Question

Balance the following chemical reaction by the half reaction method under both acidic and basic conditions. What was oxidized and what was reduced? IrO3 + V(OH)4+2  Ir(OH)5+3 + V+2

Answer 1

oxidation half reaction                                                       reduction half reaction

IrO3 --------> [Ir(OH)5]^3+                                                 [ V(OH)4]^2+   -------> V^2+

IrO3 +2H2O --------> [Ir(OH)5]^3+                                     [ V(OH)4]^2+   -------> V^2+ + 4H2O

IrO3 +2H2O +H^+ --------> [Ir(OH)5]^3+                             [ V(OH)4]^2+ +4H^+ -------> V^2+ + 4H2O

IrO3 +2H2O +H^+ --------> [Ir(OH)5]^3+ +2e^-                  [ V(OH)4]^2+ +4H^+ +4e^- -------> V^2+ + 4H2O

                  2IrO3 +4H2O +2H^+ --------> 2[Ir(OH)5]^3+ +4e^-            

     [ V(OH)4]^2+ +4H^+ +4e^- -------> V^2+ + 4H2O

           -----------------------------------------------------------------------------------------

               2IrO3 + [ V(OH)4]^2+ + 6H^+ ---------> 2[Ir(OH)5]^3+ + V^2+

Basic medium

oxidation half reaction                                                       reduction half reaction

IrO3 --------> [Ir(OH)5]^3+                                                 [ V(OH)4]^2+   -------> V^2+

IrO3 +2H2O --------> [Ir(OH)5]^3+                                     [ V(OH)4]^2+   -------> V^2+ + 4H2O

IrO3 +2H2O +H2O--------> [Ir(OH)5]^3+ +OH^-                 [ V(OH)4]^2+ +4H2O -------> V^2+ + 4H2O + 4OH^-

IrO3 +2H2O +H2O -------> [Ir(OH)5]^3+ +OH^- +2e^-          [ V(OH)4]^2+ +4H2O +4e^- -------> V^2+ + 4H2O + 4OH^-

                  2IrO3 +4H2O +2H2O -------> 2[Ir(OH)5]^3+ +2OH^- +4e^-      

      [ V(OH)4]^2+ +4H2O +4e^- -------> V^2+ + 4H2O + 4OH^-

                -----------------------------------------------------------------------------

           2IrO3 + [ V(OH)4]^2+ 6H2O -----------> 2[Ir(OH)5]^3 + V^2+ + 6OH^-

IrO3 is oxidized

[ V(OH)4]^2+ is reduced