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This equation represents the preparation of gold sol by $2 \mathrm{AuCl}_3+3 \mathrm{HCHO}+3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Au}+3 \mathrm{HCOOH}+6 \mathrm{HCl}$.
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The correct answer is:
Reduction
In the given reaction, gold $(\mathrm{Au})$ goes from +3 state in $\mathrm{AuCl}_3$ to zero in $\mathrm{Au}$.
Thus, reduction is taking place in the preparation of Au sol.
Thus, reduction is taking place in the preparation of Au sol.
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