# Need help with a redox calculationWatch

Thread starter 3 weeks ago
#1
Recently got this question and I have no clue what to do, can anyone offer any help?Calculate the masses of anhydrous Iron(II) and Iron(III) sulphate present in 1dm3 Of solution from the following results:25cm3 of the solution was oxidised by 23.0cm3 of 0.02mol dm-3 KMnO4After reduction by zinc and dilute sulphuric acid, 25.00cm3 of the solution required 40.0cm3 of the same KMnO4 solution.(Fe = 55.8, S=32.1, O=16.0)(Use MnO4^- 8H 5e- -> Mn^2 4H2O)
0
3 weeks ago
#2
in the equation given, what reacts with manganate in the titration?
how isthe second reaction different to the first
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0
3 weeks ago
#3
(Original post by johnclifford)
Recently got this question and I have no clue what to do, can anyone offer any help?Calculate the masses of anhydrous Iron(II) and Iron(III) sulphate present in 1dm3 Of solution from the following results:25cm3 of the solution was oxidised by 23.0cm3 of 0.02mol dm-3 KMnO4After reduction by zinc and dilute sulphuric acid, 25.00cm3 of the solution required 40.0cm3 of the same KMnO4 solution.(Fe = 55.8, S=32.1, O=16.0)(Use MnO4^- 8H 5e- -> Mn^2 4H2O)
I will give you a clue.

1) Fe(II) is a reducing agent and would react with an oxidising agent to form Fe(III). The oxidising agent used in the question is potassium manganate(VII). Volume required = 23.0 cm³

2) Zinc is a reducing agent and would react with the oxidising agent, Fe(III) to form Fe(II). In other words, after the reduction by zinc (sulfuric acid is just a medium and is required when there is an oxidising agent such as Fe³⁺ present), there are now more Fe(II) ions present in the original sample e.g. the original Fe(II) present plus those that were formed after the reaction with zinc. Hence, the volume required to react with Fe(II) ions = 40.0 cm³ (an increase of 40-23 = 17cm³)

With the above information, the question can then be solved. Hope it helps. Cheers.
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