what is the balanced net ionic equation for the reaction of ascorbic acid with triiodide ion
C6H8O6 ==> C6H6O6+ 2H^+ + 2e
2e + I3^- ==> 3I^-
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These are to two half equations. Add them and cancel the electrons.
To determine the balanced net ionic equation for the reaction between ascorbic acid (C6H8O6) and triiodide ion (I3^-), you can follow these steps:
1. Write the oxidation half-reaction:
C6H8O6 -> C6H6O6 + 2H^+ + 2e^-
In this equation, ascorbic acid (C6H8O6) is oxidized, losing two electrons to form dehydroascorbic acid (C6H6O6), two hydrogen ions (2H^+), and two electrons (2e^-).
2. Write the reduction half-reaction:
2e^- + I3^- -> 3I^-
In this equation, two electrons (2e^-) are used to reduce triiodide ion (I3^-) to three iodide ions (3I^-).
3. Balance the electrons:
To balance the electrons in the two half-reactions, multiply the oxidation half-reaction by 2 and the reduction half-reaction by 2:
2(C6H8O6) -> 2(C6H6O6) + 4H^+ + 4e^-
4e^- + 2I3^- -> 6I^-
4. Add the two equations together and cancel out the electrons:
2(C6H8O6) + 4e^- + 4I3^- -> 2(C6H6O6) + 4H^+ + 6I^-
This final equation represents the balanced net ionic equation for the reaction of ascorbic acid with triiodide ion.