Displacement Reaction - Practice Questions & MCQ

In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as:

$\mathrm{X}+\mathrm{YZ} \rightarrow \mathrm{XZ}+\mathrm{Y}$

Displacement reactions fit into two categories: metal displacement and non-metal displacement.

• Metal Displacement: A metal in a compound can be displaced by another metal in the uncombined state. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores. A few such  examples are:
  
  $\begin{aligned} & \mathrm{CuSO}_4(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{ZnSO}_4(\mathrm{aq}) \\ & \mathrm{V}_2 \mathrm{O}_5(\mathrm{~s})+5 \mathrm{Ca}(\mathrm{s}) \xrightarrow{\Delta} 2 \mathrm{~V}(\mathrm{~s})+5 \mathrm{CaO}(\mathrm{s}) \\ & \mathrm{TiCl}_4(\mathrm{l})+2 \mathrm{Mg}(\mathrm{s}) \xrightarrow{\Delta} \mathrm{Ti}(\mathrm{s})+2 \mathrm{MgCl}_2(\mathrm{~s})\end{aligned}$
  
  In each case, the reducing metal is a better reducing agent than the one that is being reduced which evidently shows more capability to lose electrons as compared to the one that is reduced.
• Non-metal displacement: The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement. All alkali metals and some alkaline earth metals (Ca, Sr, and Ba) which are very good reductants, will displace hydrogen from cold water.
• Less active metals such as magnesium and iron react with steam to produce dihydrogen gas:
  
  $\begin{aligned} & 2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \quad \rightarrow \quad 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_2(\mathrm{~g}) \\ & \mathrm{Ca}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{aq})+\mathrm{H}_2(\mathrm{~g})\end{aligned}$
• 
  $\mathrm{Mg}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \xrightarrow{\Delta} \mathrm{Mg}(\mathrm{OH})_2(\mathrm{~s})+\mathrm{H}_2(\mathrm{~g})$

1. Identify the oxidation and reduction half reactions and write them separately in ionic form. 
   For example:
   Mn in MnO^-₄ in acidic medium generally goes to MnSO₄ or Mn²⁺
   Write equation like this:
2. $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$
3. Balance each half-reaction separately. This is done according to the following procedure.
   1. Balance all the atoms of both reactions except 'O' and 'H'.
   2. Now balance O and H atoms depending upon the medium of reaction:
      
      Acidic Medium
   3. $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+} \quad$ (acidic medium)

$\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$
Now to balance H atoms, add as many H+ ions required to the side that is deficient in H atoms.
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$

Basic Medium

To balance O atoms, add same number of H₂O molecules to the side having excess of O atoms and add the double the number of OH^- ions to the other side (i.e., to the side deficient in O atoms).

$\mathrm{NO}_3^{-} \longrightarrow \mathrm{NH}_4^{+} \quad$ (basic medium)

$\mathrm{NO}_3^{-}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4^{+}+6 \mathrm{OH}^{-}$

Now to balance H atoms, add same number of OH^- ions to the side in excess of H atoms and then add same number of water molecules to the other side (i.e., the side deficient in H atoms).

$\begin{aligned} & \mathrm{NO}_3^{-}+3 \mathrm{H}_2 \mathrm{O}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4^{+}+6 \mathrm{OH}^{-}+4 \mathrm{OH}^{-} \\ & \mathrm{NO}_3^{-}+7 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4^{+}+10 \mathrm{OH}^{-}\end{aligned}$

Alternatively, for balancing in the basic medium, you can first balance in the acidic medium and then add as many OH^- ions on both sides such that all the H⁺ on one side is consumed to give water. Now complete the net number of water molecules after the above operation.

$\begin{aligned} & \mathrm{NO}_3^{-}+10 \mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+}+3 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{NO}_3^{-}+10 \mathrm{H}^{+}+10 \mathrm{OH}^{-} \longrightarrow \mathrm{NH}_4^{+}+3 \mathrm{H}_2 \mathrm{O}+10 \mathrm{OH}^{-} \\ & \mathrm{NO}_3^{-}+10 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4^{+}+3 \mathrm{H}_2 \mathrm{O}+10 \mathrm{OH}^{-} \\ & \mathrm{NO}_3^{-}+7 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4^{+}+10 \mathrm{OH}^{-}\end{aligned}$

3. Now add electrons to the side deficient in negative charge in order to balance the charge on both sides.

$\begin{aligned} & \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{NO}_3^{-}+7 \mathrm{H}_2 \mathrm{O}+8 \mathrm{e}^{-} \longrightarrow \mathrm{NH}_4^{+}+10 \mathrm{OH}^{-}\end{aligned}$

These are balanced half-reactions in acidic and basic medium respectively.

