Mix Examples-Redox Reactions Questions in English

(D) . $N_{2(g)} + O_{2(g)} \rightarrow 2 NO_{(g)}$ is a combination reaction where two reactants form a single product.
$B$. $2 Pb(NO_3)_{2(s)} \rightarrow 2 PbO_{(s)} + 4 NO_{2(g)} + O_{2(g)}$ is a decomposition reaction where a single reactant breaks down into multiple products.
$C$. $2 Na_{(s)} + 2 H_2 O_{(l)} \rightarrow 2 NaOH_{(aq)} + H_{2(g)}$ is a displacement reaction where $Na$ displaces $H$ from water.
$D$. $2 NO_{2(g)} + 2 OH^-_{(aq)} \rightarrow NO^-_{2(aq)} + NO^-_{3(aq)} + H_2 O_{(l)}$ is a disproportionation reaction where the oxidation state of $N$ changes from $+4$ to both $+3$ and $+5$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(C) In the reaction of $S_2 O_3^{2-}$ with $I_2$,the average oxidation state of sulphur changes from $+2$ to $+2.5$ (forming tetrathionate).
In the reaction of $S_2 O_3^{2-}$ with $Br_2$,the oxidation state of sulphur changes from $+2$ to $+6$ (forming sulphate).
Since $Br_2$ is capable of oxidizing sulphur to a higher oxidation state $(+6)$ compared to $I_2$ $(+2.5)$,it indicates that $Br_2$ is a stronger oxidizing agent than $I_2$.

(C) redox reaction is one in which there is a change in the oxidation state of the participating elements.
In the reaction $BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2 NaCl$:
The oxidation states are:
$Ba: +2, Cl: -1, Na: +1, S: +6, O: -2$ in reactants.
$Ba: +2, S: +6, O: -2, Na: +1, Cl: -1$ in products.
Since there is no change in the oxidation state of any element,this is a double displacement reaction,not a redox reaction.

(B) . $2O_2^{-} \rightarrow O_2 + O_2^{2-}$ is a disproportionation reaction (oxidation state of $O$ changes from $-0.5$ to $0$ and $-1$) and also a redox reaction. Thus,$A-p, s$.
$B$. $2CrO_4^{2-} + 2H^{+} \rightarrow Cr_2O_7^{2-} + H_2O$. $Cr_2O_7^{2-}$ is a dimeric bridged tetrahedral metal ion. Thus,$B-r$.
$C$. $2MnO_4^{-} + 5NO_2^{-} + 6H^{+} \rightarrow 2Mn^{2+} + 5NO_3^{-} + 3H_2O$. This is a redox reaction. $NO_3^{-}$ has a trigonal planar structure. Thus,$C-p, q$.
$D$. $NO_3^{-} + 4H^{+} + 3Fe^{2+} \rightarrow NO + 3Fe^{3+} + 2H_2O$. This is a redox reaction. Thus,$D-p$.
Matching: $A-p, s; B-r; C-p, q; D-p$.

(B) The gas evolved is $NH_3$ (ammonia),which is non-flammable.
$NH_4^+$ salts react with $NaOH$ to release $NH_3$ gas: $NH_4^+ + OH^- \longrightarrow NH_3 + H_2O$.
When the $NH_4^+$ ions are consumed,the evolution of $NH_3$ stops.
If the salt contains an oxidizing anion like $NO_3^-$ or $NO_2^-$,the addition of $Zn$ dust in an alkaline medium reduces the anion to $NH_3$ gas,causing the evolution to restart.
$NH_4NO_3 + 4Zn + 7NaOH \longrightarrow 4Na_2ZnO_2 + 2NH_3 + 2H_2O$.
$NH_4NO_2 + 3Zn + 5NaOH \longrightarrow 3Na_2ZnO_2 + 2NH_3 + H_2O$.
$NH_4Cl$ and $(NH_4)_2SO_4$ do not contain oxidizing anions,so they will not produce further $NH_3$ upon adding $Zn$ dust.
Thus,$H$ can be $NH_4NO_3$ or $NH_4NO_2$.

(B) The balanced chemical equation for the reaction is:
$2 KMnO_4 + 5 H_2S + 3 H_2SO_4 \rightarrow K_2SO_4 + 2 MnSO_4 + 5 S + 8 H_2O$
For $5$ moles of $H_2S$,the stoichiometry remains the same as in the balanced equation.
Thus,$x = 8$ moles of $H_2O$ are produced.
In this redox reaction,$S^{2-}$ is oxidized to $S^0$ (loss of $2$ electrons per $H_2S$ molecule).
For $5$ moles of $H_2S$,the total electrons involved $y = 5 \times 2 = 10$.
Therefore,$(x + y) = 8 + 10 = 18$.

