Redox equation and ionic form #4

Balance this equation and put it in the ionic form:
$$MnI_2+NaBiO_3+H_2SO_4 \rightarrow$$
$$\rightarrow \quad\quad NaMnO_4+NaIO_3+Bi_2(SO_4)_3+Na_2SO_4+H_2O$$

SOLUTION:
Using the inspection technique we have:
$${\color{red} a}MnI_2+{\color{red} b}NaBiO_3+{\color{red} c}H_2SO_4 \rightarrow$$
$$\rightarrow \quad\quad {\color{red} d}NaMnO_4+{\color{red} e}NaIO_3+{\color{red} f}Bi_2(SO_4)_3+{\color{red} g}Na_2SO_4+{\color{red} h}H_2O$$
$$\begin{cases}Mn:a=d\\ I:2a=e\\ Na:b=d+e+2g\\ Bi:b=2f\\ S:c=3f+g\\ H:2c=2h\\ O:3b+4c=4d+3e+12f+4g+h \end{cases}$$

This is a reduction-oxidation equation, in fact:
$$\overset{+2}{Mn}\overset{-1}{I_2}+Na\overset{+5}{Bi}O_3+H_2SO_4 \rightarrow$$
$$\rightarrow \quad\quad Na\overset{+7}{Mn}O_4+Na\overset{+5}{I}O_3+\overset{+3}{Bi_2}(SO_4)_3+Na_2SO_4+H_2O$$

- Manganese changes its oxidation number from +2 to +7: this is the first oxidation (Manganese is the reductant).
- Iodine changes its oxidation number from -1 to +5: this is the second oxidation (Iodine is the reductant).
- Bismuth changes its oxidation number from +5 to +3: this is the reduction (Bismuth is the oxidant).

Write down the three half-equations:
$$\begin{cases} \overset{+2}{Mn} \rightarrow \overset{+7}{Mn}+5e^-\\\\ \overset{-1}{I_2} \rightarrow 2\overset{+5}{I}+12e^-\\\\ 2\overset{+5}{Bi}+4e^- \rightarrow \overset{+3}{Bi_2}\end{cases}$$

Now sum the two oxidation equation:
$$\begin{cases} \overset{+2}{Mn}+\overset{-1}{I_2} \rightarrow \overset{+7}{Mn}+2\overset{+5}{I}+17e^-\\\\ 2\overset{+5}{Bi}+4e^- \rightarrow \overset{+3}{Bi_2}\end{cases}$$

Now balance the electrons and sum:
$$\begin{cases} \overset{+2}{Mn}+\overset{-1}{I_2} \rightarrow \overset{+7}{Mn}+2\overset{+5}{I}+17e^- & \quad \mbox{(x4)}\\\\ 2\overset{+5}{Bi}+4e^- \rightarrow \overset{+3}{Bi_2} & \quad \mbox{(x17)}\end{cases}+$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$= \quad\quad 4\overset{+2}{Mn}\overset{-1}{I_2}+34\overset{+5}{Bi} \rightarrow 4\overset{+7}{Mn}+8\overset{+5}{I}+17\overset{+3}{Bi_2}$$

So we have calculated the following stoichiometric coefficients:
$$\begin{cases}a=4\\ b=34\\ d=4\\ e=8\\ f=17 \end{cases}$$
The other stoichiometric coefficients are:
$$\begin{cases}g=11\\ c=62\\ h=62 \end{cases}$$

The balanced equation is:
$${\color{red} 4}MnI_2+{\color{red} 34}NaBiO_3+{\color{red} 62}H_2SO_4 \rightarrow$$
$$\rightarrow \quad\quad {\color{red} 4}NaMnO_4+{\color{red} 8}NaIO_3+{\color{red} 17}Bi_2(SO_4)_3+{\color{red} 11}Na_2SO_4+{\color{red} 62}H_2O$$

Now we write it in the ionic form:
$$ 4MnI_2++34Na^++34BiO_3^-+124H^++62SO_4^{2-} \rightarrow$$
$$\rightarrow \quad\quad 4Na^++4MnO_4^-+8Na^++8IO_3^-+34Bi^{3+}+$$
$$+ \quad\quad\quad\quad 51SO_4^{2-}+22Na^++11SO_4^{2-}+62H_2O$$
Now sum the sodium and the solfate ions and simplify:
$$4MnI_2+34BiO_3^-+124H^+ \rightarrow$$
$$\rightarrow \quad 4MnO_4^-+8IO_3^-+34Bi^{3+}+62H_2O$$
We can divide by two all the stoichiometric coefficients. The net ionic equation balanced is:
$$2MnI_2+17BiO_3^-+62H^+ \rightarrow$$
$$\rightarrow \quad 2MnO_4^-+4IO_3^-+17Bi^{3+}+31H_2O$$

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PROBLEM 17:
Balance the following redox equation and put it in the ionic form:
$$[Ni(NH_3)_4]Cl_2+(C_6H_5)NC+H_2O \rightarrow $$
$$\quad Ni[(C_6H_5)NC]_4+NH_3+NH_4Cl+(C_6H_5)NCO$$