Mix Examples-Redox Reactions Questions in English

(C) redox reaction involves a change in the oxidation state of elements.
In the reaction $2KI + Cl_2 \to 2KCl + I_2$:
The oxidation state of $I$ changes from $-1$ to $0$ (oxidation).
The oxidation state of $Cl$ changes from $0$ to $-1$ (reduction).
Therefore,this is a redox reaction.

(A) The oxidation state of $Mn$ in $KMnO_4$ is $+7$. The number of electrons gained is equal to the change in oxidation state of $Mn$.
$1$. $MnO_4^- (+7) \rightarrow MnO_4^{2-} (+6)$: Gain of $1 \ e^-$.
$2$. $MnO_4^- (+7) \rightarrow MnO_2 (+4)$: Gain of $3 \ e^-$.
$3$. $MnO_4^- (+7) \rightarrow \frac{1}{2} Mn_2O_3 (+3)$: Gain of $4 \ e^-$.
$4$. $MnO_4^- (+7) \rightarrow Mn^{2+} (+2)$: Gain of $5 \ e^-$.
Thus,the sequence is $1, 3, 4, 5$.

(D) The given reaction is $4Fe + 3O_2 \rightleftharpoons 4Fe^{3+} + 6O^{2-}$.
In this reaction,$Fe$ loses electrons to form $Fe^{3+}$,so $Fe$ acts as a reducing agent $(Fe \to Fe^{3+} + 3e^-)$.
Since $Fe$ undergoes oxidation,it cannot be reduced by $Fe^{3+}$.
Therefore,the statement 'Metallic iron is reduced by $Fe^{3+}$' is incorrect.

(D) The equivalent weight of an oxidizing agent is calculated by the formula: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}}$.
In the given redox reaction,$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,the total number of electrons gained by one mole of $K_2Cr_2O_7$ is $6$.
Therefore,the n-factor for $K_2Cr_2O_7$ is $6$.
Thus,the equivalent weight $= \frac{M}{6}$.

(B) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and converts ferrous ions $(Fe^{2+})$ to ferric ions $(Fe^{3+})$.
The reduction half-reaction for $KMnO_4$ is:
$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$
The number of electrons gained ($n$-factor) per molecule of $KMnO_4$ is $5$.
The molar mass of $KMnO_4$ is $39 + 55 + (4 \times 16) = 158 \ g/mol$.
Equivalent weight = $\frac{\text{Molar mass}}{n\text{-factor}} = \frac{158}{5} = 31.6$.

(A) In the given reaction: $2[Fe(CN)_6]^{3-} + H_2O_2 + 2OH^{-} \rightarrow 2[Fe(CN)_6]^{4-} + 2H_2O + O_2$.
$H_2O_2$ is acting as a reducing agent because it is oxidized to $O_2$.
The oxidation half-reaction is: $H_2O_2 + 2OH^{-} \rightarrow O_2 + 2H_2O + 2e^{-}$.
The change in oxidation state of oxygen in $H_2O_2$ $(-1)$ to $O_2$ $(0)$ involves a loss of $2$ electrons per molecule of $H_2O_2$.
Thus,the $n$-factor for $H_2O_2$ is $2$.
The molar mass of $H_2O_2 = (2 \times 1) + (2 \times 16) = 34 \ g/mol$.
The equivalent weight = $\frac{\text{Molar Mass}}{n\text{-factor}} = \frac{34}{2} = 17$.

(A) The balanced redox reaction in acidic medium is:
$3MnO_4^- + 5FeC_2O_4 + 24H^+ \rightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$
From the stoichiometry of the balanced equation:
$5$ moles of $FeC_2O_4$ require $3$ moles of $MnO_4^-$.
Therefore,$1$ mole of $FeC_2O_4$ requires $\frac{3}{5} = 0.6$ moles of $MnO_4^-$.
Since $KMnO_4$ provides $MnO_4^-$,the required amount is $0.6$ moles of $KMnO_4$.