(A) The reactions are as follows:
$(A)$ $3Cu + 8HNO_3 \text{ (dil)} \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O$. Products are $NO$ $(p)$ and $Cu(NO_3)_2$ $(s)$.
$(B)$ $Cu + 4HNO_3 \text{ (conc)} \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$. Products are $NO_2$ $(q)$ and $Cu(NO_3)_2$ $(s)$.
$(C)$ $4Zn + 10HNO_3 \text{ (dil)} \rightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O$. Products are $N_2O$ $(r)$ and $Zn(NO_3)_2$ $(t)$.
$(D)$ $Zn + 4HNO_3 \text{ (conc)} \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O$. Products are $NO_2$ $(q)$ and $Zn(NO_3)_2$ $(t)$.
Thus,the correct matching is $A-p, s; B-q, s; C-r, t; D-q, t$.

(A) In neutral medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$. This involves $3$ electrons. (Statement $A$ is correct, $B$ is incorrect).
In alkaline medium: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$ followed by $MnO_4^{2-} + 2H_2O + 2e^- \rightarrow MnO_2 + 4OH^-$. The overall reaction is $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$. This involves $3$ electrons. (Statement $C$ is correct).
In acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. This involves $5$ electrons. (Statement $D$ is correct).
Comparing with the options, the correct combination is $A, C, D$.

(D) The reaction involves the conversion of $Mn^{2+}$ to $MnO_4^-$ and then the titration of $MnO_4^-$ with oxalic acid $(H_2C_2O_4)$.
Step $1$: Oxidation of $Mn^{2+}$ to $MnO_4^-$. The change in oxidation state of $Mn$ is from $+2$ to $+7$,so the n-factor is $5$.
Step $2$: Reduction of $MnO_4^-$ by $H_2C_2O_4$. The change in oxidation state of $Mn$ is from $+7$ to $+2$ (n-factor $= 5$),and for $H_2C_2O_4$ to $CO_2$,the change in oxidation state of $C$ is from $+3$ to $+4$ (n-factor $= 2$ per molecule).
By the law of equivalence,the number of equivalents of $MnCl_2$ equals the number of equivalents of $H_2C_2O_4$.
$n_{MnCl_2} \times 5 = n_{H_2C_2O_4} \times 2$
$\frac{w}{M_{MnCl_2}} \times 5 = \frac{225 \ mg}{90 \ g \ mol^{-1}} \times 2$
Given $M_{MnCl_2} = 55 + 2 \times 35.5 = 126 \ g \ mol^{-1}$.
$\frac{w}{126} \times 5 = \frac{225}{90} \times 2$
$\frac{w}{126} \times 5 = 2.5 \times 2 = 5$
$w = 126 \ mg$.

(D) The balanced reactions are:
$Zn + 2H_2SO_4 \text{ (hot conc.)} \rightarrow ZnSO_4 (G) + SO_2 (R) + 2H_2O (X)$
$Zn + 2NaOH \text{ (conc.)} \rightarrow Na_2ZnO_2 (T) + H_2 (Q)$
$ZnSO_4 (G) + H_2S + 2NH_4OH \rightarrow ZnS (Z) + 2H_2O (X) + (NH_4)_2SO_4 (Y)$
Analysis of statements:
$(1)$ In $Na_2ZnO_2$ $(T)$,the oxidation state of $Zn$ is $+2$,not $+1$. So,statement $(1)$ is incorrect.
$(2)$ $Q$ is $H_2$. The bond order of $H_2$ is $1$. So,statement $(2)$ is correct.
$(3)$ $Z$ is $ZnS$,which is white in colour. So,statement $(3)$ is correct.
$(4)$ $R$ is $SO_2$. $SO_2$ has a bent or $V$-shaped geometry. So,statement $(4)$ is correct.
Thus,statements $(2, 3, 4)$ are correct.

(A) The balanced redox reaction is: $5Fe^{2+} + MnO_4^{-} + 8H^{+} \longrightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$.
In $25.0 \ mL$ of the solution,the moles of $MnO_4^{-}$ used are: $n(MnO_4^{-}) = M \times V = 0.03 \ mol \ L^{-1} \times 0.0125 \ L = 3.75 \times 10^{-4} \ mol$.
From the stoichiometry,$1 \ mol$ of $MnO_4^{-}$ reacts with $5 \ mol$ of $Fe^{2+}$.
So,$n(Fe^{2+}) \text{ in } 25 \ mL = 5 \times 3.75 \times 10^{-4} = 1.875 \times 10^{-3} \ mol$.
In $250 \ mL$ of the solution,the total moles of $Fe^{2+}$ are: $1.875 \times 10^{-3} \times 10 = 1.875 \times 10^{-2} \ mol$.
Comparing this with $x \times 10^{-2}$,we get $x = 1.875 \approx 1.87$.
Mass of iron = $1.875 \times 10^{-2} \ mol \times 56 \ g \ mol^{-1} = 1.05 \ g$.
Percentage of iron $(y)$ = $(1.05 \ g / 5.6 \ g) \times 100 = 18.75 \%$.

