91Ó°ÊÓ

The oxidation states of atoms are changed in redox reactions.Redox reactions are defined by the real or formal movement of electrons between chemical entities, with one species usually experiencing oxidation while the other experiences reduction.

${\text{ClO}}_3^{-}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)→{\text{I}}_2\left(\text{s}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)\left[\text{acidic}\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$\begin{array}{c}{\mathrm{ClO}}_3^{-}\left(\text{aq}\right)→{\mathrm{Cl}}^{-}\left(\text{aq}\right)\\ {\mathrm{I}}^{-}\left(\text{aq}\right)→{\mathrm{I}}_2\left(\text{s}\right)\end{array}$

2: Balance all the elements other than Oxygen and Hydrogen.

$\begin{array}{l}{\mathrm{ClO}}_3^{-}\left(aq\right)→{\mathrm{Cl}}^{-}\left(aq\right)\\ 2{\mathrm{I}}^{-}\left(aq\right)→{\mathrm{I}}_2(\mathrm{s})\end{array}$

3: Now add molecules to balance oxygen atoms.

$\begin{array}{l}{\mathrm{ClO}}_3^{-}\left(aq\right)→{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+3{\mathrm{H}}_2\mathrm{O}\\ 2{\mathrm{l}}^{-}\left(aq\right)→{\mathrm{I}}_2\left(s\right)\end{array}$

4: To balance hydrogen atoms we add protons,${\mathrm{H}}^{+}$

$\begin{array}{l}{\mathrm{ClO}}_3^{-}\left(aq\right)+6{\mathrm{H}}^{+}→{\mathrm{Cl}}^{-}\left(aq\right)+3{\mathrm{H}}_2\mathrm{O}\\ 2{\mathrm{l}}^{-}\left(aq\right)→{\mathrm{I}}_2(\mathrm{s})\end{array}$

5: Now balance the charge with electrons, ${\mathrm{e}}^{-}$

$\begin{array}{l}{\mathrm{ClO}}_3^{-}\left(aq\right)+6{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}→{\mathrm{Cl}}^{-}\left(aq\right)+3{\mathrm{H}}_2\mathrm{O}\\ 2{\mathrm{l}}^{-}\left(aq\right)→{\mathrm{I}}_2(\mathrm{s})+2{\mathrm{e}}^{-}\end{array}$

6: Scale the reactions to make the electron count equal.

$\begin{array}{l}{\mathrm{ClO}}_3^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}→{\mathrm{Cl}}^{-}\left(aq\right)+3{\mathrm{H}}_2\mathrm{O}\\ 6I^{-}\left(\mathrm{aq}\right)→3{\mathrm{I}}_2(\mathrm{s})+6e^{-}\end{array}$

7: Add the reactions.

${\mathrm{ClO}}_3^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}+6{\mathrm{l}}^{-}\left(\mathrm{aq}\right)→{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+3{\mathrm{H}}_2\mathrm{O}+3{\mathrm{I}}_2(\mathrm{s})+6{\mathrm{e}}^{-}$

8: Cancel out common terms.

${\mathrm{ClO}}_3^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}+6{\mathrm{ll}}^{-}\left(\mathrm{aq}\right)→{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+3{\mathrm{H}}_2\mathrm{O}+3{\mathrm{I}}_2(\mathrm{s})$

Hence, we get the following balanced reaction:

${\mathrm{ClO}}_3^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}+6{\mathrm{l}}^{-}\left(\mathrm{aq}\right)→{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+3{\mathrm{H}}_2\mathrm{O}+3{\mathrm{I}}_2(\mathrm{s})$

Oxidizing agent: ${\mathrm{ClO}}_3^{-}$

Reducing agent:${\text{I}}^{-}$

(b) ${\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)→{\mathrm{MnO}}_2(\mathrm{s})+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)\left[basic\right]$${\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)→{\mathrm{MnO}}_2(\mathrm{s})+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)\left[basic\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(aq\right)→{\mathrm{MnO}}_2\left(s\right)\\ {\mathrm{SO}}_3^{2-}\left(aq\right)→{\mathrm{SO}}_4^{2-}\left(aq\right)\end{array}$

2: Balance all the elements other than Oxygen and Hydrogen.

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)→{\mathrm{MnO}}_2\left(s\right)\\ {\mathrm{SO}}_3^{2-}\left(aq\right)→{\mathrm{SO}}_4^{2-}\left(aq\right)\end{array}$

3: Now add molecules to balance oxygen atoms.

