Standard potentials for chlorine species in 1 M acid (V): ClO3^-/HClO2 = 1.21, HClO2/HOCl = 1.65, HOCl/Cl2 = 1.61, Cl2/Cl- = 1.36. Which statements are correct?

HClO2 is thermodynamically unstable with respect to disproportionation into ClO3^- and HOCl

Cl2 is thermodynamically unstable in 1 M acid and disproportionates to HOCl and Cl-

HClO2 is thermodynamically unstable with respect to disproportionation into ClO3^- and HOCl

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ec077

Standard potentials for chlorine species in $1 M$ acid (V): $ClO X_{3} X^{-} / HClO X_{2} = 1.21$ , $HClO X_{2} / HOCl = 1.65$ , $HOCl / Cl X_{2} = 1.61$ , $Cl X_{2} / Cl X^{-} = 1.36$ . Which statements are correct?

A $E^{\circ} (HOCl / Cl X^{-}) = 1.485 V$
B $Cl X_{2}$ is thermodynamically unstable in $1 M$ acid and disproportionates to HOCl and $Cl X^{-}$
C $HClO X_{2}$ is thermodynamically unstable with respect to disproportionation into $ClO X_{3} X^{-}$ and HOCl
D $E^{\circ} (ClO X_{3} X^{-} / HOCl) = 1.43 V$

HOCl + H X^{+} + e X^{-} \frac{1}{2} Cl X_{2} + H X_{2} O

(

1.61 V

, 1 e) plus

\frac{1}{2} Cl X_{2} + e X^{-} Cl X^{-}

(

1.36 V

, 1 e) gives a 2-electron couple with

E^{\circ} = (1.61 + 1.36) /2 = 1.485 V

. B: for

Cl X_{2}

, the potential to the right (

1.36

, reduction) is less than that to the left (

1.61

, oxidation to HOCl), so

E_{disp}^{\circ} = 1.36 - 1.61 < 0

;

Cl X_{2}

is stable in acid (it disproportionates only in base). C: for

HClO X_{2}

1.65 > 1.21

, so

E_{disp}^{\circ} = 1.65 - 1.21 = + 0.44 V

and it disproportionates. D:

ClO X_{3} X^{-} HClO X_{2}

(2 e, 1.21) and

HClO X_{2} HOCl

(2 e, 1.65) combine to a 4-electron couple:

E^{\circ} = (2 \times 1.21 + 2 \times 1.65) /4 = 1.43 V