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Q.Give the necessary equations of the following with reasons:

(i) In aqueous solutions both HgCl2 and SnCl2 cannot exist together.
(ii) Ozone is an oxidizing and reducing agent.
Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2018Subjective· 2mImportance★★★★★
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(i) SnCl2 reduces HgCl2, so both cannot exist together; (ii) O3 both oxidises (liberates I2 from KI) and reduces (with H2O2), so it is both an oxidising and reducing agent.

(i) HgCl2 and SnCl2 cannot exist together:

Stannous chloride (SnCl2) is a powerful reducing agent (Sn2+ is readily oxidised to Sn4+). If both are present, SnCl2 reduces HgCl2:

2HgCl2 + SnCl2 -> Hg2Cl2 (white ppt) + SnCl4

With excess SnCl2 the mercurous chloride is further reduced to metallic mercury:

Hg2Cl2 + SnCl2 -> 2Hg (grey/black) + SnCl4

Since a chemical reaction occurs immediately, the two cannot be present together in aqueous solution.

(ii) Ozone as oxidising and reducing agent:

As an oxidising agent (it readily gives up an oxygen atom): 2KI + H2O + O3 -> 2KOH + I2 + O2 (liberates iodine). …

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