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JEE • CHEMISTRY • REDOX REACTIONS

Redox reactions practice · Medium

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121 Medium
121 of 171
Analyzing +4 | -1
The reaction; KI+I2→KI3\mathrm{KI} + {\mathrm{I}_2} \to \mathrm{K}{\mathrm{I}_3} shows :
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the reaction involves complex formation, oxidation of KI, and reduction of I2.
122 Medium
122 of 171
Analyzing +4 | -1
Assertion: Cu\mathrm{Cu} is stronger reducing agent than H2\mathrm{H}_2.

Reason: E∘\mathrm{E}^{\circ} of Cu2+/Cu\mathrm{Cu}^{2+} / \mathrm{Cu} is negative.
View answer and explanation
Correct option: D
Assertion is false, Reason is true
Option D (Assertion is false, Reason is true) is correct because copper is a weaker reducing agent than hydrogen, and the standard reduction potential of Cu2+/Cu is positive.
123 Medium
123 of 171
Application +4 | -1
Oxidation numbers of Fe inFe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4} are :
View answer and explanation
Correct option: A
+2and+3+ 2\mathrm{ }and\mathrm{ } + 3
Option A (+2 and +3) is correct because Fe3O4 is a mixed oxide containing FeO (oxidation state +2) and Fe2O3 (oxidation state +3).
124 Medium
124 of 171
Application +4 | -1
The oxidation number of nitrogen in NH2OH\mathrm{N}{\mathrm{H}_2}\mathrm{OH} is :
View answer and explanation
Correct option: B
−1- 1
Option B (-1) is correct because the oxidation number of nitrogen in NH₂OH is calculated by setting the sum of oxidation numbers to zero.
125 Medium
125 of 171
Understanding +4 | -1
Reaction of acidified KMnO4\mathrm{KMn}{\mathrm{O}_4} with ferrous oxalate gives oxidation products containing :
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because acidified KMnO4\mathrm{KMn}{\mathrm{O}_4} oxidizes ferrous oxalate to Fe3+\mathrm{F}{\mathrm{e}^{3 + }} and CO2\mathrm{C}{\mathrm{O}_2}.
126 Medium
126 of 171
Remembering +4 | -1
The most stable oxidation state of copper is :
View answer and explanation
Correct option: A
+2+ 2
Option A (+2+2) is correct because the most stable oxidation state of copper is +2 due to its higher hydration energy which compensates for the second ionization enthalpy.
127 Medium
127 of 171
Application +4 | -1
2MnO4−+10I−+16H+→A+8H2O+B,2 \mathrm{MnO}_4^{-}+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \rightarrow \mathrm{A}+8 \mathrm{H}_2 \mathrm{O}+\mathrm{B}, A and B respectively are
View answer and explanation
Correct option: D
Mn+2,I2\mathrm{Mn}^{+2}, \mathrm{I}_2
Option D (Mn+2,I2\mathrm{Mn}^{+2}, \mathrm{I}_2) is correct because it represents the products of the redox reaction where permanganate is reduced to manganese(II) and iodide is oxidized to iodine.
128 Medium
128 of 171
Application +4 | -1
Oxidation number of nitrogen is highest in
View answer and explanation
Correct option: B
N2O4{\mathrm{N}_2}{\mathrm{O}_4}
Option B (N2O4{\mathrm{N}_2}{\mathrm{O}_4}) is correct because the oxidation number of nitrogen is +4, which is the highest among the given options.
129 Medium
129 of 171
Application +4 | -1
The number of moles of Mohr’s salt required per mole of dichromate ion is :
View answer and explanation
Correct option: D
6
Option D (6) is correct because 6 moles of Mohr's salt (Fe2+Fe^{2+}) are required to reduce 1 mole of dichromate ion (Cr2O72−Cr_2O_7^{2-}).
130 Medium
130 of 171
Understanding +4 | -1
Which is not an oxidising agent?
View answer and explanation
Correct option: C
C6H12O6{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}
Option C (C6H12O6{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}) is correct because glucose is a reducing agent, meaning it donates electrons and gets oxidized, not reduced.
131 Medium
131 of 171
Analyzing +4 | -1
Which combination is odd with respect to oxidation numbers of S,Cr,N\mathrm{S},\mathrm{Cr},\mathrm{N} and H\mathrm{H} respectively:
View answer and explanation
Correct option: C
NH3,NH4+,N3H,NO2−\mathrm{N}{\mathrm{H}_3},\mathrm{NH}_4^ + ,{\mathrm{N}_3}\mathrm{H},\mathrm{NO}_2^ -