(B) The reaction between potassium dichromate $(K_2Cr_2O_7)$ and sodium sulfite $(Na_2SO_3)$ in an acidic medium $(H_2SO_4)$ is as follows:
$K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2SO_4 \rightarrow 3Na_2SO_4 + K_2SO_4 + 4H_2O + Cr_2(SO_4)_3$
In this reaction,the orange dichromate ion $(Cr_2O_7^{2-})$ is reduced to the green chromium$(III)$ ion $(Cr^{3+})$,which forms chromium$(III)$ sulfate $(Cr_2(SO_4)_3)$.
Thus,the green color is due to the formation of $Cr_2(SO_4)_3$.

(C) In option $(C)$,the reaction is $BaO_2 + H_2SO_4 \to BaSO_4 + H_2O_2$.
Checking the oxidation states:
In $BaO_2$,$Ba$ is $+2$ and $O$ is $-1$.
In $H_2SO_4$,$H$ is $+1$,$S$ is $+6$,and $O$ is $-2$.
In $BaSO_4$,$Ba$ is $+2$,$S$ is $+6$,and $O$ is $-2$.
In $H_2O_2$,$H$ is $+1$ and $O$ is $-1$.
Since the oxidation states of all elements $(Ba, O, H, S)$ remain the same on both sides,there is no change in valency.

(D) redox reaction is one in which the oxidation state of at least one element changes.
In option $(A)$,$Rb$ goes from $0$ to $+1$ and $H$ goes from $+1$ to $0$.
In option $(B)$,$Cu$ goes from $+2$ to $+1$ and $I$ goes from $-1$ to $0$.
In option $(C)$,$O$ in $H_2O_2$ goes from $-1$ to $-2$ (in $H_2O$) and $0$ (in $O_2$).
In option $(D)$,$4KCN + Fe(CN)_2 \to K_4Fe(CN)_6$ is a complex formation reaction where the oxidation states of all elements remain unchanged $(K=+1, C=+2, N=-3, Fe=+2)$.
Therefore,$(D)$ is not a redox reaction.

(D) In the reaction $CaF_2 + H_2SO_4 \rightarrow CaSO_4 + 2HF$,the oxidation states of all elements $(Ca: +2, F: -1, H: +1, S: +6, O: -2)$ remain unchanged on both sides of the equation.
Since there is no change in oxidation states,this is an acid-base reaction,not a redox reaction.
Therefore,it does not demonstrate the oxidizing behavior of concentrated $H_2SO_4$.

(C) Two compounds can exist together if they do not react with each other via a redox reaction.
$FeCl_3$ and $SnCl_2$ react as $2FeCl_3 + SnCl_2 \rightarrow 2FeCl_2 + SnCl_4$.
$HgCl_2$ and $SnCl_2$ react as $2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$.
$FeCl_3$ and $KI$ react as $2FeCl_3 + 2KI \rightarrow 2FeCl_2 + 2KCl + I_2$.
$FeCl_2$ and $SnCl_2$ do not react with each other because both are in their lower oxidation states and cannot act as oxidizing agents for each other.

(C) The balanced chemical reaction is:
$2KClO_3 + 3H_2C_2O_4 + H_2SO_4 \rightarrow K_2SO_4 + 2KCl + 6CO_2 + 4H_2O$
Calculation of oxidation states:
$1$. For $Cl$ in $KClO_3$,the oxidation state changes from $+5$ to $-1$ in $KCl$. The change is $|-1 - 5| = 6$.
$2$. For $C$ in $H_2C_2O_4$,the oxidation state changes from $+3$ to $+4$ in $CO_2$. The change is $|4 - 3| = 1$.
$3$. The oxidation states of $S$,$H$,$K$,and $O$ remain unchanged in this reaction.
Thus,$Cl$ undergoes the maximum change in oxidation number.

(A) In the formation of $Fe(CO)_5$,the reaction is: $Fe + 5CO \rightarrow Fe(CO)_5$. Here,the oxidation state of $Fe$ remains $0$ (zero).
In the rusting of iron: $4Fe + 3O_2 + 2xH_2O \rightarrow 2Fe_2O_3 \cdot xH_2O$,$Fe$ is oxidized from $0$ to $+3$.
In the reaction with steam: $3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2$,$Fe$ is oxidized from $0$ to $+8/3$.
In the reaction with $CuSO_4$: $Fe + CuSO_4 \rightarrow FeSO_4 + Cu$,$Fe$ is oxidized from $0$ to $+2$.
Therefore,the formation of $Fe(CO)_5$ does not involve the oxidation of iron.