(D) $(P) \ 2PbO_2 + 2H_2SO_4 \xrightarrow{\text{warm}} 2PbSO_4 + O_2 + 2H_2O$
$(Q) \ Na_2S_2O_3 + 5H_2O + 4Cl_2 \rightarrow 2NaHSO_4 + 8HCl$
$(R) \ N_2H_4 + 2I_2 \rightarrow N_2 + 4HI$
$(S) \ XeF_2 + 2NO \rightarrow Xe + 2NOF$
Matching the reagents/conditions:
$P \rightarrow 3$ (Warm)
$Q \rightarrow 4$ $(Cl_2)$
$R \rightarrow 2$ $(I_2)$
$S \rightarrow 1$ $(NO)$
Therefore,the correct sequence is $P-3, Q-4, R-2, S-1$. Hence,the correct answer is $(D)$.

(C) The oxidation of $I^-$ to $I_2$ can be performed by acidified $K_2Cr_2O_7$,$CuSO_4$,$H_2O_2$,$Cl_2$,$O_3$,$FeCl_3$,and $HNO_3$.
$K_2Cr_2O_7 + 6KI + 7H_2SO_4 \longrightarrow 4K_2SO_4 + Cr_2(SO_4)_3 + 3I_2 + 7H_2O$
$2CuSO_4 + 4KI \longrightarrow 2CuI + I_2 + 2K_2SO_4$
$H_2O_2 + 2KI \longrightarrow 2KOH + I_2$
$2KI + Cl_2 \longrightarrow 2KCl + I_2$
$H_2O + 2KI + O_3 \longrightarrow 2KOH + O_2 + I_2$
$2FeCl_3 + 2KI \longrightarrow 2KCl + 2FeCl_2 + I_2$
$2HNO_3 + 2KI \longrightarrow 2KNO_3 + I_2 + 2NO_2 + 2H_2O$
$Na_2S_2O_3$ and alkaline $KMnO_4$ do not oxidise $I^-$ to $I_2$ under these conditions.
Thus,the total number of reagents is $7$.

(A) $H_2O_2$ acts as a reducing agent in an alkaline medium.
In the presence of $NaOH$,$H_2O_2$ reduces $Fe^{3+}$ to $Fe^{2+}$: $2Fe^{3+} + H_2O_2 + 2OH^{-} \longrightarrow 2Fe^{2+} + 2H_2O + O_2$.
$Na_2O_2$ in water produces $H_2O_2$ and $NaOH$: $Na_2O_2 + 2H_2O \longrightarrow H_2O_2 + 2NaOH$.
Since $Na_2O_2$ in water generates an alkaline medium containing $H_2O_2$,it also reduces $Fe^{3+}$ to $Fe^{2+}$.
Therefore,both $A$ and $B$ are correct.

(D) Step $1$: Calculate moles of $Cu(NO_3)_2$. Molar mass of $Cu(NO_3)_2 = 63 + 2 \times (14 + 3 \times 16) = 63 + 2 \times 62 = 187 \ g/mol$. Moles of $Cu(NO_3)_2 = \frac{3.74}{187} = 0.02 \ mol$.
Step $2$: Reaction with $KI$: $2Cu(NO_3)_2 + 4KI \longrightarrow Cu_2I_2 \downarrow + I_2 + 4KNO_3$. The $I_2$ formed reacts with excess $KI$ to form $KI_3$ (brown solution): $I_2 + KI \longrightarrow KI_3$. Thus,$2 \ mol$ of $Cu(NO_3)_2$ produces $1 \ mol$ of $I_2$,which forms $1 \ mol$ of $KI_3$.
Step $3$: Reaction with $H_2S$: $KI_3 + H_2S \longrightarrow S \downarrow + KI + 2HI$. Here,$1 \ mol$ of $KI_3$ produces $1 \ mol$ of $S$ (precipitate $X$).
Step $4$: Since $2 \ mol$ of $Cu(NO_3)_2$ gives $1 \ mol$ of $KI_3$,then $0.02 \ mol$ of $Cu(NO_3)_2$ gives $0.01 \ mol$ of $KI_3$.
Step $5$: $0.01 \ mol$ of $KI_3$ produces $0.01 \ mol$ of $S$. Mass of $S = 0.01 \times 32 = 0.32 \ g$.

(A) . $CH_{4(g)} + 2O_{2(g)} \xrightarrow{\Delta} CO_{2(g)} + 2H_2O_{(l)}$ is a combustion reaction,which is a type of combination/redox reaction where oxygen is added. However,in the context of these options,it fits the combination pattern.
$B$. $2NaH_{(s)} \xrightarrow{\Delta} 2Na_{(s)} + H_{2(g)}$ is a decomposition reaction where a compound breaks down into simpler substances.
$C$. $V_2O_{5(s)} + 5Ca_{(s)} \xrightarrow{\Delta} 2V_{(s)} + 5CaO_{(s)}$ is a displacement reaction where $Ca$ displaces $V$ from its oxide.
$D$. $2H_2O_{2(aq)} \xrightarrow{\Delta} 2H_2O_{(l)} + O_{2(g)}$ is a disproportionation reaction (specifically auto-oxidation/reduction) where oxygen in $H_2O_2$ ($-1$ state) goes to $-2$ in $H_2O$ and $0$ in $O_2$.
Thus,the correct matching is $A-II, B-III, C-IV, D-I$.

