Q48P - Chemistry: Molecular Nature Of Matter And Change

Question

Use the following half-reactions to write three spontaneous reactions  , calculate${E^{\circ }}_{cell}$ for each reaction, and rank the strengths of the oxidizing and reducing agents:

(1)  $\mathbf{2}\mathit{H}\mathit{C}\mathit{I}\mathit{O}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{H}}^{\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathit{C}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{I}\mathbf{\right)}{\mathit{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{63}\mathit{V}$ 

(2)   $\mathit{P}{\mathit{t}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathit{P}\mathit{t}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathit{V}$

(3)    $\mathit{P}\mathit{b}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathit{P}\mathit{b}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathit{S}{{\mathit{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{31}\mathit{V}$        

 

Step-by-Step Solution

Answer

  The reaction to be spontaneous to be combined and reversed solution and the cancel common terms of oxidising agents and reducing agents.

Step 1: To Write three spontaneous reactions from the given half-reactions

 Spontaneous reactions 

(1)  $2\text{HClO}\left(\text{aq}\right)+2\text{H}+\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_2\text{O}({\text{l);E}}^{\circ }\text{=1.63V}$

(2) $P{t}^{2+}\left(aq\right)+2e^{-}Pt\left(s\right)E^{\circ }=1.20V$ 

(3)  _{$PbS{O}_4\left(s\right)+2e^{-}\to Pb\left(s\right)+S{O_4}^{2-}\left(aq\right);E^{\circ }=-0.31V,{E^{\circ }}_{cell}>0$}

(a) Combine (1) and reversed of (2):

$2\text{HClO}\left(\text{aq}\right)+2\text{H}+\left(\text{aq}\right)+2{\text{e}}^{-}+\text{Pt}\left(\text{s}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_2\text{O}\left(\text{l}\right)+{\text{Pt}}^{2+}+2{\text{e}}^{-}$

 Cancel common terms to obtain the spontaneous reaction:

$2HCIO\left(aq\right)+2H+\left(aq\right)+Pt\left(s\right)\to C{I}_2\left(g\right)+2H_{2}O\left(I\right)+P{t}^{2+}{E^{\circ }}_{cell}={E^{\circ }}_{red}-{E^{\circ }}_{pxi}=1.63-1.20=0.43V$

  (b) Combine (1) and reversed of (3):

$2\text{HClO}\left(\text{aq}\right)+2\text{H}+\left(\text{aq}\right)+2{\text{e}}^{-}+{\text{P}}^2\text{b}\left(\text{s}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_2\text{O}\left(\text{l}\right)+{\text{FbSO}}_4\left(\text{s}\right)+2\text{e}$

 Cancel common terms to obtain the spontancous reaction:

 _{$2HCIO\left(aq\right)+2H+\left(aq\right)+Pb\left(s\right)+S{O_4}^{2-}\left(aq\right)\to C{I}_2\left(g\right)+2H_{2}O\left(I\right)+PbSO\left(s\right)$}

 (c) Combine (2) and reversed of (3):

 _{${\text{Pt}}^{2+}+2{\text{e}}^{-}+\text{Pb}\left(\text{s}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)\to \text{Pt}\left(\text{s}\right)+{\text{PbSO}}_4\left(\text{s}\right)+2{\text{e}}^{-}$}

Cancel common terms to obtain the spontaneous reaction:

$P{t}^{2+}+Pb\left(s\right)+S{O_4}^{2-}\left(aq\right)\to Pt\left(s\right)+PbS{O}_4\left(s\right){E^{\circ }}_{cell}={E^{\circ }}_{red}-{E^{\circ }}_{oxi}=1.20-(-0.31)=1.52V$

 Oxidising agents : $2\text{HClO}\left(\text{aq}\right)>{\text{Pt}}^{2+}>{\text{PbSO}}_4\left(\text{s}\right)$ 

Reducing agents : $\text{Pb}\left(\text{s}\right)>\text{Pt}\left(\text{s}\right)>{\text{Cl}}_2\left(g\right)$ 

 

Therefore ,the work is done is 

(a)    $2HCIO\left(aq\right)+2H+\left(aq\right)+Pt\left(s\right)\to C{I}_2\left(g\right)+2H_{2}O\left(I\right)+P{T}^{2+};{E^{\circ }}_{cell}=0.43V$

(b)$\text{HClO}\left(\text{aq}\right)+2\text{H}+\left(\text{aq}\right)+\text{Pb}\left(\text{s}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{H}}_2\text{O}\left(\text{l}\right)+{\text{PbSO}}_4\left(\text{s}\right);{E^{\circ }}_{cell}=1.94V$ (c) _{${\text{P}}^2{\text{t}}^2+\text{Pb}\left(\text{s}\right)+{\text{SO}}_4^2\left(\text{aq}\right)\to \text{Pt}\left(\text{s}\right)+{\text{PbSO}}_4\left(\text{s}\right);{\text{K}}_{\text{coll }}^{\text{o}}=1.51\text{V}$}

Oxidising agents : $2\text{HClO}\left(\text{aq}\right)>{\text{Pt}}^{2+}>{\text{PbSO}}_4\left(\text{s}\right)$ 

Reducing agents :  _{$Pb\left(s\right)>Pt\left(s\right)>C{I}_2\left(g\right)$}