Balancing redox equation (in BR book)

[sbuxaddict]

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Hi everyone,
In the BR Physics II book (pg 263) they give an example of balancing a redox reaction via the half-cell method:

MnO4(^1-) + Zn ---> MnO2 + Zn(OH)4(^2-)

They split it into the ox and red components, and the second to last step is this (after balancing net charges and # of H's):

Reduction: 4 H2O + 2 MnO4(^1-) + 6 e- --> 2 MnO2 + 8OH(-)
Oxidation: 3 Zn + 12 OH(-) --> 3 Zn(OH)4(^2-) + 6 e-

I got everything up to here, and then they added the 2 half-reactions, giving this:

4 H2O + 2 MnO4(^1-) + 3 Zn --> 2 MnO2 + 3 Zn(OH)4(^2-) + 2 OH(-)

My question is...where did the 12 OH(-) from the oxidation reactants go (bolded above), and how did they get the 2 OH(-) instead of 8 OH(-) for the products (also bolded)?

 

[BB8730]

At first glance this problem appears to be done incorrectly in the BR book. I believe there should be 4 OH- on the left side and none on the right side. When adding two half reactions, whatever appears on both sides is cancelled out. The 6 electrons on both sides cancel out, and 8 OH- should cancel out of both sides, leaving 4 OH- on the left and none on the right. That would result in equal total charge on both sides (negative 6). The way it's done in the BR book, the left side has negative 2 and the right side has negative 8, which doesn't look right to me.

I could be wrong, but BR is known to contain lots of errors.

 

[sbuxaddict]

At first glance this problem appears to be done incorrectly in the BR book. I believe there should be 4 OH- on the left side and none on the right side. When adding two half reactions, whatever appears on both sides is cancelled out. The 6 electrons on both sides cancel out, and 8 OH- should cancel out of both sides, leaving 4 OH- on the left and none on the right. That would result in equal total charge on both sides (negative 6). The way it's done in the BR book, the left side has negative 2 and the right side has negative 8, which doesn't look right to me.

I could be wrong, but BR is known to contain lots of errors.

Ok cool, that's what I got for my final answer too, but I wasn't sure if I missed something. And you're right, the way they have it doesn't make the overall charges on each side equal. Plus, if they cancelled out the 6 electrons from each side in their answer, idk why they wouldn't have done the same for the OH's.

 

[indianjatt]

Yeah they just made a typo. There should be 4 OH- on the left side.