Question

Given : $H_2{O}_2\to O_2+2H^{\oplus }+2e^{-}$ $E^{\ominus }=-0.69V$
$H_2{O}_2+2H^{\oplus }+2e^{-}\to 2H_{2}O$
$E^{\ominus }=1.77V$
$I^{\ominus }\to +2e^{-}$
$E^{\ominus }=-0.535V$
Which of the following statements is/(are) correct ?

• A. $H_2{O}_2$ behaves as an oxidant for $I_2/I^{\ominus }$.
• B. $H_2{O}_2$ behaves as an reductant for $I_2/I^{\ominus }$.
• C. $I^{\ominus }/I_2$ behaves as an reductant for $H_2{O}_2$.
• D. None of these is correct.

Answer 1

Answer: (A), (C)

The correct options are
A $H_2{O}_2$ behaves as an oxidant for $I_2/I^{\ominus }$.
C $I^{\ominus }/I_2$ behaves as an reductant for $H_2{O}_2$.

$H_2{O}_2$ behaves as an oxidant for $I_2/I^{-}$.
$I^{-}/I_2$ behaves as an reductant for $H_2{O}_2$.
At anode :
$2I^{\ominus }\to I_2+2e^{-}$
$E_{a\left(oxid\right)}^{\ominus }=-0.535\Rightarrow E_{a\left(red\right)}^{\ominus }=+0.535V$
At cathode:
$H_2{O}_2+2H^{\oplus }+2e^{-}\to 2H_{2}O$
$E_c^{\ominus }=1.77$
$E_{cell}^{\ominus }=(E_c^{\ominus }-E_a^{\ominus }{)}_R=1.77-0.53=+ve$
Note that :
Anode :
$H_2{O}_2\to O_2+2H^{\oplus }+2e^{-}$
$E_{a\left(oxid\right)}^{\ominus }=-0.69V$
Cathode :
$I_2+2e^{-}\to 2I$
$E_{a\left(red\right)}^{\ominus }=+0.69V$
$E^{\ominus }=E_c^{\ominus }-E_a^{\ominus }=0.535-0.69=-ve$