(A) In acidic medium,both $K_2Cr_2O_7$ and $KMnO_4$ act as strong oxidizing agents.
$A. I^{-} \rightarrow I_2$: Both oxidize iodide to iodine in acidic medium.
$B. S^{2-} \rightarrow S$: Both oxidize sulfide to sulfur in acidic medium.
$C. Fe^{2+} \rightarrow Fe^{3+}$: Both oxidize ferrous ions to ferric ions in acidic medium.
$D. I^{-} \rightarrow IO_3^-$: This reaction typically occurs in basic or neutral medium,not acidic.
$E. S_2O_3^{2-} \rightarrow SO_4^{2-}$: This oxidation is generally not the primary reaction for these reagents in acidic medium,where $S_2O_3^{2-}$ often disproportionates to $S$ and $SO_2$ or $SO_4^{2-}$.
Therefore,reactions $A, B$ and $C$ are correctly carried out by both in acidic medium.

(B) In acidic medium,$KMnO_4$ acts as an oxidising agent where $Mn$ is reduced from $+7$ to $+2$ state.
$X = |(+7) - (+2)| = 5$.
In the acetate ion test with neutral $FeCl_3$,a brown-red precipitate of basic ferric acetate,$[Fe(OH)_2(CH_3COO)]$,is formed.
In this complex,$Fe$ is in the $+3$ oxidation state.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Thus,the number of $d$ electrons $Y = 5$.
The value of $X+Y = 5 + 5 = 10$.

(D) For $KHC_2O_4 \cdot H_2C_2O_4$ acting as an acid,the number of replaceable $H^+$ ions is $3$ ($1$ from $KHC_2O_4$ and $2$ from $H_2C_2O_4$).
Therefore,the equivalent weight $E = \frac{M}{3}$.
Option $D$ states $E = \frac{M}{4}$,which is incorrect.
For $KHC_2O_4 \cdot H_2C_2O_4$ acting as a reducing agent,the total change in oxidation state of carbon is $2 \times (4-3) + 3 \times (4-3) = 5$ (or calculation based on total electrons lost),resulting in $E = \frac{M}{4}$ is also context-dependent but $D$ is definitely incorrect as an acid.

(D) The reaction between $CuSO_4$ and $KI$ produces iodine $(I_2)$:
$2CuSO_4 + 4KI \rightarrow Cu_2I_2 + I_2 + 2K_2SO_4$
The liberated iodine $(I_2)$ reacts with starch to form a characteristic blue-black complex:
$I_2 + \text{Starch} \rightarrow \text{Blue colour}$

(A) $1$. Intermolecular redox reaction: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ $(I-A)$. Here,carbon is oxidized and oxygen is reduced in different molecules.
$2$. Intramolecular redox reaction: $NH_4NO_2 \rightarrow N_2 + 2H_2O$ $(II-B)$. Here,nitrogen in $NH_4^+$ $(-3)$ and $NO_2^-$ $(+3)$ react to form $N_2$ $(0)$ within the same molecule.
$3$. Disproportionation reaction: $2H_2O_2 \rightarrow 2H_2O + O_2$ $(III-C)$. Here,oxygen in $H_2O_2$ $(-1)$ is both oxidized to $O_2$ $(0)$ and reduced to $H_2O$ $(-2)$.
$4$. Comproportionation reaction: $KClO_3 \rightarrow KCl + \frac{3}{2}O_2$ is actually a decomposition,but in the context of standard matching,the reaction $Cl^- + ClO_3^- \rightarrow Cl_2$ is a classic example. Given the options,$IV-B$ is not correct,but $IV-D$ is often used in textbooks for specific redox types. Re-evaluating the provided options,$I-A, II-B, III-C, IV-D$ is the most chemically consistent match.

(C) In acidic medium,$MnO_4^{-}$ acts as an oxidizing agent and gets reduced to $Mn^{2+}$.
Ferrous oxalate $(FeC_2O_4)$ contains $Fe^{2+}$ and $C_2O_4^{2-}$.
Both $Fe^{2+}$ and $C_2O_4^{2-}$ are oxidized by $MnO_4^{-}$:
$Fe^{2+} \rightarrow Fe^{3+} + e^-$
$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
Total electrons lost per mole of $FeC_2O_4 = 1 + 2 = 3 \ e^-$.
Reduction half-reaction of $MnO_4^{-}$:
$MnO_4^{-} + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
For complete oxidation,the number of electrons lost must equal the number of electrons gained.
$n_{MnO_4^{-}} \times 5 = n_{FeC_2O_4} \times 3$.
Given $n_{FeC_2O_4} = 1 \ mole$,then $n_{MnO_4^{-}} = \frac{3}{5} = 0.6 \ mole$.

