Ncert Exercises & Solutions

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

F_{2}

can oxidize

{Cl}^{-}

{Cl}_{2}, {Br}^{-}

{Br}_{2}

, and

l^{-}

l_{2}

as:

F_{2 (aq)} + 2 C^{-} l_{(s)} \to 2 {F^{-}}_{(aq)} + {Cl}_{(g)}

F_{2 (aq)} + 2 {Br}^{-}_{(aq)} \to 2 {F^{-}}_{(aq)} + {Br}_{2 (l)}

F_{2 (aq)} + 2 {l^{-}}_{(aq)} \to 2 {F^{-}}_{(aq)} + l_{2 (g)}

On the other hand,

{Cl}_{2}, {Br}_{2}, and l_{2}

cannot oxidize

F^{-} to F_{2}

The oxidizing power of halogens increases in the order of

$l_{2} < {Br}_{2} < {Cl}_{2} < F_{2}$ .

Hence, fluorine is the best oxidant among halogens.

Hl and HBr can reduce $H_{2} {SO}_{4} to {SO}_{2}$ , but HCl and HF cannot

Therefore, Hl and HBr are stronger reductants than HCl and HF

$2 Hl + H_{2} {SO}_{4} \to l_{2} {SO}_{2} + 2 H_{2} O$
$2 HBr + H_{2} {SO}_{4} \to {Br}_{2} + {SO}_{2} + 2 H_{2} O$

Again, $l^{-}$ can reduce ${Cu}^{2 +}$ to ${Cu}^{+}$ , but ${Br}^{-}$ cannot.

$4 {l^{-}}_{(aq)} + 2 {Cu}^{2 +}_{(aq)} \to {Cu}_{2} l_{2 (g)} + l_{2 (aq)}$

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

Thus, reducing the power of hydrohalic acid increases in the order of HF<HCl<HBr<Hl.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions