Sodium reaction with Iodine(1) chloride.What is equation?my answer was 2Na+2ICl》2NaCl+I2.But it is incorrect.Correct answer is 2Na+ICl》NaCl+NaI How?

1 Answer
Feb 25, 2018

Well, sodium is a strong reducing agent....and the halogens are oxidizing agents....and so are the INTERHALOGENS....

Explanation:

#"Reduction half equation:"#

#ICl+2e^(-) rarr I^(-) + Cl^(-)# #(i)#

#"Oxidation half equation:"#

#Nararr Na^(+) + e^(-)# #(ii)#

And so we take #(i)+2xx(ii)#

#2Na+IClrarr2Na^+ +I^(-) + Cl^-#

In your answer you suggested that iodide anion was oxidized...which would be a hard thing to do in these reducing conditions....