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\mathrm{H}_2 \mathrm{O}_2 is used to restore the colour of old lead paintings. Which among the following action occurs during the restoration?

Option: 1

Converting \mathrm{PbO}_2$ to $\mathrm{Pb}


Option: 2

Oxidising \mathrm{PbS}$ to $\mathrm{PbSO}_4
 


Option: 3

Converting \mathrm{PbCO}_3$ to $\mathrm{Pb}
 


Option: 4

Oxidising \mathrm{PbSO}_3$ to $\mathrm{PbSO}_4


Answers (1)

best_answer

Old lead-based paintings are blackened by the action of \mathrm{H}_2 \mathrm{S} gas. Lead reacts with \mathrm{H}_2 \mathrm{S} and forms \mathrm{PbS} which is black precipitate. During the restoration, \mathrm{H}_2 \mathrm{O}_2 is used to oxidise this \mathrm{PbS} into \mathrm{PbSO}_4.

\mathrm{PbS}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \longrightarrow \mathrm{PbSO}_4(\mathrm{~s})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l})

Posted by

Ritika Harsh

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