Question

Electrode potential data given below

$Cl_{2}+2H_{2}O \rightarrow 2ClO^{-}+4H^{+}+2e^{-};E^\circ =-1.61volt$

$ClO^{-}+2H_{2}O \rightarrow ClO_{3}^{ -}+4H^{+}+4e^{-};E^\circ =-0.50volt$

Based on these data which is the spontaneous reaction.

• A. $Cl_{2}+ClO^{-}+ClO_{3}^{ -}$
• B. $ClO^{-} \rightarrow Cl_{2}+ClO_{3}^{ -}$
• C. $ClO_{3}^{ -} \rightarrow Cl_{2}+ClO^{-}$
• D. $ClO^{-}+Cl_{2} \rightarrow ClO_{3}^{ -}$

Answer 1

Answer: (B)

Given electron potential data is-

$Cl_{2}+2H_{2}O \rightarrow 2ClO^{-}+4H^{+}+2e^{-};E^\circ =-1.61$ Volt

$ClO^{-}+2H_{2}O \rightarrow ClO_{3}^{-}+4H^{+}+4e^{-};E^\circ =-0.50$ Volt

For reaction to be spontaneous $E^{0}$ reaction should $+ve$ , Thus substracting equation $1$ from equation $2.$