Q22.71CP - Chemistry: Molecular Nature Of Matter And Change

Question

Before the development of the Downs cell, the Castner cell was used for the industrial production of  metal. The Castner cell was based on the electrolysis of molten$\mathbf{\text{NaOH}}$  ,

(a) Write balanced cathode and anode half-reactions for this cell.

(b) A major problem with this cell was that the water produced at one electrode diffused to the other and reacted with the Na. If all the water produced reacted with  , what would be the maximum efficiency of the Castner cell expressed as moles of Na produced per mole of electrons flowing through the cell?

Step-by-Step Solution

Answer

a) Half-reactions for the Castner cell

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

b) The Castner cell's maximum efficiency will be $50\%$ .

Step 1: Concept introduction

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

Step 2: Balanced Cathode anode half reactions

Half-reactions for the Castner cell

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

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

Step 3: Calculating the amount of sodium

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:

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

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  $50\%$.