A) \[Na\]and \[B{{r}_{2}}\]
B) \[Na\] and \[{{O}_{2}}\]
C) \[{{H}_{2}},B{{r}_{2}}\] and \[NaOH\]
D) \[{{H}_{2}}\] and \[{{O}_{2}}\]
Correct Answer: C
Solution :
\[NaBr\rightleftharpoons N{{a}^{+}}+B{{r}^{-}}\] \[\left. \begin{align} & 2{{H}_{2}}O+2{{e}^{-}}\xrightarrow{{}}{{H}_{2}}+2O{{H}^{-}} \\ & N{{a}^{+}}+O{{H}^{-}}\xrightarrow{{}}NaOH \\ \end{align} \right]at\,cathode\] \[B{{r}^{-}}-{{e}^{-}}\xrightarrow{{}}Br\] \[Br+Br\xrightarrow{{}}B{{r}_{2}}\] (at anode) So, the products are \[{{H}_{2}}\], \[B{{r}_{2}}\] and \[NaOH\]. \[{{H}_{2}}\] at cathode, \[B{{r}_{2}}\] at anode and \[NaOH\] in solution.
You need to login to perform this action.
You will be redirected in
3 sec