4. Now add two half-reactions together in such a manner that electrons from both sides cancel. So multiply by coefficients so that number of electrons produced in oxidation equals the number of electrons used in reduction.

Disproportionation reactions are those reactions in which one species having some oxidation state converts into two different oxidation states, one oxidation state is higher and other is lower.

The balancing of the disproportionation reaction by ion electrode method can be understood by the following example.
The chemical reaction is as follows:

$\mathrm{Cl}_2+\mathrm{OH}^{-} \rightarrow \mathrm{ClO}_3^{-}+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$

In this reaction, Cl on the reactant side has zero oxidation state but on the product side, its oxidation states are +5(in $\mathrm{ClO}_3^{-}$) and -1 in Cl^-.

STEP 1: Write oxidation half-reaction 

$\mathrm{Cl}_2 \rightarrow \mathrm{ClO}_3^{-}$

Now balance the chlorine atoms on both sides. Thus the balance equation is as follows:

$\mathrm{Cl}_2 \rightarrow 2 \mathrm{ClO}_3^{-}$

Now chlorine atoms are changing its oxidation states from 0 to 5. Thus, there is a total exchange of 10 electrons. So, write the complete balanced equation as follows:

$
\mathrm{Cl}_2 \rightarrow 2 \mathrm{ClO}_3^{-}+10 \mathrm{e}^{-}
$

STEP 2: Write the reduction half-reaction

$\mathrm{Cl}_2 \rightarrow \mathrm{Cl}^{-}$

Now balance the chlorine atoms on both sides. Thus the balance equation is as follows:

$\mathrm{Cl}_2 \rightarrow 2 \mathrm{Cl}^{-}$

Now in this equation, chlorine atoms are changing its oxidation states from 0 to -1. Thus, there is a total exchange of 2 electrons. So, write the complete balanced equation as follows:

$
\mathrm{Cl}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}
$

Now balance the electrons exchange of equations (i) and (ii) and then add them both. Thus the final added equation is as follows:

$
6 \mathrm{Cl}_2 \rightarrow 2 \mathrm{ClO}_3^{-}+10 \mathrm{Cl}^{-}
$

STEP 3: Balance the charge
In equation(iii), there is a total of -12 charge on the product side and zero charge on the reactant side. Thus, to balance the charge on both sides, add the required number of OH^- ions on the deficient side. Thus,

$12 \mathrm{OH}^{-}+6 \mathrm{Cl}_2 \longrightarrow 2 \mathrm{ClO}_3^{-}+10 \mathrm{Cl}^{-}$

STEP 4: Balance the oxygen atoms
To balance the oxygen atoms, add the required number of H₂O molecules on the deficient side.

$12 \mathrm{OH}^{-}+6 \mathrm{Cl}_2 \longrightarrow 2 \mathrm{ClO}_3^{-}+6 \mathrm{H}_2 \mathrm{O}+10 \mathrm{Cl}^{-}$

This is the final balanced equation for the given disproportionation reaction by ion-electrode method.

While balancing a given reaction by this method, following steps are to be followed :

1. Assign oxidation state to each element (atom) on both sides of the equation and identify which element has been oxidised and which reduced.
2. Calculate the increase or decrease in the oxidation number per atom.
3. Multiply by suitable integers so as to equalise the total increase or decrease in oxidation state of the species involved.
4. Balance atoms undergoing oxidation and reduction apart o and H.
5. Balancing of the total charge in the equation using H⁺ or OH^- depending upon the medium.

• In Acidic medium, count total charge on both sides and balance it by adding H⁺ ions to the required side (i.e., to the side deficient in positive charge). Finally, add enough water molecules to balance H and O atoms to the required side.

• In a Basic medium, balance the charge by adding OH^- ions to the side with excess of positive charge and finally add required number of H₂O molecules to the appropriate side to balance O and H.