Question

Which of the following is the strongest oxidising agent?
A.$C{l^ - }$
B.$M{n^{2 + }}$
C.$Mn{O_4}^ -$
D.$C{r^{3 + }}$

Answer 1

Hint:
An electrochemical series is the arrangement of various electrode systems in the increasing order of their individual standard reduction potentials. Standard reduction potential is the electrode potential at $298K$the concentration of the ions used in the assembly where the reduction process is taking place is $1mol/L$and the pressure considered is $1atm$.

Step-by-step explanation:
Step 1:
The important features of electrochemical series are:
The electrode system having positive values acts as a cathode whereas the electrode system having a negative value acts as an anode.
On moving down, the series the tendency to gain electrons i.e. the tendency for reduction to occur increases gradually.
On moving upward in the series, the tendency to lose electrons i.e. the tendency for oxidation to occur increases gradually.
Step 2:
The oxidising power of the oxidised form increases and the reducing power of the reduced form decreases, in moving down from the top to bottom in the series.
Step 3:
The standard reduction potential of the following ions is as follows:
${E^0}_{Cl/C{l^ - }} = 1.36V$
${E^0}_{M{n^{ + 2}}/Mn{O_4}^ - } = - 1.51V$
${E^0}_{Mn{O_4}^ - /M{n^{2 + }}} = 1.51V$
${E^0}_{C{r^{3 + }}/Cr} = - 0.74V$
Step 4:
We know that the oxidising power increases as the positive value for the standard reduction potential increases of the electrode system. Thus, $Mn{O_4}^ -$ having the highest value of standard reduction potential of $1.51V$ is the strongest oxidising agent in the given ions.

Hence, $Mn{O_4}^ -$ having the highest value of standard reduction potential of $1.51V$ is the strongest oxidising agent in the given ions.

Note: The strongest oxidising agent is one which is reduced most easily.They accept electrons easily.The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential.