91Ó°ÊÓ

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.

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

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

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,

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

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

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

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,

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

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

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

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,

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

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

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

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,

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

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

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

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,

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

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

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

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,

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