Redox equation and ionic form #3

Balance the following equation and put it in the ionic form:
$$Zn+NaNO_3+NaOH+H_2O \rightarrow Na_2[Zn(OH)_4]+NH_3$$

SOLUTION:
First we write down the relationships between the stoichiometric coefficients for the balance of mass:
$${\color{red} a}Zn+{\color{red} b}NaNO_3{\color{red} c}+NaOH+{\color{red} d}H_2O \rightarrow {\color{red} e}Na_2[Zn(OH)_4]+{\color{red} f}NH_3$$
$$\begin{cases} Zn:a=e\\ Na:b+c=2e\\ N:b=f\\ H:c+2d=4e+3f\\ O:3b+c+d=4e\end{cases}$$

To verify if this is a redox equation, write down all oxidation number:
$$\overset{0}{Zn}+Na\overset{+5}{N}O_3+NaOH+H_2O \rightarrow Na_2[\overset{+2}{Zn}(OH)_4]+\overset{-3}{N}H_3$$

- Zinc changes its oxidation number from 0 to +2: this is the oxidation (Zinc is the reductant).
- Nitrogen changes its oxidation number from +5 to -3: this is the reduction (Nitrogen is the oxidant).

Now write down the two half-equations:
$$\begin{cases} \overset{0}{Zn} \rightarrow \overset{+2}{Zn}+2e^-\\\\ \overset{+5}{N}+8e^- \rightarrow \overset{-3}{N}\end{cases}$$

Balance the electrons and sum:
$$\begin{cases} \overset{0}{Zn} \rightarrow \overset{+2}{Zn}+2e^- & \quad \mbox{(x4)}\\\\ \overset{+5}{N}+8e^- \rightarrow \overset{-3}{N}\end{cases} \quad+$$
$$\rule[0.3cm]{6.5cm}{0.01cm} $$
$$=\quad\quad 4\overset{0}{Zn}+\overset{+5}{N} \rightarrow 4\overset{+2}{Zn}+\overset{-3}{N}$$

The stoichiometric coefficients found are:
$$\begin{cases}a=4\\ b=1\\e=4\\f=1 \end{cases}$$

From the relationships previously take, we have:
$$\begin{cases}Na:b+c=2e\\ O:3b+c+d=4e \end{cases} \Rightarrow \begin{cases}c=2e-b=7\\ d=4e-3b-c=6\end{cases}$$

The balanced redox equation is:
$${\color{red} 4}Zn+NaNO_3{\color{red} 7}+NaOH+{\color{red} 6}H_2O \rightarrow {\color{red} 4}Na_2[Zn(OH)_4]+NH_3$$

Now put it in the ionic form:
$$4Zn+Na^++NO_3^-+7Na^++7OH^-+6H_2O \rightarrow 8Na^++4[Zn(OH)_4]^{2-}+NH_3$$
Now simplify the sodium:
$$4Zn+NO_3^-+7OH^-+6H_2O \rightarrow 4[Zn(OH)_4]^{2-}+NH_3$$
Note that the charges are balanced.

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PROBLEM 16:
Balance the following redox equation and put it in the ionic form:
$$As_2O_3+KIO_4 \rightarrow K_3AsO_4+KIO_2+H_2O$$