Chapter 18, Problem 18.135CP

Interpretation Introduction

Interpretation:

The cathode half reaction for anodizing titanium is same as the one for anodizing aluminium. While the anode half reaction is as follows

$Ti\left(s\right) + 2H_{2}O\left(l\right) \to Ti{O}_2\left(s\right)+ 4H^{+}\left(aq\right)+4e^{-} E^{\circ }=+1.066 V$

Find out the overall cell reaction and cell potential for anodizing titanium.

Concept Introduction: Anodizing of titanium proceeds in the similar fashion as of aluminium, but the resultant coat of $Ti{O}_2$ is very much thinner $\left({10}^{-4}mm\right)$ as compared to that of coat of $A{l}_2{O}_3$. Also, the iridescent colours of anodized titanium result not from the absorption of organic dyes but from the interference of light as it is reflected by the anodized surface.

According to the Nernst equation for the overall cell reaction, cell potential is given as follows:

    $E=E^{\circ }-\frac{0.0592V}{n}log(P_{H_2}{}^2)$

    Where, $E$ = Cell Potential

    $E^{\circ }$ = Standard Cell Potential

    $n$ = Number of electrons involved in overall cell reaction

    $P_{H_2}$ = Pressure of hydrogen Gas