Learn Redox Reaction MCQ questions & answers in Physical Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
161.
The oxidation number of nitrogen in $${\left( {{N_2}{H_5}} \right)^ + }$$ is
162.
In which of the following transition metal complexes does the metal exhibit zero oxidation
state ?
A
$$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$$
B
$$\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]S{O_4}$$
C
$$Ni{\left( {CO} \right)_4}$$
D
$$\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]{X_3}$$
Answer :
$$Ni{\left( {CO} \right)_4}$$
In metal carbonyls, metal always has zero $$O.N.$$
163.
The solution in a beaker turns blue if
A
$$Cu$$ electrode is placed in $$ZnS{O_4}$$ solution
B
$$Cu$$ electrode is placed in $$AgN{O_3}$$ solution
C
$$Cu$$ electrode is placed in $$A{l_2}{\left( {S{O_4}} \right)_3}$$ solution
D
$$Cu$$ electrode is placed in $$FeS{O_4}$$ solution
Answer :
$$Cu$$ electrode is placed in $$AgN{O_3}$$ solution
$$Cu$$ being more reactive than $$Ag$$ due to lower electrode potential, displaces $$Ag$$ from $$AgN{O_3}$$ solution. It is dissolved in the solution in the form of $$C{u^{2 + }}$$ ions.
164.
In the reaction shown below, oxidation state of the carbon in reactant and product are (i) and (ii) respectively? Is the given reaction a redox reaction?
$$N{a_2}C{O_3}\left( {aq} \right) + HCl\left( {aq} \right) \to $$ $$N{a^ + }\left( {aq} \right) + C{l^ - }\left( {aq} \right) + {H_2}O\left( l \right) + C{O_2}\left( g \right)$$
A
(i) 6, (ii) 4, yes
B
(i) 6, (ii) 6, No
C
(i) 4, (ii) 4, No
D
(i) 4, (ii) 4, yes
Answer :
(i) 4, (ii) 4, No
The redox reaction involve loss or gain of electron$$(s)$$ i.e. change in oxidation state.
Given reaction is not a redox reaction as this reaction involves no change in oxidation state of reactant or product.
165.
Which of the following is the strongest oxidizing agent?
A
$${F_2}$$
B
$$C{l_2}$$
C
$$B{r_2}$$
D
$${I_2}$$
Answer :
$${F_2}$$
$${F_2}$$ is the strongest oxidizing agent with highest reduction potential.
166.
In the redox reaction, $$P{b_3}{O_4} + 8HCl \to $$ $$3PbC{l_2} + C{l_2} + 4{H_2}O$$
A
three $$P{b^{2 + }}$$ ions get oxidised to $$P{b^{4 + }}$$ state
B
one $$P{b^{4 + }}$$ ion gets reduced to $$P{b^{2 + }}$$ and two $$P{b^{2 + }}$$ ions remain unchanged in their oxidation state
C
one $$P{b^{2 + }}$$ ion gets oxidised to $$P{b^{4 + }}$$ and two $$P{b^{4 + }}$$ ions remain unchanged in their oxidation states
D
three $$P{b^{4 + }}$$ ions get reduced to $$P{b^{2 + }}$$ state
Answer :
one $$P{b^{4 + }}$$ ion gets reduced to $$P{b^{2 + }}$$ and two $$P{b^{2 + }}$$ ions remain unchanged in their oxidation state
$$P{b_3}{O_4}$$ is a mixture of $$PbO$$ and $$Pb{O_2}.$$
$$P{b_3}{O_4} \equiv 2PbO \cdot Pb{O_2} + 8HCl$$ $$ \to 3PbC{l_2} + C{l_2} + 4{H_2}O$$
In the reaction one $$P{b^{4 + }}$$ ion is reduced to $$P{b^{2 + }}$$ and two $$P{b^{2 + }}$$ ions remain unchanged.
167.
When $$KMn{O_4}$$ is reduced with oxalic acid in acidic solution, the oxidation number of $$Mn$$ changes from
B
In atmosphere, formation of $${O_3}$$ from $${O_2}$$ by lightening
C
Formation of oxides of nitrogen from nitrogen and oxygen by lightening
D
Evaporation of $${H_2}O$$
Answer :
Formation of oxides of nitrogen from nitrogen and oxygen by lightening
(A) Neutralisation reaction
(B) $$3{O_2} \to 2{O_3},$$ allotropic formation
(C) $${N_2} + {O_2} \to 2NO$$
$${N_2}$$ is reduced and $${O_2}$$ is oxidised.
(D) In evaporation of $${H_2}O,$$ state changes only.
169.
The number of electrons involved in the conversion of $$MnO_4^ - $$ to $$Mn{O_2}$$ is
A
3
B
4
C
1
D
2
Answer :
3
$$MnO_4^ - \to Mn{O_2}$$
Oxidation number of $$MnO_4^ - :x - 8 = - 1 \Rightarrow x = + 7$$
Oxidation number of $$Mn{O_2}:x - 4 = 0 \Rightarrow x = + 4$$
Change in oxidation number = 3
170.
For decolourisation of $$1\,mole$$ of acidified $$KMn{O_4}$$ the moles of $${H_2}{O_2}$$ required are
A
$$\frac{1}{2}$$
B
$$\frac{3}{2}$$
C
$$\frac{5}{2}$$
D
$$\frac{7}{2}$$
Answer :
$$\frac{5}{2}$$
$$2KMn{O_4} + 3{H_2}S{O_4} \to $$ $${K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5\left[ O \right]$$
$$\left[ {{H_2}{O_2} + O \to {H_2}O + {O_2}} \right] \times 5$$
Hence, $$1\,mole$$ of $$KMn{O_4}$$ requires $$\frac{5}{2}$$ $$moles$$ of $${H_2}{O_2}.$$