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


F2 can oxidise Cl- to Cl2, Br- to Br2and I- to I2 as:


F2(aq) + 2Cl-(s) 2F-(aq) + Cl2(g)


F2(aq) + 2Br-(aq) 2F-(aq) + Br2(l)


F2(aq) + 2I-(aq) 2F-(aq) + I2(s)


On the other hand, Cl2, Br2and I2 cannot oxidise F- to F2. The oxidising power of halogens increases in the order of I2˂Br2˂Cl2˂F2. Hence, fluorine is the best oxidant among halogens. 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 I2 + SO2 + 2H2O


2HBr + H2SO4 Br2 + SO2 + 2H2O


Again, I- can reduce Cu + 2 to Cu+, but Br-cannot.


4I-(aq) + 2Cu2+(aq) Cu2I2(s) + I2(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.


15
1