Balance ionic reaction

[Mary]

I am trying to balance the following ionic reaction but nothing works, either the atoms don't balance or the charges on both sides of the reaction don't balance. Can you please show me the steps how to balance it. The charges are in curly brackets.
IO₃^-(aq) + I^-(aq) = I₃^- (aq)
I first balanced the I atoms
IO₃^-(aq) + 2I^-(aq) = I₃^- (aq)
Then tried to balance O atoms by adding water
IO₃^-(aq) + 2I^-(aq) = I₃^- (aq) + 3H₂O
Then balance H atoms
IO₃^-(aq) + 2I^-(aq) + 6H⁺ = I₃^- (aq) + 3H₂O
Now all atoms balance but I have +3 charge on the left and -1 charge on the right
What am I doing wrong?

Thank you very much,
Mary

[Vitalii]

I am trying to balance the following ionic reaction but nothing works, either the atoms don't balance or the charges on both sides of the reaction don't balance. Can you please show me the steps how to balance it. The charges are in curly brackets.
IO₃^-(aq) + I^-(aq) = I₃^- (aq)
I first balanced the I atoms
IO₃^-(aq) + 2I^-(aq) = I₃^- (aq)
Then tried to balance O atoms by adding water
IO₃^-(aq) + 2I^-(aq) = I₃^- (aq) + 3H₂O
Then balance H atoms
IO₃^-(aq) + 2I^-(aq) + 6H⁺ = I₃^- (aq) + 3H₂O
Now all atoms balance but I have +3 charge on the left and -1 charge on the right
What am I doing wrong?

Thank you very much,
Mary

For red-ox reactions you should start with balancing half-cell reactions:
1) 2 IO₃^- + 12 H⁺ + 10 ^- + I^- = I₃^- + 6 H₂O
2) 3 I^- = I₃^- + 2 ^-
and then add them together with appropriate coefficients (you should get an equation without free electrons):
2 IO₃^- + 16 I^- + 12 H⁺ = 6 I₃^- + 6 H₂O
Last step is to get rid of common multiplier in the coefficients:
IO₃^- + 8 I^- + 6 H⁺ = 3 I₃^- + 3 H₂O

You may always check with the online equation balancer :
IO₃^- + I^- + H⁺ = I₃^- + H₂O

[bobbybeck]

Hello. I stumbled across this problem when looking for some help in solving a practice problem for school. How would one balance this equation using the ion-electron method of redox. Thanks!

[GrahamKemp]

Hello. I stumbled across this problem when looking for some help in solving a practice problem for school. How would one balance this equation using the ion-electron method of redox. Thanks!

Take the equation stripped of spectator ions, water, and its components. As given:

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IO3{-} + I{-} = I3{-}

Split into two half equation. Balance the Iodine while there.

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3 IO3{-} = I3{-}

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3 I{-} = I3{-}

Balance oxygen with water

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3 IO3{-} = I3{-} + 9 H2O

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3 I{-} = I3{-}

Balance hydrogen with protons

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3 IO3{-} + 18 H{+} = I3{-} + 9 H2O

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3 I{-} = I3{-}

Balance charges with electrons

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3 IO3{-} + 18 H{+} + 16 e = I3{-} + 9 H2O

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3 I{-} = I3{-} + 2 e

Multiply to equate electron counts .

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3 IO3{-} + 18 H{+} + 16 e = I3{-} + 9 H2O

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24 I{-} = 8 I3{-} + 16 e

Merge and cancel common terms

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3 IO3{-} + 24 I{-} + 18 H{+} = 9 I3{-} + 9 H2O

Simplfy

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IO3{-} + 8 I{-} + 6 H{+} = 3 I3{-} + 3 H2O

Check everything balances
Check sums pass okay