Chapter 17, Problem 17.AE

Interpretation Introduction

Interpretation:

The voltage required to electrolyze sodium sulfate with given current and current density has to be determined.  And also the voltage when gold electrodes replaces platinum electrodes has to be calculated.

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

$\text{E = E(cathode)}\text{-}\text{E(anode)}$

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Overpotential: The activation energy of a reaction at an electrode can be overcome by voltage.  The required voltage to apply is called overpotential.

Ohmic potential:  In electrochemical cell, the electrical resistance of a solution while current I flows can be overcome by voltage.  The required voltage to apply is called ohmic potential.

${\text{E}}_{\text{ohmic}}\text{=}\text{IR}$

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

Answer to Problem 17.AE

The voltage required to electrolyze sodium sulfate with given current and current density is $\text{-}\text{2}\text{.35 V}$

The voltage when gold electrodes replaces platinum electrodes is $\text{-2}\text{.78 V}\text{.}$

Explanation of Solution

To determine: The voltage required to electrolyze sodium sulfate with given current and current density.

At cathode: ${\text{2H}}^{\text{+}}\text{ +}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftarrows }{\text{ H}}_{\text{2}}{}_{\text{(g)}}{\text{ E}}^{\text{o}}\text{=}0\text{V}$

At anode: $\text{0}\text{.5}{\text{O}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{2H}}^{\text{+}}\text{ +}{\text{ 2e}}^{\text{-}}\stackrel{}{\rightleftarrows }{\text{H}}_{\text{2}}\text{O}{\text{E}}^{\text{o}}\text{=}\text{1}\text{.229}\text{V}$

$\begin{array}{l}\text{E(cathode) =}\text{0 -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{{\text{P}}_{\text{H}}{}_{{}_{\text{2}}}}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}\\ \\ \text{E(anode)}\text{= 1}\text{.229 }\text{-}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{\text{1}}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}{\text{P}}_{{\text{o}}_{\text{2}}}{}^{\text{0}\text{.5}}}\\ \\ \text{E(cell) =}\text{E(cathode) -}\text{ E(anode)}\\ \\ \text{=}\text{-1}\text{.229 -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}{\text{P}}_{\text{H}}{}_{{}_{\text{2}}}{\text{P}}_{{\text{o}}_{\text{2}}}{}^{\text{0}\text{.5}}\text{=}\text{-1}\text{.299V}\\ \\ \text{E =}\text{ E(cell) - I R - overpotentials}\\ \\ \text{=}\text{1}\text{.229 - (0}\text{.100 A) (2}\text{.00 }\Omega \text{)}\text{-}\text{0}\text{.85 V - 0}\text{.068 V =}\text{2}\text{.35 V}\end{array}$

For gold electrodes, the voltage is,

$\begin{array}{l}\text{E =}\text{ E(cell) - I R - overpotentials}\\ \\ \text{=}\text{1}\text{.229 - (0}\text{.100 A) (2}\text{.00 Ω)}\text{-}\text{0}\text{.963 V - 0}\text{.390 V =}\text{-2}\text{.78 V}\end{array}$

Conclusion

The voltage required to electrolyze sodium sulfate with given current and current density was determined.  And also the voltage when gold electrodes replaces platinum electrodes was calculated.