Question

The products formed when an aqueous solution of $NaBr$ electrolysed in a cell having inert electrodes are:
A. $Na$ and $B{r_2}$
B. $Na$ and ${O_2}$
C. ${H_2},B{r_2}$ and $NaOH$
D. ${H_2}$ and ${O_2}$

Answer 1

The aqueous solution of $NaBr$ is subjected to electrolysis. Hence the first thing that needs to be checked is the position of $Na$ in the activity series. The inert electrodes are placed and hence look for the positive ions which move towards the cathode and negative ions which move towards the anode.

Complete step by step answer:
At first the aqueous solution is added with $NaBr$ in the electrophoresis process. Hence there is the presence of $NaBr$ and ${H_2}O$ in the electrophoresis apparatus.
Under the action of current both the chemical elements are converted to their ionic state. Hence the resultant ions will be:
$NaBr \rightleftharpoons N{a^ + } + B{r^ - }$
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$
Here the two positive ions are produced which are $N{a^ + }$ and ${H^ + }$ .
${H^ + }$ has a lower discharge potential as compared to that of $N{a^ + }$ according to the trends of activity series, which makes the hydrogen ion move towards the cathode. At the cathode the reaction that takes place: $2{H^ + } + 2{e^ - } \to {H_2}$
Hence ${H_2}$ is released at the cathode.
There are two negative ions produced which are $B{r^ - }$ and $O{H^ - }$.
The discharge potential is lower in $B{r^ - }$ here as compared to that of the $O{H^ - }$ ion. The $B{r^ - }$ ion moves towards the anode and hence the reaction taking place at the anode is: $2B{r^ - } - 2e \to B{r_2}$
Hence the $B{r_2}$ is released at the anode.
Therefore, the residual ions form a compound in the electrolysis solution. The reaction taking place here will be: $N{a^ + } + O{H^ - } \to NaOH$
Therefore $NaOH$ is produced inside the solution. Along with it $B{r_2}$ is produced at the anode and ${H_2}$ is released at the cathode. The correct answer is option “C” .

Note: In the process of electrolysis always focus on the activity series first. The elements involved have the trend of moving towards the electrode according to the nature of activity series. Here as it is visible that discharge potential plays an important role in the process. Therefore, using the ions moving towards cathode or anode can be determined.