Q45 / 120 · NEET 2012 (Mains) · Chemistry

Standard reduction potentials of the half reactions are given below $$ \begin{array}{ll} \mathrm{F}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{~F}^{-}(a q) ; & E^{\circ}=+2.85 \mathrm{~V} \\ \mathrm{Cl}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q) ; & E^{\circ}=+1.36 \mathrm{~V} \\ \mathrm{Br}_2(l)+2 e^{-} \longrightarrow 2 \mathrm{Br}^{-}(a q) ; & E^{\circ}=+1.06 \mathrm{~V} \\ \mathrm{I}_2(s)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) ; & E^{\circ}=+0.53 \mathrm{~V} \end{array} $$ The strongest oxidising and reducing agents respectively are

  • Correct answer
    A
    $\mathrm{F}_2$ and $\mathrm{I}^{-}$
  • B
    $\mathrm{Br}_2$ and $\mathrm{Cl}^{-}$
  • C
    $\mathrm{Cl}_2$ and $\mathrm{Br}^{-}$
  • D
    $\mathrm{Cl}_2$ and $\mathrm{I}_2$

Solution

Higher the value of standard reduction potential, stronger will be the oxidising agent. Therefore, $\mathrm{F}_2$ will act as strongest oxidising agent. Similarly lower the value of standard reduction potential, stronger will be the reducing agent. Therefore, $\mathrm{I}^{-}$will act as strongest reducing agent.

Hence, the correct answer is option: 'A'.

Relevant Questions