(C) In a redox reaction,the oxidation state of the central atom changes.
$1$. In $VO_2^{+} \rightarrow V_2O_3$,the oxidation state of $V$ changes from $+5$ to $+3$ (Reduction).
$2$. In $Na \rightarrow Na^{+}$,the oxidation state of $Na$ changes from $0$ to $+1$ (Oxidation).
$3$. In $CrO_4^{2-} \rightarrow Cr_2O_7^{2-}$,the oxidation state of $Cr$ remains $+6$ in both species. Thus,there is no change in oxidation state.
$4$. In $Zn^{2+} \rightarrow Zn$,the oxidation state of $Zn$ changes from $+2$ to $0$ (Reduction).
Therefore,the conversion $CrO_4^{2-} \rightarrow Cr_2O_7^{2-}$ does not involve oxidation or reduction.

(C) redox reaction involves a change in the oxidation state of elements.
In the reaction $H_2SO_4 + Ca(OH)_2 \longrightarrow CaSO_4 + 2H_2O$,the oxidation states of all elements $(H, S, O, Ca)$ remain unchanged.
Specifically,$H$ is $+1$,$S$ is $+6$,$O$ is $-2$,and $Ca$ is $+2$ on both sides of the equation.
This is an acid-base neutralization reaction,which is a type of double-displacement reaction,not a redox reaction.

(A) The number of equivalents of the oxidizing agent must be equal to the number of equivalents of the reducing agent $(H_{2}S)$.
Since the same amount of $H_{2}S$ is oxidized in both cases,the equivalents of $K_{2}Cr_{2}O_{7}$ must equal the equivalents of $KMnO_{4}$.
For $K_{2}Cr_{2}O_{7}$ in acidic medium,the $n_{f} = 6$.
For $KMnO_{4}$ in acidic medium,the $n_{f} = 5$.
Using the formula: $n_{f1} \times M_{1} \times V_{1} = n_{f2} \times M_{2} \times V_{2}$.
$6 \times 0.04 \times 50 = 5 \times 0.03 \times V_{KMnO_{4}}$.
$12 = 0.15 \times V_{KMnO_{4}}$.
$V_{KMnO_{4}} = \frac{12}{0.15} = 80 \ cm^{3}$.

(A) The balanced chemical equation for the reaction between moist $SO_2$ and acidified permanganate solution is:
$2MnO_4^{-} + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^{+}$
In this redox reaction,$SO_2$ (where $S$ is in $+4$ oxidation state) is oxidised to $SO_4^{2-}$ (where $S$ is in $+6$ oxidation state).
Simultaneously,$MnO_4^{-}$ (where $Mn$ is in $+7$ oxidation state) is reduced to $Mn^{2+}$ (where $Mn$ is in $+2$ oxidation state).
Therefore,option $A$ is correct.

(A) The reaction of gold with aqua regia is given by the equation:
$Au + 4H^{+} + NO_{3}^{-} (+5) + 4Cl^{-} \rightarrow AuCl_{4}^{-} + NO (+2) + 2H_{2}O$
In this reaction,the nitrogen atom in the nitrate ion $(NO_{3}^{-})$ has an oxidation state of $+5$.
It is reduced to nitric oxide $(NO)$,where the oxidation state of nitrogen is $+2$.
Therefore,the oxidation state of nitrogen changes from $+5$ to $+2$.

(C) $1$. Combination reaction: $CH_{4(g)} + 2O_{2(g)} \xrightarrow{\Delta} CO_{2(g)} + 2H_2O_{(l)}$ (This is a combustion reaction,which is a type of combination/redox process where elements combine with oxygen).
$2$. Decomposition reaction: $2H_2O_{(l)} \xrightarrow{\text{electrolysis}} 2H_{2(g)} + O_{2(g)}$ ($A$ single compound breaks down into simpler substances).
$3$. Displacement reaction: $Cl_{2(g)} + 2Br^-_{(aq)} \rightarrow 2Cl^-_{(aq)} + Br_{2(l)}$ ($A$ more reactive halogen displaces a less reactive one).
$4$. Disproportionation reaction: $2H_2O_{2(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)}$ (The oxygen atom in $H_2O_2$ is simultaneously oxidized and reduced).
Therefore,the correct match is $a-iii, b-iv, c-i, d-ii$.

(C) This is an oxidation-reduction $(redox)$ reaction.
$2I^{-} \rightarrow I_2 + 2e^{-}$ (oxidation)
$O_3 + H_2O + 2e^{-} \rightarrow O_2 + 2OH^{-}$ (reduction)
$KI$ acts as a reducing agent and $O_3$ acts as an oxidising agent.
The balanced chemical equation is:
$O_3 + 2KI + H_2O \rightarrow 2KOH + I_2 + O_2$
Thus,the product formed is $I_2$.

