Question

Which of the following will be possible in electrochemical cell obtained from ${E^o}_{C{l_2}l2C{l^ - }} = 1.36V$ and ${E^o}_{B{r_2}l2B{r^ - }} = 1.09V$
A.$B{r_2} + 2C{l^ - } \to 2B{r^ - } + C{l_2}$
B.$2B{r^ - } + C{l_2} \to B{r_2} + 2C{l^ - }$
C.$2C{l^ - } + 2B{r^ - } \to C{l_2} + B{r_2}$
D.$C{l_2} + B{r_2} \to 2C{l^ - } + 2B{r^ - }$

Answer 1

We must have to know that the electrochemical cell is an apparatus which is used to bring out the electricity by using a chemical reaction which is conducted in that cell. And here, the chemical energy is converted to electrical energy and also electrical energy is converted back into chemical energy. An electrochemical cell contains cathode and anode and the reduction reaction takes place in cathode and oxidation reaction takes place in anode.

Complete answer:
Here, $B{r_2} + 2C{l^ - } \to 2B{r^ - } + C{l_2}$ Reaction does not take place in this electrochemical cell. Because, the ionized bromine is reacted with chlorine atoms. Hence, option (A) is incorrect.
In this reaction, both oxidation and reduction reaction takes place. The oxidation reaction takes place in anode and the number of oxidation states increases. And the reduction reaction occurs in the cathode and there is a decrease in the number of oxidation states.
Consider the cell reaction which is takes place in the electrochemical cell, and that is,
$2NaB{r_{(aq)}} + C{l_{2(g)}} \to 2NaC{l_{(aq)}} + B{r_{2(g)}}$
Let’s see the ionic form of the reaction,
$2N{a^ + }_{(aq)} + 2B{r^ - }_{(aq)} + C{l_{2(g)}} \to 2N{a^ + }_{(aq)} + 2Cl_{(aq)}^ - + B{r_{2(aq)}}$
Here, the sodium ions terms will cancels each other and the net equation can be written as,
$2B{r^ - }_{(aq)} + C{l_{2(g)}} \to 2Cl_{(aq)}^ - + B{r_{2(aq)}}$
The half- cell reaction takes place at the anode and the oxidation state of bromine increased here, the reaction can be written as,
$2B{r^ - }_{(aq)} \to B{r_{2(g)}} + 2{e^ - }$
The half- cell reaction takes place at the cathode and the oxidation state of chlorine decreased here, the reaction can be written as,
$C{l_{2(g)}} + 2{e^ - } \to 2Cl_{(aq)}^ -$
Therefore, the possible in electrochemical cell obtained from ${E^o}_{C{l_2}l2C{l^ - }} = 1.36V$and ${E^o}_{B{r_2}l2B{r^ - }} = 1.09V$ is equal to $2B{r^ - }_{(aq)} + C{l_{2(g)}} \to 2Cl_{(aq)}^ - + B{r_{2(aq)}}$. Hence, option (B) is correct.
Here, $2C{l^ - } + 2B{r^ - } \to C{l_2} + B{r_2}$ reaction is not possible in an electrochemical cell. Hence, option (C) is incorrect.
The oxidation state of bromine is increasing in the oxidation reaction. Therefore,$C{l_2} + B{r_2} \to 2C{l^ - } + 2B{r^ - }$ reaction is not possible in an electrochemical cell. Hence, the option (D) is incorrect.

Note:
As we know, the electrochemical cell contains both cathode and anode. The cathode is denoted as a positive sign and the anode is denoted as negative sign. In anode, oxidation reaction occurs and in cathode, reduction reaction occurs. Hence, in the case of cathode, the electrons are moved towards cathode and in anode, the electrons are moved towards anode. The electrochemical cell is used for the preparation of high purity $Zn$, $Al$ etc.