Question 17.10: Balance this equation: CrO4^2- + Fe(OH)2→Cr(OH)3 + Fe(OH)3 ...

Use the Ion–electron Problem-Solving strategy

1. Write the two half-reactions:

$Fe(OH)_2 \rightarrow Fe(OH)_3$      (oxidation)

$CrO^{2-}_4 \rightarrow Cr(OH)_3$        (reduction)

2. Balance elements other than H and O (accomplished in Step 1).

3. Remember the solution is basic. Balance O and H as though the solution were acidic. Use $H_2O    and    H^+$ . To balance O and H in the

oxidation equation, add 1$H_2O$  on the left and $1   H^+$ on the right side:

Add $1 OH^-$ to each side:

Combine $H^+    and     OH^-    as    H_2O$ and rewrite, canceling $H_2O$  on each side:

$Fe(OH)_2 + OH^- \rightarrow Fe(OH)_3$  (oxidation)

To balance O and H  in the reduction equation, add $1  H_2O$  on the right and  5 $H^+$ on the left:

Add 5  $OH^-$ to each side:

Combine $5 H^+ + 5 OH^- \rightarrow  5 H_2O$:

Rewrite, canceling $1  H_2O$  from each side:

$CrO_4^{2-} + 4 H_2O \rightarrow Cr(OH)_3  + 5 OH^-$    (reduction)

4. Balance each half-reaction electrically with electrons:

$Fe(OH)_2 + OH^- \rightarrow Fe(OH)_3 + e^-$    (balanced oxidation equation)

$CrO^{2-}_4 + 4 H_2O + 3 e^- \rightarrow Cr(OH)_3 + 5 OH^-$    (balanced reduction equation)

5. Equalize the loss and gain of electrons. Multiply the oxidation reaction by 3:

6. Add the two half-reactions together, canceling the $3e^-$ and  $3   OH^-$ from each side of the equation:

$\underset{CrO_4^{2-} + 3Fe(OH)_2+ 4 H_2O \rightarrow Cr(OH)_3+ 3Fe(OH)_3 +2 OH^-} {\underline {CrO_4^{2-} + 4 H_2O + \cancel{3e^-} \rightarrow Cr(OH)_3 + 5 OH^-}}$                          (balanced)

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