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:
$$Cu(CNS) \rightarrow Cu_2S+CS_2+(CN)_2+N_2$$