Chapter 8: Redox Reactions

Q1Written answer

Which of the following is not an example of redox reaction?

\[(i)CuO + H_2 \rightarrow Cu + H_2O\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{F}{{\mathbf{e}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}\text{ }+\text{ }\mathbf{3CO}\text{ }\to \text{ }\mathbf{2Fe}\text{ }+\text{ }\mathbf{3C}{{\mathbf{O}}_{\mathbf{2}}}\]

\[(iii) 2K + F_2 \rightarrow 2KF \]

\[(iv) BaCl_{2} + H_{2} SO_{4} \rightarrow BaSO_{4}+2HCl \]

Show solutionHide solution

Worked solution

Ans: Redox reaction is defined as the simultaneous oxidation and reduction of reacting species. Thus, change in oxidation state will decide whether a reaction is redox or not. Thus, assigning the oxidation states as:

\[\left( \text{i} \right)\text{ CuO + }{{\text{H}}_{\text{2}}}\to \text{ Cu + }{{\text{H}}_{\text{2}}}\text{O}\]

Here, oxidation of H and reduction of Cu is taking place.

\[\left( \text{ii} \right)\text{ F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{ + 3CO}\to \text{ 2Fe + 3C}{{\text{O}}_{\text{2}}}\]

Here, oxidation of C and reduction of Fe is taking place.

\[\left( \text{iii} \right)\text{ 2K + }{{\text{F}}_{\text{2}}}\to \text{2KF}\]

Here, the oxidation of K and reduction of F is taking place.

\[\left( \text{iv} \right)\text{ BaC}{{\text{l}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\to \text{ BaS}{{\text{O}}_{\text{4}}}\text{ + 2HCl}\]

This is not a redox reaction, but it is a double displacement reaction.

Hence, option(iv) is the correct answer.

Q2Written answer

The more positive the value of Ee, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent. Ee values:

Fe3+/ Fe2+ = +0.77; I2 (s) / I- = +0.54; Cu2+ / Cu = +0.34; Ag+ / Ag = +0.80V

(i) \[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}\]

(ii) \[{{\mathbf{I}}_{\mathbf{2}}}\] (s)

(iii) \[\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+}}\]

(iv) \[\mathbf{A}{{\mathbf{g}}^{+}}\]

Show solutionHide solution

Worked solution

Ans: The strongest oxidising agent means it has greater tendency to oxidise other species and itself gets easily reduced. So higher the \[{{\text{E}}^{\odot }}\]values, stronger is the oxidising agent it is. Thus, Ag+ having highest positive \[{{\text{E}}^{\odot }}\] value among the given systems, is the strongest oxidising agent. Hence, option (iv) is the correct answer.

Q3Written answer

\[{{\mathbf{E}}^{\odot }}\]values of some redox couples are given below. On the basis of these values choose the correct option. \[{{\mathbf{E}}^{\odot }}\] values:

\[\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}\text{}/\mathbf{Br}\text{}=\text{}+\text{}\mathbf{1}.\mathbf{90};\text{}\mathbf{A}{{\mathbf{g}}^{\mathbf{+}}}\text{}/\mathbf{Ag}\left(\mathbf{s}\right)\text{}=\text{}+\text{}\mathbf{0}.\mathbf{80}\text{}\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+}}/\mathbf{Cu}\left(\mathbf{s}\right)\text{}=\text{}+\text{}\mathbf{0}.\mathbf{34};\text{}{{\mathbf{I}}_{\mathbf{2}}}\text{}\left(\mathbf{s}\right)/{{\mathbf{I}}^{\text{ }}}=\text{ }+\text{ }\mathbf{0}.\mathbf{54}\]

(i) Cu will reduce \[\mathbf{Br}\]

(ii) Cu will reduce Ag

(iii) Cu will reduce \[{{\mathbf{I}}^{\text{ }}}\]

(iv) Cu will reduce \[\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}\].

Show solutionHide solution

Worked solution

Ans: Reduction potential is defined as the tendency of the specie to get reduced. More positive the value of \[{{\text{E}}^{\odot }}\text{ }\], greater is the tendency of the species to get reduced and stronger is the oxidising agent.

On the basis of the given \[{{\text{E}}^{\odot }}\text{ }\] values, the order of getting reduced is:

\[\text{B}{{\text{r}}_{\text{2}}}\text{A}{{\text{g}}^{\text{+}}}\text{}{{\text{I}}_{^{\text{2}}}}\text{C}{{\text{u}}^{\text{2+}}}\]

Hence, Cu has the least tendency to get reduced and will itself gets oxidise and reduce other species as: \[\text{B}{{\text{r}}_{\text{2}}}\], \[\text{A}{{\text{g}}^{\text{+}}}\] and \[{{\text{I}}_{^{\text{2}}}}\]. Hence, option (iv) is the correct answer.

