Chapter 14, Problem 57QP

Complete and balance the following oxidation-reduction reactions, assuming they occur in basic solution.

$\begin{array}{l}\left(\text{a}\right){\text{ NH}}_{\text{3}}\left(aq\right)+{\text{ClO}}^{-}\left(aq\right)\to {\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\\ \left(\text{b}\right)\text{ Cr}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+{\text{HO}}_2^{-}\left(aq\right)\to {\text{CrO}}_4^{2-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ \left(\text{c}\right){\text{ Br}}_{\text{2}}\left(aq\right)\to {\text{Br}}^{-}\left(aq\right)+{\text{BrO}}^{-}\left(aq\right)\end{array}$

Interpretation Introduction

Interpretation:

The given redox chemical equation, which takes place in basic medium, is to be completed and balanced.

Explanation of Solution

The skeleton chemical equation is shown as follows,

${\text{NH}}_3\left(aq\right){\text{+ClO}}^{-}\left(aq\right)\stackrel{}{\to }{\text{N}}_2{\text{H}}_4\left(aq\right){\text{+Cl}}^{-}\left(aq\right)$ …(1)

The oxidation states are assigned in the above skeleton chemical equation which are shown as follow,

$\stackrel{-3}{\text{N}}{\text{H}}_3\left(aq\right)\text{+}\stackrel{+1}{\text{Cl}}{\text{O}}^{-}\left(aq\right)\stackrel{}{\to }{\stackrel{-2}{\text{N}}}_2{\text{H}}_4\left(aq\right){\text{+Cl}}^{-}\left(aq\right)$ …(2)

Two electrons are added on the right of the half-reaction.

$2\stackrel{-3}{\text{N}}{\text{H}}_3\left(aq\right)\stackrel{}{\to }{\stackrel{-2}{\text{N}}}_2{\text{H}}_4\left(aq\right)\text{+2}{e}^{-}$ …(3)

Two electrons are added on the left side of the reduction-half reaction of the one given.

$\stackrel{+1}{\text{Cl}}{\text{O}}^{-}\left(aq\right)+2e^{-}\stackrel{}{\to }{\text{Cl}}^{-}\left(aq\right)$ …(4)

Further, equation (3) and equation (4) are added.

$2\stackrel{-3}{\text{N}}{\text{H}}_3\left(aq\right)+\stackrel{+1}{\text{Cl}}{\text{O}}^{-}\left(aq\right)\stackrel{}{\to }{\stackrel{-2}{\text{N}}}_2{\text{H}}_4\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$ …(5)

To balance the hydrogen atom $1$ of the water molecules is added to the right side of chemical equation (5).

$2\stackrel{-3}{\text{N}}{\text{H}}_3\left(aq\right)+\stackrel{+1}{\text{Cl}}{\text{O}}^{-}\left(aq\right)\stackrel{}{\to }{\stackrel{-2}{\text{N}}}_2{\text{H}}_4\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$ …(6)

The chemical equation (6) represents the balanced chemical equation for the given reaction.

Interpretation Introduction

Interpretation:

The given redox chemical equation, which takes place in basic medium, is to be completed and balanced.

Explanation of Solution

The skeleton chemical equation is shown as follows,

$\text{Cr}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+{\text{HO}}_2^{-}\left(aq\right)\stackrel{}{\to }{\text{CrO}}_4^{-2}\left(aq\right)+{\text{H}}_2\text{O}\left(aq\right)$ …(7)

The oxidation states are assigned in the above skeleton chemical equation which are shown as follow,

$\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)\text{+H}\stackrel{-1}{{\text{O}}_2^{-}}\left(aq\right)\stackrel{}{\to }\stackrel{+6}{\text{Cr}}{\text{O}}_4^{-2}\left(aq\right){\text{+H}}_2\stackrel{-2}{\text{O}}\left(aq\right)$ …(8)

Three electrons are added on the right side in the oxidation half-reaction.

$\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)\stackrel{}{\to }\stackrel{+6}{\text{Cr}}{\text{O}}_4^{-2}\left(aq\right)+3e^{-}$ …(9

Five electrons are added on the left side of the reduction-half reaction of the one given.

$\text{H}\stackrel{-1}{{\text{O}}_2^{-}}\left(aq\right)+2e^{-}\stackrel{}{\to }{\text{2H}}_2\stackrel{-2}{\text{O}}\left(aq\right)$ …(10)

To balance the electrons on both sides of the overall reaction, equation (9) is multiplied with $2$ .

