How do you balance lead (iv) nitrate reacting with a solution of sodium iodide?

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Answer 1 · 2018-05-12T14:41:10

Well, you are combining an oxidant, i.e. #Pb^(4+)# with a potential reductant....#I^(-)#

And so a proposed sequence....

#Pb^(4+) + 2e^(-)rarr Pb^(2+) # #(i)#

#I^(-) rarr 1/2I_2(s) + e^(-)# #(ii)#

And so we could propose #(i)+2xx(ii)#

#Pb^(4+) +2I^(-)+ 2e^(-)rarr Pb^(2+) +I_2(s)darr + 2e^(-)#

And for completeness, we could add #2xxI^-# to each side....

#Pb^(4+) +4I^(-) rarr underbrace(PbI_2(s)darr)_"yellow salt" +underbrace(I_2(s)darr)_"dark solid" #

And so plumbic ion is reduced by iodide ion to give a #Pb^(2+)# salt...and the oxidation product is iodine....