Question

In cell representation, oxidation half-cell is represented on the:
a.) Left-hand side
b.) Right-hand side
c.) Salt bridge
d.) None of the above

Answer 1

$Zn\left( s \right) + C{u^{ + 2}}\left( {aq} \right)\xrightarrow{{}}Z{n^ + }\left( {aq} \right) + Cu\left( s \right)$
– Consider the following reactions -$Zn\left( s \right) + C{u^{ + 2}}\left( {aq} \right)\xrightarrow{{}}Z{n^ + }\left( {aq} \right) + Cu\left( s \right)$ . The electrochemical cell representation of this cell will be $Zn\left( s \right)|Z{n^{ + 2}}\left( {aq} \right)||C{u^{ + 2}}\left( {aq} \right)|Cu\left( s \right)$ . Since the reaction occurring in the oxidation half-cell is $Zn\left( s \right)\xrightarrow{{}}Z{n^{ + 2}}\left( {aq} \right) + 2{e^ - }$ , we can see that the oxidation half-cell is represented on the left-hand side of the electrochemical cell representation.

Complete step by step answer:
In this problem, we have to deal with electrochemical cell representation, so let’s learn about electrochemical cell representation first.
Electrochemical (also called voltaic or galvanic) cell is made up of two half cells. In one half-cell, the oxidation reaction takes place and in one half-cell reduction reaction takes place. Both the half-cells are connected by a salt bridge (usually made of a strong electrolyte such as $AgN{O_3}$ , $KCl$ ) and by connecting electrical wires.
You can also say that the electrochemical cell is made up of an anode and a cathode. The half-cell in which oxidation reaction occurs acts as the anode and the half-cell in which reduction reaction occurs acts as the cathode.
Let’s consider the following electrochemical cell
Overall reaction - $Zn\left( s \right) + C{u^{ + 2}}\left( {aq} \right)\xrightarrow{{}}Z{n^ + }\left( {aq} \right) + Cu\left( s \right)$
Oxidation reaction in one half-cell (anode) - $Zn\left( s \right)\xrightarrow{{}}Z{n^{ + 2}}\left( {aq} \right) + 2{e^ - }$
Reduction reaction in one half-cell (cathode) - $C{u^{ + 2}}\left( {aq} \right) + 2{e^ - }\xrightarrow{{}}Cu\left( s \right)$
We can see in the above reactions that the two free electrons that are formed in the oxidation reaction are utilized in the reduction reaction, this flow of electrons results in the flow of current in the electrochemical cell.
The electrochemical cell representation for the following example is
$Zn\left( s \right)|Z{n^{ + 2}}\left( {aq} \right)||C{u^{ + 2}}\left( {aq} \right)|Cu\left( s \right)$

Here, $Zn\left( s \right)$ represents the anode electrode
$Z{n^{ + 2}}\left( {aq} \right)$ represents the anode electrolyte
$C{u^{ + 2}}\left( {aq} \right)$ represents cathode electrolyte
$Cu\left( s \right)$ represents cathode electrode
And the symbol $||$ represents the salt bridge
As we can see that the oxidation half-cell (anode) is represented by the left-hand side of the electrochemical cell representation. So, the correct answer is “Option A”.

Note: In the solution above we discussed how the electrons move from the anode to the cathode. This movement of electrons (hence current) takes place because of a difference in the electric potential (called cell potential) of the anode and the cathode, i.e. anode has a higher potential than the cathode.