How do you solve #I^- + ClO^- rarr I_3^- +Cl# using the redox reaction method in a base solution?

1 Answer
Jan 26, 2017

Answer:

#"Hypochlorite"# is reduced to #"chloride"#..........

Explanation:

And #"iodide"# is oxidized to #"triiodide"#.............

#"Reduction:"#

#ClO^(-) +2H^(+) + 2e^(-) rarrCl^(-) + H_2O# #(i)#, yes I know you asked for basic solution, but attend........

#"Oxidation:"#

#3I^(-) rarrI_3^(-) + 2e^(-)# #(ii)#

For each half equation, mass and charge are conserved as is required. Am I right?

We add #(i)# and #(ii)# to remove the electrons.

#ClO^(-) +2H^(+) + 3I^(-) rarrI_3^(-) + Cl^(-) +H_2O#

But you specified basic solution. So all we have to do is add #2xxHO^-# to each side:

#ClO^(-) +cancel2H_2O(l) + 3I^(-) rarrI_3^(-) + + Cl^(-) +2HO^(-)+cancel(H_2O)#

To give me finally,

#ClO^(-) +H_2O(l) + 3I^(-)rarrI_3^(-) + Cl^(-) + 2HO^(-)#

Which I think is balanced with respect to mass and charge.