Balance this equation using the change-in-oxidation-number method:

$CrO_4^{2-} + Fe(OH)_2 → Cr(OH)_3 + Fe(OH)_3$ (basic solution)

Question

Balance this equation using the change-in-oxidation-number method:

[latex]CrO_4^{2-} + Fe(OH)_2 → Cr(OH)_3 + Fe(OH)_3[/latex] (basic solution)

Accepted Answer

1. and 2. Assign oxidation numbers and balance the charges with electrons:

[latex]Cr^{6+} + 3 e^-[/latex] → [latex]Cr^{3+}[/latex] (reduction)

[latex]Cr^{6+} gains 3 e^-[/latex]

[latex]Fe^{2+}[/latex] → [latex]Fe^{3+} + e^-[/latex] (oxidation)

[latex]Fe^{2+} loses 1 e^-[/latex]

3. Equalize the loss and gain of electrons, by multiplying the oxidation step by 3:

[latex]Cr^{6+} + 3 e^-[/latex] → [latex]Cr^{3+}[/latex] (reduction)

[latex]Cr^{6+} gains 3 e^-[/latex]

[latex]3 Fe^{2+}[/latex] → [latex]3 Fe^{3+} + 3 e^-[/latex] (oxidation)

[latex]3 Fe^{2+} loses 3 e^-[/latex]

4. Transfer coefficients back to the original equation:

[latex]CrO_4^{2-} + 3 Fe(OH)_2 → Cr(OH)_3 + 3 Fe(OH)_3[/latex]

5. Balance electrically. Because the solution is basic, use [latex]OH^-[/latex] to balance charges. The charge on the left side is -2 and on the right side is 0. Add [latex]2 OH^-[/latex] ions to the right side of the equation:

[latex]CrO_4^{2-} + 3 Fe(OH)_2 → Cr(OH)_3 + 3 Fe(OH)_3 + 2 OH^-[/latex]

Adding [latex]4 H_2O[/latex] to the left side balances the equation:

[latex]CrO_4^{2-} + 3 Fe(OH)_2 + 4 H_2O[/latex]→

[latex]Cr(OH)_3 + 3 Fe(OH)_3 + 2 OH^-[/latex] (balanced)

Check: Each side of the equation has a charge of -2 and contains the same number of atoms of each element.

Question 17.11: Balance this equation using the change-in-oxidation-number m...

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