SELECT * FROM question_mgmt as q WHERE id=3252 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=49 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Answer:

F2 can oxidize Cl- to Cl2, Br-  to Br2, and I- to I2 as:

F2(aq) + S2cl-(s)  →  2F-(aq)  +  Cl (g)

F2(aq) + 2Br-(aq)  →  2F-(aq)  + Br 2(l)

F2(aq) + 2l-(aq)  →  2F-(aq)  + 12(s)

 

On the other hand, Cl2, Br2, and I2 cannot oxidize F- to F2. The oxidizing power of halogens increases in the order of I2 < Br2 < Cl2 < F2. Hence, fluorine is the best oxidant.

HI and HBr can reduce H2SO4 to SO2, but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF.

2HI + H2SO4  →  l2  +  SO2  +  2H2O

2HBr + H2SO4  →  Br2 + SO2  +  2H2O

Again, I- can reduce Cu2+ to Cu+, but Br- cannot.

4l-(aq)  +   2Cu2+(aq)  →  Cu2l2(s)  + l2(aq)

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.

Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.


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