Redox equation and ionic form #2

Balance the stoichiometric coefficients of this redox equation, and rewrite it in the ionic form:
$$Sn+HNO_3+H_2O \rightarrow H_2SnO_3+NO$$

SOLUTION:
For the principle of conservation of mass we have:
$${\color{red} a}Sn+{\color{red} b}HNO_3+{\color{red} c}H_2O \rightarrow {\color{red} d}H_2SnO_3+{\color{red} e}NO$$
$$\begin{cases} Sn:a=d\\ N:b=e\\ H:b+2c=2d\\ O:3b+c=3d+e\end{cases}$$

This is a redox equation, infact Sn and N change their oxidation number:
$$\overset{0}{Sn}+H\overset{+5}{N}O_3+H_2O \rightarrow H_2\overset{+4}{Sn}O_3+\overset{+2}{N}O$$

- Tin changes its oxidation number from 0 to +4: this is the oxidation (it is the reductant).
- Nitrogen changes its oxidation number from +5 to +2: this is the reduction (it is the oxidant).

So write down the two half-equations:
$$\begin{cases} \overset{0}{Sn} \rightarrow \overset{+4}{Sn}+4e^-\\\\ \overset{+5}{N}+3e^- \rightarrow \overset{+2}{N}\end{cases}$$

Now balance the electrons and sum:
$$\begin{cases} \overset{0}{Sn} \rightarrow \overset{+4}{Sn}+4e^- & \quad \mbox{(x3)}\\\\ \overset{+5}{N}+3e^- \rightarrow \overset{+2}{N} & \quad \mbox{(x4)}\end{cases} \quad+$$
$$\rule[0.3cm]{6.5cm}{0.01cm} $$
$$=\quad\quad 3\overset{0}{Sn}+4\overset{+5}{N} \rightarrow 3\overset{+4}{Sn}+4\overset{+2}{N}$$

From this equation we have found 4 stoichiometric coefficients:
$$\begin{cases} a=3\\ b=4\\ d=3\\ e=4\end{cases}$$
From the hydrogen mass relationship, we calculate $c=1$.

The balanced equation is:
$${\color{red} 3}Sn+{\color{red} 4}HNO_3+H_2O \rightarrow {\color{red} 3}H_2SnO_3+{\color{red} 4}NO$$

Now put it in the ionic form:
$$3Sn+4H^++4NO_3^-+2H_2O \rightarrow 6H^++3SnO_3^{2-}+4NO$$
We can simplify the $4H^++H_2O$ in reagents with the $6H^+$ in products. The net ionic equation is:
$$3Sn+4NO_3^- \rightarrow 3SnO_3^{2-}+4NO$$

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PROBLEM 15:
Balance the following redox equation and put it in the ionic form, if possible:
$$Al+K_2SiF_6 \rightarrow AlF_3+Si+KF$$