Q4Written answer

Using the standard electrode potential, find out the pair between which redox reaction is not feasible. \[{{\mathbf{E}}^{\odot }}\]values:

\[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}/\mathbf{F}{{\mathbf{e}}^{\mathbf{2}+}}\text{}=\text{}+\text{}\mathbf{0}.\mathbf{77};\text{}{{\mathbf{I}}_{\mathbf{2}}}\text{}/{{\mathbf{I}}^{}}\text{}=\text{}+\text{}\mathbf{0}.\mathbf{54};\text{}\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+}}/\mathbf{Cu}\text{}=\text{}+\text{}\mathbf{0}.\mathbf{34};\text{}\mathbf{A}{{\mathbf{g}}^{+}}\text{}/\mathbf{Ag}\text{}=\text{}+\text{ }\mathbf{0}.\mathbf{80}\text{ }\mathbf{V}\]

(i) \[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}\] and \[{{\mathbf{I}}^{}}\]

(ii) Ag+ and Cu

(iii) \[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}\] and Cu

(iv) Ag and \[\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}\].

Show solutionHide solution

Worked solution

Ans: \[{{\text{E}}^{\odot }}\] will be negative for the pair Ag and \[\text{F}{{\text{e}}^{\text{3+}}}\]. Hence the reaction is not feasible. Hence, option(iv) is the correct answer.

Q5Written answer

Thiosulphate reacts differently with iodine and bromine in the reactions given below:

\[\mathbf{2}{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}}+{{\mathbf{I}}_{\mathbf{2}}}\text{ }\to \text{ }{{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}\text{ }}}^{\mathbf{2}}+\mathbf{2}{{\mathbf{I}}^{}}\text{ }\]

\[{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}\text{ }}}^{\mathbf{2}}+\text{ }\mathbf{2B}{{\mathbf{r}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{5}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\text{ }\to \text{ }\mathbf{2S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\text{ }+\text{ }\mathbf{2B}{{\mathbf{r}}^{}}+\mathbf{10}\text{ }{{\mathbf{H}}^{+}}\]

Which of the following statements justifies the above dual behaviour of thiosulphate?

(i) Bromine is a stronger oxidant than iodine.

(ii) Bromine is a weaker oxidant than iodine.

(iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.

(iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.

Show solutionHide solution

Worked solution

Ans: Standard reduction potential of bromine is higher than Iodine, hence Bromine is a stronger oxidant than iodine. Therefore, option (i) is the correct answer.

Q6Written answer

The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?

(i) The oxidation number of hydrogen is always +1.

(ii) The algebraic sum of all the oxidation numbers in a compound is zero.

(iii) An element in the free or the uncombined state bears oxidation number zero.

(iv) In all its compounds, the oxidation number of fluorine is – 1.

Show solutionHide solution

Worked solution

Ans: In ionic hydrides hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion. Hence, option(i) is the correct answer.

Q7Written answer

In which of the following compounds, an element exhibits two different oxidation states.

\[\left( \mathbf{i} \right)\text{ }\mathbf{N}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{OH}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{N}{{\mathbf{H}}_{\mathbf{4}}}\mathbf{N}{{\mathbf{O}}_{\mathbf{3}}}\]

\[\left( \mathbf{iii} \right)\text{ }{{\mathbf{N}}_{\mathbf{2}}}{{\mathbf{H}}_{\mathbf{4}}}\]

\[\left( \mathbf{iv} \right)\text{ }{{\mathbf{N}}_{\mathbf{3}}}\mathbf{H}\]

Show solutionHide solution

Worked solution

Ans: The oxidation states are given below-

In NH2OH oxidation of N is -1.

NH4NO3 exists as \[\text{N}{{\text{H}}_{\text{4}}}^{\text{+}}\text{.N}{{\text{O}}_{\text{3}}}^{\text{-}}\text{, }\!\!~\!\!\text{ }\]

Thus, the oxidation state of N in \[\text{NH}_{4}^{+}\] is -3 while in \[\text{NO}_{3}^{-}\] is +5.

In N2H4 oxidation of N is -2

In N3H oxidation of N is \[\frac{\text{-1}}{3}\]

Hence, option (ii) is the correct answer.

Q8Written answer

Which of the following arrangements represent increasing oxidation number of the central atom?

\[\left( \mathbf{i} \right)\mathbf{ Cr}{{\mathbf{O}}_{3}}^{\mathbf{ }}\mathbf{ , Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\mathbf{ , Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}\]

\[\left( \mathbf{ii} \right)\mathbf{ Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\mathbf{ , Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{}}\]

\[\left( \mathbf{iii} \right)\mathbf{ Cr}{{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{}}\mathbf{ , Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{}}\mathbf{ , Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\]

\[\left( \mathbf{iv} \right)\mathbf{ Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\mathbf{ , Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}\mathbf{ , Cr}{{\mathbf{O}}_{\mathbf{2 }}}^{\mathbf{}}\mathbf{ , Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{}}\]

Show solutionHide solution

Worked solution

Ans: Let the oxidation number of central atom be 'y'.

