Please balance this equation. : IO3- + I- + H+ = H2O + I2

IO−3(aq]+5I−(aq]+6H+(aq]→3I2(aq]+3H2O(l]

The iodate, IO−3, and iodide, I−, ions will react in acidic medium to form iodine, I2.

Your starting equation will be

IO−3(aq]+I−(aq]+H+(aq]→I2(aq]+H2O(l]

Right from the start, you can probably tell that this is a disproportionation raction, which a redox reaction in which the same chemical species is being reduced and oxidized at the same time.

Assign oxidation numbers to all the atoms that take part in the reaction

+5I−2O−3+−1I−++1H+→0I2++1H2−2O

Some of the iodine atoms are being reduced from an oxidation state of +5 to an oxidation state of 0, while other are being oxidized from an oxidation state of -1 to an oxidation state of 0.

The two half-reactions will be

• oxidation half-reaction

2−1I−→0I2+2e−

Each iodine atom loses one electrons, so two iodine atoms will lose 2 electrons.

• reduction half-reaction

2+5IO−3+10e−→0I2

Each iodine atom will gain 5 electrons, which means that two iodine atoms will gain a total of 10 electrons* to form I2.

Since you're in acidic solution, use water to balance the oxygen atoms and protons to balance the hydrogen atoms

12H++2+5IO−3+10e−→0I2+6H2O

In any redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

This means that you need to multiply the oxidation half-reaction by 5 to get

10I−→5I2+10e−

The two half-reactions will now be

{10I−→5I2+10e−12H++2IO−3+10e−→I2+6H2O

Add the two half-reaction to get

10I−+12H++2IO−3+10e−→5I2+10e−+I2+6H2O

2IO−3(aq]+10I−(aq]+12H+(aq]→6I2(aq]+6H2O(l]

The balanced net ionic equation will thus be

IO−3(aq]+5I−(aq]+6H+(aq]→3I2(aq]+3H2O(l]