NCERT Solutions for Class 11 Science Chemistry Chapter 8 - Redox Reactions

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Answer:

Reactions that involve changes in the oxidation number of the interacting species are called redox reactions. In option (iv), there is no change in the oxidation number of interacting species. 

Hence, the correct answer is option (iv).

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Oxidising agent is the species which oxidises other species and itself gets reduced. From the above data, the species which has the greatest tendency to get reduced is Ag⁺. So, it is also the strongest oxidising agent.

Hence, the correct answer is option (iv).

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Br₂ has the highest reduction potential value and thus, the highest tendency to get reduced among the given options. While Cu²⁺ has the least reduction potential value, so Cu will easily get oxidised to Cu²⁺ and reduce Br₂ to Br⁻.

Hence, the correct answer is option (iv).

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A redox reaction is feasible if the $E^{⊝}$ value for overall reaction is positive. In option (iv), Ag is getting oxidised and Fe³⁺ is getting reduced.
$$\begin{gathered} \frac{Ag\to A{g}^{+} + e^{-}; E^{⊝}=-0.80 V (sign of E^{⊝}is reversed) \\ F{e}^{3+} + e^{-}\to F{e}^{2+}; E^{⊝}=0.77 V}{Ag + F{e}^{3+}\to A{g}^{+} + F{e}^{2+}; E^{⊝}=-0.80 + 0.77=-0.03 V} \end{gathered}$$
The overall $E^{⊝}$ value is negative.

Hence, the correct answer is option (iv).

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The standard reduction potential of Br₂ is more than I₂. So, it acts as a stronger oxidizing agent and oxidizes thiosulphate to SO₄²⁻ while I₂ oxidizes thiosulphate to S₄O₆²⁻ only.

Hence, the correct answer is option (i). 

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The oxidation number of hydrogen is −1 when it combines with metals as metal hydrides.

Hence, the correct answer is option (i).

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In NH₄NO₃, nitrogen exhibits two different oxidation states. NH₄NO₃ is an ionic compound consisting of  NH₄⁺ and NO₃⁻. In NH₄⁺,  oxidation state of nitrogen is −3 while in NO₃⁻, the oxidation state is +5.

Hence, the correct answer is option (ii).

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The oxidation numbers of central atom in CrO₂⁻, ClO₃⁻, CrO₄²⁻ and MnO₄⁻ are +3, +5, +6 and +7 respectively.

Hence, the correct answer is option (i).

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The oxidation number of an element is equal to number of (n−1)d electrons and ns electrons. By this formula, option (iv) has 7 (5+2) electrons and thus, can exhibit highest oxidation number.

Hence, the correct answer is option (iv).

Answer:

 In a disproportionation reaction, an element in one oxidation state is simultaneously oxidised and reduced. In option (iv), nitrogen in NO₂ in +4 oxidation state is oxidised to +5 in NO₃⁻ as well as reduced to +3 in NO₂⁻.

Hence, the correct answer is option (iv).

Answer:

Fluorine is the most electronegative element and thus, can be reduced only. Also, it does not have d-orbitals to get oxidised and therefore, does not positive oxidation states. It always exists in −1 oxidation state. So, it does not show disproportionation tendency.

Hence, the correct answer is option (iii).

Answer:

The oxidation state of potassium (K) is same i.e. +1 throughout the reaction while chlorine (Cl) gets reduced from +5 oxidation state to −1. Oxygen (O) gets oxidised from −2 oxidation state to 0. So, none of the statements are true.

Hence, the correct answers are options (i), (ii), (iii) and (iv).

Answer:

Zinc (Zn) is oxidised from 0 oxidation state to +2, thus, acting as a reductant while hydrogen ion (H⁺) is reduced to H₂, acting as an oxidant. The oxidation state of chlorine (Cl) remains same i.e. −1 throughout the reaction.

Hence, the correct answers are options (iii) and (iv). 

Answer:

Variable oxidation states are characteristics of elements with more than 1 electron in d-orbitals. So, titanium (Ti) with outer orbital electronic configuration of 3d²4s² shows +2, +3 and +4 oxidation states. Also, phosphorus (P) with outer orbital electronic configuration of  3s²3p³ shows +3 and +5 oxidation states due to presence of d-orbitals.

Hence, the correct answers are options (iii) and (iv).

