Q22.57CP - Chemistry: Molecular Nature Of Matter And Change

Question

Balance the following redox reactions:

(a) In acidic solution, xenon trioxide reacts with iodide ion to produce xenon gas, triiodide ion $\left({I_3}^{-}\right)$ ,and water.

(b) In basic solution, the hydrogen xenate ion $\left(\mathrm{HXe}{O_4}^{-}\right)$  disproportionate to xenon gas, perxenate ion $\left(\mathrm{Xe}{O_6}^{4-}\right)$ , water, and oxygen gas.

(c) In basic solution, bismuthate ion $\left(\mathrm{Bi}{O_3}^{-}\right)$ reacts with manganese(II) ion to produce bismuth(III) and permanganate ions.

(d) In basic solution, oxygen difluoride reacts to produce fluoride ion, oxygen gas, and water.

(e) In acidic solution, manganese(IV) oxide reacts with sulfite ion to form manganese(II) ion, water, and dithionate ion $\left(S_2{O_6}^{2-}\right)$.

(f) In one of the few reactions known for astatine, the oxidation of actinide ion by chlorine gas in acid solution produces astinate ion $\left(\mathrm{At}{O_3}^{-}\right)$  and chloride ion.

Step-by-Step Solution

Answer

(a) The redox reaction for (a) is ${\mathrm{XeO}}_{3\left(aq\right)}+9{\mathrm{I}}_{\left(aq\right)}^{-}+6{\mathrm{H}}_{\left(aq\right)}^{+}\to {\mathrm{Xe}}_{\left(g\right)}+3{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}+3{\mathrm{I}}_{3\left(aq\right)}^{-}$.

(b) The redox reaction for (b) is $2{\mathrm{HXeO}}_{4\left(aq\right)}^{-}+2{\mathrm{OH}}_{\left(aq\right)}^{-}+\to {\mathrm{Xe}}_{\left(g\right)}+{\mathrm{XeO}}_{6\left(g\right)}^{4-}+{\mathrm{O}}_{2\left(g\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}$.

(c) The redox reaction for (c) is $5{\mathrm{BiO}}_{3\left(aq\right)}^{-}+2{\mathrm{Mn}}_{\left(aq\right)}^{2+}+7{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}\to 5{\mathrm{Bi}}_{\left(aq\right)}^{3+}+2{\mathrm{MnO}}_{4\left(aq\right)}^{-}+14O{\mathrm{H}}_{\left(aq\right)}^{-}$.

(d) The redox reaction for (d) is $2{\mathrm{OH}}_{\left(aq\right)}^{-}+{\mathrm{OF}}_{2\left(aq\right)}\to 2 {\mathrm{F}}_{\left(aq\right)}^{-}+{\mathrm{O}}_{2\left(g\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}$.

(e) The redox reaction for (e) is ${\mathrm{MnO}}_{2\left(aq\right)}+2{\mathrm{SO}}_{3\left(aq\right)}^{2-}+4{\mathrm{H}}_{\left(aq\right)}^{+}\to {\mathrm{Mn}}_{\left(aq\right)}^{2+}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}+{\mathrm{S}}_2{\mathrm{O}}_{6\left(aq\right)}^{2-}$.

(f) The redox reaction for (f) is ${\mathrm{At}}_{\left(aq\right)}^{-}+{\mathrm{Cl}}_{2\left(g\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}\to {\mathrm{AtO}}_{\left(aq\right)}^{-}+2{\mathrm{Cl}}_{\left(aq\right)}^{-}+2{\mathrm{H}}_{\left(aq\right)}^{+}$.

Step 1: Introduction

Redox is a chemical reaction that involves changing the oxidation states of atoms. Redox reactions are defined as the actual or formal movement of electrons between chemical entities, with one species often experiencing oxidation while the other experiences reduction.

Step 2: Subpart (a)

Divide each reaction into half reactions and use the half-reaction method to balance the redox.

${\mathrm{XeO}}_{3\left(aq\right)}+I_{\left(aq\right)}^{-}\to Xe\left(g\right)+I_{3\left(aq\right)}^{-}+H_2{O}_{\left(l\right)}\left[\mathrm{acidic}\right]$

Split the reactions in half.

$\begin{array}{l}OHR:I^{-}\to +I_3^{-}\\ RHR:X_3\to Xe\end{array}$

Electrons and non  atoms must be balanced.

