Question

Balance the following redox equations in basic medium by oxidation number method :
a) ${\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}\underset{}{\to }{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+\mathrm{Zn}(\mathrm{OH}{)}_{6\left(\mathrm{aq}\right)}^{2-}$
b) ${\mathrm{MnCl}}_{2\left(\mathrm{aq}\right)}+{\mathrm{HO}}_{2\left(\mathrm{aq}\right)}^{-}\underset{}{\to }\mathrm{Mn}{\left(\mathrm{OH}\right)}_{3\left(\mathrm{s}\right)}+{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}$
c) ${\mathrm{Mn}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}\underset{}{\to }{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$
d) $\mathrm{Cu}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{s}\right)}+{\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{aq}\right)}\underset{}{\to }{\mathrm{Cu}}_{\left(\mathrm{s}\right)}+{\mathrm{N}}_{2\left(\mathrm{g}\right)}$
e) $\mathrm{Fe}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{s}\right)}+{\mathrm{CrO}}_{4\left(\mathrm{aq}\right)}^{2-}\underset{}{\to }\mathrm{Fe}{\left(\mathrm{OH}\right)}_{3\left(\mathrm{s}\right)}+\mathrm{Cr}{\left(\mathrm{OH}\right)}_{4\left(\mathrm{aq}\right)}^{-}$
f) ${\mathrm{TeO}}_{3\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}\underset{}{\to }{\mathrm{TeO}}_{4\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}$

Answer 1

(a) ${\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}\underset{}{\to }{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+\mathrm{Zn}(\mathrm{OH}{)}_{6\left(\mathrm{aq}\right)}^{2-}$

Step-1:- Oxidation number is assigned to each atom.

Step-2:- The atoms whose oxidation number changes are identified.

Step-3:- The total increase and decrease of oxidation number of atoms are calculated.

Zn(0) → Zn(+4)
Net increase = +4
N(+5) → N(−3)
Net decrease = −8

Step-4:- The factor that makes the total increase and decrease of oxidation number equal is chosen.

The net increase in oxidation number is multiplied by 2. Hence, the coefficient 2 is required for Zn on both sides of the equation.

$2{\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}\underset{}{\to }{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+2\mathrm{Zn}(\mathrm{OH}{)}_{6\left(\mathrm{aq}\right)}^{2-}$

Step-5:- The oxygen atoms on the two sides are counted, and then appropriate number of water molecules are added on the side of the equation deficient in oxygen.

9H₂O is added on the left side of equation.

$2{\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}+9{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+2\mathrm{Zn}(\mathrm{OH}{)}_{6\left(\mathrm{aq}\right)}^{2-}$

The equation is balanced for H-atoms by adding H⁺ ions to the side with less H-atoms.

$2{\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}+9{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+2\mathrm{Zn}(\mathrm{OH}{)}_{6\left(\mathrm{aq}\right)}^{2-}+3{\mathrm{H}}^{+}\left(\mathrm{aq}\right)$

Step-6:- As the reaction occurs in a basic medium, 3OH⁻ is added to both sides of the equation.

$2{\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}+9{\mathrm{H}}_2\mathrm{O}+3{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\underset{}{\to }{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+2\mathrm{Zn}(\mathrm{OH}{)}_{6\left(\mathrm{aq}\right)}^{2-}+3{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+3{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Hence, the balanced redox equation is

$2{\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}+6{\mathrm{H}}_2\mathrm{O}+3{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\underset{}{\to }{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}+2\mathrm{Zn}(\mathrm{OH}{)}_{6\left(\mathrm{aq}\right)}^{2-}$

(b) ${\mathrm{MnCl}}_{2\left(\mathrm{aq}\right)}+{\mathrm{HO}}_{2\left(\mathrm{aq}\right)}^{-}\underset{}{\to }\mathrm{Mn}{\left(\mathrm{OH}\right)}_{3\left(\mathrm{s}\right)}+{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}$

Using the procedure of step (a), the resultant balanced redox equation is

$2{\mathrm{MnCl}}_{2\left(\mathrm{aq}\right)}+{\mathrm{HO}}_{2\left(\mathrm{aq}\right)}^{-}+3{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }2\mathrm{Mn}{\left(\mathrm{OH}\right)}_{3\left(\mathrm{s}\right)}+4{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}$

(c) ${\mathrm{Mn}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}\underset{}{\to }{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

Step-1:- Oxidation number is assigned to each atom.

Step-2:- The atoms whose oxidation number change are identified.

Step-3:- The total increase and decrease of oxidation number of atoms are calculated.

Mn(+2) → Mn(+4)
Net increase = +2
2O(−1) → 2O(−2)
Net decrease = −2

Step-4:- The net increase in oxidation number of Mn and decrease in oxidation number of O is equal. Thus, no coeffecient is required.
${\mathrm{Mn}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}\underset{}{\to }{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

Step-5:- The oxygen atoms on the two sides are counted and then appropriate number of water molecules are added on the side of the equation deficient in oxygen.

H₂O is added on the left side of equation.

