Question

Given,
$E_{C{r^{3 + }}/Cr}^0$= - 0.74 V ; $E_{MnO_4^ - /M{n^{2 + }}}^0$= 1.51 V
$E_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^0$= 1.33 V ; $E_{Cl/C{l^ - }}^0$= 1.36 V
Based on the data given above, the strongest oxidising agent will be :
a.) $C{r^{3 + }}$
b.) $M{n^{2 + }}$
c.) $MnO_4^ - $
d.) $C{l^ - }$

Answer 1

Hint: The oxidising agent is one that helps in oxidation of other species. This means it will help in removal of electrons of other species and itself get reduced by accepting these electrons.
The Standard reduction potential values given above tell the measure of species to get reduced. Higher the SRP, better will be the oxidising agent.

Complete step by step answer:
First, we should understand what an oxidising agent is.
The oxidising agent is the one that helps any other species to get reduced by helping it in removal of electrons in redox reaction. The reagent takes the electrons and thus let others oxidise and itself get reduced.
The Standard reduction potential is the measure of tendency of a chemical species to get reduced. The SRP values help us in measuring oxidising and reducing agents.
The species which can easily donate the electrons, it can be easily oxidised. The species with high Standard reduction potential will be the stronger oxidising agent.
If we see the data given to us, we see that the reduction potential of $MnO_4^ - /M{n^{2 + }}$ has the high value. This means it can easily get reduced and accept electrons. Thus, it can easily oxidise the other species. So, it will be the best oxidising agent.
So, the correct answer is “Option C”.

Note: It must be noted that we say the answer is option c.) and not option b.) while both of these have Manganese atoms. This is because in option b.) we have $M{n^{2 + }}$. The Mn is in +2 state already. It can accept two electrons to reach the neutral state. But in c.) $MnO_4^ - $. The Mn is in +7 oxidation state. It can accept more electrons. So, it will be a better oxidising agent.