Question

Which of the following redox couples is a stronger oxidizing agent?
A. $C{l_2}({E^0} = 1.36V)$ and $B{r_2}({E^0} = 1.09V)$
B. $Mn{O_4}^ - ({E^0} = 1.51V)$ and $C{r_2}{O_7}^{2 - }({E^0} = 1.33V)$

Answer 1

Hint: Higher the positive${E^0}$ value stronger the oxidizing agent. The substance having the highest positive ${E^0}$ potential will always get reduced and will undergo reduction reaction and will be a stronger oxidizing agent.

Complete step by step answer:
A. As we have given a redox couple:
$C{l_2}({E^0} = 1.36V)$ and $B{r_2}({E^0} = 1.09V)$
The substance having higher positive ${E^0}$potential value will get easily reduced and will be a stronger oxidizing agent.
Here, $C{l_2}$ has a higher positive ${E^0}$ potential value as compared to $B{r_2}$ that means it will get easily reduced than $B{r_2}$. So, $C{l_2}$ is a stronger oxidizing agent than $B{r_2}$.
Hence, In the redox couple $C{l_2}({E^0} = 1.36V)$ is a stronger oxidizing agent than $B{r_2}({E^0} = 1.09V)$.
B. As we have given a redox couple:
$Mn{O_4}^ - ({E^0} = 1.51V)$ and $C{r_2}{O_7}^{2 - }({E^0} = 1.33V)$
As the substance having higher positive ${E^0}$ potential value will get easily reduced and will be a stronger oxidizing agent.
Here, $Mn{O_4}^ - $ has higher positive ${E^0}$ potential value as compared to $C{r_2}{O_7}^{2 - }$ that means it will get easily reduced than $C{r_2}{O_7}^{2 - }$. So, $Mn{O_4}^ - $ is a stronger oxidizing agent as compared to $C{r_2}{O_7}^{2 - }$.
Hence, In the redox couple $Mn{O_4}({E^0} = 1.51V)$is stronger oxidizing agent that $C{r_2}{O_7}^{2 - }({E^0} = 1.33V)$.

Additional Information:
A substance that can cause combustion of other materials are called the dangerous goods of an oxidizing agent.
Oxidizing agents are the substances that oxidize other substances and reduce themselves. In simple words oxidizing agents accept electrons and get reduced by oxidizing others.
Some common oxidizing agents are oxygen, ozone, fluorine, chlorine, bromine, iodine, hypochlorite, chlorate, nitric acid and so on.

Note:
Higher the positive value of ${E^0}_{(oxidation)}$ then the compound will get easily oxidized and will be a stronger reducing agent.
And greater the negative value of ${E^0}_{(reduction)}$ then the compound will get easily reduced and will be a stronger oxidizing agent.