Oxidation number of O = −2

\[\text{CrO}_{2}^{-}\]: y + 2 × (−2) = −1

y − 4 = −1 or y = +3

\[\text{CrO}_{4}^{2-}\] : y + 4 × (−2) = −2

y − 8 = −2 or y = +6

\[\text{ClO}_{3}^{-}\]: y + 3 × (−2) = −1

y − 6 = −1 or y = +5

\[\text{MnO}_{4}^{-}\]: y + 4 × (−2) = −1

y − 8 = −1 or y = +7

increasing order of oxidation number of the central atom is:

\[\text{Cr}{{\text{O}}_{\text{2}}}^{\text{--}}\mathbf{ }<\text{Cl}{{\text{O}}_{\text{3}}}^{\text{--}}\mathbf{ }<\text{Cr}{{\text{O}}_{\text{4}}}^{\text{2--}}\text{Mn}{{\text{O}}_{\text{4}}}^{\text{--}}\].

Hence, option (i) is the correct answer.

Q9Written answer

The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?

\[\left( \mathbf{i} \right)\text{ }\mathbf{3}{{\mathbf{d}}^{\mathbf{1}\text{ }}}\mathbf{4}{{\mathbf{s}}^{\mathbf{2}}}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{3}{{\mathbf{d}}^{\mathbf{3}}}\mathbf{4}{{\mathbf{s}}^{\mathbf{2}}}\]

\[\left( \mathbf{iii} \right)\text{ }\mathbf{3}{{\mathbf{d}}^{\mathbf{5}}}\mathbf{4}{{\mathbf{s}}^{\mathbf{1}}}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{3}{{\mathbf{d}}^{\mathbf{5}}}\mathbf{4}{{\mathbf{s}}^{\mathbf{2}}}\]

Show solutionHide solution

Worked solution

Ans: For \[\text{3}{{\text{d}}^{\text{1 }}}\text{4}{{\text{s}}^{\text{2}}}\] can exhibit highest oxidation state as 2+1 = +3,

For \[\text{3}{{\text{d}}^{\text{3}}}\text{4}{{\text{s}}^{\text{2}}}\]can exhibit highest oxidation state as 2+3 = +5

For \[\text{3}{{\text{d}}^{\text{5}}}\text{4}{{\text{s}}^{\text{1}}}\]can exhibit the highest oxidation state as 1+5 = +6

For \[\text{3}{{\text{d}}^{\text{5}}}\text{4}{{\text{s}}^{\text{2}}}\]can exhibit highest oxidation state as 2+5 = +7.

Hence, option(iv) is the correct answer.

Q10Written answer

Identify disproportionate reaction

\[\left( \mathbf{i} \right)\text{ }\mathbf{C}{{\mathbf{H}}_{\mathbf{4}}}+\mathbf{2}{{\mathbf{O}}_{\mathbf{2}}}\text{ }\to \text{ }\mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{2}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{C}{{\mathbf{H}}_{\mathbf{4}}}+\mathbf{4C}{{\mathbf{l}}_{\mathbf{2}}}\text{ }\to \text{ }\mathbf{CC}{{\mathbf{l}}_{\mathbf{4}}}\text{ }+\text{ }\mathbf{4HCl}\]

\[\left( \mathbf{iii} \right)\text{ }\mathbf{2}{{\mathbf{F}}_{\mathbf{2}}}+\mathbf{2O}{{\mathbf{H}}^{}}\text{ }\to \text{ }\mathbf{2}{{\mathbf{F}}^{}}\text{ }+\text{ }\mathbf{O}{{\mathbf{F}}_{\mathbf{2}}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{2N}{{\mathbf{O}}_{\mathbf{2}}}+\mathbf{2O}{{\mathbf{H}}^{}}\text{ }\to \text{ }\mathbf{N}{{\mathbf{O}}_{\mathbf{2}\text{ }}}^{\text{ }}+\text{ }\mathbf{N}{{\mathbf{O}}_{\mathbf{3}}}^{}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Show solutionHide solution

Worked solution

Ans: Disproportionate reactions are defined as the reactions in which the same substance is oxidized as well as reduced. Here, the below reaction is given as-

\[\text{2N}{{\text{O}}_{\text{2}}}\text{+2O}{{\text{H}}^{\text{--}}}\to \text{N}{{\text{O}}_{\text{2 }}}^{\text{-- }}\text{+ N}{{\text{O}}_{\text{3}}}^{\text{--}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\]

In this reaction, N is both oxidized as well as reduced since O.N. of N increases from +4 in \[\text{NO}_{3}^{-}\] to +5 in NO2 and decreases from +4 in NO to +3 in \[\text{NO}_{2}^{-}\].

Hence, option(iv) is the correct answer.

My Cart

Your Cart 0

Multiple Choice Questions (Type-I)