4

(iii) From the balanced redox equation, 1 mol MnO₄^– oxidises 5 mol of iron(II) ions

mol Fe²⁺ = 5 x 4.772 x 10^–4 = 2.386 x 10^–3

(iv) You can assume mass of Fe = mass of Fe²⁺ ion, (2 electrons don't weigh much!)

mol = mass/atomic mass, so mass = mol x atomic mass

mass Fe = 2.386 x 10^–3 x 55.8 = 0.1331 g

(v) Total mass of iron in original sample = 10 x 0.1331 = 1.331 g (scaling up by factor of 250/25)

(vi) % iron in the original salt = 1.331 x 100 / 8.25 = 16.1% (1dp, 3sf)

Question 2:(a) 2S₂O₃^2–_(aq)  +  I_2(aq)  ==>  S₄O₆^2–_(aq) + 2I^–_(aq)

(b) mol S₂O₃^2– = 0.0120 x 23.5 / 1000 = 0.000282, mole iodine as I₂ = mol  S₂O₃^2– / 2 (1 : 2 in equation) = 0.000141,

A_r(I) = 126.9, so M_r(I₂) = 2 x 126.9 = 253.8

therefore: mass of iodine = 0.000141 x 126.9 x 2 = 0.0358 g

(c) mol S₂O₃^2– = 0.095 x 26.5 / 1000 = 0.002518,

mol of iodine = mole 'thio' / 2 = 0.002518 / 2 = 0.001259 in 25.0 cm³,

scaling up to 1 dm³ gives 0.001259 x 1000 /25 = 0.0504 mol dm^–3 of molecular iodine I₂.

mass concentration of I₂ = 0.0504 x 2 x 126.9 = 12.8 g dm^–3 of iodine

Question 3: (a) (i) Sn²⁺_(aq) + 2Fe³⁺_(aq) ==> Sn⁴⁺_(aq) + 2Fe²⁺_(aq)

(ii) Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) ==> 2Cr³⁺_(aq) + 6Fe³⁺_(aq) + 7H₂O_(l)

(b) for a 25.0 cm³ aliquot titrated

mol Cr₂O₇^2– = 0.0200 x 26.4 / 1000 = 0.000528,

mol Fe²⁺ titrated = 6 x Cr₂O₇^2– = 0.003168

(from 1 mol Cr₂O₇^2– : 6 mol Fe²⁺ in equation),

mol Fe₂O₃ = mol Fe²⁺ / 2 = 0.003168 / 2 =  0.001584

(because of 2Fe in Fe₂O₃)

M_r(Fe₂O₃) = 159.8 (Fe == 55.9, 0 = 16.0)

so mass of Fe₂O₃ = 0.001584 x 159.8 = 0.2531 g.

Total mass of Fe₂O₃ = 0.2531 x 10 (¹/₁₀th titrated) = 2.531 g

Therefore % Fe₂O₃ = 2.531 x 100 / 2.83 = 89.4%

(c) Potassium manganate(VII) isn't used for this titration because it is strong enough to oxidise chloride ions (from the hydrochloric acid) to form chlorine, giving a completely false titration.

(The half-cell potential proves the feasibility of this reaction, with a greater positive potential than chlorine/chloride, the manganate(VII) ion can oxidise the chloride ion to chlorine. Further note, that the dichromate(VI) ion cannot oxides the chloride ion, which is why it would used in this context.)

Note: There is actually a flaw in this question. In order to ensure all the Fe³⁺ is reduced, you would need excess Sn²⁺ solution, BUT, any excess Sn²⁺ would be oxidised by the Cr₂O₇^2– giving a false titration value. Never-the-less, it is a legitimate problem to solve!

Question 4: mol Fe²⁺ = 0.100 x 25.0 / 1000 = 0.0025,

mol MnO₄^– = mol Fe²⁺ / 5 (from equation 1 : 5) = 0.0005 in 24.15 cm³,

scaling up to 1 dm³, molarity of MnO₄^– = 0.0005 x 1000 / 24.15 = 0.0207 mol dm^–3.

Question 5: mol Cr₂O₇^2– = 0.0200 x 21.25 / 1000 = 0.000425,

mol of Fe salt = mol Fe²⁺ titrated = 6 x Cr₂O₇^2– = 6 x 0.000425 = 0.00255,

BUT only ¹/₁₀th of Fe²⁺ salt used in titration,

so 1 g of FeSO₄.(NH₄)₂S0₄ .xH₂O is equal to 0.00255 mol.