(A) The reaction between acidified $K_2Cr_2O_7$ and $Na_2SO_3$ is a redox reaction where $Cr_2O_7^{2-}$ acts as an oxidizing agent and $SO_3^{2-}$ acts as a reducing agent.
The balanced chemical equation is:
$K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2SO_4 \rightarrow K_2SO_4 + 3Na_2SO_4 + Cr_2(SO_4)_3 + 4H_2O$
In this reaction,the orange-colored dichromate ion $(Cr_2O_7^{2-})$ is reduced to the green-colored chromium$(III)$ sulfate $(Cr_2(SO_4)_3)$,which causes the solution to turn green.

(B) The total energy change involved in the conversion of $\frac{1}{2} Cl_{2(g)}$ to $Cl^{-}_{(aq)}$ is given by the sum of the enthalpy changes of the individual steps:
$\Delta H = \frac{1}{2} \Delta_{diss} H_{Cl_2}^{\Theta} + \Delta_{eg} H_{Cl}^{\Theta} + \Delta_{Hyd} H_{Cl}^{\Theta}$
Substituting the given values:
$\Delta H = (\frac{1}{2} \times 240) + (-349) + (-381) \ kJ \ mol^{-1}$
$\Delta H = 120 - 349 - 381 \ kJ \ mol^{-1}$
$\Delta H = - 610 \ kJ \ mol^{-1}$

(D) The sodium salt of oxalic acid,$Na_2C_2O_4$ $(X)$,reacts with conc. $H_2SO_4$ to produce $CO$ and $CO_2$ gases,causing effervescence.
$Na_2C_2O_4 + H_2SO_4 \to Na_2SO_4 + H_2C_2O_4$
$H_2C_2O_4 \xrightarrow{Conc. H_2SO_4} H_2O + CO \uparrow + CO_2 \uparrow$
$X$ reacts with $CaCl_2$ to form a white precipitate of calcium oxalate $(CaC_2O_4)$:
$Na_2C_2O_4 + CaCl_2 \to CaC_2O_4 \downarrow + 2NaCl$
Oxalate ions $(C_2O_4^{2-})$ decolourise acidic $KMnO_4$ solution due to the redox reaction:
$5C_2O_4^{2-} + 2MnO_4^{-} + 16H^{+} \to 10CO_2 \uparrow + 2Mn^{2+} + 8H_2O$
Thus,$'X'$ is $Na_2C_2O_4$.

(C) redox reaction is one in which oxidation and reduction occur simultaneously,involving a change in the oxidation states of the elements.
In option $C$,the reaction is $2F_{2(g)} + 2H_2O_{(l)} \rightarrow 4H_{(aq)}^+ + 4F_{(aq)}^- + O_{2(g)}$.
Here,the oxidation state of $F$ changes from $0$ to $-1$ (reduction) and the oxidation state of $O$ in $H_2O$ changes from $-2$ to $0$ (oxidation).
Therefore,this is a redox reaction involving water.
In options $A$,$B$,and $D$,there is no change in the oxidation states of any of the elements involved.

(D) substance can act as both an oxidizing and a reducing agent when the central atom is in an intermediate oxidation state.
In $SO_2$,the oxidation state of sulfur is $+4$. Since sulfur can range from $-2$ to $+6$,it can be oxidized to $+6$ or reduced to lower states,thus acting as both.
In $H_2O_2$,the oxidation state of oxygen is $-1$. Since oxygen can range from $-2$ to $0$,it can be reduced to $-2$ or oxidized to $0$,thus acting as both.
Therefore,both $H_2O_2$ and $SO_2$ can act as both oxidizing and reducing agents.

