Chapter 20.3, Problem 28SSC

Interpretation Introduction

Interpretation:

Voltage for Down’s cell is to be calculated.

Concept introduction:

Electrode potential is capacity of electrode to gain or lose electron when it is dipped in solution of its own ions. The absolute magnitude of cell potential of an electrode cannot be determined as oxidation half reaction or reduction half reaction cannot occur alone.

Down’s cell is the cell in which electrolysis of molten $\text{NaCl}$ occurs which results into $\text{Na}$ metal and ${\text{Cl}}_2$ gas.

Answer to Problem 28SSC

Voltage for Down’s cell that is ${\text{2Na}}^{+}\left(\text{l}\right)+2{\text{Cl}}^{-}\left(\text{l}\right)\to 2\text{Na}\left(\text{l}\right)+{\text{Cl}}_2\left(\text{g}\right)$ is $4.06827\text{V}$.

Explanation of Solution

When molten $\text{NaCl}$ undergo electrolysis, ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions are generated. ${\text{Na}}^{+}$ moves to cathode and undergo reduction to $\text{Na}$ and ${\text{Cl}}^{-}$ move to anode and oxidizes to ${\text{Cl}}_2$ gas.

The overall reaction for Down’s cell is:

${\text{2Na}}^{+}\left(\text{l}\right)+2{\text{Cl}}^{-}\left(\text{l}\right)\to 2\text{Na}\left(\text{l}\right)+{\text{Cl}}_2\left(\text{g}\right)$

The half cell reactions are as follows:

${\text{Na}}^{+}\left(\text{l}\right)+{\text{e}}^{-}\to \text{Na}\left(\text{l}\right)$

${\text{2Cl}}^{-}\left(\text{l}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{e}}^{-}$

As per the latest convention of sign, the electrode at which reduction occurs with respect to standard hydrogen electrode is assigned positive sign or has higher reduction potential and the electrode at which oxidation occurs with respect to standard hydrogen electrode is assigned negative sign or has lower reduction potential.

As per table 20.1, standard potential for the half cell reactions are as follows:

$\begin{array}{l}{\text{Na}}^{+}\left(\text{l}\right)+{\text{e}}^{-}\to \text{Na}\left(\text{l}\right);-2.71\text{V}\\ {\text{2Cl}}^{-}\left(\text{l}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{e}}^{-};1.35827\text{V}\end{array}$

Since chlorine has positive electrode potential then sodium so reduction of chlorine and oxidation of sodium occur.

The formula for cell potential is as follows:

$E_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}$

Where,

$E_{\text{cell}}^{\text{0}}$ is standard cell potential.

$E_{\text{Reduction}}^{\text{0}}$ is half cell potential for reduction.

$E_{\text{oxidation}}^{\text{0}}$ is half cell potential for oxidation.

Substitute $-2.71\text{V}$ for $E_{\text{oxidation}}^{\text{0}}$ and $1.35827\text{V}$ for $E_{\text{Reduction}}^{\text{0}}$ in above formula.

$\begin{array}{c}{E}_{\text{cell}}^{\text{0}}=E_{\text{Reduction}}^{\text{0}}-E_{\text{oxidation}}^{\text{0}}\\ =1.35827\text{V}-\left(-2.71\text{V}\right)\\ =4.06827\text{V}\end{array}$

Voltage for Down’s cell that is ${\text{2Na}}^{+}\left(\text{l}\right)+2{\text{Cl}}^{-}\left(\text{l}\right)\to 2\text{Na}\left(\text{l}\right)+{\text{Cl}}_2\left(\text{g}\right)$ is $4.06827\text{V}$.

Conclusion

Voltage for Down’s cell that is ${\text{2Na}}^{+}\left(\text{l}\right)+2{\text{Cl}}^{-}\left(\text{l}\right)\to 2\text{Na}\left(\text{l}\right)+{\text{Cl}}_2\left(\text{g}\right)$ is $4.06827\text{V}$.