Answer:

It is a disproportionation reaction in which phosphorus (P) in 0 oxidation state in P₄ undergoes oxidation to +1 oxidation state in H₂PO₂⁻ as well as reduction to −3 oxidation state in PH₃. The oxidation state of hydrogen (H) remains same i.e. +1 throughout the reaction.

Hence, the correct answers are options (iii) and (iv).

Answer:

The electrodes having negative electrode potential values will undergo oxidation and act as anodes when connected to Standard Hydrogen Electrode.

Hence, the correct answers are options (i) and (ii).

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The species which bleaches substances due to its oxidising action is ClO⁻ and it is known as hypochlorite ion.

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Mn is in the highest oxidation state i.e. +7 in MnO₄⁻. Hence, it does not undergo disproportionation. MnO₄²⁻ undergoes disproportionation because in this anion, Mn is in +6 oxidation state and it can be oxidised as well as reduced. The following reaction takes place : 
${3Mn{O}_4}^{2-}+ 4H^{+}\to {2Mn{O}_4}^{-} + Mn{O}_{2 }+ 2H_{2}O$

Answer:

In the first reaction, lead (Pb) is in its stable oxidation state of +2 and it undergoes no further change in oxidation state. It is a simple acid-base reaction. But in the second reaction, oxidation state of lead (Pb) is +4 which is unstable and thus, undergoes reduction to +2 oxidation state. Chlorine is oxidised from −1 oxidation state to 0 oxidation state. Hence, it is a redox reaction. 

Answer:

Nitric acid reacts with PbO (basic oxide) because lead (Pb) is in +2 oxidation state. A simple acid-base reactions takes place resulting in formation of Pb(NO₃)₂ and H₂O. But in PbO₂, lead (Pb) is in +4 oxidation state and can not be oxidised further. Therefore, nitric acid does not react with PbO₂.

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$\text{(i)} \stackrel{+2}{F{e}^{2+}}+\stackrel{+6}{C{r}_2}{O_7}^{2-}\to 2\stackrel{+3}{C{r}^{3+}}+\stackrel{+3}{F{e}^{3+}}$
Increase in oxidation number of 1 per Fe atom and decrease in oxidation number of 3 per Cr atom.
Cross multiply the total change in oxidation number on product side and balance Fe and Cr atoms.

$$\begin{gathered} 6F{e}^{2+}+C{r}_2{O_7}^{2-}\to 2C{r}^{3+}+6F{e}^{3+} \\ \text{Add }{H}^{+}\text{ to balance the charges on both sides.} \\ 6F{e}^{2+}+C{r}_2{O_7}^{2-}+14H^{+}\to 2C{r}^{3+}+6F{e}^{3+} \\ \text{Add }{H}_{2}O\text{ to balance H and O atoms.} \\ \mathbf{6}{\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathbf{Cr}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{7}}}^{\mathbf{2}\mathbf{-}}\mathbf{+}\mathbf{14}{\mathbf{H}}^{\mathbf{+}}\mathbf{\to }\mathbf{2}{\mathbf{Cr}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{6}{\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{7}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

$\text{(ii) }\stackrel{0}{I_2}+\stackrel{+5}{N}{O_3}^{-}\to \stackrel{+4}{N}{O}_2+\stackrel{+5}{I}{O_3}^{-}$
Increase in oxidation number of 5 per I atom and decrease in oxidation number of 1 per N atom.
Cross multiply the total change in oxidation number on product side and balance I and N atoms.

$$\begin{gathered} I_2+10N{O_3}^{-}\to 10N{O}_2+2I{O_3}^{-} \\ \text{Add }{H}^{+}\text{ to balance the charges on both sides.} \\ {\mathrm{I}}_2+10\mathrm{N}{{\mathrm{O}}_3}^{-}+8{\mathrm{H}}^{+}\to 10{\mathrm{NO}}_2+2\mathrm{I}{{\mathrm{O}}_3}^{-} \\ \text{Add }{\mathrm{H}}_{2}O\text{ to balance H and O atoms.} \\ {\mathbf{I}}_{\mathbf{2}}\mathbf{+}\mathbf{10}\mathbf{N}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{+}\mathbf{8}{\mathbf{H}}^{\mathbf{+}}\mathbf{\to }\mathbf{10}{\mathbf{NO}}_{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{I}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

$\text{(iii)} \stackrel{0}{I_2}+{\stackrel{+2}{S}}_2{O_3}^{2-}\to \stackrel{-1}{I^{-}}+\stackrel{+2.5}{S_4}{O_6}^{2-}$
Increase in oxidation number of 0.5 per S atom and decrease in oxidation number of 1 per I atom.
Cross multiply the total change in oxidation number on product side and balance I and S atoms.

