Question

Question: Titanium can be produced by electrolytic reduction from an anhydrous molten salt electrolyte that contains titanium(IV) chloride and a spectator salt that furnishes ions to make the electrolyte conduct electricity. The standard enthalpy of formation ofis, and the standard entropies of${\mathbf{\text{TiCl}}}_{\mathbf{4}}\mathbf{\left(}\mathit{\ell }\mathbf{\right)}\mathbf{,}\mathbf{\text{Ti}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{}\mathbf{,}\mathbf{25330}\mathbf{,}\mathbf{223}{\mathbf{\text{JK}}}^{\mathbf{-}\mathbf{1}}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$, arespectively. What minimum applied voltage will be necessary at 100⁰?

Short answer

Answer

The minimum applied voltage will be -1.73V

Step-by-step solution

 Step 1: ∆H°and ∆S° of a given system.

$\mathbf{∆}{\mathit{H}}^{\mathbf{0}}$is the enthalpy change accompanying a reaction.. It can be calculated by taking the difference between the enthalpy of formation of the products and reactants.

$\mathbf{∆}{\mathit{H}}^{\mathbf{0}}\mathbf{=}\mathbf{∆}{\mathit{H}}_{\mathbf{\text{f}}}^{\mathbf{0}}\mathbf{\left(}\mathbf{\text{products}}\mathbf{\right)}\mathbf{-}\mathbf{∆}{\mathit{H}}_{\mathbf{\text{f}}}^{\mathbf{0}}\mathbf{\left(}\mathbf{\text{reactants}}\mathbf{\right)}$

$\mathbf{∆}{\mathit{S}}^{\mathbf{0}}$is a used to denote the total change in entropy. Entropy is a measure of the degree of randomness or the degree of disorder in a system

${\mathit{S}}^{\mathbf{0}}\mathbf{=}{\left(S_{\text{f}}^0\right)}_{\mathbf{\text{products}}}\mathbf{-}{\left(S_{\text{f}}^0\right)}_{\mathbf{\text{reactants}}}$

Calculate in ∆H0a given system.

Value of is $∆H°$calculated as

role="math" localid="1663666573412" $$\begin{gathered} {\text{TiCl}}_4\left(l\right)\to \text{Ti}\left(s\right)+2{\text{Cl}}_2\left(g\right) \\ ∆H°=H_{\text{f}}^0\left(\text{products}\right)-H_{\text{f}}^0\left(\text{reactants}\right) \\ ∆H°=\left[H_{\text{f}}^0(Ti(s\left)\right)+2∆H_{\text{f}}^0\left({\text{Cl}}_2\left(g\right)\right)\right]-\left[∆H_{\text{f}}^0\left({\text{TiCl}}_4\left(l\right)\right)\right] \\ =0+0-\left(1\text{mol}\right)\left(-750{\text{kJmol}}^{-1}\right) \\ 750\text{kJ} \end{gathered}$$

Calculate ∆S° in a given system.

Value of$∆S°$ is calculated as

$∆S°=∆{\left(S_{\text{f}}^{°}\right)}_{\text{products}}-∆{\left(S_{\text{f}}^{°}\right)}_{\text{reactants}}$

Substitute and solve the problem

$$\begin{gathered} \Delta S^{°}=\left[\Delta S_{\text{f}}^{°}\left(\text{Ti}\right(s\left)\right)+2S_{\text{f}}^{°}\left({\text{Cl}}_2\left(g\right)\right)\right]-\left[\Delta S_{\text{f}}^{°}\left({\text{TiCl}}_4\left(l\right)\right)\right] \\ =\left[\left(1\text{mol}\right)\left(30{\text{JK}}^{-1}{\text{mol}}^{-1}\right)+\left(2\text{mol}\right)\left(223{\text{JK}}^{-1}{\text{mol}}^{-1}\right)\right]-\left[\left(1\text{mol}\right)\left(253{\text{JK}}^{-1}{\text{mol}}^{-1}\right)\right] \\ =30{\text{JK}}^{-1}+446{\text{JK}}^{-1}-253{\text{JK}}^{-1} \\ =223{\text{JK}}^{-1} \end{gathered}$$

Find the value of∆G .

The given temperature is

$$\begin{gathered} T={100}^0\text{C} \\ =373\text{K} \end{gathered}$$

The value of,

$$\begin{gathered} \text{∆G =∆H -∆S} \\ \text{= 750000} \text{J - 373K×233J/K} \\ =668KJ/Mol \end{gathered}$$

The minimum applied voltage.

Use Gibbs free energy,

$$\begin{gathered} \Delta G=-nFE \\ 1{\text{JC}}^{-1}=1\text{V} \end{gathered}$$

The value of E is,

$$\begin{gathered} E=-\frac{\Delta G}{nF} \\ E=-\frac{668\times {10}^3{\text{Jmol}}^{-}}{4\text{mol}\times 96,485{\text{Cmol}}^{-1}} \\ E=-1.73 V \end{gathered}$$