Question

Question: Consider a galvanic cell for which the anode reaction is

$\mathbf{\text{Pb}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{Pb}}}^{\mathbf{2}\mathbf{+}}\left(1.0\times {10}^{-2}\text{M}\right)\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}$

and the cathode reaction is 0.640 V

${\mathbf{\text{VO}}}^{\mathbf{2}\mathbf{+}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{\text{M}}\mathbf{)}\mathbf{+}\mathbf{2}{\mathbf{\text{H}}}_{\mathbf{3}}{\mathbf{\text{O}}}^{\mathbf{+}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{\text{M}}\mathbf{)}\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }{\mathbf{\text{V}}}^{\mathbf{3}\mathbf{+}}\left(1.0\times {10}^{-5}\text{M}\right)\mathbf{+}\mathbf{3}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathit{\ell }\mathbf{\right)}$

The measured cell potential is.

(a) Calculate${\mathit{E}}^{\mathbf{0}}$for${\mathbf{\text{VO}}}^{\mathbf{2}\mathbf{+}}\mathbf{\mid }{\mathbf{\text{V}}}^{\mathbf{3}\mathbf{+}}$ thehalf-reaction, using${\mathit{E}}^{\mathbf{0}}\left(P{b}^{2+}|Pb\right)$from Appendix E.

(b) Calculate the equilibrium constant Kat 25⁰Cfor the reaction

Short answer

Answer

for the ${\text{VO}}^{2+}\mid {\text{V}}^{3+}$half-reaction is $E^0\left(\left({\text{VO}}^{2+}\mid {\text{V}}^{3+}\right)\right)=0.514\text{V}$

The equilibrium constant K at 25⁰C for the reaction $K=4.2\times {10}^{21}$

Step-by-step solution

The cell potential.

In an electrochemical cell, the cell potential is defined as the difference in potential between two half cells.

The difference between the potential of anode become reduced and the cathode's potential to become reduced is the cell potential.

$\left({E^0}_{cell}\right)\mathbf{=}{{\mathit{E}}^{\mathbf{0}}}_{\mathbf{c}\mathbf{a}\mathbf{t}\mathbf{h}\mathbf{o}\mathbf{d}\mathbf{e}}\mathbf{-}{{\mathbf{E}}^{\mathbf{0}}}_{\mathbf{a}\mathbf{n}\mathbf{o}\mathbf{d}\mathbf{e}}$

Reduction potential of half- cell anode reaction 

$$\begin{gathered} {\text{E}}_{{\text{Pb}}^{\text{2 +}}\text{/Pb}}{\text{= E}}^0\text{-}\frac{\text{0.0592}}{\text{2}}\text{log}\frac{\text{1}}{{\text{Pb}}^{\text{+ 2}}} \\ \text{= - 0.1263 -}\frac{0.0592}{2}\text{log}\frac{1}{1\times {10}^{-2}} \\ \text{= - 0.1855} \text{V} \end{gathered}$$

Reduction potential for half-cell cathode reaction   

$$\begin{gathered} {\text{E}}_{\text{cell}}{\text{=E}}_{\text{cathode}}{\text{-E}}_{\text{anode}} \\ \text{0.640} {\text{V =E}}_{\text{cathode}}\text{-}\left(\text{- 0.1855V}\right) \\ {\text{E}}_{\text{cathode}}\text{= 0.4545} \text{V} \end{gathered}$$

Reduction potential for standard half-cell cathode reaction

The reduction potential for the cathode is not in standard conditions. So we have to calculate the standard reduction potential for the cathode.

The cathode reaction

${\text{VO}}^{2+}(0.10\text{M})+2{\text{H}}_3{\text{O}}^{+}(0.10\text{M})+e^{-}\to {\text{V}}^{3+}\left(1.0\times {10}^{-5}\text{M}\right)+3{\text{H}}_2\text{O}\left(l\right)$

$$\begin{gathered} {\text{E}}_{cell}{\text{=E}}^{\text{o}}\text{-}\frac{\text{0.0592}}{1}\text{log}\frac{\left[V^{+3}\right]}{\left[V{O_2}^{+}\right]{\left[H_3{O}^{+}\right]}^2} \\ {\text{0.4545 =E}}^{\text{o}}\text{-}\frac{\text{0.0592}}{1}\text{log}\frac{\text{1}\times {\text{10}}^{-5}}{\left(0.10\right){\left(0.10\right)}^2} \\ {\text{E}}^{\text{o}}\text{= - 0.366} \text{V} \end{gathered}$$

Calculate the equilibrium constant

$\text{Pb}\left(s\right)+2{\text{VO}}^{2+}(0.10\text{M})+4{\text{H}}_3{\text{O}}^{+}(0.10\text{M})\to {\text{Pb}}^{2+}\left(1.0\times {10}^{-2}\text{M}\right)+2{\text{V}}^{3+}\left(1.0\times {10}^{-5}\text{M}\right)+6{\text{H}}_2\text{O}\left(l\right)$

$$\begin{gathered} n=2 \\ {E}_{cell}°=E_{cathode}°-E_{anode}° \\ {E}_{cell}°=0.336 V-\left(-0.126V\right) \\ =0.462 \\ lo{g}_{10}K=\frac{n}{0.0592\text{V}}E° \\ =\frac{2}{0.0592\text{V}}\times 0.462\text{V} \\ =15.6 \end{gathered}$$

Solve the steps further:\

$$\begin{gathered} K={\log }^{-1}(15.6) \\ =4\times {10}^{15} \end{gathered}$$