${\mathit{I}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathit{S}}_{\mathbf{2}}{{\mathit{O}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}\mathbf{\to }\mathbf{2}{\mathit{I}}^{\mathbf{-}}\mathbf{+}{\mathit{S}}_{\mathbf{4}}{{\mathit{O}}_{\mathbf{6}}}^{\mathbf{2}\mathbf{-}}$

$\text{(iv)} \stackrel{+4}{Mn}{O}_2+\stackrel{+3}{C_2}{O_4}^{2-}\to \stackrel{+2}{M{n}^{2+}}+\stackrel{+4}{C}{O}_2$
Increase in oxidation number of 1 per C atom and decrease in oxidation number of 2 per Mn atom.
Cross multiply the total change in oxidation number on product side and balance C and Mn atoms.

$$\begin{gathered} Mn{O}_2+C_2{O_4}^{2-}\to M{n}^{2+}+2C{O}_2 \\ \text{Add} H^{+}\text{ to balance the charges on both sides.} \\ {\mathrm{MnO}}_2+{\mathrm{C}}_2{{\mathrm{O}}_4}^{2-}+4{\mathrm{H}}^{+}\to {\mathrm{Mn}}^{2+}+2{\mathrm{CO}}_2 \\ \text{Add }{H}_{2}O\text{ to balance H and O atoms.} \\ {\mathbf{MnO}}_{\mathbf{2}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\mathbf{+}\mathbf{4}{\mathbf{H}}^{\mathbf{+}}\mathbf{\to }\mathbf{2}{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$
 

Answer:

Reactions (i), (iii) and (v) are redox reactions because they involve change in oxidation number of interacting species.
 

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Fluorine is the best oxidant as it has the highest tendency to get reduced. This is due to its high standard electrode potential.

Hence, the correct answer is option (ii).

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The following reaction takes place:
$6\stackrel{-1}{I^{-}}+2\stackrel{+7}{Mn}{O_4}^{-}+ 4H_{2}O\to 3\stackrel{0}{I_2}+2\stackrel{+4}{Mn}{O}_2+8O{H}^{-}$
Permanganate ions act as oxidising agent and manganese gets reduced as its oxidation number changes from +7 to +4. Iodide ions get oxidised to molecular iodine.

Hence, the correct answer is option (iii).

Answer:

The following reaction takes place:
$2H_2{O}_2\to 2H_{2}O+O_2$
It is an example of disproportionation reaction because oxygen in –1 oxidation state on reactant side gets reduced to –2 oxidation state as well as gets oxidized to zero oxidation state on the product side.

Hence, the correct answer is option (i).

Answer:

In the representation ${E^{⊝}}_{F{e}^{3+}/F{e}^{2+}}, F{e}^{3+}$ is the oxidised form and $F{e}^{2+}$ is the reduced form. Similarly, in the representation ${E^{⊝}}_{C{u}^{2+}/Cu}, C{u}^{2+}$ is the oxidised form and $Cu$ is the reduced form.

Hence, the correct answer is option (ii).

Answer:

Redox reactions are reactions which are composed of two half cell reactions-oxidation and reduction simultaneously. The half cell reaction in which loss of electrons takes place is called oxidation reactions and the half cell reactions in which gain of electrons takes place is known as reduction reactions. But the total gain or loss of electrons in a redox reaction remains same. For example,
 

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For a reaction to take place, the overall  value of cell reaction should be positive.


Hence, reactions (ii) and (iv) will take place because they have positive  value.
 

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 Reaction in which an element in one oxidation state is simultaneously oxidised and reduced is called disporportionation reaction. Fluorine is the most electronegative element of the periodic table and has the tendency to gain electron. It has the most positive standard electrode potential value , so, it always tends to get reduced and attain −1 oxidation state. It does not exhibit any positive oxidation state as it can not be oxidised. Hence, it does not show disporportionation reaction.

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The following redox couples are:

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The method which can be used to find out strength of reductant/oxidant in a solution is the titration method. This method is used for reagents like Cu(II) which can oxidise iodide ions.

When starch is added to the solution, the colour of the solution becomes intense blue due to presence of iodine liberated in the above reaction. But the colour fades away as soon as sodium thiosulphate is added to the solution. This is because thiosulphate ions react with iodine.

In this way, end point can be determined and strength of reductant/oxidant can be find out.