jazz_xox_
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Iron(II) ethanedioate dihydrate can be analysed by titration using potassium manganate(VII) in acidic solution. In this reaction, manganate(VII) ions oxidise iron(II) ions and ethanedioate ions.A 1.381 g sample of impure FeC2O4.2H2O was dissolved in an excess of dilute sulfuric acid and made up to 250 cm3 of solution.
25.0 cm3 of this solution decolourised 22.35 cm3 of a 0.0193 mol dm–3 solution of potassium manganate(VII).(i) Use the half-equations given below to calculate the reacting ratio of moles of manganate(VII) ions to moles of iron(II) ethanedioate. MnO4 + 8H+ + 5e Image Mn2+ + 4H2O Fe2+ Image Fe3+ + e C2O42– Image 2CO2 + 2e



please could someone help me find the reacting ratio?


I was wanting to write an overall equation, but I could only do this for MnO4- with either Fe2+ or C2O42-? The answer is 3:5


the examiners report says They realised that the reduction of manganate(VII) ions is a 5–electron change reaction. The oxidation of FeC2O4 is a 3–electron change reaction. Therefore, the reacting ratio is 3:5... but I don't understand how you know the oxidation of FeC2O4 is a 3 electron change?

Thanks
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charco
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(Original post by jazz_xox_)
Iron(II) ethanedioate dihydrate can be analysed by titration using potassium manganate(VII) in acidic solution. In this reaction, manganate(VII) ions oxidise iron(II) ions and ethanedioate ions.A 1.381 g sample of impure FeC2O4.2H2O was dissolved in an excess of dilute sulfuric acid and made up to 250 cm3 of solution.
25.0 cm3 of this solution decolourised 22.35 cm3 of a 0.0193 mol dm–3 solution of potassium manganate(VII).(i) Use the half-equations given below to calculate the reacting ratio of moles of manganate(VII) ions to moles of iron(II) ethanedioate. MnO4 + 8H+ + 5e Image Mn2+ + 4H2O


Fe2+ Image Fe3+ + e
C2O42– Image 2CO2 + 2e



please could someone help me find the reacting ratio?


I was wanting to write an overall equation, but I could only do this for MnO4- with either Fe2+ or C2O42-? The answer is 3:5


the examiners report says They realised that the reduction of manganate(VII) ions is a 5–electron change reaction. The oxidation of FeC2O4 is a 3–electron change reaction. Therefore, the reacting ratio is 3:5... but I don't understand how you know the oxidation of FeC2O4 is a 3 electron change?

Thanks
You've shown it in the question (red above).

What is the formula of iron(II) ethandioate?
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jazz_xox_
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(Original post by charco)
You've shown it in the question (red above).

What is the formula of iron(II) ethandioate?
Thank you, got it now
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username4441362
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can someone explain this to me, im still lost about the ratio thingy
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