$2\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{Cr}}{\text{O}}_4^{-2}\left(aq\right)+6e^{-}$ …(11)

To balance the electrons on both sides of the overall reaction, equation (10) is multiplied with $3$ .

$\text{3H}\stackrel{-1}{{\text{O}}_2^{-}}\left(aq\right)+6e^{-}\stackrel{}{\to }{\text{6H}}_2\stackrel{-2}{\text{O}}\left(aq\right)$ …(12)

Further, equation (11) and equation (12) are added.

$\begin{array}{l}\underset{¯}{\begin{array}{l}2\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{Cr}}{\text{O}}_4^{-2}\left(aq\right)+6e^{-}\\ \text{3H}\stackrel{-1}{{\text{O}}_2^{-}}\left(aq\right)+6e^{-}\stackrel{}{\to }{\text{6H}}_2\stackrel{-2}{\text{O}}\left(aq\right)\end{array}}\\ \underset{¯}{2\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+\text{3H}\stackrel{-1}{{\text{O}}_2^{-}}\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{Cr}}{\text{O}}_4^{-2}\left(aq\right)+{\text{6H}}_2\stackrel{-2}{\text{O}}\left(aq\right)}\end{array}$ …(13)

To balance ionic charge on both sides of the chemical reaction, $1$ hydroxyl ion is added on the right side of equation (13).

$2\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+\text{3H}\stackrel{-1}{{\text{O}}_2^{-}}\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{Cr}}{\text{O}}_4^{-2}\left(aq\right)+{\text{6H}}_2\stackrel{-2}{\text{O}}\left(aq\right)+{\text{OH}}^{-}$ …(14)

To balance the hydrogen atom $1$ of the water molecules is added to the left side of chemical equation (14).

$2\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+\text{3H}\stackrel{-1}{{\text{O}}_2^{-}}\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{Cr}}{\text{O}}_4^{-2}\left(aq\right)+5{\text{H}}_2\stackrel{-2}{\text{O}}\left(aq\right)+{\text{OH}}^{-}$ …(15)

The chemical equation (15) represents the balanced chemical equation for the given reaction.

Interpretation Introduction

Interpretation:

The given redox chemical equation, which takes place in basic medium, is to be completed and balanced.

Explanation of Solution

The skeleton chemical equation is shown as follows,

${\text{Br}}_2\left(aq\right)\stackrel{}{\to }{\text{Br}}^{-}\left(aq\right)+{\text{BrO}}^{-}\left(aq\right)$ …(16)

The oxidation states are assigned in the above skeleton chemical equation which are shown as follow,

${\stackrel{0}{\text{Br}}}_2\left(aq\right)\stackrel{}{\to }{\text{Br}}^{-}\left(aq\right)+\stackrel{+1}{\text{Br}}{\text{O}}^{-}\left(aq\right)$ …(17)

Two electrons are added on the left side of the reduction half-reaction.

${\stackrel{0}{\text{Br}}}_2\left(aq\right)+2e^{-}\stackrel{}{\to }{\text{2Br}}^{-}\left(aq\right)$ …(18)

Two electrons are added on the right side of the oxidation-half reaction of the one given.

${\stackrel{0}{\text{Br}}}_2\left(aq\right)\stackrel{}{\to }2\stackrel{+1}{\text{Br}}{\text{O}}^{-}\left(aq\right)+2e^{-}$ …(19)

Further, equation (18) and equation (19) are added.

$4{\stackrel{0}{\text{Br}}}_2\left(aq\right)\stackrel{}{\to }{\text{2Br}}^{-}\left(aq\right)+2\stackrel{+1}{\text{Br}}{\text{O}}^{-}\left(aq\right)$ …(20)

To balance ionic charge on both sides of the chemical reaction, $4$ hydroxyl ions are added on the left side of equation (20).

$4{\stackrel{0}{\text{Br}}}_2\left(aq\right)+4{\text{OH}}^{-}\stackrel{}{\to }{\text{2Br}}^{-}\left(aq\right)+2\stackrel{+1}{\text{Br}}{\text{O}}^{-}\left(aq\right)$ …(21)

To balance the hydrogen atom $2$ of the water molecules are added to the right side of chemical equation (21).

$4{\stackrel{0}{\text{Br}}}_2\left(aq\right)+4{\text{OH}}^{-}\stackrel{}{\to }{\text{2Br}}^{-}\left(aq\right)+2\stackrel{+1}{\text{Br}}{\text{O}}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}$ …(22)

The chemical equation (22) represents the balanced chemical equation for the given reaction.