Option D is correct because electrolytes increase conductivity when dissolved in water by furnishing ions, making the assertion incorrect.
132 Medium
132 of 171
Application +4 | -1
For the reaction: NH3+OCl−⟶N2H4+Cl−{NH}_3+{OCl}^{-} \longrightarrow {N}_2 {H}_4+{Cl}^{-}in basic medium, the coefficients of NH3,OCl−{NH}_3, {OCl}^{-}and N2H4{N}_2 {H}_4 for the balanced equation are respectively
View answer and explanation
Correct option: C
2,1,12,1,1
Option C (2,1,1) is correct because it represents the correct coefficients for the balanced redox reaction in a basic medium.
133 Medium
133 of 171
Understanding +4 | -1
FeX2Fe{X_2} and FeY3Fe{Y_3} are known when xx and yy are
View answer and explanation
Correct option: A
x = F, Cl, Br, l  and  y = F, Cl  and  Brx\,\mathrm{ = }\,F,\,Cl,\,Br,\,l\,\,\mathrm{and}\,\,y\, = \,F,\,Cl\,\,and\,\,Br
Option A (x = F,Cl,Br,l  and  y = F,Cl  and  Brx\,\mathrm{ = }\,F,Cl,Br,l\,\,\mathrm{and}\,\,y\, = \,F,Cl\,\,and\,\,Br) is correct because FeI3FeI_3 does not exist due to I−I^- being a strong reducing agent.
134 Medium
134 of 171
Understanding +4 | -1
A group 15 element, which is a metal and forms a hydride with strongest reducing power among group 15 hydrides. The element is:
View answer and explanation
Correct option: A
BiBi
The correct answer is A (BiBi) because Bismuth is a metal in group 15 and forms BiH3BiH_3, which has the weakest Bi−HBi-H bond and thus the strongest reducing power.
135 Medium
135 of 171
Application +4 | -1
How many electrons are involved in oxidation of KMnO4\mathrm{KMn}{\mathrm{O}_4} in basic medium?
View answer and explanation
Correct option: A
1
Option A (1) is correct because in a basic medium, KMnO4 is reduced to K2MnO4, involving a change of +7 to +6 in oxidation state, which means only 1 electron is involved.
136 Medium
136 of 171
Application +4 | -1
The oxidation number of ‘ P ’ in Mg2P2O7M{g_2}{P_2}{O_7}^{^{}} is
View answer and explanation
Correct option: B
+5+ 5
The correct answer is +5 because the oxidation number of P in Mg2P2O7Mg_2P_2O_7 is calculated based on the known oxidation numbers of Mg and O.
137 Medium
137 of 171
Application +4 | -1
In which of the following compounds iron has lowest oxidation number?
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Correct option: C
Fe2O\mathrm{Fe}_2 \mathrm{O}
Option C (Fe2O\mathrm{Fe}_2 \mathrm{O}) is correct because the oxidation state of iron in this compound is +1, which is the lowest among the given options.
138 Medium
138 of 171
Analyzing +4 | -1
Which of the following is disproportionation reaction?
View answer and explanation
Correct option: D
2NO2+2OH−→NO2−+NO3−+H2O2 \mathrm{NO}_2+2 \mathrm{OH}^{-} \rightarrow \mathrm{NO}_2^{-}+\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{O}
Option D (2NO2+2OH−→NO2−+NO3−+H2O2 \mathrm{NO}_2+2 \mathrm{OH}^{-} \rightarrow \mathrm{NO}_2^{-}+\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{O}) is correct because nitrogen in NO2NO_2 is both oxidized (to NO3−NO_3^-) and reduced (to NO2−NO_2^-).
139 Medium
139 of 171
Understanding +4 | -1
What would happen when a small quantity of H2O2{\mathrm{H}_2}{\mathrm{O}_2} is added to a solution of FeSO4\mathrm{FeS}{\mathrm{O}_4}?
View answer and explanation
Correct option: D
An electron is lost by Fe2+\mathrm{F}{\mathrm{e}^{2 + }}
Option D (An electron is lost by Fe2+\mathrm{F}{\mathrm{e}^{2 + }}) is correct because H2O2{\mathrm{H}_2}{\mathrm{O}_2} acts as an oxidizing agent, oxidizing Fe2+\mathrm{F}{\mathrm{e}^{2 + }} to Fe3+\mathrm{F}{\mathrm{e}^{3 + }}.
140 Medium
140 of 171
Remembering +4 | -1
Which of the following reactions leads to the breakdown of a compound into two or more components?
View answer and explanation
Correct option: C
Decomposition reactions
Option C is correct because decomposition reactions involve the breakdown of a compound into two or more components.
141 Medium
141 of 171
Analyzing +4 | -1
Assertion (A): Addition of water to a solution containing solute and solvent cannot change it's normality or molarity.

