Both Ti and V are reactive enough to displace ${\mathbf{H}}_{\mathbf{2}}$ from water. The difference in their role="math" localid="1663897644113" ${{\mathbf{E}}^{\mathbf{o}}}_{\mathbf{half}\mathbf{-}\mathbf{cell}}$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.

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.

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.

role="math" localid="1663898895162" $\begin{array}{c}0{.43\mathrm{V}=\mathrm{E}}_{\mathrm{V}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{Ti}}^{\mathrm{o}}\\ =(-1{.20\mathrm{V})-\mathrm{E}}_{\mathrm{Ti}}^{\mathrm{o}}\\ {\mathrm{E}}_{\mathrm{Ti}}^{\mathrm{o}}=-(0.43\mathrm{V}+1.20\mathrm{V})\\ =-1.63\mathrm{V}\end{array}$

Therefore, we get:

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