E deg values (V): Fe^3+/Fe^2+ = 0.77, I2/I- = 0.54, Br2/Br- = 1.09, Cl2/Cl- = 1.36. An acidic Fe^3+ solution is added to a solution containing KI, KBr and KCl (all 1 M). Which halide ion(s) is/are oxidised?

JEE Main levelSingle correct optionDifficulty 2 of 3ec041

$E^{\circ}$ values (V): $Fe X^{3 +} / Fe X^{2 +} = 0.77$ , $I X_{2} / I X^{-} = 0.54$ , $Br X_{2} / Br X^{-} = 1.09$ , $Cl X_{2} / Cl X^{-} = 1.36$ . An acidic $Fe X^{3 +}$ solution is added to a solution containing KI, KBr and KCl (all $1 M$ ). Which halide ion(s) is/are oxidised?

Fe X^{3 +}

can oxidise a halide only if

E^{\circ} (Fe X^{3 +} / Fe X^{2 +})

exceeds

E^{\circ}

of the halogen couple, so that

E_{cell}^{\circ} > 0

. For

I X^{-}

0.77 - 0.54 = + 0.23 V

, spontaneous. For

Br X^{-}

0.77 - 1.09 = - 0.32 V

and for

Cl X^{-}

0.77 - 1.36 = - 0.59 V

, both non-spontaneous. Only iodide is oxidised, giving

I X_{2}