Scaling up to 1 mol gives a molar mass for the salt in g mol^–1 of 1 x 1 /0.00255 = 392.2.

So the formula mass for FeSO₄.(NH₄)₂S0₄.xH₂O is 392.2.

Now the formula mass of FeSO₄.(NH₄)₂S0₄ = 284.1, this leaves 392.2 – 284.1 = 108.1 mass units.

M_r(H₂O) = 18, so 108.1 / 18 = 6.005 mol of water, so x = 6 in the salt formula, FeSO₄.(NH₄)₂S0₄.6H₂O.

Question 6: (a) 2MnO₄^–_(aq) + 16H⁺_(aq)  + 5C₂O₄^2–_(aq) ==> 2Mn²⁺_(aq)  + 8H₂O_(l) + 10CO_2(g)

or  2MnO₄^–_(aq) + 6H⁺_(aq)  + 5H₂C₂O_4(aq) ==> 2Mn²⁺_(aq)  + 8H₂O_(l) + 10CO_2(g)

(b)  M_r(H₂C₂O₄.2H₂O) = 126.0,

total mol H₂C₂O₄.2H₂O (or C₂O₄^2–)  = 1.52 / 126 = 0.01206,

but mol of C₂O₄^2– in titration = 0.001206 (¹/₁₀th used, 25 of 250 cm³),

mol MnO₄^– = mol of C₂O₄^2– / 2.5 (2:5 or 1:2.5 ratio),

mol MnO₄^– = 0.001206 / 2.5 = 0.0004824 (in 24.55 cm³),

scaling up to 1 dm³ the molarity of MnO₄^– = 0.0004824 x 1000 / 24.55 = 0.0196 mol dm^–3.

M_r(KMnO₄) = 158, so in terms of mass concentration = 0.0196 x 158 = 3.10 g dm^–3

Question 7: mol KHC₂O₄.H₂C₂O₄.2H₂O (M_r = 254.1) = 0.15 / 254.1 = 0.0005903,

ratio of tetroxalate to manganate(VII) is 2:2.5 or 1:1.25 (note equiv of 2 C₂O₄^2– in salt),

so mol MnO₄^– in titration = 0.0005903 / 1.25 = 0.0004723 in 23.2 cm³,

scaling up to 1 dm³ gives for [MnO₄^–] = 0.0004723 x 1000 / 23.2 = 0.0204 mol dm^–3.

Question 8: (a)  2MnO₄^–_(aq) + 6H⁺_(aq)  + 5H₂O_2(aq)  ==> 2Mn²⁺_(aq) + 8H₂O_(l) + 5O_2(g)  

(b)  in titration mol MnO₄^– =  0.0200 x 20.25 / 1000 = 0.000405,

MnO₄^–:H₂O₂ ratio is 2:5 or 1:2.5, so mol H₂O₂ in titration = 0.000405 x 2.5 = 0.0010125,

scaling up for total mol H₂O₂ in diluted solution (of 1 dm³ or 1000 cm³) = 0.0010125 x 1000 / 25.0 = 0.0405 mol,

but in the original 50 cm³ solution,

therefore scaling up to 1 dm³, the original molarity of H₂O₂ is 0.0405 x 1000 / 50 = 0.810 mol dm^–3.

Question 9: (a) Zn_(s) + 2Fe³⁺_(aq) ==> Zn²⁺_(aq) +  2Fe²⁺_(aq)

(b) mol MnO₄^– in  titration = 0.0100 x 26.5 / 1000 =  0.000265,

mol Fe (Fe²⁺) = mol MnO₄^– x 5 = 0.001325 in 20.o cm³ of the alum solution,

scaling up gives total mol Fe = 0.001325 x 500 / 20 = 0.033125,

total mass Fe in the 13.2 g of alum = 0.033125 x 55.9 = 1.852,

so % Fe = 1.852 x 100 / 13.2 = 14.0%

Question 10: mol MnO₄^– in titration = 0.05 x 24.5 / 1000 = 0.001225,

ratio MnO₄^–:Na₂C₂O₄ is 2:5 or 1:2.5, so mol Na₂C₂O₄ titrated = 0.001225 x 2.5 = 0.003063 in 5 cm³,

scaling up to 1 dm³, molarity Na₂C₂O₄ = 0.003063 x 1000 / 5 = 0.613  mol dm^–3

M_r(Na₂C₂O₄) = 134, so concentration = 0.613 x 134 = 82.1 g dm^–3

Question 11: mol KMnO₄ = 0.0100 x 43.85 / 1000 = 0.0004385, mol Fe (Fe²⁺) = mol KMnO₄ x 5,

