Consider the half equation IO3^-+xH2O+ye^-》IO^-+zOH^-

Question

X. Y. Z

A. 1. 2. 2

B. 2. 4. 4

C. 3. 6. 6

D. 4. 8. 8

Answer 1

I'm guessing you are asking how to balance this? Correct?

Since there are OH- ions on the right side of the equation, this indicates the reaction is taking place (and being balanced) in alkaline conditions.

IO₃^- ===> IO^- initial equation

IO₃^- ===> IO^- + 2H₂O to balance the O atoms

IO₃^- + 4OH^- + 4H⁺ ===> IO^- + 2H₂O + 4OH^- to balance the H atoms using base

IO₃^- + 4OH^- + 4H⁺ +4e- ===> IO^- + 2H₂O + 4OH^- to balance the charge

Canceling and reducing we end up with...

IO₃^- + 2H₂O + 4e- ==> IO^- + 4OH^-

So answer would be (B) 2 H2O, 4e- and 4 OH^-