Question

Balance the following net ionic reaction that occurs in acid using the half-reaction method of balancing.
Cr2O72- + H2AsO3 --> Cr3+ + H3AsO4

Must use the half reaction method to answer this problem. Please state the unbalanced equation to start the problem.

Answer 1

In this case Cr₂O₇^2- change to Cr³⁺, and H₂AsO₃ change to H₃AsO₄

+++++++++++++++++++++++

But the reaction occurs in acid, so we need to add H⁺/H₂O to balance each half reaction:

Cr₂O₇^2- + H⁺ --> Cr³⁺ + H₂O

(we add water as a product because there was no O atoms, so we will have to add H⁺ as a reactant)

H₂AsO₃ + H₂O --> H₃AsO₄ + H⁺

(we add water as a reactant due there was more O atoms in the right of the reaction, therefore we add H⁺ as a product)

+++++++++++++++++++++++++++++

Now we balance each half reaction:

Cr₂O₇^2- + 14H⁺ --> 2Cr³⁺ + 7H₂O

H₂AsO₃ + H₂O --> H₃AsO₄ + H⁺

+++++++++++++++++++++++++++++

-We add electrons to balance the charges of each half reaction:

Charges	-2		+14		6+		0
	Cr2O72-	+	14H+	-->	2Cr3+	+	7H2O

So we have +12 on the left and +6 on the right, so we add 6 electrons on the left to balance the equation;

Cr₂O₇^2- + 14H⁺ +6e^- --> 2Cr³⁺ + 7H₂O

************

Charges	0		0		0		+1
	H2AsO3	+	H2O	-->	H3AsO4	+	H+

So we add one electron on the right of the equation:

H₂AsO₃ + H₂O --> H₃AsO₄ + H⁺ + e^-

+++++++++++++++++++++++++++++

we have to match the numbers of electrons in both equations so that they can be canceled, so we multiply by 6 the equation: (H₂AsO₃ + H₂O ->  H₃AsO₄ + H⁺ + e^-) = 6H₂AsO₃ + 6H₂O --> 6H₃AsO₄ + 6H⁺ + 6e^-

++++++++++++

we add both half reactions:

6H₂AsO₃ + 6H₂O --> 6H₃AsO₄ + 6H⁺ + 6e^-

Cr₂O₇^2- + 14H⁺ +6e^- --> 2Cr³⁺ + 7H₂O

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Cr₂O₇^2- + 8H⁺ + 6H₂AsO₃ --> 6H₃AsO₄ + 2Cr³⁺ + H₂O