Q21.151CP - Chemistry: Molecular Nature Of Matter And Change

Question

Both Ti and V are reactive enough to displace ${\mathbf{H}}_{\mathbf{2}}$  from water. The difference in their  ${{\mathbf{E}}^{\mathbf{o}}}_{\mathbf{half}\mathbf{-}\mathbf{cell}}$ values is  $\mathbf{0}\mathbf{.}\mathbf{42}$ V.

Given:

$\mathbf{V}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{C}{\mathbf{u}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }{\mathbf{V}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Cu}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ ${\mathbf{\Delta G}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{298}$kJ/mol 

use Appendix D to calculate the ${{\mathbf{E}}^{\mathbf{o}}}_{\mathbf{half}\mathbf{-}\mathbf{cell}}$  values for V and Ti.

Step-by-Step Solution

Answer

The values of ${\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}$  for V and Ti is obtained as:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{V}}^{\mathrm{o}}=-1.20\mathrm{V}\\ {\mathrm{E}}_{\mathrm{Ti}}^{\mathrm{o}}=-1.63\mathrm{V}\end{array}$

Step 1: Chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate a new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

Step 2: Evaluate the value of E o half - cell for V and Ti

To determine the number of electrons participating in the process, divide the reaction into half-reactions:

$\begin{array}{l}\mathrm{Oxidation}:{\mathrm{V}}_{\left(\mathrm{s}\right)}\to {\mathrm{V}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\\ {\mathrm{Reduction}:\mathrm{Cu}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Cu}}_{\left(\mathrm{s}\right)}\end{array}$

In this procedure, two electrons are engaged. The equation  ${\mathrm{\Delta G}}^{\mathrm{o}}=-\mathrm{nF}{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}$ may be used to calculate the ${\mathrm{E}}^{\mathrm{o}}$  given the ${\mathrm{\Delta G}}^{\mathrm{o}}$ .

$\begin{array}{c}{\mathrm{\Delta G}}^{\mathrm{o}}=-298\mathrm{kJ}\\ {=-\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}}\\ =-2{\mathrm{e}}^{-}\times 96,500\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}{\times \mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}\end{array}$

$\begin{array}{c}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=\frac{-{\mathrm{\Delta G}}^{\mathrm{o}}}{\mathrm{nF}}\\ =\frac{298,000 \mathrm{J}}{2{\mathrm{e}}^{-}\times 96,500\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}}\\ =1.54 \mathrm{V}\end{array}$

$\begin{array}{c}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}{=\mathrm{E}}_{\mathrm{reduction}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}\\ {=\mathrm{E}}_{\mathrm{Cu}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{V}}^{\mathrm{o}}1.54\mathrm{V}\\ =0{.34\mathrm{V}-\mathrm{E}}_{\mathrm{V}}^{\mathrm{o}}\\ {\mathrm{E}}_{\mathrm{V}}^{\mathrm{o}}=-1.54 \mathrm{V}+0.34 \mathrm{V}\\ =-1.20\mathrm{V}\end{array}$

As both V and Ti can remove   from water, their standard reduction potentials must be lower than hydrogen's, which is  $-0.83$ V. The difference between the two metals' half-cell value is $0.43$ . Ti will be able to displace hydrogen from water if V has a higher positive reduction potential than Ti.

$\begin{array}{c}0{.43\mathrm{V}=\mathrm{E}}_{\mathrm{V}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{Ti}}^{\mathrm{o}}\\ =(-1{.20\mathrm{V})-\mathrm{E}}_{\mathrm{Ti}}^{\mathrm{o}}\\ {\mathrm{E}}_{\mathrm{Ti}}^{\mathrm{o}}=-(0.43\mathrm{V}+1.20\mathrm{V})\\ =-1.63\mathrm{V}\end{array}$

Therefore, we get:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{V}}^{\mathrm{o}}=-1.20\mathrm{V}\\ {\mathrm{E}}_{\mathrm{Ti}}^{\mathrm{o}}=-1.63\mathrm{V}\end{array}$