$\begin{array}{l}OHR:3I^{-}\to +I_3^{-}+2e^{-}\\ \mathrm{RHR}:Xe{O}_3+6e^{-}\to Xe\end{array}$

$H^{+}$

is used to balance reaction charges (acidic environment).

$\begin{array}{l}OHR:3I^{-}\to +I_3^{-}+2e^{-}\\ RHR:{\mathrm{XeO}}_3+6e^{-}+6H^{+}\to Xe\end{array}$

Add ${\mathrm{H}}_2\mathrm{O}$ to the opposite side to balance $O$atoms.

$\begin{array}{l}OHR:3I^{-}\to +I_3^{-}+2e^{-}\\ RHR:Xe{O}_3+6e^{-}+6H^{+}\to Xe+3{\mathrm{H}}_{2}O\end{array}$

Combine half reactions and cancel common ions/ compounds on both sides by multiplying reactions by the least common factor.

$\begin{array}{l}3xOHR:9I^{-}\to +3I_3^{-}+6e^{-}\\ \mathrm{RHR}:X_3+6e^{-}+6H^{+}\to Xe+3H_{2}O\\ \mathrm{Reaction}:Xe{O}_{3\left(aq\right)}+9I_{\left(aq\right)}^{-}+6H_{\left(aq\right)}^{+}\to X{e}_{\left(g\right)}+3H_2{O}_{\left(l\right)}+3I_{3\left(aq\right)}^{-}\end{array}$

Therefore, xenon trioxide reacts with iodide ion to produce xenon gas, triiodide ion $\left({{\mathrm{I}}_3}^{-}\right)$ ,and water is ${\mathrm{XeO}}_{3\left(aq\right)}+9{\mathrm{I}}_{\left(aq\right)}^{-}+6{\mathrm{H}}_{\left(aq\right)}^{+}\to {\mathrm{Xe}}_{\left(g\right)}+3{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}+3{\mathrm{I}}_{3\left(aq\right)}^{-}$

Step 3: Subpart (b)

Partition each reaction into half reactions and use the half-reaction approach to balance the redox.

$HX{\mathrm{eO}}_{4\left(aq\right)}^{-}\to X{e}_{\left(g\right)}+Xe{O}_{6\left(g\right)}^{4-}+H_2{O}_{\left(l\right)}+O_{2\left(g\right)}\left[\mathrm{basic}\right]$

Divide each reaction in half and use the half-reaction approach to balance the redox.

$\begin{array}{l}OHR:HXe{O}_4^{-}+H_{2}O\to Xe{O}_6^{4-}+O_2\\ RHR:HXe{O}_4^{-}\to Xe\end{array}$

Electrons and non ${}^{-}O$ atoms must be balanced.

$\begin{array}{l}OHR:HXe{O}_4^{-}+H_{2}O\to Xe{O}_6^{4-}+O_2+6e^{-}\\ RHR:6e^{-}+HXe{O}_4^{-}\to Xe\end{array}$

$O{H}^{-}$ is used to balance reaction charges.

$\begin{array}{l}\mathrm{OHR}:{\mathrm{HXeO}}_4^{-}+{\mathrm{H}}_2\mathrm{O}+9{\mathrm{OH}}^{-}\to {\mathrm{XeO}}_6^{4-}+{\mathrm{O}}_2+6e^{-}\\ \mathrm{RHR}:6e^{-}+\mathrm{HXeO}\end{array}$

Balance $\mathrm{O}$ by adding ${\mathrm{H}}_2\mathrm{O}$ to the contrary side.

$\begin{array}{l}\mathrm{OHR}:HXe{O}_4^{-}+H_2\mathrm{O}+9{\mathrm{HH}}^{-}\to {\mathrm{XeO}}_6^{4-}+{\mathrm{O}}_2+6e^{-}+6{\mathrm{H}}_2\mathrm{O}\\ RH\mathrm{H}:3{\mathrm{H}}_2\mathrm{O}+6e^{-}+{\mathrm{HXO}}_4^{-}\to \mathrm{Xe}+7{\mathrm{OH}}^{-}\end{array}$

Combine half reactions and cancel common ions/ compounds on both sides by multiplying reactions by the least common factor.