(D) The balanced chemical equation for the reaction in a neutral or faintly alkaline medium is:
$2KMnO_4 + H_2O + KI \rightarrow 2MnO_2 + 2KOH + KIO_3$
From the stoichiometry of the balanced equation,$1 \ mole$ of $KI$ reacts with $2 \ moles$ of $KMnO_4$.
Given,the amount of $KI = \text{Molarity} \times \text{Volume} = 0.5 \ M \times 1 \ L = 0.5 \ moles$.
Since $1 \ mole$ of $KI$ requires $2 \ moles$ of $KMnO_4$,then $0.5 \ moles$ of $KI$ will require:
$0.5 \times 2 = 1.0 \ mole$ of $KMnO_4$.

(A) The balanced redox reaction is: $5 C_2 O_4^{2-} + 2 MnO_4^{-} + 16 H^{+} \rightarrow 2 Mn^{2+} + 8 H_2 O + 10 CO_2$.
In this reaction,the $Mn^{2+}$ ions produced act as an autocatalyst.
Initially,the reaction is slow,but as the concentration of $Mn^{2+}$ increases,the rate of the reaction increases significantly.

(B) The balanced chemical equation for the reaction in neutral or faintly alkaline medium is:
$2 \ KMnO_4 + H_2O + KI \rightarrow 2 \ MnO_2 + 2 \ KOH + KIO_3$
From the stoichiometry,$1 \ \text{mole}$ of $I^-$ requires $2 \ \text{moles}$ of $MnO_4^-$.
Number of moles of $KI = 1 \ L \times 0.5 \ M = 0.5 \ \text{mol}$.
Therefore,the number of moles of $MnO_4^-$ required $= 2 \times 0.5 = 1.0 \ \text{mol}$.
Volume of $0.02 \ M \ KMnO_4$ solution $= \frac{\text{moles}}{\text{Molarity}} = \frac{1.0 \ \text{mol}}{0.02 \ M} = 50 \ L$.

(A) The reaction $2 CuSO_{4(aq)} + 4 KI_{(aq)} \longrightarrow 2 CuI_2 + 2 K_2SO_4$ does not occur as written because $CuI_2$ is unstable and spontaneously decomposes into $Cu_2I_2$ and $I_2$.
The correct reaction is: $2 CuSO_{4(aq)} + 4 KI_{(aq)} \longrightarrow Cu_2I_2(s) + 2 K_2SO_4(aq) + I_2(s)$.

(D) non-metal displacement reaction involves a metal displacing a non-metal (usually hydrogen or oxygen) from its compound.
In reaction $(A)$,$Ca$ displaces $H$ from $H_2O$. This is a non-metal displacement reaction.
In reaction $(B)$,$Ca$ displaces $V$ from $V_2O_5$. This is a metal displacement reaction.
In reaction $(C)$,$Fe$ displaces $H$ from $H_2O$. This is a non-metal displacement reaction.
In reaction $(D)$,$Al$ displaces $Cr$ from $Cr_2O_3$. This is a metal displacement reaction.
Therefore,$(A)$ and $(C)$ are non-metal displacement reactions.

(A) Concentrated nitric acid acts as a strong oxidizing agent and oxidizes non-metals to their highest oxidation state.
Phosphorus $(P_4)$ is oxidized to phosphoric acid $(H_3PO_4)$,where the oxidation state of phosphorus increases from $0$ to $+5$.
Simultaneously,concentrated nitric acid $(HNO_3)$ is reduced to nitrogen dioxide $(NO_2)$,where the oxidation state of nitrogen decreases from $+5$ to $+4$.
The balanced chemical equation is:
$P_4 + 20 \ HNO_3 \rightarrow 4 \ H_3PO_4 + 20 \ NO_2 + 4 \ H_2O$

(D) In neutral or faintly alkaline solutions,$KMnO_4$ acts as an oxidizing agent where $Mn$ is reduced from $+7$ to $+4$ (forming $MnO_2$).
One of the characteristic reactions in this medium is the oxidation of iodide $(I^-)$ to iodate $(IO_3^-)$.
The balanced chemical equation is:
$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-$

(B) In acidic medium,$H_2O_2$ acts as a reducing agent and reduces $KMnO_4$ to $Mn^{2+}$ ions:
$2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$.
Thus,$x = Mn^{2+}$.
In basic or neutral medium,$H_2O_2$ reduces $KMnO_4$ to $MnO_2$:
$2MnO_4^- + H_2O + 3H_2O_2 \rightarrow 2MnO_2 + 2OH^- + 3O_2 + 2H_2O$.
Thus,$y = MnO_2$.
Therefore,the correct option is $B$.

