Q21.38 P - Chemistry: Molecular Nature Of Matter And Change

Question

In basic solution${\mathbf{\text{Se}}}^{\mathbf{2}\mathbf{-}}$,  and ${\mathbf{\text{SO}}}_{\mathbf{\text{3}}}^{\mathbf{\text{2 - }}}$ ions react spontaneously

   $$\begin{gathered} {\mathbf{\text{2Se}}}^{\mathbf{\text{2 - }}}{\mathbf{\text{(aq) + 2SO}}}_{\mathbf{\text{3}}}^{\mathbf{\text{2 - }}}{\mathbf{\text{(aq) + 3H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)n}} \\ {\mathbf{\text{2Se(s) + 6OH}}}^{\mathbf{\text{ - }}}{\mathbf{\text{(aq) + S}}}_{\mathbf{\text{2}}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{2 - }}}\mathbf{\text{(aq)}}\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{\text{E}}}_{\mathbf{\text{cell}}}^{\mathbf{\text{o}}}\mathbf{\text{ = 0.35V}} \end{gathered}$$

(a) Write balanced half-reactions for the process. 

(b) If   ${\text{E}}_{\text{sulfite}}^{\text{o}}$is $\text{ - 0.57V}$ , calculate  ${\text{E}}_{\text{selenium.}}^{\text{o}}$

Step-by-Step Solution

Answer

a. $$\begin{gathered} {\text{Reduction:2SO}}_{\text{3(aq)}}^{\text{2 - }}{\text{ + 3H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + 4e}}^{\text{ - }}\to {\text{6OH}}_{\text{(aq)}}^{\text{ - }}{\text{ + S}}_{\text{2}}{\text{O}}_{\text{3(aq)}}^{\text{2 - }} \\ {\text{Oxidation: 2Se}}^{\text{2 - }}\to {\text{2Se}}_{\text{(s)}}{\text{ + 4e}}^{\text{ - }} \end{gathered}$$

b. $E_{selenium}^0\text{ = - 0.92}$

Step 1: Meaning of half-reaction

Redox reactions are frequently balanced using half reactions. After balancing the atoms and oxidation numbers in acidic oxidation-reduction processes, one must add H + ions to balance the hydrogen ions in the half reaction.

Step 2: Determining the balanced half-reaction for the process

${\text{Se}}^{\text{2 - }}$ is oxidised to $\text{Se}$ in basic solution. Whereas  ${\text{SO}}_{\text{3}}^{\text{ - }}$is reduced to  ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 - }}$

 $$\begin{gathered} {\text{Reduction:2SO}}_{\text{3(aq)}}^{\text{2 - }}{\text{ + 3H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + 4e}}^{\text{ - }}{\text{6OH}}_{\text{(aq)}}^{\text{ - }}{\text{ + S}}_{\text{2}}{\text{O}}_{\text{3(aq)}}^{\text{2 - }} \\ {\text{Oxidation:2Se}}^{\text{2 - }}{\text{2Se}}_{\text{(s)}}{\text{ + 4e}}^{\text{ - }} \end{gathered}$$

Step 3: Calculating the E s u l f i t e 0

Given that $E_{\$sulfite}^0$,$\text{ - 0.57V}$   and $E_{cell}^0\text{ = 0.35V}$ , altering the equation used to obtain the  $E_{cell}^0$may be utilised to obtain Selenium's standard reduction potential.

 $E_{Cell}^0=E_{reduced}^0{\text{ - E }}_{oxidized}^0$

Given the above responses,

 $E_{cell}^0=E_{sulfite}^0-E_{selenium}^0$

$E_{selenium}^0$ can be isolated by substituting the known values for data-custom-editor="chemistry" ${\mathrm{E}}_{\mathrm{cell}}^0$ and  $E_{sulfite}^0$

$$\begin{gathered} E_{cell}^0=E_{sulfite}^0-E_{selenium}^0 \\ {E}_{selenium}^0=E_{sulfite}^0-E_{cell}^0 \\ {E}_{selenium}^0\text{ = - 0.57 - 0.35} \\ {E}_{selenium}^0\text{ = - 0.92} \end{gathered}$$