Q4.122CP - Chemistry: Molecular Nature Of Matter And Change

Question

Use the oxidation number method to balance the following equations by placing coefficients in the blanks. Identify the reducing and oxidizing agents:

 $\begin{array}{l}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\_}\mathbf{KOH}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{\_}{\mathbf{K}}_{{}_{\mathbf{2}}}{\mathbf{CrO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{\_}{\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{ClO}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{\_}{\mathbf{MnO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{ClO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{\_}{\mathbf{KMnO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{\_}{\mathbf{MnO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}\mathbf{KOH}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{\_}{\mathbf{CrO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{HSnO}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{\_}{{\mathbf{CrO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{HSnO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{e}\mathbf{\right)}\mathbf{\_}{\mathbf{KMnO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{NaNO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{\_}{\mathbf{MnO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{NaNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}\mathbf{KOH}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{\_}{\mathbf{I}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{\_}{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{\_}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\end{array}$

Step-by-Step Solution

Answer

The reactions are balanced by oxidation number method.

Step 1: Meaning of oxidation and reduction

Oxidation: In this process, electrons are lost by the element and the oxidation number of the element increases. The element that undergoes oxidation are called Reducing agent.

 

Reduction: In this process, the electrons are gained by the element and the oxidation number of the element decreases. The element that undergoes reduction are called Oxidising agent.

 

Step 2: Reaction (a)

The oxidation state for each element present in the given reaction are shown below,

 $\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{K}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}_{\mathbf{2}}{\stackrel{\mathbf{-}\mathbf{1}}{\mathbf{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{3}}{\mathbf{Cr}}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{K}}}_{{}_{\mathbf{2}}}\stackrel{\mathbf{+}\mathbf{6}}{\mathbf{Cr}}{\stackrel{\mathbf{-}\mathbf{6}}{\mathbf{O}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}_{\mathbf{2}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

 

From the reaction, it can be concluded that the oxidation state of chromium is increased from +3 to +6 and the oxidation state of oxygen is decreased from -1 to -2. Thus, the oxidising agent is H₂O₂ and the reducing agent is Cr(OH)₃.

 

Now, the balanced equation for the given chemical reaction is,

 $\mathbf{4}\mathbf{KOH}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{Cr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{K}}_{{}_{\mathbf{2}}}{\mathbf{CrO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

Step 3: Reaction (b)

The oxidation state for each element present in the given reaction are shown below,

 $\stackrel{\mathbf{+}\mathbf{7}}{\mathbf{Mn}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{3}}{\mathbf{Cl}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}_{\mathbf{2}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\stackrel{\mathbf{+}\mathbf{4}}{\mathbf{Mn}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{7}}{\mathbf{Cl}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}{\stackrel{\mathbf{+}}{\mathbf{H}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

 

From the reaction, it can be concluded that the oxidation state of chlorine is increased from +3 to +7 and the oxidation state of manganese is decreased from +7 to +4. Thus, the oxidising agent is MnO₄^- and the reducing agent is ClO₂^-.

 

Now, the balanced equation for the given chemical reaction is,

 $\mathbf{4}{\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{ClO}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{4}{\mathbf{MnO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{ClO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

Step 4: Reaction (c)

The oxidation state for each element present in the given reaction are shown below,

 

 $\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{K}}\stackrel{\mathbf{+}\mathbf{7}}{\mathbf{Mn}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{Na}}}_{\mathbf{2}}\stackrel{\mathbf{+}\mathbf{4}}{\mathbf{S}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}_{\mathbf{2}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\stackrel{\mathbf{+}\mathbf{4}}{\mathbf{Mn}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{Na}}}_{\mathbf{2}}\stackrel{\mathbf{+}\mathbf{6}}{\mathbf{S}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{K}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

From the reaction, it can be concluded that the oxidation state of sulfur is increased from +4 to +6 and the oxidation state of manganese is decreased from +7 to +4. Thus, the oxidising agent is KMnO₄ and the reducing agent is Na₂SO₄.

 

Now, the balanced equation for the given chemical reaction is,

 $\mathbf{2}{\mathbf{KMnO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{MnO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{KOH}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

Step 5: Reaction (d)

The oxidation state for each element present in the given reaction are shown below,

 $\stackrel{\mathbf{+}\mathbf{6}}{\mathbf{Cr}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}\stackrel{\mathbf{+}\mathbf{2}}{\mathbf{Sn}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}_{\mathbf{2}}\stackrel{\mathbf{=}\mathbf{2}}{\mathbf{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\stackrel{\mathbf{+}\mathbf{3}}{\mathbf{Cr}}{{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}\stackrel{\mathbf{+}\mathbf{4}}{\mathbf{Sn}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

 

From the reaction, it can be concluded that the oxidation state of Sn is increased from +2 to +4 and the oxidation state of chromium is decreased from +6 to +3. Thus, the oxidising agent is CrO₄^2- and the reducing agent is HSnO₂^-.

 

Now, the balanced equation for the given chemical reaction is,

 $\mathbf{2}{\mathbf{CrO}}_{\mathbf{4}}^{\mathbf{-}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{HSnO}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}{{\mathbf{CrO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{HSnO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

Step 6: Reaction (e)

The oxidation state for each element present in the given reaction are shown below,

 

 $\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{K}}\stackrel{\mathbf{+}\mathbf{7}}{\mathbf{Mn}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{Na}}\stackrel{\mathbf{+}\mathbf{3}}{\mathbf{N}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}_{\mathbf{2}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\stackrel{\mathbf{+}\mathbf{4}}{\mathbf{Mn}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{Na}}\stackrel{\mathbf{+}\mathbf{5}}{\mathbf{N}}{\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{K}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

From the reaction, it can be concluded that the oxidation state of nitrogen is increased from +3 to +5 and the oxidation state of manganese is decreased from +7 to +4. Thus, the oxidising agent is KMnO₄ and the reducing agent is NaNO₂.

 

Now, the balanced equation for the given chemical reaction is,

 $\mathbf{2}{\mathbf{KMnO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{NaNO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{MnO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{NaNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{KOH}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

Step 7: Reaction (f)

The oxidation state for each element present in the given reaction are shown below,

 ${\stackrel{\mathbf{-}\mathbf{1}}{\mathbf{I}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{0}}{\mathbf{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}_{\mathbf{2}}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }{\stackrel{\mathbf{0}}{\mathbf{I}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\stackrel{\mathbf{-}\mathbf{2}}{\mathbf{O}}{\stackrel{\mathbf{+}\mathbf{1}}{\mathbf{H}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

 

From the reaction, it can be concluded that the oxidation state of iodine is increased from -1 to 0 and the oxidation state of oxygen is decreased from 0 to -2. Thus, the oxidising agent is O₂ and the reducing agent is I^-.

 

Now, the balanced equation for the given chemical reaction is,

 $\mathbf{4}{\mathbf{I}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$