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Balance ionic reaction
Posted: Mon Nov 06, 2006 8:46 am
by 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
Re: Balance ionic reaction
Posted: Tue Nov 07, 2006 2:04 am
by Vitalii
Mary wrote: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 IO
3- + 12 H
+ + 10
- + I
- = I
3- + 6 H
2O
2) 3 I
- = I
3- + 2
-
and then add them together with appropriate coefficients (you should get an equation without free electrons):
2 IO
3- + 16 I
- + 12 H
+ = 6 I
3- + 6 H
2O
Last step is to get rid of common multiplier in the coefficients:
IO
3- + 8 I
- + 6 H
+ = 3 I
3- + 3 H
2O
You may always check with the
online equation balancer :
IO3- + I- + H+ = I3- + H2O
Posted: Sat Apr 16, 2016 5:06 pm
by 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!
Posted: Thu Aug 18, 2016 9:43 pm
by GrahamKemp
bobbybeck wrote: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