Q21.43P - Chemistry: Molecular Nature Of Matter And Change

Question

Balance each skeleton reaction, calculate $\mathbf{E}{\mathbf{°}}_{\mathbf{cell}}$, and state whether the reaction is spontaneous:

(a)  ${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + Fe}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) Cl}}}^{\mathbf{\text{ - }}}{\mathbf{\text{(aq) + Fe}}}^{\mathbf{\text{3 + }}}\mathbf{\text{(aq)}}$

(b)  ${\mathbf{\text{Mn}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) + Co}}}^{\mathbf{\text{3 + }}}{\mathbf{\text{(aq) MnO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s) + Co}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq)[acidic]}}$

(c) ${\mathbf{\text{AgCl(s) + NO(g) Ag(s) + Cl}}}^{\mathbf{\text{ - }}}{\mathbf{\text{(aq) + NO}}}_{\mathbf{\text{3}}}^{\mathbf{\text{ - }}}\mathbf{\text{(aq)[acidic]}}$

Step-by-Step Solution

Answer

The required work uses the reactions of electrons and atoms on both side andredox through the half-reaction method use spontaneous and non-spontaneous reaction

Step 1: Given the Half reaction and redox reaction

(a) To balance each reaction, divide into half reactions and balance the electrons and atoms on both sides.

${\text{ OHR :2Fe}}_{\text{(aq)}}^{\text{2 + }}\to {\text{2Fe}}_{\text{(aq)}}^{\text{3 + }}{\text{ + e}}^{\text{ - }}$

${\text{RHR:Cl}}_{\text{2(g)}}{\text{ + 2e}}^{\text{ - }}\to {\text{2Cl}}_{\text{(aq)}}^{\text{ - }}$

${\text{Reaction :Cl}}_{\text{2(g)}}{\text{ + 2Fe}}_{\text{(aq)}}^{\text{2 + }}\to {\text{2Fe}}_{\text{(aq)}}^{\text{3 + }}{\text{ + 2Cl}}_{\text{(aq)}}^{\text{ - }}$

${\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{Fe}}^{\text{3 + }}}^{\text{o}}\text{ = 0.77V}$

${\text{E}}_{\text{reduction }}^{\text{o}}{\text{ = E}}_{{\text{Cl}}_{\text{2}}}^{\text{o}}\text{ = 1.36V}$

${\text{E}}_{\text{cell }}^{\text{o}}{\text{ = E}}_{\text{reduction }}^{\text{o}}{\text{ - E}}_{\text{oxidation }}^{\text{o}}$

${\text{ = E}}_{{\text{Cl}}_{\text{2}}}^{\text{o}}{\text{ - E}}_{{\text{Fe}}_{\text{((q) )}}}^{\text{o + }}$$=1.36-0.77=0.59V$

$\mathrm{E}{°}_{\mathrm{cell}}=0.59\mathrm{V}$, spontaneous reaction

(b) to balance each reaction, divide into half reactions and balance the redox through the half-reaction method.

${\text{Mn}}_{\text{(aq)}}^{\text{2 + }}{\text{ + Co}}_{\text{(aq)}}^{\text{3 + }}\to {\text{MnO}}_{\text{2(s)}}{\text{ + Co}}_{\text{(aq)}}^{\text{2 + }}\text{[ acidic ]}$

Separate the half reactions.

${\text{OHR:M}}^{\text{2 + }}\to {\text{MnO}}_{\text{2}}$

Balance electrons and non- O atoms.

${\text{ OHR :Mn}}^{\text{2 + }}\to {\text{MnM}}_{\text{2}}{\text{ + 2e}}^{\text{ - }}$ 

Balance reaction charges with  ${\text{H}}^{\text{ + }}{\text{OHR:Mn}}^{\text{2 + }}\to {\text{MnM}}_{\text{2}}{\text{ + 2e}}^{\text{ - }}{\text{ + 4H}}^{\text{ + }}{\text{RHR:Co}}^{\text{3 + }}{\text{ + e}}^{\text{ - }}\to {\text{Co}}^{\text{2 + }}$

Balance O atoms by ${\text{H}}_{\text{2}}{\text{O OHR:Mn}}^{\text{2 + }}{\text{ + 2H}}_{\text{2}}\text{O}\to {\text{MnO}}_{\text{2}}{\text{ + 2e}}^{\text{ - }}{\text{ + 4H}}^{\text{ + }}{\text{RHR:Co}}^{\text{3 + }}{\text{ + e}}^{\text{ - }}\to {\text{Co}}^{\text{2 + }}$

 

 

 

Step 2: Define multiple reactions

Multiply reactions by least common factor and combine half reactions, and cancel common ions/ compounds on both sides accordingly. 

