Question

Given :

Oxidation :

$H_2{O}_2\to O_2+2H^{\oplus }+2e^{-}$

$E^{\ominus }=-0.69V$;

$2F^{\ominus }\to F_2+2e^{-}$

$E^{\ominus }=-2.87V$;

Reduction :

$H_2{O}_2+2H^{\oplus }+2e^{-}\to 2H_{2}O$

$E^{\ominus }=-1.77V$;

$2I^{\ominus }\to I_2+2e^{-}$ $E^{\ominus }=-0.54V$;

Which of the following statements is/are correct ?

• A. $H_2{O}_2$ behaves as an oxidant for $I^{\ominus }$.
• B. $H_2{O}_2$ behaves as a reductant for $I_2$.
• C. $H_2{O}_2$ behaves as an oxidant for $F^{\ominus }$.
• D. $H_2{O}_2$ behaves as a reductant for $F_2$.

Answer 1

Answer: (A), (D)

$H_2{O}_2$ behaves as an oxidant for $I^{\ominus }$.
$H_2{O}_2$ behaves as a reductant for $F_2$.

Create a cell with required equation (cell reaction) and find its $E_{cell}^{\ominus }$.

$\left(A\right):H_2{O}_2+I^{\ominus }\to I_2+H_{2}O$ $E_{cell}^{\ominus }=\left(1.77\right)-\left(0.54\right)>0$

$\left(B\right):H_2{O}_2+I_2\to I^{\ominus }+O_2$ $E_{cell}^{\ominus }=\left(0.54\right)-\left(0.69\right)<0$

(Not possible to occur spontaneously) And so on.

$\left(C\right)H_2{O}_2+F^{\ominus }\to F_2+H_{2}O$ $E_{cell}^{\ominus }=\left(1.77\right)-\left(2.87\right)<0$

$\left(D\right):H_2{O}_2+F_2\to F^{\ominus }+O_2$ $E_{cell}^{\ominus }=\left(2.87\right)-\left(0.69\right)>0$

The reaction for which $E_{cell}>0$ will have $\Delta G<0$. So, options A and D are correct.