Reason (R): The milliequivalent and millimoles of the solute are changed on dilution.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D (Both (A) and (R) are false) is correct because adding water changes normality and molarity, and the milliequivalents and millimoles of solute remain constant upon dilution.
142 Medium
142 of 171
Application +4 | -1
MnO4−\mathrm{MnO}_4^ - is a good oxidising agent in different medium changing to MnO4−→Mn2+\mathrm{MnO}_4^ - \to \mathrm{M}{\mathrm{n}^{2 + }},→MnO42−,→MnO2,→Mn2O3\to \mathrm{MnO}_4^{2 - }, \to \mathrm{Mn}{\mathrm{O}_2}, \to \mathrm{M}{\mathrm{n}_2}{\mathrm{O}_3} Changes in oxidation number respectively are
View answer and explanation
Correct option: C
5,1,3,4
Option C (5,1,3,4) is correct because it accurately reflects the changes in oxidation number of Mn in the given transformations.
143 Medium
143 of 171
Analyzing +4 | -1
The redox reaction among the following is
View answer and explanation
Correct option: C
combination of dinitrogen with dioxygen at 200 K
The correct answer is C (combination of dinitrogen with dioxygen at 200 K) because the oxidation states of nitrogen and oxygen change during the reaction.
144 Medium
144 of 171
Analyzing +4 | -1
Assertion (A): Oxygen has oxidation state of -2 in both O2\mathrm{O}_2 and O3.\mathrm{O}_3.

Reason (R): Oxygen is assigned an oxidation state of -2 in almost all its compounds.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D is correct because the oxidation state of oxygen in O2O_2 is 0, in O3O_3 is 0, and the general rule of -2 has exceptions like peroxides and superoxides.
145 Medium
145 of 171
Application +4 | -1
Among NH3,HNO3,NaN3\mathrm{N}{\mathrm{H}_3},\mathrm{HN}{\mathrm{O}_{3,}}\mathrm{Na}{\mathrm{N}_3} and Mg3N2;\mathrm{M}{\mathrm{g}_3}{\mathrm{N}_2}; the number of molecules having nitrogen in negative oxidation state is
View answer and explanation
Correct option: C
3
Option C (3) is correct because NH3, NaN3 and Mg3N2 have nitrogen in negative oxidation state.
146 Medium
146 of 171
Analyzing +4 | -1
Assertion: In alkaline medium, KMnO4\mathrm{KMnO}_4 acts as powerful oxidising agent.
Reason: KMnO4\mathrm{KMnO}_4 reduces to give Mn2+\mathrm{Mn}^{2+} in alkaline medium.
View answer and explanation
Correct option: C
If Assertion is correct but Reason is incorrect
The assertion is correct; KMnO4 is a powerful oxidizing agent in alkaline medium. However, the reason is incorrect; KMnO4 reduces to Mn+6 in alkaline medium, not Mn+2.
147 Medium
147 of 171
Analyzing +4 | -1
Identify the disproportionation reaction :
View answer and explanation
Correct option: D
2NO2+2OH−→NO2−+NO3−+H2O2N{O_2} + 2O{H^ - } \to NO_2^ - + NO_3^ - + {H_2}O
Option D (2NO2+2OH−→NO2−+NO3−+H2O2N{O_2} + 2O{H^ - } \to NO_2^ - + NO_3^ - + {H_2}O) is correct because nitrogen undergoes both oxidation and reduction in the same reaction.
148 Medium
148 of 171
Application +4 | -1
Which of the following is the cell reaction that occurs when the following half - cells are combined?

I2+2e−→2I−(1 M);E∘=+0.54 VBr2+2e−→2Br−(1 M);E∘=+1.09 V\begin{aligned} & \mathrm{I}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-}(1\,\mathrm{M}) ; \mathrm{E}^{\circ}=+0.54 \mathrm{~V} \\ & \mathrm{Br}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(1\,\mathrm{M}) ; \mathrm{E}^{\circ}=+1.09 \mathrm{~V} \end{aligned}
View answer and explanation
Correct option: C
2I−+Br2→I2+2Br−2 \mathrm{I}^{-}+\mathrm{Br}_2 \rightarrow \mathrm{I}_2+2 \mathrm{Br}^{-}
Option C (2I−+Br2→I2+2Br−2 \mathrm{I}^{-}+\mathrm{Br}_2 \rightarrow \mathrm{I}_2+2 \mathrm{Br}^{-}) is correct because Bromine has a higher reduction potential and will be reduced, while Iodide will be oxidized.
149 Medium
149 of 171
Application +4 | -1
The value of 'n' in the reaction Cr2O72−+14H++nFe2+→2Cr3++nFe3++7H2O\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - } + 14{\mathrm{H}^ + } + n\mathrm{F}{\mathrm{e}^{2 + }} \to 2\mathrm{C}{\mathrm{r}^{3 + }} + n\mathrm{F}{\mathrm{e}^{3 + }} + 7{\mathrm{H}^2}\mathrm{O} will be
View answer and explanation
Correct option: C
6
Option C (6) is correct because balancing the redox reaction shows that 6 moles of Fe2+ are required to react with 1 mole of dichromate.
150 Medium
150 of 171
Remembering +4 | -1
Starch iodide paper is used to test for the presence of:
View answer and explanation
Correct option: A
Iodine
Option A (Iodine) is correct because starch iodide paper turns blue in the presence of iodine.
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