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

Asked by Pragya Singh | 1 year ago |  72

1 Answer

Solution :-

F2​ can oxidize \( Cl^{ – } \;to\; Cl_{ 2 }, Br^{ – }\; to\; Br_{ 2 }​,\; and\; I^{ – } to\; I_{ 2 }\) as:

But, \( Cl_{ 2 }, Br_{ 2 }​,\; and\; I_{ 2 }\)​ cannot oxidize \( F^{ – }\; to\; F_{ 2 }\)​. 

The oxidizing power of halogens increases in the order as given below:

\( I_{ 2 }​< Br_{ 2 }< Cl_{ 2 }​

Therefore, fluorine is the best oxidant among halogens.

HI and HBr can reduce \( H_{ 2 }SO_{ 4 }\)​ to \( SO_{ 2 }\), but HCI and HF cannot. 

Hence, HI and HBr are stronger reductants compared to HCl and HF.

\( H_2 SO_4\) can be converted to SO  using HI and HBr , but not with HCl or HF. HI and
HBr are thus more effective reductants than HCl and HF

Therefore, hydrochloric acid is the best reductant among hydrohalic compounds.

Hence, the reducing power of hydrohalic acids increases as given below:

HF< HCl< HBr

Answered by Abhisek | 1 year ago

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