From the following reduction potentials

$$\begin{gathered} l_2\left(s\right)+2e^{-}\rightleftharpoons 2l{E}^{°}=0.535V \\ {l}_2\left(aq\right)+2c^{-}\rightleftharpoons 2l^{-}{E}^{°}=0.620V \\ {l}_3+2e^{-}\rightleftharpoons 3l3l{E}^{°}=0.535V \end{gathered}$$

(a) Calculate the equilibrium constant for${\mathrm{l}}_2\left(\mathrm{aq}\right)+{\mathrm{l}}^{-}\rightleftharpoons {\mathrm{l}}_3^{-}.$

(b)The equilibrium constant for ${\mathrm{l}}_2\left(\mathrm{s}\right)+{\mathrm{l}}^{-}\rightleftharpoons {\mathrm{l}}_3^{-}.$

(c)The solubility (g/L.) of ${\mathrm{l}}_2\left(\mathrm{s}\right)$in water is.

• A chemical reaction's equilibrium constant is the value of its reaction quotient at chemical equilibrium, a state attained by a dynamic chemical system after a sufficient amount of time has passed in which its composition has no measurable tendency to change further.
• The equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture for a given set of reaction conditions. As a result, given the initial composition of a system, known equilibrium constant values can be used to determine the system's composition at equilibrium.
• $K_{°}=K^{!}/l^{-}=\frac{{\left[R\right]}^P{\left[S\right]}^{\sigma }...}{{\left[A\right]}^{\alpha }{\left[B\right]}^{\beta }....}$
  

We have to reverse the reactants and products in third reaction so we can get the reaction from task

a)By reversing the reaction, the standard reduction potential must be multiplied with -1. Now the reactions are:

$$\begin{gathered} l_2\left(s\right)+2e\rightleftharpoons 2l^{-}{E}^{°}=0.620V \\ 3l^{-}\rightleftharpoons l_3^{-}+2-e{E}^{°}=-0.535V \end{gathered}$$

The net reaction is:

$l_2\left(s\right)+3l^{-}\rightleftharpoons 2l^{-}+l_3^{-}$

The standard reduction potential is then:

${\mathrm{E}}^{°}=(0.620-0.535)\mathrm{V}=0.085\mathrm{V}$

By using equation for equilibrium constant, we can calculate it:

$$\begin{gathered} \mathrm{K}=10\frac{\mathrm{n}.{\mathrm{E}}^{°}}{0.05916} \\ \mathrm{k}=10\frac{2.0.085\mathrm{V}}{0.05916} \\ \mathrm{k}=10\frac{2.0.085\mathrm{V}}{0.05916} \\ \mathrm{k}=747,42 \end{gathered}$$

Therefore the equilibrium constant for $l_2\left(aq\right)+l^{-}\rightleftharpoons l_3^{-}$is 747, 42.

We have to reverse the reactants and products in third reaction so we can get the reaction from task

b). By reversing the reaction, the standard reduction potential must be multiplied with -l. Now the reactions are:

$$\begin{gathered} l_2\left(s\right)+2-e\rightleftharpoons 2l^{-}{E}^{°}=0.535V \\ 3l^{-}\rightleftharpoons l_3^{-}+2-e{E}^{°}=0.535V \end{gathered}$$

The net reaction is:

$l_2\left(aq\right)+3l^{-}\rightleftharpoons 2l^{-}+l_3^{-}$

The standard reduction potential is then:

${\mathrm{E}}^{°}=(0.535-0.535)\mathrm{V}=0.000\mathrm{V}$

By using equation 14-25 for equilibrium constant, we can calculate it:

Therefore the equilibrium constant for$l_2\left(S\right)+l^{-}\rightleftharpoons l_3^{-}is1$

c) We can describe solubility of iodine as its ability to mix with water. In other words, its transition from solid state to aqueous state. Solubility of iodine can be calculated by using this reaction:

$l_2\left(s\right)\rightleftharpoons l_2\left(aq\right)$

In order to get the reaction from step 9, we can use the first and second reaction from the task:

$$\begin{gathered} I_2\left(s\right)+2e^{-}\rightleftharpoons 2I^{-}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{E}^{°}=0.535V \\ {I}_2("aq")+2e^{-}\rightleftharpoons 2I^{-}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{E}^{°}=0.620V/\cdot \left(1\right) \end{gathered}$$

We have to reverse the reactants and products in second reaction so we can get the reaction for solubility. By reversing the reaction, the standard reduction potential must be multiplied with -1. Now the reactions are:

$$\begin{gathered} I_2+2e\rightleftharpoons 2I^{-}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{E}^{°}=0.535V \\ 2I^{-}\rightleftharpoons I_2("aq")+2-e\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{E}^{°}=-0.620V \end{gathered}$$

The net reaction is:

$I_2\left(s\right)\rightleftharpoons I_2("aq")$

The standard reduction potential is then:

${\mathrm{E}}^{°}=(0.620-0.535)\mathrm{V}=-0.085\mathrm{V}$

The solubility of solid is equal to the equilibrium constant:

$$\begin{gathered} K=\frac{\left[I_2\right(aq\left)\right]}{\left[I_2\right(s\left)\right]}=\frac{\left[I_2\right(aq\left)\right]}{1}=\left[I_2\right(aq\left)\right]=10\frac{n.E^{°}}{0.05916} \\ \left[I_2\right(aq\left)\right]=10\frac{2.(-0.085V)}{0.05916} \\ \left[I_2\right(aq\left)\right]=1.338\times {10}^{-3}M \end{gathered}$$

To calculate solubility in we need to multiply molarity with the molar mass of iodine (M=253.8g/mol):

$$\begin{gathered} \mathrm{solubility}=\mathrm{\gamma }=\mathrm{c}.\mathrm{M}=1.338\times {10}^{-3}\frac{\mathrm{mot}}{\mathrm{L}}.253.8\frac{\mathrm{g}}{\mathrm{mroT}} \\ \mathrm{solubility}=0.340\mathrm{g}/\mathrm{L} \end{gathered}$$

Therefore the solubility (g/L.) of ${\mathrm{l}}_2\left(\mathrm{s}\right)$in water is 0.340 g /L..