Question

Balance the following equations by the oxidation number method.

$A.F{e}^{2+}+H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$

$B.I_2+N{O}_3^{-}\to N{O}_2+I{O}_3^{-}$

$C.I_2+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$

$D.Mn{O}_2+C_2{O}_4^{2-}\to M{n}^{2+}+C{O}_2$

Answer 1

$A.$
Identifying species which are getting oxidized and reduced
$F{e}^{2+}+H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$
$Fe=+2H=+1Cr=+6Cr=+3Fe=+3H=+1$
$O=-2$ $O=-2$
This indicates that $F{e}^{2+}$ is getting oxidized and
$C{r}_2{O}_7^{2-}$ is getting reduced.

Calculating the increase and decrease of oxidation number to make equal exchange of electrons
$F{e}^{2+}+H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$
$Fe=+2H=+1Cr=+6Cr=+3Fe=+3H=+1$
$O=-2$ $O=-2$
$Cr$ is changing its oxidation state from $+6$ to $+3$, so $C{r}^{3+}$ in the product side must be multiplied by a factor of $3$ and $Fe$ is changing its oxidation state from $+2$ to $+3$, so multiply $F{e}^{2+}$ and $F{e}^{3+}$ by a factor of $6$. Now, the equation becomes:
$6F{e}^{2+}+H^{+}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+}+6F{e}^{3+}+H_{2}O$

Balancing the charges
To balance the charges in acidic medium as in this reaction, $13H^{+}$ ions need to be added on the left side of the reaction. Now, the equation becomes:
$6F{e}^{2+}+14H^{+}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+}+6F{e}^{3+}+H_{2}O$

Balancing hydrogen and oxygen atoms by adding $H_{2}O$
To balance hydrogen and oxygen atoms, adding $7H_{2}O$ molecules to the right side of the reaction. Now, the final balanced reaction is:
$6F{e}^{2+}+14H^{+}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+}+6F{e}^{3+}+7H_{2}O$

$B.$
Identifying species which are getting oxidized and reduced
$I_2+N{O}_3^{-}\to N{O}_2+I{O}_3^{-}$
$I=0N=+5\to N=+4I=+5$
$O=-2$ $O=-2$
This indicates that $I_2$ is getting oxidized and $N{O}_3^{-}$ is getting reduced.

Calculating the increase and decrease of oxidation number to make equal exchange of electrons
$I_2+N{O}_3^{-}\to N{O}_2+I{O}_3^{-}$
$I=0N=+5N=+4I=+5$
$O=-2$ $O=-2$
$I$ is changing its oxidation state from $0$ to $+5$, so $I{O}_3^{-}$ in the product side must be multiplied by a factor of $2$ and $N$ is changing its oxidation state from $+5$ to $+4$, so multiplying bothe $N{O}_3^{-}$ and $N{O}_2$ by a factor of $10$. So now equation becomes:
$I_2+10N{O}_3^{-}\to 10N{O}_2+2I{O}_3^{-}$

Balancing the charges
To balance the charges in acidic medium as in this reaction, $8H^{+}$ ions need to be added on the left side of the reaction. Now, the equation becomes:
$I_2+10N{O}_3^{-}+8H^{+}\to 10N{O}_2+2I{O}_3^{-}$

Balancing hydrogen and oxygen atoms by adding $H_{2}O$
To balance hydrogen and oxygen atoms, adding $4H_{2}O$ molecules to the right side of the reaction. Now, the final balanced reaction is:
$I_2+10N{O}_3^{-}+8H^{+}\to 10N{O}_2+2I{O}_3^{-}+4H_{2}O$

$C.$
Identifying species which are getting oxidized and reduced
$I_2+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$
$I=0S=+2I=-1S=+5/2$
$O=-2$ $O=-2$
This indicates that $I_2$ is getting reduced and $S_2{O}_3^{2-}$ is getting oxidized.

Calculating the increase and decrease of oxidation number to make equal exchange of electrons
$I_2+S_2{O}_3^2-\to I^{-}+S_4{O}_6^{2-}$
$I=0S=+2I=-1S=+2.5$
$O=-2$ $O=-2$
Iodine is changing its oxidation state from $0$ to $-1$, so $I^{-}$ in the product side must be multiplied by a factor of $2$ and sulphur is changing its oxidation state from $+2$ to $+2.5$, so multiplying $S_2{O}_3^{2-}$ by a factor of $2$ and $S_4{O}_6^{2-}$ by a factor of $1$. So, now equation becomes:
$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$

Balancing the charges
To balance the charges in acidic medium as in this reaction, $H^{+}$ ions need to be added but in this reaction charges are already balanced so, no need to add any $H^{+}$ ion. Now, the equation becomes:
$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$

Balancing hydrogen and oxygen atoms by adding $H_{2}O$
To balance hydrogen and oxygen atoms, $H_{2}O$ molecules are added but here all oxygen atoms are already balanced so, no need to add any $H_{2}O$ molecule. Now, the final balanced reaction is:
$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$

$D.$
Identifying species which are getting oxidized and reduced
$Mn{O}_2+C_2{O}_4^{2-}\to M{n}^{2+}+C{O}_2$
$Mn=+4C=+3Mn=+2C=+4$
$O=-2$ $O=-2$
This indicates that $C_2{O}_4^{2-}$ is oxidized and $Mn{O}_2$ is reduced in above reaction.

Calculating the increase and decrease of oxidation number to make equal exchange of electrons
$Mn{O}_2+C_2{O}_4^{2-}\to M{n}^{2+}+C{O}_2$
$Mn=+4C=+3Mn=+2C=+4$
$O=-2$ $O=-2$
Carbon is changing its oxidation state from $+3$ to $+4$, so $C{O}_2$ in the product side must be multiplied by a factor of $2$ and $Mn$ is changing its oxidation state from $+4$ to $+2$, so multiplying both $Mn{O}_2$ and $M{n}^{2+}$ by a factor of $1$. So now equation becomes:
$Mn{O}_2+C_2{O}_4^{2-}\to M{n}^{2+}+2C{O}_2$

Balancing the charges
To balance the charges in acidic medium as in this reaction, $4H^{+}$ ions need to be added on the left side of the reaction. Now, the equation becomes:
$Mn{O}_2+4H^{+}+C_2{O}_4^{2-}\to M{n}^{2+}+2C{O}_2$

Balancing hydrogen and oxygen atoms by adding $H_{2}O$
To balance hydrogen and oxygen atoms, adding $2H_{2}O$ molecules to the right side of the reaction. Now, the final balanced reaction is:
$Mn{O}_2+4H^{+}+C_2{O}_4^{2-}\to M{n}^{2+}+2C{O}_2+2H_{2}O$