Question

Complete and balance the following equations for reactions taking place in basic solution.

(a)$\mathbf{Cr}{\left(\mathbf{OH}\right)}_{\mathbf{3}}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{Br}}_{\mathbf{2}}\left(\mathbf{aq}\right)\to {\mathbf{CrO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{Br}}^{\mathbf{-}}\left(\mathbf{aq}\right)$

(b)$\mathbf{ZrO}{\left(\mathbf{OH}\right)}_{\mathbf{2}}\left(\mathbf{s}\right){\mathbf{+}\mathbf{SO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\left(\mathbf{aq}\right)\to \mathbf{Zr}\left(\mathbf{s}\right){\mathbf{+}\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\left(\mathbf{aq}\right)$

(c)${\mathbf{HPbO}}_{\mathbf{2}}^{\mathbf{-}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{Re}\left(\mathbf{s}\right)\to \mathbf{Pb}\left(\mathbf{s}\right){\mathbf{+}\mathbf{ReO}}_{\mathbf{4}}^{\mathbf{-}}\left(\mathbf{aq}\right)$

(d)${\mathbf{HXeO}}_{\mathbf{4}}^{\mathbf{-}}\left(\mathbf{aq}\right)\to {\mathbf{XeO}}_{\mathbf{6}}^{\mathbf{4}\mathbf{-}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{Xe}\left(\mathbf{g}\right)$

(e)${\mathbf{N}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\left(\mathbf{aq}\right){\mathbf{+}\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\left(\mathbf{aq}\right)\to {\mathbf{N}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}\mathbf{CO}\left(\mathbf{g}\right)$

Short answer

(a) $\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{+ 3 Br}}_{\text{2}}{\text{+ 10 OH}}^{\text{-}}\to {\text{CrO}}_{\text{4}}^{\text{2 -}}\left(\text{aq}\right){\text{+ 3 Br}}^{\text{-}}\left(\text{aq}\right){\text{+ 5H}}_{\text{2}}\text{O}$

(b)$\text{2 ZrO}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right){\text{+ 4 SO}}_{\text{3}}^{\text{-}}\left(\text{aq}\right)\to \text{2 Zr}\left(\text{s}\right){\text{+ 4 SO}}_{\text{4}}^{\text{2 -}}\left(\text{aq}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{l}\right)$

(c)${\text{7 HPbO}}_{\text{2}}^{\text{-}}\left(\text{aq}\right)\text{+ 2 Re}\left(\text{s}\right){\text{+ OH}}^{\text{-}}\to \text{7 Pb}\left(\text{s}\right){\text{+ 2 ReO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}$

(d)${\text{14 HXeO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right)\to {\text{6 XeO}}_{\text{6}}^{\text{4 -}}\left(\text{aq}\right)\text{+ 8 Xe}\left(\text{g}\right){\text{+ 7H}}_{\text{2}}{\text{O + 26 OH}}^{\text{-}}$

(e)${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{aq}\right){\text{+ CO}}_{\text{3}}^{\text{2 -}}\left(\text{aq}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)\text{+ CO}\left(\text{g}\right){\text{+ 2H}}_{\text{2}}\text{O}$

Step-by-step solution

Balancing redox reactions

Redox reactions comprise reactions where processes like oxidation and reduction happen together. There are several methods to balance redox reactions. They include the oxidation method and ion-electron method.

Balancing the equations for the reactions taking place in the basic solution

The steps of balancing the redox reaction in the basic medium is:

Note down the skeletal equations of all the reactants and products of the reaction.

Specify the oxidation number of each element and identify the element that undergoes a change in the oxidation state.

Calculate the increase or decrease in oxidation number per atom and identify the oxidizing and reducing agents.

Multiply the formulae of oxidizing and reducing agent by a suitable coefficient so as to equalize the total increase or decrease in oxidation number as calculated in step 3.

The atoms other than H or O must be balanced.

Finally balance the oxygen atoms by incorporating water molecules by hit and trial method.

In case of ionic reactions (Basic medium) , first balance the oxygen atoms by adding water molecule to the side deficient in oxygen atoms. The H atoms are then balanced by adding water molecules equal in number to the deficiency of H atoms an equal number of ions to the opposite side of the equation.

The balanced reactions in the basic medium are:

(a) $\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{+ 3 Br}}_{\text{2}}{\text{+ 10 OH}}^{\text{-}}\to {\text{CrO}}_{\text{4}}^{\text{2 -}}\left(\text{aq}\right){\text{+ 3 Br}}^{\text{-}}\left(\text{aq}\right){\text{+ 5H}}_{\text{2}}\text{O}$

(b)$\text{2 ZrO}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right){\text{+ 4 SO}}_{\text{3}}^{\text{-}}\left(\text{aq}\right)\to \text{2 Zr}\left(\text{s}\right){\text{+ 4 SO}}_{\text{4}}^{\text{2 -}}\left(\text{aq}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{l}\right)$

(c)${\text{7 HPbO}}_{\text{2}}^{\text{-}}\left(\text{aq}\right)\text{+ 2 Re}\left(\text{s}\right){\text{+ OH}}^{\text{-}}\to \text{7 Pb}\left(\text{s}\right){\text{+ 2 ReO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}$

(d)${\text{14 HXeO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right)\to {\text{6 XeO}}_{\text{6}}^{\text{4 -}}\left(\text{aq}\right)\text{+ 8 Xe}\left(\text{g}\right){\text{+ 7H}}_{\text{2}}{\text{O + 26 OH}}^{\text{-}}$

(e)${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{aq}\right){\text{+ CO}}_{\text{3}}^{\text{2 -}}\left(\text{aq}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)\text{+ CO}\left(\text{g}\right){\text{+ 2H}}_{\text{2}}\text{O}$