Question

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

(a)role="math" localid="1663418747793" ${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + Fe}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{(aq) Cl}}}^{\mathbf{\text{-}}}{\mathbf{\text{(aq) + Fe}}}^{\mathbf{\text{3 +}}}\mathbf{\text{(aq)}}$

(b)role="math" localid="1663418761168" ${\mathbf{\text{Mn}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{(aq) + Co}}}^{\mathbf{\text{3 +}}}{\mathbf{\text{(aq) MnO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s) + Co}}}^{\mathbf{\text{2 +}}}\mathbf{\text{(aq)[acidic]}}$

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

Short 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-by-step solution

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.

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

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

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.