Balancing Redox reaction

[cc609]

For balancing this half reaction

3I- --> I3-
is shown as 3I- ---> I3- + 2e-

How does this answer come about?? How does 2e- balance this equation out?! If anything I thought it would be 3e-...

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[FeralisExtremum]

3 I- [for clarity: this is three I atoms, each with a -1 charge] --> I3 - [for clarity: this is an I3 molecule with a -1 charge overall).

Now, look at the e- transfer between individual I atoms (do this by comparing oxidation states of individual atoms). We go from a -1 charge to a -1/3 charge (an I3 molecule with -1 charge overall has -1/3 charge on each atom). That means in this reaction, each I atom is losing 2/3 e-. But we are forming I3, which means this reaction has to occur 3 times to form a single I3 molecule. (2/3) * (3) = 2 e- lost overall.

 

[DAT Destroyer]

For balancing this half reaction

3I- --> I3-
is shown as 3I- ---> I3- + 2e-

How does this answer come about?? How does 2e- balance this equation out?! If anything I thought it would be 3e-...

When balancing a redox reaction or half reaction.........Mass and Charge must BOTH balance. In this example, Mass is balanced by the coefficients of 3. Now for charge......the Left side has a MINUS 3....the right side only a Minus 1.......ADD electrons to the more positive side.......thus,,,,,2 electrons added to the right side does the trick. To check it......mass is balanced,,,,,and both sides now have Minus 3.

Hope this helps.

Dr. Jim Romano