${\mathrm{Mn}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

The equation is balanced for H atoms by adding H⁺ ions to the side with less H-atoms.

${\mathrm{Mn}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)$

Step-6:- Since, the reaction occurs in a basic medium, 2OH⁻ is added to both sides of the equation.

${\mathrm{Mn}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Hence, the balanced redox equation is

${\mathrm{Mn}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\underset{}{\to }{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

(d) $\mathrm{Cu}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{s}\right)}+{\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{aq}\right)}\underset{}{\to }{\mathrm{Cu}}_{\left(\mathrm{s}\right)}+{\mathrm{N}}_{2\left(\mathrm{g}\right)}$

Step-1:- Oxidation number is assigned to each atom.

Step-2:- The atoms whose oxidation number change are identified.

Step-3:- The total increase and decrease of oxidation number of atoms are calculated.

Cu(+2) → Cu(0)
Net decrease = +2
N₂(−2) → N₂(0)
Net increase = −4

Step-4:- The factor that makes the total increase and decrease of oxidation number equal is chosen.

The net increase in oxidation number is multiplied by 2. Hence, the coefficient 2 is required for Cu on both sides of the equation.

$2\mathrm{Cu}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{s}\right)}+{\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{aq}\right)}\to 2{\mathrm{Cu}}_{\left(\mathrm{s}\right)}+{\mathrm{N}}_{2\left(\mathrm{g}\right)}$

Step-5:- The oxygen atoms on the two sides are counted and then appropriate number of water molecules are added on the side of the equation deficient in oxygen.

4H₂O is added on the right side of equation.

$2\mathrm{Cu}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{s}\right)}+{\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{aq}\right)}\to 2{\mathrm{Cu}}_{\left(\mathrm{s}\right)}+{\mathrm{N}}_{2\left(\mathrm{g}\right)}+4{\mathrm{H}}_2\mathrm{O}$

Since, the number of H atoms on the two sides on the reaction are already balanced, thus, the resultant redox reaction is

$2\mathrm{Cu}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{s}\right)}+{\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{aq}\right)}\to 2{\mathrm{Cu}}_{\left(\mathrm{s}\right)}+{\mathrm{N}}_{2\left(\mathrm{g}\right)}+4{\mathrm{H}}_2\mathrm{O}$

(e) $\mathrm{Fe}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{s}\right)}+{\mathrm{CrO}}_{4\left(\mathrm{aq}\right)}^{2-}\underset{}{\to }\mathrm{Fe}{\left(\mathrm{OH}\right)}_{3\left(\mathrm{s}\right)}+\mathrm{Cr}{\left(\mathrm{OH}\right)}_{4\left(\mathrm{aq}\right)}^{-}$

Using the procedure of part (d), the resultant redox reaction is

$3\mathrm{Fe}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{s}\right)}+{\mathrm{CrO}}_{4\left(\mathrm{aq}\right)}^{2-}+4{\mathrm{H}}_2\mathrm{O}\underset{}{\to }3\mathrm{Fe}{\left(\mathrm{OH}\right)}_{3\left(\mathrm{s}\right)}+\mathrm{Cr}{\left(\mathrm{OH}\right)}_{4\left(\mathrm{aq}\right)}^{-}+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

(f) ${\mathrm{TeO}}_{3\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}\underset{}{\to }{\mathrm{TeO}}_{4\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}$

Step-1:- Oxidation number is assigned to each atom and atoms other than O are balanced.

Step-2:- The atoms whose oxidation number change are identified.

Step-3:- The total increase and decrease of oxidation number of atoms are calculated.

Te(+4) → Te(+6)
Net increase = +2
Cl₂(0) → 2Cl⁻(−1)
Net decrease = (−2)

Step-4:- The net increase in oxidation number of Te and decrease in oxidation number of Cl is equal. Thus, no coeffecient is required.
${\mathrm{TeO}}_{3\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}\underset{}{\to }{\mathrm{TeO}}_{4\left(\mathrm{aq}\right)}^{2-}+2{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}$

Step-5:- The oxygen atoms on the two sides are counted and then appropriate number of water molecules are added on the side of the equation deficient in oxygen.

H₂O is added on the left side of equation.

${\mathrm{TeO}}_{3\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{TeO}}_{4\left(\mathrm{aq}\right)}^{2-}+2{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}$

The equation is balanced for H atoms by adding H⁺ ions to the side with less H-atoms.

${\mathrm{TeO}}_{3\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{TeO}}_{4\left(\mathrm{aq}\right)}^{2-}+2{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)$

Step-6:- Since, the reaction occurs in a basic medium, 2OH⁻ is added to both sides of the equation.

${\mathrm{TeO}}_{3\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\underset{}{\to }{\mathrm{TeO}}_{4\left(\mathrm{aq}\right)}^{2-}+2{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Hence, the balanced redox equation is

${\mathrm{TeO}}_{3\left(\mathrm{aq}\right)}^{2-}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\underset{}{\to }{\mathrm{TeO}}_{4\left(\mathrm{aq}\right)}^{2-}+2{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}+{\mathrm{H}}_2\mathrm{O}$