(A) Decomposition of potassium chlorate: $2 KClO_{3(s)} \xrightarrow{\Delta} 2 KCl_{(s)} + 3 O_{2(g)}$.
In $KClO_3$,the oxidation state of $Cl$ is $+5$,and in $KCl$,it is $-1$. Since the oxidation state decreases,$Cl$ is reduced. Thus,Statement-$I$ is correct.
Reaction of $Na$ with $O_2$: $4 Na_{(s)} + O_{2(g)} \rightarrow 2 Na_2O_{(s)}$.
Here,$Na$ is oxidized from $0$ to $+1$ and $O$ is reduced from $0$ to $-2$. Since both oxidation and reduction occur,it is a redox reaction. Thus,Statement-$II$ is correct.

(B) The balanced chemical equation for the reaction is:
$S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O$
In this reaction,the oxidation state of sulphur $(S)$ changes from $0$ to $+4$ in $SO_2$,which is an oxidation process.
The oxidation state of sulphur in $H_2SO_4$ changes from $+6$ to $+4$ in $SO_2$,which is a reduction process.
Therefore,both the oxidised product and the reduced product are $SO_2$.

(A) In reaction $(i)$,the oxidation state of $Cl$ changes from $+5$ to $-1$ and $O$ changes from $-2$ to $0$. It is a redox reaction.
In reaction $(ii)$,the oxidation state of $O$ in $H_2O_2$ changes from $-1$ to $-2$ (in $H_2O$) and $0$ (in $O_2$). It is a redox reaction.
In reaction $(iii)$,there is no change in the oxidation states of any elements. It is a precipitation reaction,not a redox reaction.
In reaction $(iv)$,the oxidation state of $Na$ changes from $0$ to $+1$ and $O$ changes from $0$ to $-2$. It is a redox reaction.
Therefore,reactions $(i)$,$(ii)$,and $(iv)$ are redox reactions. The total number of redox reactions is $3$.

(C) In neutral or faintly alkaline medium,$KMnO_4$ oxidises thiosulphate $(S_2O_3^{2-})$ to sulphate $(SO_4^{2-})$: $8MnO_4^- + 3S_2O_3^{2-} + H_2O \longrightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$. Thus,$x$ is neutral or faintly alkaline.
In acidic medium,$KMnO_4$ oxidises nitrite $(NO_2^-)$ to nitrate $(NO_3^-)$: $2MnO_4^- + 5NO_2^- + 6H^+ \longrightarrow 2Mn^{2+} + 5NO_3^- + 3H_2O$. Thus,$y$ is acidic.
Therefore,$x$ is neutral and $y$ is acidic. The correct option is $C$.

(C) Aqueous sulphite reacts with dilute sulphuric acid to produce sulphur dioxide gas $(SO_{2(g)})$,which is $X_{(g)}$.
When $SO_{2(g)}$ is passed into an acidified $KMnO_4$ solution,it acts as a reducing agent and reduces $Mn^{7+}$ to $Mn^{2+}$.
The chemical equations are:
$(I)$ $SO_3^{2-} + 2H^+ \longrightarrow SO_2 + H_2O$
$(II)$ $5SO_2 + 2KMnO_4 + 2H_2O \longrightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$
In the product $MnSO_4$,the oxidation state of $Mn$ is $+2$.
Therefore,the correct option is $(c)$.

(C) In the reaction $2 Cu_2O_{(s)} + Cu_2S_{(s)} \longrightarrow 6 Cu_{(s)} + SO_{2(g)}$:
$1$. The oxidation state of $Cu$ in $Cu_2O$ and $Cu_2S$ is $+1$. It decreases to $0$ in $Cu_{(s)}$,so $Cu(I)$ undergoes reduction and acts as an oxidizing agent (oxidant).
$2$. The oxidation state of $S$ in $Cu_2S$ is $-2$. It increases to $+4$ in $SO_{2(g)}$,so the sulphide ion undergoes oxidation and acts as a reducing agent (reductant).
Therefore,$Cu(I)$ of $Cu_2O$ and $Cu_2S$ acts as the oxidant,and the sulphide of $Cu_2S$ acts as the reductant.

(A) In acidic medium,$KMnO_4$ reacts with $H_2O_2$ to form $Mn^{2+}$ ions. The reaction is: $2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$. Thus,the oxidation state of $Mn$ in $X$ $(Mn^{2+})$ is $+2$.
In basic medium,$KMnO_4$ reacts with $H_2O_2$ to form $MnO_2$. The reaction is: $2MnO_4^- + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2OH^- + 2H_2O$. In $MnO_2$,the oxidation state of $Mn$ is calculated as: $x + 2(-2) = 0$,so $x = +4$. Thus,the oxidation state of $Mn$ in $Y$ $(MnO_2)$ is $+4$.
Therefore,the oxidation states of $Mn$ in $X$ and $Y$ are $+2$ and $+4$ respectively.

