The Old Nassau (Gold and Black) Reaction

Last updated
Save as PDF

Page ID: 222030

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Chemical Concept Demonstrated

Chemical kinetics

Demonstration

Three solutions are combined together in alphabetical order.:

Solution A: starch solution
Solution B: HgCl₂ solution
Solution C: KIO₃ solution

Video 1: The attractiveness of Landolt reaction is mainly due to spectacular, even more unusual than in other reactions clock, the time course of processes, complicated its course, and the myriad possibilities in modifying the show.

Observations

The resulting solution turns yellow, then turns black.

Explanation (including important chemical equations)

This demonstration is a variant of the Landolt or iodine clock reaction.

\[IO^-_{3(aq)} + 3 HSO^-_{3(aq)} \rightarrow I^-_{(aq)} + 3 SO^{2-}_{4(aq)} + 3 H^+_{(aq)} \tag{1}\]

\[IO^-_{3(aq)} + 5 I^-_{(aq)} + 6 H^+_{(aq)} \rightarrow 3 I_{2 (aq)} + 3 H_2O_{(l)} \tag{2}\]

\[I_{2 (aq)} + HSO^-_{3(aq)} + H_2O_{(l)} \rightarrow 2 I^-_{(aq)} + SO^{2-}_{4(aq)} + 3 H^+_{(aq)} \tag{3}\]

\[ I_{2 (aq)} + I^-_{(aq)} + \text{starch}_{(aq)} \rightarrow \text{starch}-I^{3-}_{(aq)} \tag{4}\]

The presence of HgCl₂ means that a fifth (and perhaps sixth) reaction can occur.

\[Hg^{2+}_{(aq)} + 2 I^-_{(aq)} \rightarrow HgI_{2 (s)} \tag{5}\]

\[HgI_{2 (s)} + 2 I^-_{(aq)} \rightarrow HgI^{2-}_{4(aq)} \tag{6}\]

The yellow HgI₂ precipitate forms as soon as the I^{^-} concentration is large enough for the ion product to exceed the solubility product for HgI₂. Using the quantities specified above, the HgI₂ precipitate does not dissolve. Instead, the yellow suspension of the HgI₂ precipitate is masked by the appearance of the deep-blue-black starch-I^{^3-} complex when the HSO₃^{^-} is exhausted.

An alternative form of the demonstration can be prepared by doubling the concentration of HSO₃^{^-} in solution A. There is now enough HSO₃^{^-} present initially to consume all of the IO₃^{^-} ion. The HSO₃^{^-} is never exhausted and the I^{^-}concentration can become high enough to redissolve the HgI₂ precipitate. The solution, therefore, turns from clear to yellow and then back to clear again.

Contributors

George M. Bodner (Purdue University)