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(aq\right)→{\mathrm{MnO}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{SO}}_3^{2-}\left(aq\right)+{\mathrm{H}}_2\mathrm{O}→{\mathrm{SO}}_4^{2-}\left(aq\right)\end{array}$

4: To balance hydrogen atoms we add protons,

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(aq\right)+4{\mathrm{H}}^{+}→{\mathrm{MnO}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{SO}}_3^{2-}\left(aq\right)+{\mathrm{H}}_2\mathrm{O}→{\mathrm{SO}}_4^{2-}\left(aq\right)+2{\mathrm{H}}^{+}\end{array}$

5: Now balance the charge with electrons,${\mathrm{e}}^{-}$

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(aq\right)+4{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}→{\mathrm{MnO}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{SO}}_3^{2-}\left(aq\right)+{\mathrm{H}}_2\mathrm{O}→{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\end{array}$

6: Scale the reactions to make the electron count equal.

$\begin{array}{l}2{\mathrm{MnO}}_4^{-}\left(aq\right)+8H^{+}+6e^{-}→2{\mathrm{MnO}}_2(\mathrm{s})+4H_{2}O\\ 3S{O}_3^{2-}\left(aq\right)+3H_{2}O→3S{O}_4^{2-}\left(aq\right)+6H^{+}+6e^{-}\end{array}$

7: Add the reactions.

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}+3{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)+3{\mathrm{H}}_2\mathrm{O}→2{\mathrm{MnO}}_2(\mathrm{s})+4{\mathrm{H}}_2\mathrm{O}+3{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}$

8: Cancel out common terms.

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}^{+}+3{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)→2{\mathrm{MnO}}_2(\mathrm{s})+{\mathrm{H}}_{2}O+3{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$

9: Since the reaction is in the basic medium we add ${\mathrm{OH}}^{-}$to balance the ${\mathrm{H}}^{+}$ions.

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}^{+}+2{\mathrm{OH}}^{-}+3{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)→2{\mathrm{MnO}}_2(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}+3{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}$

10: Combine ${\mathrm{OH}}^{-}\text{ionsand}{\mathrm{H}}^{+}$ions present on the same side to form water.

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}+3{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)→2{\mathrm{MnO}}_2(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}+3{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}$

11: Cancel out common terms.

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}+3{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)→2{\mathrm{MnO}}_2(\mathrm{s})+3{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}$

Hence, we get the following balanced reaction:

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}+3{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)→2{\mathrm{MnO}}_2(\mathrm{s})+3{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}$

Oxidizing agent:${\mathrm{MnO}}_4^{-}$

Reducing agent:${\mathrm{SO}}_3^{2-}$

(c)${\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)→{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{O}}_2(\mathrm{g})\left[acidic\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(aq\right)→{\mathrm{Mn}}^{2+}\left(aq\right)\\ {\mathrm{H}}_2{\mathrm{O}}_2\left(aq\right)→{\mathrm{O}}_2\left(g\right)\end{array}$

2: Balance all the elements other than Oxygen and Hydrogen.

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(aq\right)→{\mathrm{Mn}}^{2+}\left(aq\right)\\ {\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)→{\mathrm{O}}_2\left(g\right)\end{array}$

3: Now add ${\mathrm{H}}_2\mathrm{O}$molecules to balance oxygen atoms.

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(aq\right)→{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)→{\mathrm{O}}_2\left(g\right)\end{array}$

4: To balance hydrogen atoms we add protons,

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}→{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)→{\mathrm{O}}_2(\mathrm{g})+2{\mathrm{H}}^{+}\end{array}$

5: Now balance the charge with electrons,${\mathrm{e}}^{-}$

$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}→{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{H}}_2{\mathrm{O}}_2\left(aq\right)→{\mathrm{O}}_2(\mathrm{g})+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\end{array}$

6: Scale the reactions to make the electron count equal.

$\begin{array}{l}2{\mathrm{MnO}}_4^{-}\left(aq\right)+16H^{+}\left(aq\right)+10e^{-}→2{\mathrm{Mn}}^{2+}\left(aq\right)+8H_{2}O\\ 5H_2{O}_2\left(aq\right)→5{\mathrm{O}}_2(\mathrm{g})+10H^{+}+10e^{-}\end{array}$

7: Add the reactions.

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+16{\mathrm{H}}^{+}+10{\mathrm{e}}^{-}+5{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)→2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2(\mathrm{g})+10{\mathrm{H}}^{+}+10{\mathrm{e}}^{-}$

8: Cancel out common terms.

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)→2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2(\mathrm{g})$

Hence, we get the following balanced reaction:

$2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)→2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2(\mathrm{g})$

Oxidizing agent:${\mathrm{MnO}}_4^{-}$

Reducing agent:${\mathrm{H}}_2{\mathrm{O}}_2$