$\begin{array}{l}OHR:HXe{O}_4^{-}+H_{2}O+9O{H}^{-}\to Xe{O}_6^{4-}+O_2+6e^{-}+6H_2\mathrm{O}RHR\\ :3{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}+{\mathrm{HXeOH}}_4^{-}\to \mathrm{Xe}+7{\mathrm{OH}}^{-}\\ Reaction\\ :2{\mathrm{HXeO}}_{4\left(aq\right)}^{-}+2{\mathrm{OH}}_{\left(aq\right)}^{-}\to {\mathrm{Xe}}_{\left(g\right)}+{\mathrm{XeO}}_{6\left(g\right)}^{4-}+{\mathrm{O}}_{2\left(g\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}\end{array}$

Hence, the hydrogen xenate ion $\left(\mathrm{HXe}{{\mathrm{O}}_4}^{-}\right)$ disproportionate to xenon gas, perxenate ion $\left(\mathrm{Xe}{{\mathrm{O}}_6}^{4-}\right)$ , water, and oxygen gas is

$2{\mathrm{HXeO}}_{4\left(aq\right)}^{-}+2{\mathrm{OH}}_{\left(aq\right)}^{-}+\to {\mathrm{Xe}}_{\left(g\right)}+{\mathrm{XeO}}_{6\left(g\right)}^{4-}+{\mathrm{O}}_{2\left(g\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}$

Step 4: Subpart (c)

Divide each reaction into half reactions and use the half-reaction approach to balance the redox.

${\mathrm{BiO}}_{3\left(aq\right)}^{-}+M{n}_{\left(aq\right)}^{2+}\to B{i}_{\left(aq\right)}^{3+}+Mn{O}_{4\left(aq\right)}^{-}\left[\mathrm{basic}\right]$

Split the reactions in half.

$\begin{array}{l}OHR:M{n}^{2+}\to Mn{O}_4^{-}\\ RHR:Bi{O}_3^{-}\to B{i}^{3+}\end{array}$

Electrons and non ${}^{-}O$ atoms must be balanced.

$\begin{array}{l}OHR:M{n}^{2+}\to Mn{O}_4^{-}+5e^{-}\\ RHR:Bi{O}_3^{-}+2e^{-}\to B{i}^{3+}\end{array}$

${\mathrm{OH}}^{-}$ is used to balance reaction charges (basic environment).

$\begin{array}{l}OHR:M{n}^{2+}+8O{H}^{-}\to Mn{O}_4^{-}+5e^{-}\\ RHR:Bi{O}_3^{-}+2e^{-}\to B{i}^{3+}+6O{H}^{-}\end{array}$

Add ${\mathrm{OH}}^{-}$ to the other side to balance the $O$atoms.

$\begin{array}{l}OHR:M{n}^{2+}+8O{H}^{-}\to Mn{O}_4^{-}+5e^{-}+4H_{2}O\\ RHR:Bi{O}_3^{-}+2e^{-}+3H_{2}O\to B{i}^{3+}+6O{H}^{-}\end{array}$

Combine half reactions and cancel common ions/ compounds on both sides by multiplying reactions by the least common factor.

$\begin{array}{l}2xOHR:2{\mathrm{Mn}}^{2+}+16O{\mathrm{H}}^{-}\to 2{\mathrm{MnO}}_4^{-}+10e^{-}+8{\mathrm{H}}_2\mathrm{O}\\ 5xRHR:5{\mathrm{BiO}}_3^{-}+10e^{-}+15{\mathrm{H}}_2\mathrm{O}\to 5{\mathrm{Bi}}^{3+}+30{\mathrm{OH}}^{-}\\ Reaction\\ :5{\mathrm{BiO}}_{3\left(aq\right)}^{-}+2{\mathrm{Mn}}_{\left(aq\right)}^{2+}+7{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}\to 5{\mathrm{Bi}}_{\left(aq\right)}^{3+}+2{\mathrm{MnO}}_{4\left(aq\right)}^{-}+14{\mathrm{OH}}_{\left(aq\right)}^{-}\end{array}$

Therefore, bismuth ate ion  $\left(\mathrm{Bi}{{\mathrm{O}}_3}^{-}\right)$ reacts with manganese(II) ion to produce bismuth(III) and permanganate ions is $5{\mathrm{BiO}}_{3\left(aq\right)}^{-}+2{\mathrm{Mn}}_{\left(aq\right)}^{2+}+7{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}\to 5{\mathrm{Bi}}_{\left(aq\right)}^{3+}+2{\mathrm{MnO}}_{4\left(aq\right)}^{-}+14O{\mathrm{H}}_{\left(aq\right)}^{-}$.

