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Topic: ionic equation  (Read 3289 times)

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Offline jalen

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ionic equation
« on: March 29, 2008, 08:17:10 PM »
I3- ---> I- + IO3-
I3- + H2O ----> IO3-
I3- + H2O ----> 3IO3-
I3- + 9H2O ----> 3IO3-
I3- + 9H2O ----> 3IO3- + 18H+
I3- + 9H2O ----> 3IO3- + 18H+ + 6e-

I- ---> I3-
I- + 1e- ---> I3- (x6)
6I- + 6e----> 6I3-

I3- + 9H2O + 6I----> 3IO3- + 18H+ + 6I3-

Where did I go wrong?? Can't seem to find my mistake.
 

Offline boostar

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Re: ionic equation
« Reply #1 on: March 29, 2008, 08:55:15 PM »
I'm not saying I know the answer and I may be making a fool out of myself but I'll give it a shot:

I3- ---> I- + IO3-
I3- + H2O ----> IO3-... consider the charge on I, remembering that the overall compound is -1
I3- + H2O ----> 3IO3-
I3- + 9H2O ----> 3IO3-
I3- + 9H2O ----> 3IO3- + 18H+
I3- + 9H2O ----> 3IO3- + 18H+ + 6e-

I- ---> I3-
I- + 1e- ---> I3- (x6)
6I- + 6e----> 6I3-... this definitely is not balanced. Rethink the charges on both Iodines in the equation.


Sorry I did not use the sup and sub scripts but it just got very confusing :p

« Last Edit: March 30, 2008, 05:48:20 AM by boostar »

Offline Borek

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Re: ionic equation
« Reply #2 on: March 30, 2008, 05:11:33 AM »
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