Question

The strongest oxidising agent amongst $F_2,C{l}_2,B{r}_2$ and $I_2$ is

• A. $F_2$
• B. $I_2$
• C. $C{l}_2$
• D. $B{r}_2$

Answer 1

Answer: (A)

Since all the halogens have strong tendency to accept electrons, they act as strong oxidising agent.
$F_2+2e^{-}\to 2F^{-};E^o=+2.87V$
$C{l}_2+2e^{-}\to 2C{l}^{-};E^o=+1.36V$
$B{r}_2+2e^{-}\to 2B{r}^{-};E^o=+1.09V$
$I_2+2e^{-}\to 2I^{-};E^o=+0.54V$
The electrode potential of $F$; is the maximum while that of $I_2$ is the minimum.
So, $F_2$ is the strongest oxidising agent while $I_2$ is the weakest oxidising agent.