Question

Estimate E^o for the half reaction

$2H_{2}O+2e^{-}\to H_2+2O{H}^{-}$

Given the following values of $\Delta {G^o}_f$

$$\begin{gathered} H_{2}O\left(l\right)=-237 kJ/mol \\ {H}_2\left(g\right)=0.0 \\ O{H}^{-}\left(aq\right)=-157 kJ/mol \\ {e}^{-}=0.0 \end{gathered}$$

Compare the values of E^o with the value of E^o given in Table 11.1.

Short answer

The reduction potential of the cell in the given table is -0.83 V. The result obtained for the given problem matches the value given in the table.

Step-by-step solution

Cell potential of an electrochemical cell

The measure of the potential difference between anode and cathode is the cell potential of an electrochemical cell. The overall cell potential can be calculated using,

${E^o}_{cell}={E^o}_{cathode}-{E^o}_{anode}$

Calculation of free energy for the reaction

Given that the half-reaction,

$2H_{2}O\left(I\right)+2e^{-}\to H_2\left(g\right)+2O{H}^{-}\left(aq\right)$

On calculating the free energy using the reaction:

$$\begin{gathered} \Delta G^o=\sum {n_p\Delta G^o}_{\left(products\right)}-\sum {n_f\Delta G^o}_{\left(reactants\right)} \\ =\Delta {G^o}_{f\left(H_{2\left(g\right)}\right)}+2\Delta {G^o}_{f\left(H_{2}O\right(l\left)\right)}-2\Delta {G^o}_{f\left(H_{2}O\right(l\left)\right)}-2\Delta {G^o}_{f\left(e^{-}\right)} \\ =160 KJ \end{gathered}$$

Calculation to find the potential of the cell

On using the other equation to find potential of cell,

$$\begin{gathered} \Delta G^o=-nF{E}^o \\ {E}^o=\frac{-\Delta G^o}{nF} \end{gathered}$$

Faraday’s constant is $F=96500 C mo{l}^{-1}$. On substituting the values in the reaction,

$$\begin{gathered} E^o=\frac{-160kJ}{(2 mol{e}^{-1})(96485 C mo{l}^{-1})}\times \frac{1000 J}{1 kJ} \\ {E}^o=-0.829 J/C \\ {E}^o=-0.83 V \end{gathered}$$