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Redox Reactions

Long Answer Type

91.

Why do the following reactions proceed differently?

Pb₃O₄ +8HCl --->3PbCl₂ +Cl₂ +4H₂O
Pb₃O₄ +4HNO₃ --->2Pb(NO₃)₂ +PbO₂ +2H₂O

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92. While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agent is their reactions, ozone and nitric acid act only an oxidants. Why?

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93.

Consider the reaction:

Why does the same reductant—thiosulphate react differently with iodine and bromine?

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Short Answer Type

94. Why does the following reaction occur?

bold XeO subscript bold 6 superscript bold 4 bold minus end superscript bold left parenthesis bold aq bold right parenthesis bold space bold plus bold 2 bold F to the power of bold minus bold left parenthesis bold aq bold right parenthesis bold space bold plus bold 6 bold H to the power of bold plus bold left parenthesis bold aq bold right parenthesis bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold rightwards arrow bold XeO subscript bold 3 bold left parenthesis bold g bold right parenthesis bold space bold plus bold F subscript bold 2 bold left parenthesis bold g bold right parenthesis bold plus bold 3 bold H subscript bold 2 bold O bold left parenthesis bold l bold right parenthesis

conclusion about the compound Na₄XeO₆ (of which XeO₆^4– is a part) can be drawn from the reaction ?

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Long Answer Type

95. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compunds, hydroiodic acid is the best reductant.

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96. When a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed when the oxidising agent is in excess. Justify this statement giving three illustrations.

The Fact given in the question is clear from the following illustration.

First illustration: Carbon is a reducing agent while oxygen is an oxidising agent; Let carbon be taken in excess and allowed to burn in a limited supply of O₂. CO has formed in which carbon has +2 oxidation state (lower oxidation state).

Also, let O₂ be taken in excess, then initially formed CO gets oxidised to CO₂ in which carbon has +4 oxidation state (higher oxidation state).

Second illustration. Sodium is a reducing agent while oxygen is an oxidising agent. Let sodium be taken in excess and allowed to burn in a limited supply of O₂. Sodium oxide (Na₂O) is formed in which oxygen has -2 oxidation state.

On the other hand, let O₂ be taken in excess, then Na₂O₂ is formed in which has -1 oxidation state which is higher than -2.

Third illustration P₄ is a reducing agent while Cl₂ is an oxidising agent. Let P₄ be taken in excess and allow to react it with Cl₂, then PCl₃ is formed in which P has +3 oxidation state.

On the other hand, let Cl₂ be taken in excess, then initially formed PCl₃ reacts further to form PCl₅in which P has +5 oxidation state.

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97. How do you account for the following observations?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants; yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox for the equation reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapours of bromine. Why?

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98.

Consider the reactions:

$bold left parenthesis bold a bold right parenthesis bold space bold space bold 6 bold CO subscript bold 2 bold left parenthesis bold g bold right parenthesis bold space bold plus bold space bold 6 bold H subscript bold 2 bold O bold left parenthesis bold l bold right parenthesis bold space bold space bold rightwards arrow bold space bold space bold space bold C subscript bold 6 bold H subscript bold 12 bold O subscript bold 6 bold left parenthesis bold aq bold right parenthesis bold space bold plus bold space bold 6 bold O subscript bold 2 bold left parenthesis bold g bold right parenthesis$
Why is it more appropriate to write these reactions as:
$bold left parenthesis bold a bold right parenthesis bold space bold 6 bold CO subscript bold 2 bold left parenthesis bold g bold right parenthesis bold space bold plus bold space bold 12 bold H subscript bold 2 bold O bold left parenthesis bold l bold right parenthesis bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold rightwards arrow bold space bold space bold space bold C subscript bold 6 bold H subscript bold 12 bold O subscript bold 6 bold left parenthesis bold aq bold right parenthesis bold space bold plus bold space bold 6 bold H subscript bold 2 bold O bold left parenthesis bold I bold right parenthesis bold space bold plus bold space bold 6 bold O subscript bold 2 bold left parenthesis bold g bold right parenthesis bold left parenthesis bold b bold right parenthesis bold space bold O subscript bold 3 bold left parenthesis bold g bold right parenthesis bold space bold plus bold space bold H subscript bold 2 bold O subscript bold 2 bold left parenthesis bold l bold right parenthesis bold space bold space bold rightwards arrow bold space bold space bold space bold space bold space bold H subscript bold 2 bold O bold left parenthesis bold l bold right parenthesis bold space bold plus bold space bold O subscript bold 2 bold left parenthesis bold g bold right parenthesis bold space bold plus bold space bold O subscript bold 2 bold left parenthesis bold g bold right parenthesis$

Also, suggest a technique to investigate the path of the above (a) and (b) redox reactions.

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99.

What is the oxidation number of sulphur in:
(i)H₂SO₄ (ii) Na₂S₂O₃?

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100.

Determine the oxidation number of:
$left parenthesis straight i right parenthesis space straight S space in space straight S subscript 2 straight O subscript 3 superscript 2 minus end superscript space straight S subscript 2 straight O subscript 3 superscript 2 minus end superscript space space space space space left parenthesis ii right parenthesis space Br space in space BrO subscript 4 superscript minus space space space space space space left parenthesis iii right parenthesis space straight P space in space HPO subscript 4 superscript 2 minus end superscript$

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