What is the balanced ionic equation for the reaction of IO3-(aq) with I- in the acidic solution, H+ ?

##”IO”_text(3(aq])^(-) + 5″I”_text((aq])^(-) + 6″H”_text((aq])^(+) -> 3″I”_text(2(aq]) + 3″H”_2″O”_text((l])##

The iodate, ##”IO”_3^(-)##, and iodide, ##”I”^(-)##, ions will react in acidic medium to form iodine, ##”I”_2##.

Your starting equation will be

##”IO”_text(3(aq])^(-) + “I”_text((aq])^(-) + “H”_text((aq])^(+) -> “I”_text(2(aq]) + “H”_2″O”_text((l])##

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

Assign to all the atoms that take part in the reaction

##stackrel(color(blue)(+5))(“I”) stackrel(color(blue)(-2))(“O”_3^(-)) + stackrel(color(blue)(-1))(“I”^(-)) + stackrel(color(blue)(+1))(“H”^(+)) -> stackrel(color(blue)(0))(“I”_2) + stackrel(color(blue)(+1))(“H”_2) stackrel(color(blue)(-2))(“O”)##

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

##2stackrel(color(blue)(-1))(“I”^(-)) -> stackrel(color(blue)(0))(“I”_2) + 2e^(-)##

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

##2stackrel(color(blue)(+5))(“I”)”O”_3^(-) + 10e^(-) -> stackrel(color(blue)(0))(“I”_2)##

Each iodine atom will gain 5 electrons, which means that two iodine atoms will gain a total of 10 electrons* to form ##”I”_2##.

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

##12″H”^(+) + 2stackrel(color(blue)(+5))(“I”)”O”_3^(-) + 10e^(-) -> stackrel(color(blue)(0))(“I”_2) + 6″H”_2″O”##

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

##10″I”^(-) -> 5″I”_2 + 10e^(-)##

The two half-reactions will now be

##{(10″I”^(-) -> 5″I”_2 + 10e^(-)), (12″H”^(+) + 2″IO”_3^(-) + 10e^(-) -> “I”_2 + 6″H”_2″O”) :}##

Add the two half-reaction to get

##10″I”^(-) + 12″H”^(+) + 2″IO”_3^(-) + color(red)(cancel(color(black)(10e^(-)))) -> 5″I”_2 + color(red)(cancel(color(black)(10e^(-)))) + “I”_2 + 6″H”_2″O”##

##2″IO”_text(3(aq])^(-) + 10″I”_text((aq])^(-) + 12″H”_text((aq])^(+) -> 6″I”_text(2(aq]) + 6″H”_2″O”_text((l])##

The balanced net ionic equation will thus be

##”IO”_text(3(aq])^(-) + 5″I”_text((aq])^(-) + 6″H”_text((aq])^(+) -> 3″I”_text(2(aq]) + 3″H”_2″O”_text((l])##

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