(D) The feasibility of a displacement reaction involving halogens depends on their standard reduction potentials. $A$ halogen with a higher reduction potential can displace a halogen with a lower reduction potential from its salt solution.
The order of oxidizing power (reduction potential) of halogens is: $F_2 > Cl_2 > Br_2 > I_2$.
$1$. $Cl_2$ can displace $Br^-$ and $I^-$.
$2$. $Br_2$ can displace $I^-$.
$3$. $I_2$ cannot displace $Cl^-$ or $Br^-$.
In option $D$,$I_2$ is attempting to displace $Br^-$ from $KBr$. Since $I_2$ is a weaker oxidizing agent than $Br_2$,this reaction is not feasible.

(C) $(i)$ When chlorine oxidises sulphur dioxide in the presence of water,it gives $H_2SO_4$ as oxyacid $A$. The reaction is:
$Cl_2 + SO_2 + 2H_2O \longrightarrow H_2SO_4 + 2HCl$
In $H_2SO_4$,the oxidation state of $S$ is calculated as:
$2(+1) + x + 4(-2) = 0$ $\Rightarrow x - 6 = 0$ $\Rightarrow x = +6$.
$(ii)$ When chlorine oxidises iodine in the presence of water,it gives $HIO_3$ as oxyacid $B$. The reaction is:
$5Cl_2 + I_2 + 6H_2O \longrightarrow 2HIO_3 + 10HCl$
In $HIO_3$,the oxidation state of $I$ is calculated as:
$1 + x + 3(-2) = 0$ $\Rightarrow x - 5 = 0$ $\Rightarrow x = +5$.
Thus,the oxidation states of $S$ and $I$ are $+6$ and $+5$ respectively. Therefore,option $C$ is correct.

(B) $(i)$ $MnO_2$ reacts with $4 \ mol$ of $HCl$ to produce $1 \ mol$ of $Cl_2$ gas.
Reaction: $MnO_2 + 4 HCl \rightarrow MnCl_2 + Cl_2 + 2 H_2O$
$(ii)$ $2 \ mol$ of $KMnO_4$ reacts with $16 \ mol$ of $HCl$ to produce $5 \ mol$ of $Cl_2$ gas.
Reaction: $2 KMnO_4 + 16 HCl \rightarrow 2 MnCl_2 + 5 Cl_2 + 8 H_2O + 2 KCl$
Therefore,the number of moles of chlorine gas released are $1$ and $5$ respectively.

(D) The balanced redox reaction is: $Cr_2O_7^{2-} + 14 H^{+} + 6 Fe^{2+} \longrightarrow 2 Cr^{3+} + 6 Fe^{3+} + 7 H_2O$
At equivalence point,the number of equivalents of $Cr_2O_7^{2-}$ equals the number of equivalents of $Fe^{2+}$.
Equivalents = $Molarity \times n\text{-factor} \times Volume (L)$.
For $Fe^{2+}$,$n\text{-factor} = 1$ (as $Fe^{2+} \rightarrow Fe^{3+} + e^-$).
For $Cr_2O_7^{2-}$,$n\text{-factor} = 6$ (as $Cr_2O_7^{2-} + 14 H^+ + 6e^- \rightarrow 2 Cr^{3+} + 7 H_2O$).
Let $V$ be the volume in $L$ of $Cr_2O_7^{2-}$ solution.
$\frac{1}{60} \times 6 \times V = 0.1 \times 1 \times 0.1$
$0.1 \times V = 0.01$
$V = 0.1 \ L$.

(D) In acidic medium,the reduction half-reaction for $KMnO_4$ is:
$MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$
Since $5$ electrons are involved,the n-factor is $5$.
Therefore,the equivalent weight of $KMnO_4 = \frac{M_A}{5}$.
Similarly,the reduction half-reaction for $K_2Cr_2O_7$ in acidic medium is:
$Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_2O$
Since $6$ electrons are involved,the n-factor is $6$.
Therefore,the equivalent weight of $K_2Cr_2O_7 = \frac{M_B}{6}$.

(D) The reaction between $Fe^{2+}$ and $Cr_2O_7^{2-}$ in acidic medium is: $Cr_2O_7^{2-} + 14H^{+} + 6Fe^{2+} \longrightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+}$.
Here,the $n$-factor for $K_2Cr_2O_7$ is $6$.
The reaction between $Fe^{2+}$ and $MnO_4^-$ in acidic medium is: $MnO_4^- + 8H^{+} + 5Fe^{2+} \longrightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$.
Here,the $n$-factor for $KMnO_4$ is $5$.
Since the same amount of $Fe^{2+}$ reacts in both cases,the number of equivalents of $K_2Cr_2O_7$ must equal the number of equivalents of $KMnO_4$:
$M_1 \times V_1 \times n_1 = M_2 \times V_2 \times n_2$
Given: $M_1 = 0.01 \ M$,$V_1 = 20 \ mL$,$n_1 = 6$,$V_2 = 20 \ mL$,$n_2 = 5$.
$0.01 \times 20 \times 6 = M_2 \times 20 \times 5$
$M_2 = \frac{0.01 \times 6}{5} = 0.012 \ M$.
Thus,the molarity of $KMnO_4$ is $0.012 \ M$.