A) \[{{I}_{2}}+2B{{r}^{-}}\to B{{r}_{2}}+2{{I}^{-}}\]
B) \[B{{r}_{2}}+2{{I}^{-}}\to 2B{{r}^{-}}+{{I}_{2}}\]
C) \[{{I}_{2}}+2B{{r}^{-}}\to IB{{r}_{2}}+{{I}^{-}}\]
D) \[{{I}_{2}}+B{{r}_{2}} & \to 2IBr\]
Correct Answer: B
Solution :
This reaction will be occurred (that cell reaction is possible in which cathode is of higher reduction potential). \[B{{r}_{2}}+2{{I}^{-}}\xrightarrow{{}}2B{{r}^{-}}+{{I}_{2}}\] because,\[E_{Cell}^{o}=\text{positive}\] \[=E_{B{{r}_{2}}/B{{r}^{-}}}^{o}+E_{{{I}^{-}}/{{I}_{2}}}^{o}\] \[=1.09-0.54=0.55\]You need to login to perform this action.
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