Demonstrate that step 8 of the example in basic solution is balanced with respect to both mass andcharge. That is, for mass balance, show that the number of each type of atom is the same on thereactant and product side, and for charge balance, show that the total charge of all species is thesame on the reactant and product side. 8Add the half reactions together, cancel out anyspecies that occur on both sides of theequation, and double-check for balanced atomsand charge.2Cr(OH)3 + 10 OH + 3C1O+ 3H₂O + 6e* →2CrO4²- + 8H₂O + 6e + 3Cl + 60H*The simplified equation becomes:► 2Cr(OH)3 + 4OH + 3C10 ➜ 2CrO4²- +5H₂O + 3C1-This is balanced for all atoms and with respectto charge.

Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms…

Answered: 8 Add the half reactions together,…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Demonstrate that step 8 of the example in basic solution is balanced with respect to both mass and charge. That is, for mass balance, show that the number of each type of atom is the same on the reactant and product side, and for charge balance, show that the total charge of all species is the same on the reactant and product side.

8
Add the half reactions together, cancel out any
species that occur on both sides of the
equation, and double-check for balanced atoms
and charge.
2Cr(OH)3 + 10 OH + 3C1O+ 3H₂O + 6e* →
2CrO4²- + 8H₂O + 6e + 3Cl + 60H*
The simplified equation becomes:
► 2Cr(OH)3 + 4OH + 3C10 ➜ 2CrO4²- +
5H₂O + 3C1-
This is balanced for all atoms and with respect
to charge.

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