Question

Given : $H_2{O}_2\to O_2+2H^{\oplus }+2e^{-}$
$e^{\oplus }=-0.69V$
$H_2{O}_2+2H^{\oplus }+2e^{-}\to 2H_{2}O$
$E^{\oplus }=1.77V$
$2I^{\ominus }\to I_2+2e^{-}$
$E^{\ominus }=-0.535V$
Assertion: $H_2{O}_2$ acts as an oxidizing agent for $I^{\ominus }$.

Reason: $E^{\ominus }$ is positive.

• A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion
• B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
• C. Assertion is true but Reason is false
• D. Assertion is false but Reason is true
• E. Assertion is false but Reason is true
  Both Assertion and Reason are false

Answer 1

Answer: (A)

Both Assertion and Reason are true and Reason is the correct explanation of Assertion

Cathode : $I_2+2e^{-}\to 2I^{\oplus }$
$E^{\ominus }=0.535V\left(red\right)$
Anode : $H_2{O}_2\to O_2+2H^{\oplus }+2e^{-}$
$E^{\oplus }=0.69V\left(red\right)$
As $E^{\ominus }<0$, the cell is not spontaneous.
Hence $H_2{O}_2$ is not a reducing agent.
[$H_2{O}_2$ is an oxidizing agent (if it is reduced)]
cathode : $H_2{O}_2+2H^{\oplus }+2e^{-}\to 2H_{2}O;E^{\ominus }=1.77V\left(red\right)$
Anode : $2I^{\ominus }\to I_2+2e^{-}$;
$E^{\ominus }=+0.535V\left(red\right)$
As $E^{\ominus }>0$, hence cell reaction is spontaneous.
$\therefore H_2{O}_2$ acts as an oxidizing agent for $I^{\ominus }$.