Question

Amounts of iodine dissolved in aqueous solution ${\mathbf{I}}_{\mathbf{2}}\left(\mathbf{aq}\right)$ can be determined by titration with thiosulphate ion ${\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}^{\mathbf{-}\mathbf{2}}$. The thiosulpate ion is oxidized to ${\mathbf{S}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{6}}^{\mathbf{-}\mathbf{2}}$ while the iodine is reduced to iodide ion . Starch is used as an indicator because it has a strong blue colors in the presence of dissolve iodine .

1. Write a balance equation for this reaction .
2. If$\mathbf{56}\mathbf{.}\mathbf{40} \mathbf{ml}$ of $\mathbf{0}\mathbf{.}\mathbf{100} \mathbf{M}$${\mathit{S}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}$ solution is used to reach the endpoint of a titration of an unknown amount of iodine , calculate the number of moles of iodine originally present .
3. Combine the appropriate half cell potential from appendix E with thermodynamic data from appendix D for the equilibrium

${\mathit{I}}_{\mathbf{2}}\left(s\right)\underset{}{\mathbf{\leftrightarrows }}{\mathit{I}}_{\mathbf{2}}\left(aq\right)$

to calculate the equilibrium constant at ${\mathbf{25}}^{\mathbf{o}}\mathbf{C}$ for the reaction in part (a).

Short answer

1. The overall balanced equation is :${\text{I}}_{\text{2}}\left(\text{aq}\right){\text{+ 2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{- 2}}\left(\text{aq}\right)\to {\text{2I}}^{\text{-}}\left(\text{aq}\right){\text{+S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{- 2}}\left(\text{aq}\right)$
2. The number of moles of iodine titrated are $\text{2.82}\times {\text{10}}^{\text{- 3}} \text{moles}$ .

1. Equilibrium constant =$9\times {10}^{17}$

Step-by-step solution

To find the overall balance equation by adding two half-cell reaction

The anode half -cell reaction for the oxidation of$S_2{O_3}^{2-}$is given below.

$2S_2{O_3}^{2-}\left(aq\right)\to S_4{O_6}^{2-}\left(aq\right)+2e^{-} {E^0}_{anode}=0.0895$

The cathode half reaction for the reduction of$I_2 to I^{-}$is given below.

$I_2\left(aq\right)+2e^{-}\to 2I^{-}\left(aq\right)$${E^0}_{cathode}=0.535V$

The overall balanced equation is obtained by adding two half reaction.

${\text{I}}_{\text{2}}\left(\text{aq}\right){\text{+ 2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{- 2}}\left(\text{aq}\right)\to {\text{2I}}^{\text{-}}\left(\text{aq}\right){\text{+S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{- 2}}\left(\text{aq}\right)$

calculate of the number of moles of iodine present .

One mole of iodine is reduced by two moles of thiosulfate ion.

Therefore,$\text{Mole} \text{ratio =}\frac{\text{1} \text{mol} {\text{I}}_{\text{2}}}{\text{2} \text{mol} {\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{- 2}}}$

Number of moles of$S_2{O_4}^{2-}$=$56.40ml\times \frac{1L}{1000ml}\times 0.100mol/L=0.00564 mol$

Number of moles of ${\text{I}}_{\text{2}}\text{=}\frac{\text{1}}{\text{2}}\times \text{0.0564 = 0.00282} \text{mole}$

calculation of the equilibrium constant

Since,

$2S_2{O_3}^{2-}\left(aq\right)\to S_4{O_6}^{2-}\left(aq\right)+2e^{-} {E^0}_{anode}=0.0895$

$I_2\left(aq\right)+2e^{-}\to 2I^{-}\left(aq\right)$ ${E^0}_{cathode}=0.535V$

The cell potential is :

$$\begin{gathered} {E^0}_{cell}={E^0}_{cathode}-{E^0}_{anode} \\ {E^0}_{cell}=0.535V-0.0895V \\ {E^0}_{cell}=0.446V \end{gathered}$$

For the reaction,

${\text{I}}_{\text{2}}\left(\text{aq}\right){\text{+ 2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{- 2}}\left(\text{aq}\right)\to {\text{2I}}^{\text{-}}\left(\text{aq}\right){\text{+S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{- 2}}\left(\text{aq}\right) ∆{\text{E}}_{\text{cell}}\text{= 0.446V}$

The equilibrium constant is given by :

$\begin{array}{rcl}lo{g}_{10}{K}_1& =& \frac{n}{0.0592}\times {E^0}_{cell}\\ lo{g}_{10}{K}_1& =& \frac{n}{0.0592}\times 0.446V\\ lo{g}_{10}{K}_1& =& 15.06V\\ K_1& =& 1.2\times {10}^{15}\end{array}$

For, ${\text{I}}_{\text{2}}\text{(aq)}\rightleftharpoons {\text{I}}_{\text{2}}\text{(s)}$ , we calculate $∆{\text{G}}^{\text{0}}$ :

$\begin{array}{rcl}∆{\text{G}}^{\text{0}}& =& ∆{{\text{G}}_{\text{products}}}^{\text{0}}\text{-∆}{{\text{G}}_{\text{reactants}}}^{\text{0}}\\ & =& \text{0 - 16.40} \text{KJ}\\ & =& \text{- 16.40} \text{KJ}\end{array}$

$\begin{array}{rcl}{\text{K}}_{\text{2}}& =& {\text{e}}^{-\left(\frac{∆{\text{G}}^{\text{0}}}{\text{RT}}\right)}\\ & =& {\text{e}}^{\text{-}\left(\frac{\text{- 16.40} \text{KJ}}{\text{8.314J/Kmol}\times \text{298K}}\right)}\\ & =& \text{7.5}\times {\text{10}}^{\text{- 2}}\end{array}$

$$\begin{gathered} {\text{I}}_{\text{2}}\left(\text{s}\right){\text{+ 2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{- 2}}\left(\text{aq}\right)\to {\text{2I}}^{\text{-}}\left(\text{aq}\right){\text{+S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{- 2}}\left(\text{aq}\right) \\ \frac{{\text{I}}_{\text{2}}\text{(aq)}\rightleftharpoons {\text{I}}_{\text{2}}\text{(s)}}{{\text{I}}_{\text{2}}\left(\text{aq}\right){\text{+ 2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{- 2}}\left(\text{aq}\right)\to {\text{2I}}^{\text{-}}\left(\text{aq}\right){\text{+S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{- 2}}\left(\text{aq}\right)} \end{gathered}$$

Therefore the equilibrium constant for reaction a is calculated below.

$\begin{array}{rcl}\text{K}& =& {\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\\ & =& \text{9}\times {\text{10}}^{\text{17}}\end{array}$