91Ó°ÊÓ

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.

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

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

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