91Ó°ÊÓ

On the anode, where oxidation occurs, the half-reaction is $\mathit{Z}\mathit{n}\left(s\right)\mathbf{→}\mathit{Z}{\mathit{n}}^{\mathbf{2}\mathbf{+}}\left(aq\right)\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}$. The zinc loses two electrons to create Zn²⁺.$\mathit{C}{\mathit{u}}^{\mathbf{2}\mathbf{+}}\left(aq\right)\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathbf{→}\mathit{C}\mathit{u}\left(s\right)$ is the half-reaction on the cathode where reduction happens (s). Copper ions gain electrons and solidify at this point.

Half-reactions for the Castner cell

$$\begin{gathered} \frac{\text{Cathode:}    {\text{Na}}^{+}\left(l\right)+e^{-}→\text{Na}\left(s\right) \\ Anode:4O{\text{H}}^{-}\left(l\right)→{\text{O}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)+4e^{-} \\ }{4N{a}^{+}\left(l\right)+4O{\text{H}}^{-}\left(l\right)→4\text{Na}\left(\text{s}\right)+{\text{O}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)} \end{gathered}$$

Therefore, the above balanced cathode anode half reactions is used to solve problem

We may calculate how much sodium would be squandered if all of the water created in the above stated cell reacted with the obtained Na:

$$\begin{gathered} 4{\text{Na}}^{+}\left(l\right)+4{\text{OH}}^{-}\left(l\right)→4\text{Na}\left(\text{s}\right)+{\text{O}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(g\right) \\ 2\text{Na}\left(\text{s}\right)+2{\text{H}}_2\text{O}\left(g\right)→2\text{NaOH}\left(aq\right)+{\text{H}}_2\left(g\right) \end{gathered}$$

From equation (1), we see that for every 4 moles of sodium obtained, 2 moles of water are obtained as well. We can see from equation (2) that water and sodium react in a$1:1$ ratio, therefore we can deduce that $50\%$of the sodium will be "wasted," and the Castner cell's maximum efficiency will be$50\%$ .

Hence, the Castner cell's maximum efficiency will be $50\%$.