# Redox Titration Calculation Help A-Level Chemistry AQA

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#1
A 25.0cm^3 sample of a solution containing iron (II) and iron (III) sulfates together with dilute sulfuric acid required 21.9cm^3 of 0.0200 mol dm^-3 potassium manganate (VII) for complete oxidation. A further sample of the same solution was treated with tin (II) chloride solution to potassium manganate (VII) for complete oxidation. Calculate the concentrations of the iron (II) and iron (III) in grams per dm^3.

No idea where to go with this, can someone talk me through it step by step?

Thanks for any help 0
3 years ago
#2
(Original post by Jessinoch)
A 25.0cm^3 sample of a solution containing iron (II) and iron (III) sulfates together with dilute sulfuric acid required 21.9cm^3 of 0.0200 mol dm^-3 potassium manganate (VII) for complete oxidation. A further sample of the same solution was treated with tin (II) chloride solution to potassium manganate (VII) for complete oxidation. Calculate the concentrations of the iron (II) and iron (III) in grams per dm^3.

No idea where to go with this, can someone talk me through it step by step?

Thanks for any help Hi, I'm not a chemistry expert, but I'm going to move your thread over to the chemistry section to help you get more replies! For future reference, here is the subject forum directory.
0
3 years ago
#3
(Original post by Jessinoch)
A further sample of the same solution was treated with tin (II) chloride solution to potassium manganate (VII) for complete oxidation.
Is there a typo in this sentence? I'm having difficulty making sense from this
0
#4
(Original post by Metanoia)
Is there a typo in this sentence? I'm having difficulty making sense from this
Sorry yes it’s supposed to say: a further sample of the same solution was treated with tin (II) chloride solution to reduce the iron (III) to iron (II): this solution then needed 36.6cm^3 of 0.0200 mol dm^-3 potassium manganate for complete oxidation.
0
3 years ago
#5
(Original post by Jessinoch)
A 25.0cm^3 sample of a solution containing iron (II) and iron (III) sulfates together with dilute sulfuric acid required 21.9cm^3 of 0.0200 mol dm^-3 potassium manganate (VII) for complete oxidation. A further sample of the same solution was treated with tin (II) chloride solution to potassium manganate (VII) for complete oxidation. Calculate the concentrations of the iron (II) and iron (III) in grams per dm^3.

No idea where to go with this, can someone talk me through it step by step?

Thanks for any help First step: what’s the balanced half equations for Fe2+ -> Fe3+ and for [MnO4]- -> Mn2+

Second: combine them to get the testing mole ratios
0
#6
First step: what’s the balanced half equations for Fe2+ -> Fe3+ and for [MnO4]- -> Mn2+

Second: combine them to get the testing mole ratios
MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 5Fe3+ + 4H2O
0
3 years ago
#7
(Original post by Jessinoch)
MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 5Fe3+ + 4H2O
Then using the info you’ve provided, you can calculate the number of moles of MnO4- used, and therefore the moles of Fe2+ present in the sample (5x that of MnO4-)
0
#8
Then using the info you’ve provided, you can calculate the number of moles of MnO4- used, and therefore the moles of Fe2+ present in the sample (5x that of MnO4-)
Yep got that
0
3 years ago
#9
I have included the main steps below while leaving out the actual calculations, let me know if you need further clarification after trying them out

(Original post by Jessinoch)
A 25.0cm^3 sample of a solution containing iron (II) and iron (III) sulfates together with dilute sulfuric acid required 21.9cm^3 of 0.0200 mol dm^-3 potassium manganate (VII) for complete oxidation.
The first part of the question should be used to find the moles of Fe2+ present in the mixture, since only the Fe2+ (not Fe3+) will react with MnO4-

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

Step 1) moles of MnO4- required = 0.0219 x 0.02
Step 2) moles of Fe2+ in 1st reaction =
Step 3) concentration of Fe2+ in mixture =

(Original post by Jessinoch)
Sorry yes it’s supposed to say: a further sample of the same solution was treated with tin (II) chloride solution to reduce the iron (III) to iron (II): this solution then needed 36.6cm^3 of 0.0200 mol dm^-3 potassium manganate for complete oxidation.
For the 2nd part, the Fe3+ portion was reduced to Fe2+. In addition to the Fe2+ already present (calculated in step 2), Fe2+ (old and new) was then oxidised by MnO4-.

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

Step 4) moles of MnO4- required = 0.0366 x 0.02
Step 5) moles of Fe2+ in 2nd reaction =
Step 6) moles of Fe3+ originally present = moles of Fe2+ calculated in step 5 - moles of Fe2+ calculated in step 2
Step 7) concentration of Fe3+ =
1
#10
(Original post by Metanoia)
I have included the main steps below while leaving out the actual calculations, let me know if you need further clarification after trying them out

For the 2nd part, the Fe3+ portion was reduced to Fe2+. In addition to the Fe2+ already present (calculated in step 2), Fe2+ (old and new) was then oxidised by MnO4-.

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

Step 4) moles of MnO4- required = 0.0366 x 0.02
Step 5) moles of Fe2+ in 2nd reaction =
Step 6) moles of Fe3+ originally present = moles of Fe2+ calculated in step 5 - moles of Fe2+ calculated in step 2
Step 7) concentration of Fe3+ =
Got you. Thank you!
0
3 years ago
#11
(Original post by Jessinoch)
Yep got that
Then do the exact same thing for the Fe3+ that was converted into Fe2+, then knowing the number of moles of Fe2+ and Fe3+, you can calculate the gdm^-3
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