Step 5: Subpart (d)

Divide each reaction into half reactions and use the half-reaction approach to balance the redox.

$O{F}_{2\left(aq\right)}\to F_{\left(aq\right)}^{-}+O_{2\left(g\right)}+H_2{O}_{\left(l\right)}\left[\mathrm{basic}\right]$

Split the reactions in half.

$\begin{array}{l}OHR:H_{2}O\to O_2\\ RHR:O_2\to F^{-}+H_{2}O\end{array}$

Electrons and non ${}^{-}O$ must be balanced.

$\begin{array}{l}OHR:H_{2}O\to O_2+4e^{-}\\ RHR:O{F}_2+4e^{-}\to 2F^{-}+H_{2}O\end{array}$

$H^{+}$ is used to balance the reaction charges (acidic environment).

$\begin{array}{l}OHR:4O{H}^{-}+H_{2}O\to O_2+4e^{-}\\ RHR:O{F}_2+4e^{-}\to 2F^{-}+H_{2}O+2O{H}^{-}\end{array}$

Add ${\mathrm{H}}_2\mathrm{O}$ to the other side to balance the $O$ atoms.

$\begin{array}{l}OHR:2H_2\mathrm{O}+4O{H}^{-}+H_2\mathrm{O}\to O_2+4e^{-}\\ RHR:O{F}_2+4e^{-}\to 2F^{-}+H_2\mathrm{O}+2O{H}^{-}+3H_{2}O\end{array}$

Combine half reactions and cancel common ions/ compounds on both sides by multiplying reactions by the least common factor.

$\begin{array}{l}\mathrm{OHR}:2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{OH}}^{-}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4e^{-}\\ \mathrm{RHR}:{\mathrm{OF}}_2+4e^{-}\to 2 {\mathrm{F}}^{-}+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{OH}}^{-}+3{\mathrm{H}}_2\mathrm{O}\\ \mathrm{Reaction}:2{\mathrm{OH}}_{\left(aq\right)}^{-}+{\mathrm{OF}}_{2\left(aq\right)}\to 2 {\mathrm{F}}_{\left(aq\right)}^{-}+{\mathrm{O}}_{2\left(g\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}\end{array}$

As a result, oxygen difluoride reacts to produce fluoride ion, oxygen gas, and water is 

$2{\mathrm{OH}}_{\left(aq\right)}^{-}+{\mathrm{OF}}_{2\left(aq\right)}\to 2 {\mathrm{F}}_{\left(aq\right)}^{-}+{\mathrm{O}}_{2\left(g\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}$

Step 6: Subpart (e)

Divide each reaction into half reactions and use the half-reaction approach to balance the redox.

${\mathrm{MnO}}_{2\left(aq\right)}+{\mathrm{SO}}_{3\left(aq\right)}^{2-}\to {\mathrm{Mn}}_{\left(aq\right)}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}+{\mathrm{S}}_2{\mathrm{O}}_{6\left(\mathrm{aq}\right)}^{2-}\left[\mathrm{acidic}\right]$

Split the reactions in half.

$\begin{array}{l}OHR:{\mathrm{SO}}_3^{2-}\to {\mathrm{S}}_2{\mathrm{O}}_6^{2-}\\ \mathrm{RHR}:{\mathrm{MnO}}_2\to {\mathrm{Mn}}^{2+}\end{array}$

Equilibrium of electrons and non ${}^{-}O$ atoms.

$\begin{array}{l}OHR:2S{O}_3^{2-}\to S_2{O}_6^{2-}+2e^{-}\\ RHR:2e^{-}+{\mathrm{MnO}}_2\to {\mathrm{Mn}}^{2+}\end{array}$

$H^{+}$is used to balance reaction charges (acidic environment).