mol Fe = 0.0004385 x 5 = 0.0021925, so mol  FeSO₄.xH₂O is also 0.0021925,

in the titration ¹/₂₀th of the salt was used (²⁵/₅₀₀), so ¹/₂₀th of 12.18 g = 0.0021925 mol of the salt = 0.609 g,

scaling up the mass of 1 mole of the salt is 0.609 x 1 / 0.0021925 = 277.8,

so formula mass of FeSO₄.xH₂O is 277.8, now the formula mass of FeSO₄ is 152.0,

so the formula mass of xH₂O = 277.8 – 152.0 = 125.8,

M_r(H₂O) = 18, so x = 125.8 / 18 = 6.989,

so x = 7 and the formula of the salt is  FeSO₄.7H₂O, i.e. seven molecules of water of crystallisation. 

Question 12:  (a) 2MnO₄^–_(aq) + 6H⁺_(aq)  + 5NO₂^–_(aq) ==> Mn²⁺_(aq) + 5NO₃^–_(aq) +  3H₂O_(l)

(b) mol KMnO₄ in titration = 0.0250 x 25 / 1000 = 0.000625,

mol ratio MnO₄^–:NO₂^– is 2:5 or 1:2.5, so mol NO₂^– in titration = 0.000625 x 2.5 = 0.0015625 in 24.2 cm³,

scaling up to 1 dm³ gives a molar concentration of NaNO₂ of  0.0015625 x 1000 / 24.2 = 0.0646 mol dm^–3

M_r(NaNO₂) = 69, so in terms of mass concentration = 0.0646 x 69 = 4.46 g dm^–3

Question 13: M_r(FeC₂O₄) = 143.9, mol FeC₂O₄ in original solution = 2.68 / 143.9 = 0.01862,

scaling down the mol FeC₂O₄ in the titration = 0.01862 x 25 / 500 = 0.000931,

mol KMnO₄ in titration = 0.0200 x 28.0 / 1000 = 0.00056,

so ratio KMnO₄:FeC₂O₄ is 0.00056:0.000931 = giving the 'not so easy to spot' 3:5 the reacting mole ratio.

FeC₂O₄ is made up of a  Fe²⁺ ion and a  C₂O₄^2– ion, and the full redox equation is:

3MnO₄^–_(aq)+ 5FeC₂O_4(aq) + 24H⁺_(aq)  ==> 3Mn²⁺_(aq) + 5Fe³⁺_(aq)+ 12H₂O_(l) + 10CO_2(g)

or 3MnO₄^–_(aq)+ 5Fe²⁺_(aq) + 5C₂O₄^2–_(aq) + 24H⁺_(aq)  ==> 3Mn²⁺_(aq) + 5Fe³⁺_(aq)+ 12H₂O_(l) + 10CO_2(g)

Question 14: (a) IO₃^–_(aq) + 5I^–_(aq) + 6H⁺_(aq) ==> 3I_2(aq) + 3H₂O_(l)

(b) mol I^– titrated = 0.100 x 20.0 / 1000 = 0.002, mole ratio IO₃^–:I^– is 1:5,

so mole IO₃^– reacted = 0.002 / 5 = 0.0004,

so 0.0004 = 0.012 x (volume IO₃^– required) / 1000,

volume IO₃^– required = 0.0004 x 1000 / 0.012 = 33.3 cm³

(c)(i) mole S₂O₃^2– ('thio') = 0.0500 x 24.1 / 1000 = 0.001205,

I₂:S₂O₃^2– ratio is 1:2 in the titration reaction, so mol I₂ = mole S₂O₃^2– / 2 = 0.001205 / 2 = 0.0006025,

now the  IO₃^–:I₂ reaction ratio is 1:3,

so mol IO₃^– reacting to give iodine = mole I₂ formed / 3 = 0.0006025 / 3 = 0.000201 in 25 cm³,

so scaling up to 1 dm³ the molarity of the KIO₃ (IO₃^–) = 0.000201 x 1000 / 25 = 0.00804 mol dm^–3,

 M_r(KIO₃) = 214.0, so in terms of mass, concentration = 0.00804 x 214 = 1.72 g dm^–3.

A quicker approach if confident! – ratios from all equations involved are: 2S₂O₃^2– : I₂ : ¹/₃IO₃^–, means that the overall mole iodate(V) = mole thiosulphate / 6, so you can 'jump' from line '1' to the last 'few' lines. However in exams these days all the stages (i.e. , to , !) are often 'broken down' for you and it might be best you work through the problem thoroughly and methodically.

