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← IIT-JEE 2007 · Paper 2
ChemistrySingle correct+4 / -1Question 38 of 66

IIT-JEE 2007 · Paper 2, Chemistry question 38

Passage

Paragraph for Question Nos. 36 to 38

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E∘E^\circ) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their E∘E^\circ (V with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 36-38.

Half-cell reaction E∘E^\circ (V)
IX2+2 eX−→2 IX−\ce{I2 + 2e- -> 2I-} 0.54
ClX2+2 eX−→2 ClX−\ce{Cl2 + 2e- -> 2Cl-} 1.36
MnX3++eX−→MnX2+\ce{Mn^{3+} + e- -> Mn^{2+}} 1.50
FeX3++eX−→FeX2+\ce{Fe^{3+} + e- -> Fe^{2+}} 0.77
OX2+4 HX++4 eX−→2 HX2O\ce{O2 + 4H+ + 4e- -> 2H2O} 1.23

(The paragraph is not printed in the official archive copy (which reproduces only the questions with their answers); its text is restored from the JEE 2007 question paper.)

Sodium fusion extract, obtained from aniline, on treatment with iron(II) sulphate and HX2SOX4\ce{H2SO4} in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of

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Answer and worked solution

Official answer: A

Lassaigne's test. The fusion converts the nitrogen and carbon of aniline into NaCN, which forms [Fe(CN)X6]X4−\ce{[Fe(CN)6]^{4-}} with FeX2+\ce{Fe^{2+}}. Air (or FeX3+\ce{Fe^{3+}}) oxidises some iron to FeX3+\ce{Fe^{3+}}, which gives Prussian blue, iron(III) hexacyanoferrate(II):
4 FeX3++3 [Fe(CN)X6]X4−→FeX4[Fe(CN)X6]X3.\ce{4Fe^{3+} + 3[Fe(CN)6]^{4-} -> Fe4[Fe(CN)6]3}.

(A)