$\begin{array}{l}\mathrm{OHR}:2{\mathrm{SO}}_3^{2-}\to {\mathrm{S}}_2{\mathrm{O}}_6^{2-}+2e^{-}\\ \mathrm{RHR}:4{\mathrm{H}}^{+}+2e^{-}+{\mathrm{MnO}}_2\to {\mathrm{Mn}}^{2+}+2{\mathrm{H}}_2\mathrm{O}\text{'}\end{array}$

Combine half reactions and cancel common ions/ compounds on both sides by multiplying reactions by the least common factor.

$\begin{array}{l}OHR:2{\mathrm{SO}}_3^{2-}\to {\mathrm{S}}_2{\mathrm{O}}_6^{2-}+2e^{-}\\ RHR:4H^{+}+2e^{-}+{\mathrm{MnO}}_2\to {\mathrm{Mn}}^{2+}+2{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{ReactionMnO}}_{2\left(aq\right)}+2{\mathrm{SO}}_{3\left(aq\right)}^{2-}+4{\mathrm{H}}_{\left(aq\right)}^{+}\to {\mathrm{Mn}}_{\left(aq\right)}^{2+}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}+{\mathrm{S}}_2{\mathrm{O}}_{6\left(aq\right)}^{2-}\end{array}$

Therefore, manganese(IV) oxide reacts with sulfite ion to form manganese(II) ion, water, and dithionate ion $\left({\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}\right)$ is

${\mathrm{MnO}}_{2\left(aq\right)}+2{\mathrm{SO}}_{3\left(aq\right)}^{2-}+4{\mathrm{H}}_{\left(aq\right)}^{+}\to {\mathrm{Mn}}_{\left(aq\right)}^{2+}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}+{\mathrm{S}}_2{\mathrm{O}}_{6\left(aq\right)}^{2-}$

Step 7: Subpart (f)

Divide each reaction into half reactions and use the half-reaction approach to balance the redox.

$A{t}_{\left(aq\right)}^{-}+C{l}_{2\left(g\right)}\to At{O}_{3\left(aq\right)}^{-}+C{l}_{\left(aq\right)}^{-}\left[\mathrm{acidic}\right]$

Split the reactions in half.

$\begin{array}{l}OHR:A{t}^{-}\to At{O}^{-}\\ RHR:{\mathrm{Cl}}_2\to {\mathrm{Cl}}^{-}\end{array}$

Electrons and non ${}^{-}O$ atoms must be balanced.

$\begin{array}{l}OHR:A{t}^{-}\to At{O}^{-}+2e^{-}\\ RHR:C{l}_2+2e^{-}\to 2C{l}^{-}\end{array}$

$H^{+}$is used to balance reaction charges (acidic environment).

$\begin{array}{l}OHR:A{t}^{-}\to At{O}^{-}+2e^{-}+2H^{+}\\ RHR:C{l}_2+2e^{-}\to 2C{l}^{-}\end{array}$

Add ${\mathrm{H}}_2\mathrm{O}$ to the other side to balance the $O$ atoms.

$\begin{array}{l}OHR:A{t}^{-}+H_2\mathrm{O}\to {\mathrm{AtO}}^{-}+2e^{-}+2{\mathrm{H}}^{+}\\ RHR:l_2+2e^{-}\to 2{\mathrm{Cl}}^{-}\end{array}$

Combine half reactions and cancel common ions/ compounds on both sides by multiplying reactions by the least common factor.

$\begin{array}{l}OHR:A{t}^{-}+H_{2}O\to At{O}^{-}+2e^{-}+2H^{+}\\ RHR:C{l}_2+2e^{-}\to 2C{l}^{-}\\ \mathrm{Reaction}:A{t}_{\left(aq\right)}^{-}+C{l}_{2\left(g\right)}+H_2{O}_{\left(l\right)}\to At{O^{-}}_{\left(aq\right)}+2C{l}_{\left(aq\right)}^{-}+2H_{\left(aq\right)}^{+}\end{array}$

Hence, astatine, the oxidation of astinide ion by chlorine gas in acid solution produces astinate ion $\left({\mathrm{AtO}}^{-}\right)$ and chloride ion.

${\mathrm{At}}_{\left(aq\right)}^{-}+{\mathrm{Cl}}_{2\left(g\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}\to {\mathrm{AtO}}_{\left(aq\right)}^{-}+2{\mathrm{Cl}}_{\left(aq\right)}^{-}+2{\mathrm{H}}_{\left(aq\right)}^{+}$