Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 15: Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic comp....
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