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:
$$As_2O_3+KIO_4 \rightarrow K_3AsO_4+KIO_2+H_2O$$
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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:
$$Al+K_2SiF_6 \rightarrow AlF_3+Si+KF$$
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Redox equation and ionic form #1

Rewrite the following equation showing all substances in their ionic form in solution (the ionic equation) and balance it:
$$Hg(NO_3)_2+KI \rightarrow K_2HgI_4+KNO_3$$

SOLUTION:
The ionic equation is:
$$Hg^{2+}+2NO_3^-+K^++I^- \rightarrow 2K^++HgI_4^{2-}+K^++NO_3^-$$

Note that in reactants there are two $NO_3^-$ and only one $NO_3^-$ in products. So multiply by two $KNO_3$ in products:
$$Hg^{2+}+2NO_3^-+K^++I^- \rightarrow 2K^++HgI_4^{2-}+2K^++2NO_3^-$$

Now sum the same species -$K^+$- and simplify -$2NO_3^-$-:
$$Hg^{2+}+K^++I^- \rightarrow 4K^++HgI_4^{2-}$$

Now there are $4K^+$ in products. So multiply by four the $KI$ in reagents:
$$Hg^{2+}+4K^++4I^- \rightarrow 4K^++HgI_4^{2-}$$

Now simplify $K^+$:
$$Hg^{2+}+4I^- \rightarrow HgI_4^{2-}$$
This is the net ionic equation. Note that che charges are balanced.

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PROBLEM 14:
Balance the following redox equation and put it in the ionic form:
$$Na+H_2O \rightarrow NaOH+H_2$$
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How to balance redox equations #4

Balance the following chemical equation:
$$CrI_3+NaNO_3+Na_2CO_3 \rightarrow Na_2CrO_4+NaIO_3+NaNO_2+CO_2$$

SOLUTION:
Using the inspection technique, we have:
$${\color{red} a}CrI_3+{\color{red} b}NaNO_3+{\color{red} c}Na_2CO_3 \rightarrow {\color{red} d}Na_2CrO_4+{\color{red} e}NaIO_3+{\color{red} f}NaNO_2+{\color{red} g}CO_2$$
$$\begin{cases} I:3a=e\\ Cr:a=d\\ Na:b+2c=2d+e+f\\ C:c=g\\ O:3b+3c=4d+3e+2f+2g\end{cases}$$
We have more variable than equation. So we can balance the chemical equation using the only inspection technique.

Is this a redox equation? Find out the oxidation number:

$$\overset{+3}{Cr}\overset{-1}{I_3}+Na\overset{+5}{N}O_3+Na_2CO_3 \rightarrow Na_2\overset{+6}{Cr}O_4+Na\overset{+5}{I}O_3+Na\overset{+3}{N}O_2+CO_2$$

Here there are one elements that is reducting (N), and TWO elements that are oxidating (Cr and I). Write down the three equations:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N}\\\\ \overset{+3}{Cr} \rightarrow \overset{+6}{Cr}+3e^-\\\\ \overset{-1}{I_3} \rightarrow 3\overset{+5}{I}+18e^-\end{cases}$$

Now simply sum the two oxidation reaction:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N}\\\\ \overset{+3}{Cr}+\overset{-1}{I_3} \rightarrow \overset{+6}{Cr}+3\overset{+5}{I}+21e^-\end{cases}$$

Now balance the electrons and sum the two half-equations:

$$\begin{cases} \overset{+5}{N}+2e^- \rightarrow \overset{+3}{N} & \quad \mbox{(x21)}\\\\ \overset{+3}{Cr}+\overset{-1}{I_3} \rightarrow \overset{+6}{Cr}+3\overset{+5}{I}+21e^- & \quad \mbox{(x2)}\end{cases}\quad+$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$=\quad\quad 21\overset{+5}{N}+2\overset{+3}{Cr}+2\overset{-1}{I_3} \rightarrow 21\overset{+3}{N}+2\overset{+6}{Cr}+6\overset{+5}{I}$$

We now have:
$$\begin{cases} b=21\\ a=2\\ f=21\\ d=2\\ e=6\end{cases}$$

From the relationships previously found, we can obtain:

$$\begin{cases} c=5\\ g=5\end{cases}$$

The balanced equation is:
$${\color{red} 2}CrI_3+{\color{red} 21}NaNO_3+{\color{red} 5}Na_2CO_3 \rightarrow {\color{red} 2}Na_2CrO4+{\color{red} 6}NaIO3+{\color{red} 21}NaNO_2+{\color{red} 5}CO_2$$

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PROBLEM 13:
Balance the following reaction:
sodium perchlorate reacts with ammonium chloride giving sodium chloride, nitrogen gas, hydrochloric acid and water.
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How to balance redox equations #3

Balance the following chemical equation:
$I_2+KOH \rightarrow KIO_3+KI+H_2O$

SOLUTION:
Using the principle of conservation of mass, we have:
${\color{red} a}I_2+{\color{red} b}KOH \rightarrow {\color{red} c}KIO_3+{\color{red} d}KI+{\color{red} e}H_2O$
$\begin{cases}K:b=c+d\\ I:2a=c+d\\ O:b=3c+e\\ H:b=2e\end{cases}$
There are 4 equations and 5 variables. We havo to find more relationships.

Is this a reduction-oxidation equation? Look at the oxidation number:
$\overset{0}{I_2}+KOH \rightarrow K\overset{+5}{I}+K\overset{-1}{I}+H_2O$
The iodine element $I_2$ changes its oxidation number from 0 to +5 (it is oxidating) and to -1 (it is reducing). At the same time, the iodine is the oxidant and the reductant. This is a disproportionation or dismutation equation.
But don't worry. We can use the same method to balance.

The two half-equations are:
$\begin{cases} \overset{0}{I_2} +2e^-\rightarrow 2\overset{-1}{I}\\\\ \overset{0}{I_2} \rightarrow 2\overset{+5}{I}+10e^-\end{cases}$
Now balance the electrons and sum the two half-equations:
$$\begin{cases} \overset{0}{I_2} +2e^-\rightarrow 2\overset{-1}{I} & \quad \mbox{(x5)}\\\\ \overset{0}{I_2} \rightarrow 2\overset{+5}{I}+10e^-\end{cases} \quad\quad+$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$= \quad\quad 6\overset{0}{I_2}+10e^- \rightarrow 10\overset{-1}{I}+2\overset{+5}{I}+10e^-$$

So we have found that:
$\begin{cases} a=6\\ c=2\\ d=10\end{cases}$
From the relationships previously found, we can calculate:
$\begin{cases} b=12\\ e=6\end{cases}$

This dismutation reaction balanced is:
${\color{red} 6}I_2+{\color{red} 12}KOH \rightarrow {\color{red} 2}KIO_3+{\color{red} 10}KI+{\color{red} 6}H_2O$
All the stoichiometric coefficients we calculated are even. We can divide by two all of this:
${\color{red} 3}I_2+{\color{red} 6}KOH \rightarrow KIO_3+{\color{red} 5}KI+{\color{red} 3}H_2O$

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PROBLEM 12:
Balance the following chemical equation:
$$Cu(CNS) \rightarrow Cu_2S+CS_2+(CN)_2+N_2$$
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