(ii) Starch indicator is used for the titration, when the last of the iodine reacts with the thiosulphate, the blue colour from the starch–iodine 'complex' is discharged and the solution becomes colourless.

Question 15: (i) mol KMnO₄ = 0.0200 x 22.5 / 1000 = 0.00045,

mol Fe²⁺ = mol KMnO₄ x 5 = 0.00225 in 25 cm³,

scaling up to 1 dm³, molarity of the original Fe²⁺ = 0.00225 x 1000 / 25.0 = 0.090 mol dm^–3

(ii) the 2nd titration gives the total concentration of Fe²⁺ + Fe³⁺ because any Fe³⁺ has been reduced to Fe²⁺,

mol KMnO₄ = 0.0200 x 37.6 / 1000 = 0.000752, total mol Fe²⁺ titrated = mol KMnO₄ x 5 = 0.00376 in 25 cm³,

scaling up to 1 dm³, total molarity of  Fe²⁺ + Fe³⁺ in original solution = 0.00376 x 1000 / 25.0 = 0.150 mol dm ^–3,

so using the result from (a) the  Fe³⁺ concentration = 'Fe' total – Fe²⁺ = 0.150 – 0.090 = 0.060 mol dm^–3.

Question 16: you can ignore the 25 cm³ of the solution because you use the same volume in each titration and you can work on the ratio of the moles of 'Fe' out of the (a) and (b) titration calculations.

(a) mol Fe²⁺ = 5 x MnO₄^– = 5 x 0.0200 x 16.9 / 1000 = 0.00169 mol = unreacted iron (which dissolved in the acid to form Fe²⁺).

(b) mol Fe³⁺ = EDTA^4– = 0.100 x 17.6 / 1000 = 0.00176 mol = total mol iron in the sample titrated.

(c) calculation (a) gives the relative moles of unreacted iron Fe, as it dissolved to form the titratable Fe²⁺.

Calculation (b) gives the total of unreacted Fe + the rust i.e. Fe³⁺, because any Fe²⁺ formed from Fe has been oxidised to Fe³⁺.

So from the original mixture (in terms of the 25 cm³ sample), mol unreacted Fe = 0.00169,

mol of reacted iron = 0.00176 – 0.00169 = 0.00007.

Therefore the % rusted iron = 0.00007 x 100 / 0.00176 = 3.98 % rusted iron.

Question 17: (a) I_2(aq) + 2S₂O₃^2–_(aq) ==> S₄O₆^2–_(aq) + 2I^–_(aq)

 or I_2(aq) + 2Na₂S₂O_3(aq) ==> Na₂S₄O_6(aq) + 2NaI_(aq)

(b) Starch indicator is used, starch gives a blue/black colour with iodine, this colour disappears when the last of the iodine is titrated, so a blue to colourless sharp end–point is observed.

(c) mole 'thio' = 0.100 x 17.6/1000 = 0.00176,

mol I₂ = 0.00176 ÷ 2 = 0.00088 in 25 cm³,

scaling up gives 0.00088 x 1000 ÷ 25 = 0.0352 mol dm^–3 for molarity of iodine,

formula mass I₂ = 2 x 127 = 254, so concentration = 0.0352 x 254 = 8.94 g dm^–3

Question 18: (a) Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6I^–_(aq) ==> 2Cr³⁺_(aq) + 3I_2(aq) + 7H₂O_(l)

(b) 2S₂O₃^2–_(aq)  +  I_2(aq)  ==>  S₄O₆^2–_(aq) + 2I^–_(aq)

(c) mol 'thio' = 20.0 x 0.100/1000 = 0.002,

therefore from equation (b), mol iodine = mol 'thio'/2 = 0.001

(d) From equation (a) mol dichromate(VI) reacting = mol iodine liberated/3 = 0.000333 (3sf)

(e) M_r(K₂Cr₂O₇) = 294.2

mass K₂Cr₂O₇ titrated = 0.000333 x 294.2 =  0.0980 g (3 sf)

(f) Since the aliquot of 25.0 cm³ is 1/10th of the total solution in the flask, the total mass of the K₂Cr₂O₇ in original sample dissolved in the flask solution = 10 x 0.0980g = 0.98g

and the % purity of the K₂Cr₂O₇ = 0.98 x 100/1.01 = 97.0 % (3 sf)

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