Question

Question: An ambitious chemist discovers an alloy electrode that is capable of catalytically converting ethanol reversibly to carbon dioxide at 25 ⁰C according to the half-reaction

${\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{5}}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{+}\mathbf{15}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{12}{\mathit{H}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{12}{\mathit{e}}^{\mathbf{-}}$

Believing that is discovery is financially important, the chemist patents its composition and designs a fuel-cell that may be represented as

$\mathit{A}\mathit{l}\mathit{l}\mathit{o}\mathit{y}\left|C_2{H}_{5}OH\left(l\right)\right|\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{H}_3{O}^{+}\left(1M\right)||{\mathit{H}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}}\mathbf{\left(}\mathbf{1}\mathit{M}\mathbf{\right)}\left|O_2\right|\mathit{N}\mathit{i}$

1. Write the half-reaction occurring at the cathode
2. Using data from appendix D, Calculate $\mathbf{∆}\mathit{E}$ for the cell at 25⁰C
3. What is the E⁰ value for the ethanol half-cell?

Short answer

Answer:

$$\begin{gathered} 1.O_2\left(g\right)+4H_3{O}^{+}\left(aq\right)+4e^{-}\to 6H_{2}O\left(l\right) \\ 2.E_{cell}^o=1.14V \\ 3.E_{anode}^o=-0.089V \end{gathered}$$

Step-by-step solution

Step-1: Analysis the electron addition part of the reaction.

The given cell reaction can be written as

$Alloy\left|C_2{H}_{5}OH\left(l\right)\right|C{O}_2\left(g\right)+H_3{O}^{+}\left(1M\right)||H_3{O}^{+}\left(1M\right)\left|O_2\left(g\right)\right|Alloy$

The anode of the electrochemical cell is :

$||H_3{O}^{+}\left(1M\right)\left|O_2\left(g\right)\right|Alloy$

Step-2: Obtaining the cathode reaction

Reduction takes place at the cathode. So, the complete cathode reaction is:

$O_2\left(g\right)+4H_3{O}^{+}\left(aq\right)+4e^{-}\to 6H_{2}O\left(l\right)$

The reduction potentialfrom appendix D is $E^o=1.229V$

Hence, the half-reaction at the cathode is:

$O_2\left(g\right)+4H_3{O}^{+}\left(aq\right)+4e^{-}\to 6H_{2}O\left(l\right)$

Step-3: Finding the anode reaction

For calculating the $E_{cell}^o$ at 25⁰Cwe have to calculate $\Delta G_{anode}^o$ occurring in the reaction.

The reaction occurring at the anode is as:

$C_2{H}_{5}OH\left(l\right)+15H_{2}O\left(l\right)\to 2C{O}_2\left(g\right)+12H_3{O}^{+}\left(aq\right)+12e^{-}$

Step-4: Calculating the free energy for the anode reaction

The free energy change for the reaction at anode can be given as:

$$\begin{gathered} \Delta G^{\text{o}}=\Delta G_f^o\left(products\right)-\Delta G_f^o\left(reactants\right) \\ \Delta G^{\text{o}}=\left[2\Delta G_f^o\left(C{O}_2\left(g\right)\right)+3\Delta G_f^o\left(H_{2}O\left(l\right)\right)\right]-\left[\Delta G_f^o\left(C_2{H}_{5}OH\left(l\right)\right)+3\Delta G_f^o\left(O_2\left(g\right)\right)\right] \\ \Rightarrow \Delta G^o=\left[2\times (-394.36KJ)+3\times (-237.1KJ)\right]-\left[-174.89KJ\right] \\ \Rightarrow \Delta G^o=-788.72KJ-711.54KJ+174.89KJ \\ \Rightarrow \Delta G^o=-1325.37KJ \end{gathered}$$

Here, all the values of $∆{G_f}^0$ are obtained from appendix D

Step-5: To calculate the Ecello   at  250C

We know that, $\Delta G^o=-nF{E}_{cell}^o$, where $F=96485 Cmo{l}^{-1}$ and n is the number of moles of electrons.

$$\begin{gathered} E_{cell}^o=\frac{\Delta G^o}{-nF} \\ \Rightarrow E_{cell}^o=\frac{-1325.37KJ}{\left(12mol\right)\left(96485cmo{l}^{-1}\right)} \\ \Rightarrow E_{cell}^o=1.14J{C}^{-1} \\ \Rightarrow E_{cell}^o=1.14V \end{gathered}$$

Hence, the potential of the $E_{cell}^o$ comes out to be 1.14 V

Step-6: The reaction occurring at the anode.

The value for the ethanol half-cell is the reduction potential at the anode.

The reaction involved at the anode is:

$C_2{H}_{5}OH\left(l\right)+15H_{2}O\left(l\right)\to 2C{O}_2\left(g\right)+12H_3{O}^{+}\left(aq\right)+12e^{-}$

Step-7: Calculation of the E0  value for the ethanol half-cell

We know, $E_{cell}^o=E_{cathode}^o-E_{anode}^o$

We have, $E_{cell}^o=1.14V$ and $E_{cathode}^o=1.229V$

Thus,

$$\begin{gathered} E_{anode}^o=1.14V-1.229V \\ \Rightarrow E_{anode}^o=-0.089V \end{gathered}$$

Hence, the reduction potential at anode for the ethanol half-cell comes out of be $-0.089V$