Redox Reactions MCQs for NEET — Chemistry Questions with Answers

(A). Silver nitrate solution becomes bluish when copper rod is placed in it because copper $\left(E_{C{u}^{2+}/Cu}^0=0.34 V\right)$is more electropositive than silver $\left(E_{A{g}^{2+}/Ag}^0=0.8 V\right)$ Therefore, copper is able to displace silver from silver nitrate solution. The dissolution of copper in the solution as $C{u}^{2+}$ions make it bluish.

In the extraction of copper from its sulphide ore, when ore is subjected to roasting, some of it is oxidised to Cu₂O which reacts with the remaining Cu₂S (sulphide ore) to give copper metal.

Cu₂S + 2Cu₂O $\to$6Cu + SO₂ $\uparrow$

In this process Cu₂S behaves as reducing agent.

$\stackrel{+3}{\mathrm{N}}{\mathrm{O}}_2^{-}\stackrel{\mathrm{Oxidation}}{\to }\stackrel{+5}{\mathrm{N}}{\mathrm{O}}_3^{-}$

 Bleaching action of: Bleaching by sulphur dioxide is due to the process of reduction in the presence of moisture.
SO₂​+2H_2​O→H₂​SO₄​+2[H] (nascent)

The nascent hydrogen liberated in the reaction is responsible for bleaching flower but when the flower comes in contact with air.

It get oxidized and be come coloured.

Colour material $+\left[H\right]\to$ Colourless air​ Colour

The reaction of CaF2 with H2SO4 does not involve any oxidation process. It is an acid-base reaction where HF is formed. The other reactions given involve the oxidation of Cu, S, and C by hot concentrated H2SO4 acting as an oxidizing agent.

(d) Nitric acid acts as an oxidising agent while reacting with copper. When dil. HNO₃ reacts, reaction proceeds as:

$3\mathrm{Cu} + 8{\mathrm{HNO}}_3\left(\mathrm{dil}.\right) \stackrel{ }{\to }3\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2 +\hspace{0.17em}2\mathrm{NO} + {\mathrm{H}}_2\mathrm{O}$

and when conc. HNO₃ is used, reaction proceeds as

$\mathrm{Cu} + 4{\mathrm{HNO}}_3\left(\mathrm{conc}.\right) \stackrel{ }{\to }\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2 +\hspace{0.17em}2\mathrm{NO}2 + 2{\mathrm{H}}_2\mathrm{O}$

(c) ${\mathrm{I}}_2+10{\mathrm{HNO}}_3\underset{}{\to }2{\mathrm{HIO}}_3+10{\mathrm{NO}}_2+4{\mathrm{H}}_2\mathrm{O}$

(d)    $6{\mathrm{CaOCl}}_2 \underset{\mathbf{oxidation}}{\overset{\mathbf{Auto}}{\to }} \mathrm{Ca}({\mathrm{CIO}}_3{)}_2+5{\mathrm{CaCl}}_2$

H₂SO₄ cannot act as reducing agent, because when a substance acts as reducing agent, it is oxidised itself and there is increase in O.N. In H₂SO₄ the O.N. of sulphur is +6,which is maximum.

Fluorine is most electronegative element and its size is very small, so it does not have a tendency to lose electrons. Hence, never behave as a reducing agent. HF is a good oxidising agent.