${\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}{=\mathrm{E}}_{{\mathrm{MnO}}_2}^{\mathrm{o}}=1.23$

${\mathrm{E}}_{\mathrm{reduction}}^{\mathrm{o}}{=\mathrm{E}}_{{\mathrm{Co}}^{3+}}^{\mathrm{o}}=1.82$

${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}{=\mathrm{E}}_{\mathrm{reduction}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}$

${=\mathrm{E}}_{{\mathrm{Co}}^{3+}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{MnO}}^{\mathrm{o}}=1.82-1.23=0.59V$

(c) to balance each reaction, divide into half reactions and balance the redox through the half-reaction method.

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

Separate the half reactions.

$OHR:NO\to {\mathrm{NO}}_3^{-}RHR:AgCl\to Ag+C{l}^{-}$

 Balance electrons and non- O atoms.

$OHR:NO\to N{O}_3^{-}+3e^{-}RHR:AgCl+2e^{-}\to Ag+C{l}^{-}$

 Balance reaction charges with $H^{+}$ (acidic environment).

$OHR:NO\to N{O}_3^{-}+3e^{-}+4H^{+}RHR:H^{+}+AgCl+2e^{-}\to Ag+C{l}^{-}$

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

$OHR:NO+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NO}}_3^{-}+3e^{-}+4{\mathrm{H}}^{+}$

$RHR:AgCl+e^{-}\to Ag+C{l}^{-}$

Step 3: Finding the combine half reactions, and cancel common ions/ compounds on both sides

$OHR:NO+2H_2\mathrm{O}\to {\mathrm{NO}}_3^{-}+3e^{-}+4H^{+}3xRHR:3\mathrm{AgCl}+3e^{-}\to 3\mathrm{Ag}+3{\mathrm{Cl}}^{-}$

${\mathrm{ReactionNO}}_{\left(g\right)}+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+3{\mathrm{AgCl}}_{\left(s\right)}\to {\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}+3{\mathrm{Ag}}_{\left(s\right)}+3{\mathrm{Cl}}_{\left(aq\right)}^{-}+4{\mathrm{H}}_{\left(aq\right)}^{+}$

${\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}{=\mathrm{E}}_{{\mathrm{NO}}_3^{-}}^{\mathrm{o}}=0.96$

${=\mathrm{E}}_{\mathrm{AgCl}}^{\mathrm{o}}{-\mathrm{E}}_{{\mathrm{NO}}_3^{-}}^{\mathrm{o}}=0.22-0.96=0.74V$

${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=-0.74$,non-spontaneous reaction

Therefore, the solution is 

$\left(a\right)C{l}_{2\left(g\right)}+2F{e}_{\left(aq\right)}^{2+}\to 2F{e}_{\left(aq\right)}^{3+}+2C{l}_{\left(aq\right)}^{-};E_{cell}^o=0.59 \mathrm{V},$spontaneous reaction.

$\left(b\right){\mathrm{Mn}}_{\left(aq\right)}^{2+}+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{Co}}_{\left(aq\right)}^{3+}\to {\mathrm{MnO}}_{2\left(s\right)}+2{\mathrm{Co}}_{\left(aq\right)}^{2+}+4{\mathrm{H}}_{\left(aq\right)}^{+};E_{cell}^o=0.59 \mathrm{V},$spontaneous reaction.

$\left(c\right)N{O}_{\left(g\right)}+2H_2{O}_{\left(l\right)}+3AgC{l}_{\left(s\right)}\to N{O}_{3\left(aq\right)}^{-}+3A_{\left(s\right)}+3C{l}_{\left(aq\right)}^{-}+4H_{\left(aq\right)}^{+};E_{cell}^o=-0.74,$non-spontaneous reaction.