Q47P - 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) ${\mathbf{\text{Au}}}^{\mathbf{\text{ + }}}{\mathbf{\text{(aq) + e}}}^{\mathbf{\text{ - }}}\mathbf{\text{nAu(s)}}\mathbf{ }{\mathbf{\text{E}}}^{\mathbf{\text{0}}}\mathbf{\text{ = 1.69V}}$

(2)  ${\mathbf{\text{N}}}_{\mathbf{\text{2}}}{\mathbf{\text{O(g) + 2H}}}^{\mathbf{\text{ + }}}{\mathbf{\text{(aq) + 2e}}}^{\mathbf{\text{ - }}}{\mathbf{\text{nN}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)}}$

(3)$\mathit{C}{\mathit{r}}^{\mathbf{3}}\mathbf{+}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathit{e}}^{\mathbf{-}}\mathit{n}{\mathit{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{77}\mathit{V}{\mathit{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{74}\mathit{V}$

Step-by-Step Solution

Answer

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

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

(1)${\text{Au}}^{+}\left(\text{aq}\right)+{\text{e}}^{-}\to \text{Au}\left(\text{s}\right) E^{\circ }=1.69V$${\text{Au}}^{+}\left(\text{aq}\right)+{\text{e}}^{-}\to \text{Au}\left(\text{s}\right) E^{\circ }=1.69V$

 (2)  ${\text{N}}_2\text{O}\left(\text{g}\right)+2{\text{H}}^{+}+2{\text{e}}^{-}\to {\text{N}}_2\left(\text{g}\right)+{\text{H}}_2{\text{O(I)E}}^{\circ }\text{=1.77V}$

(3) ${\text{Cr}}^{3+}\left(\text{aq}\right)+3{\text{e}}^{-}\to \text{Cr}\left(\text{s}\right)E^{\circ }=-0.74V$

For a reaction to be spontaneous, ${{E^{\circ }}_{cell}}^{>0}$ (a) Combine  ${}^{3\times \left(1\right)}$and reversed of ${}^{\left(3\right)}$:

$3{\text{Au}}^{+}\left(\text{aq}\right)+3{\text{e}}^{-}+\text{Cr}\left(\text{s}\right)\to 3\text{Au}\left(\text{s}\right)+{\text{Cr}}^{3+}\left(\text{aq}\right)+3{\text{e}}^{-}$

Cancel common terms to obtain the spontaneous reaction:

${\text{3Au}}^{\text{ + }}\text{(aq) + Cr(s)}\to {\text{3Au(s) + Cr}}^{\text{3 + }}{{\text{(aq)E}}^{\circ }}_{cell}{{\text{=E}}^{\circ }}_{red}{{\text{-E}}^{\circ }}_{oxi}\text{=1.69-(-0.74)=2.43V}$

 (b) Combine$3\times \left(2\right)$ and reversed of $2\times \left(3\right)$ :

$3{\text{N}}_2\text{O}\left(\text{g}\right)+6{\text{H}}^{+}+6{\text{e}}^{-}+2Cr\left(\text{s}\right)\to 3{\text{N}}_2\left(\text{g}\right)+3{\text{H}}_2\text{O}\left(\text{l}\right)+2{\text{Cr}}^3+\left(\text{aq}\right)+6{\text{e}}^{-}$

Cancel common terms to obtain the spontaneous reaction:

${\text{3N}}_{\text{2}}{\text{O(g) + 6H}}^{\text{ + }}\text{ + 2Cr(s)}\to {\text{3N}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}{\text{O(l) + 2Cr}}^{\text{3 + }}{\text{(aq) E}}_{\text{cell }}^{\text{a}}{\text{ = E}}_{\text{red }}^{\text{a}}{\text{ - E}}_{\text{oxi }}^{\text{a}}\text{ = 1.77 - ( - 0.74) = 2.51V}$

 (c) Combine (2) and reversed of $2\times \left(1\right)$:

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

Cancel common terms to obtain the spontaneous reaction:

${\text{N}}_{\text{2}}{\text{O(g) + 2H}}^{\text{ + }}\text{ + 2Au(s)}\to {\text{N}}_{\text{2}}{\text{(g) + H}}_{\text{2}}{\text{O(l) + 2Au}}^{\text{ + }}{{\text{(aq) E}}^a}_{cell}{{\text{E}}^a}_{red}{{\text{E}}^a}_{oxi}\text{=1.77-(-0.74)=2.52V}$

Oxidising agents :${\text{N}}_2\text{O}\left(\text{g}\right)>{\text{Au}}^{+}\left(\text{aq}\right)>{\text{Cr}}^{3+}\left(aq\right)$ 

Reducing agents : $\text{Cr}\left(\text{s}\right)>\text{Au}\left(\text{s}\right)>{\text{N}}_2\left(\text{g}\right)$

Therefore, the work done is

(a)  $3{\text{Au}}^{+}\left(\text{aq}\right)+\text{Cr}\left(\text{s}\right)\to 3\text{Au}\left(\text{s}\right)+{\text{Cr}}^{3+}{{\text{(aq);E}}^{\circ }}_{cell}\text{=2.43V}$

(b) $3{\text{N}}_2\text{O}\left(\text{g}\right)+6{\text{H}}^{+}+2\text{Cr}\left(\text{s}\right)\to 3{\text{N}}_2\left(\text{g}\right)+3{\text{H}}_2\text{O}\left(\text{l}\right)+2{\text{Cr}}^{3+}{\text{(aq);E}}^{\circ }\text{=2.51V}$

(c) $N_{2}O\left(g\right)+2H^{+}+2Au\left(s\right)\to N_2\left(g\right)+H_{2}O\left(I\right)+2A{u}^{+}\left(aq\right);{E^{\circ }}_{cell}=0.08V$

Oxidising agents: ${\text{N}}_2\text{O}\left(\text{g}\right)>{\text{Au}}^{+}\left(\text{aq}\right)>{\text{Cr}}^{3+}\left(\text{aq}\right)$

Reducing agents : $\text{Cr}\left(\text{s}\right)>\text{Au}\left(\text{s}\right)>{\text{N}}_2\left(\text{g}\right)$