A) \[Na\,\,and\,\,B{{r}_{2}}\]
B) \[Na\,\,and\,\,{{O}_{2}}\]
C) \[{{H}_{2}},\,B{{r}_{2}},\,\,NaOH\]
D) \[{{H}_{2}}\,and\,\,{{O}_{2}}\]
Correct Answer: C
Solution :
[c] \[~{{H}_{2}}\text{, }B{{r}_{2}},\text{ }NaOH\] \[NaBr\text{ }\xrightarrow{iorises\text{ }}N{{a}^{+}}(aq.)+\text{ }B{{r}^{-}}(aq.)\] At cathode: \[2{{H}_{2}}O\,(l)+2e-\to \text{ }{{H}_{2}}\text{(g)}+20{{H}^{-}}(aq.)\] Since the electrode potential of \[{{H}_{2}}O\,(-\,0.83V)\]is higher than that of\[N{{a}^{+}}(-2.71)\], therefore \[{{H}_{2}}O\] is reduced in preference to \[N{{a}^{+}}\text{ }ion\]. At anode: \[2B{{r}^{-}}(aq.)\to \,B{{r}_{2}}(g)+2{{e}^{-}}\] Thus, \[{{H}_{2}}\] is liberated at the cathodes, \[B{{r}_{2}}\]at the anode and NaOH is left in the solution.
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