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 role="math" localid="1659507265051" ${\text{E}}_{\text{sulfite}}^{\text{o}}$is $\text{- 0.57V}$, calculate ${\text{E}}_{\text{selenium.}}^{\text{o}}$

Short 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-by-step solution

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.

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 ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}$

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

Calculating the  Esulfite0

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 $E_{sulfite}^0$

role="math" localid="1659509375311" $$\begin{gathered} E_{cell}^0=E_{sulfite}^0-E_{selenium}^0 \\ {E}_{selenium}^0=E_{sulfite}^0-E_{cell}^0 \\ {E}_{selenium}^0\text{= - 0.57 - 0.35} \\ {E}_{selenium}^0\text{= - 0.92} \end{gathered}$$