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Published on: 22/06/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
Explain the bleaching action of Chlorine.
2.
How is chlorine manufactured by the electrolysis of brine.
3.
How does sulphuric acid react with metals at various conditions.
4.
Explain the oxidising property of sulphuric acid.
5.
Explain the oxidising and reducing property of SO2·
1.
Oxidising and bleaching action:
Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen.
\({ H }_{ 2 }O+{ Cl }_{ 2 }\longrightarrow Hcl+\underset { Hypo\ chlorous \ acid }{ HOCl } \)
HOCI \(\longrightarrow \) HCI + (0)
Colouring matter + Nascent oxygen - 7 Colourless oxidation product.
The bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to sulphates and hydrogen sulphide to sulphur.
2FeCl2 + Cl2 \(\longrightarrow \) 2FeCl3
Cl2 + H2O \(\longrightarrow \)HCI + HOCI
2FeSO4 + H2SO4 + HOCI \(\longrightarrow \)Fe2 (SO4)3 + HCI + H2O
Overall reaction
2FeSO4 + H2SO4 + Cl2 \(\longrightarrow \) Fe2(SO4)3 + 2HCI
Cl2 + H2O \(\longrightarrow \) HCI + HOCI
Na2SO3 + HOCI\(\longrightarrow \) Na2SO4 + HCI
Overall reaction
Na2SO3 + H2O+Cl2 \(\longrightarrow \) Na2SO4 + 2HCI
Cl2 + H2S\(\longrightarrow \)2HCI + S
2.
(i) When a solution of brine (NaCl) is electrolysed, Na+ and Cl ions are formed.
(ii) Na+ ion reacts with OH- ions of water and forms sodium hydroxide. Hydrogen and chlorine are liberated as gases.
NaCI \(\longrightarrow \) Na+ + CI-1
H2O \(\longrightarrow \) H+ + OH-
Na+ + OH- \(\longrightarrow \) NaOH
At the cathode,
H+ + e- \(\longrightarrow \) H
H + H\(\longrightarrow \)H2
At the anode,
Cl- \(\longrightarrow \) CI + e-
CI + CI \(\longrightarrow \) Cl2
3.
Reaction with metals:
(i) Sulphuric acid reacts with metals and gives different product depending on the reactants and reacting condition
(ii) Dilute sulphuric acid reacts with metals like: tin, aluminium, zinc to give corresponding: sulphates.
Zn + H2SO4\(\longrightarrow \) ZnSO4 + H2 \(\uparrow \)
2AI + 3H2SO4 \(\longrightarrow \) Al2(SO4)3+ 3H2 \(\uparrow \)
(iii) Hot concentrated sulphuric acid reacts with copper and lead to give the respective sulphates as shown below
Cu + 2H2SO4 \(\longrightarrow \) CuSO4 + 2H2O + SO2\(\uparrow \)
Pb + 2H2SO4 \(\longrightarrow \) PbSO4 + 2H2O + SO2\(\uparrow \)
(iv) Sulphuric acid doesn't react with noble metals like gold, silver and platinum.
4.
Oxidising property of H2SO4:
Sulphuric acid is an oxidising agent as it produces nascent oxygen as shown below.
\({ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { H }_{ 2 }O+\underset { nascentoxygen }{ { SO }_{ 2 } } +\left( O \right) \)
Sulphuric acid oxidises elements such as carbon, sulphur and phosphorus. It also oxides bromide and iodide to bromine and iodine respectively.
C + 2H2SO4 \(\longrightarrow \) 2SO2 + 2H2O + CO2
S + 2H2SO4 \(\longrightarrow \) 3SO2 + 2H2O
P4 + 10H2SO4 \(\longrightarrow \) 4H3PO4 + 10SO2 + 4H2O
H2S + H2SO4 \(\longrightarrow \) SO2 + 2H2O + S
H2SO4 + 2HI \(\longrightarrow \) SO2 + H2O + I2
H2SO4 + 2HBr \(\longrightarrow \) 2SO2 + 2H2O + Br2
5.
Oxidising property :
Sulphur dioxide, oxidises hydrogen sulphide to sulphur and magnesium to magnesium oxide.
\({ 2H }_{ 2 }S+{ SO }_{ 2 }\longrightarrow 3S+{ 2H }_{ 2 }O\)
\(2Mg+{ SO }_{ 2 }\longrightarrow 2MgO+S\)
Reducing property :
As it can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
\({ SO }_{ 2 }+2{ H }_{ 2 }O+{ { Cl }_{ 2 }\longrightarrow { H }_{ 2 }{ SO }_{ 4 }+2HCl }\)
It also reduces potassium permanganate and dichromate to Mn2+ and Cr3+ respectively.
\({ 2KMnO }_{ 4 }+5{ SO }_{ 2 }+2{ H }_{ 2 }O\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ 2MnSO }_{ 4 }+2{ H }_{ 2 }{ SO }_{ 4 }\)
\({ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+{ 3SO }_{ 2 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ Cr }_{ 2 }\left( SO_{ 4 } \right) _{ 3 }+{ H }_{ 2 }O\)
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