I must be wrong... or am I?

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what's wrong with this?:

This is Chapter 20, Section 1, Problem 3. This problem's ID is: 954460247

-------------------------------------------------------------------------------- Following the model below, 2H2 + 2O2 ==> H2O

-------------------------------------------------------------------------------- balance the following oxidation-reduction skeleton reaction which occurs in basic solution: 6BH4-(aq) + 7ClO3-(aq) + ==> 6H2BO3-(aq) + 7Cl-(aq) + 9H2O(l) +

-- Anonymous, March 30, 2000


I didn't go into CALM to check the problem, but I'll give it a shot. First divide up the equation into half reactions:

BH4- --> 6H2BO3- 7ClO3- --> Cl-

Now figure out how many electrons were transferred in each half reaction.

The oxidation state of B in BH4- is -5, and in H2BO3- it is +3. Thus, BH4- loses 8 electons.

The oxidation state of Cl in ClO3- is +5 and for Cl- it is -1. Thus, ClO3- gains 6e-.


6e- + ClO3- ---> Cl- BH4- --> H2BO3- + 8e-

Now add water / OH- to each half reaction to balance the number of O's and H's.

3H2O + 6e- + ClO3- --> Cl- + 6OH- BH4- + 8OH- --> H2BO3- + 8e- + 5H2O

Now multiply each half reaction by a number such that the same number of electrons are in each half reaction:

12H2O + 24e- + 4ClO3- --> 4Cl- + 24OH- 3BH4- + 24OH- --> 3H2BO3- + 24e- + 15H2O

Add these together to get your balanced equation: 3BH4- + 4ClO3- --> 4Cl- + 3H2BO3- + 3H2O

-- Anonymous, March 31, 2000

Thank you Amy. That was very helpful. I think I missed the negative charge after the BH4. Once again. Thank you.

-- Anonymous, March 31, 2000

umm... Amy, I tried entering your answer into CALM... Thanks anyway.

-- Anonymous, March 31, 2000

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