(A) The reaction involves the oxidation of $X^{n+}$ to $XO_3^-$ and the reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$.
According to the law of equivalence,the number of gram equivalents of the oxidizing agent equals the number of gram equivalents of the reducing agent.
$\text{Equivalents of } K_2Cr_2O_7 = \text{Equivalents of } X^{n+}$
$(\text{n-factor})_{K_2Cr_2O_7} \times \text{moles of } K_2Cr_2O_7 = (\text{n-factor})_{X^{n+}} \times \text{moles of } X^{n+}$
For $K_2Cr_2O_7$,the change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are $2$ atoms of $Cr$,the n-factor is $2 \times (6 - 3) = 6$.
For $X^{n+}$,the oxidation state changes from $+n$ to $+5$ (in $XO_3^-$). The n-factor is $(5 - n)$.
Substituting the values: $6 \times (6 \times 10^{-3}) = (5 - n) \times (9 \times 10^{-3})$
$36 = (5 - n) \times 9$
$4 = 5 - n$
$n = 1$.

(C) Let us analyze each reaction:
$1$. $Cr_2O_7^{2-} + H_2O_2 + H^{+} \rightarrow CrO_5 + H_2O$. $CrO_5$ is a deep blue coloured compound.
$2$. $NO_2 + NO \xrightarrow{-23^{\circ}C} N_2O_3$. $N_2O_3$ is a blue solid at low temperatures.
$3$. $Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$. This complex is blood-red in colour,not blue.
$4$. $CuSO_{4(s)} + 5H_2O_{(g)} \rightarrow CuSO_4 \cdot 5H_2O_{(s)}$. Hydrated copper$(II)$ sulfate is blue.
Therefore,the reaction that does not produce a blue colour is $Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$.

(B) The reaction between $SnCl_2$ (stannous chloride) and $HgCl_2$ (mercuric chloride) is a redox reaction.
$SnCl_2$ acts as a reducing agent and reduces $HgCl_2$ to metallic mercury $(Hg)$.
The balanced chemical equation is: $SnCl_2 + HgCl_2 \rightarrow Hg + SnCl_4$.
Therefore,'$A$' is $Hg$.

(D) The reaction between $KMnO_4$,$Na_2SO_4$,and $BaCl_2$ involves the formation of $BaSO_4$ as a white precipitate,not a pink one.
$KMnO_4$ is a strong oxidizing agent,but it does not react with $Na_2SO_4$ or $BaCl_2$ to form a pink precipitate.
$BaSO_4$ is white,$MnS$ is flesh-colored or pinkish,but it is formed by the reaction of $Mn^{2+}$ with $S^{2-}$.
In this specific mixture,no reaction occurs that produces a pink precipitate.
Therefore,the correct statement is that a pink precipitate is not obtained.

(C) The reaction of $Cu^{2+}$ ions with $KI$ is a redox reaction.
$2Cu^{2+} + 4I^-$ $\rightarrow 2CuI_2$ $\rightarrow Cu_2I_2 (white) + I_2$.
However,the reaction of $Hg_2^{2+}$ ions with $KI$ produces a green precipitate of mercury$(I)$ iodide $(Hg_2I_2)$.
$Hg_2(NO_3)_2 + 2KI \rightarrow Hg_2I_2 (green ppt) + 2KNO_3$.
Thus,the correct reagent is $Hg_2(NO_3)_2$.

(A) The thermal decomposition of ammonium dichromate is given by the reaction:
$(NH_4)_2Cr_2O_7 \rightarrow N_2(g) + 4H_2O(g) + Cr_2O_3(s)$
In this reaction,$(NH_4)_2Cr_2O_7$ decomposes to form nitrogen gas,water vapour,and chromium$(III)$ oxide $(Cr_2O_3)$.
$Cr_2O_3$ is a green-coloured solid powder that is observed during the reaction.

(A) $1$. In alkaline medium: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$. Here,$x = 1$.
$2$. In acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Here,$y = 5$.
$3$. In neutral or weakly alkaline medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$. Here,$z = 3$.
Thus,the values are $x = 1, y = 5, z = 3$.

(D) When $Na_2O_2$ is added to an acidified dichromate solution,it reacts to form chromium pentoxide $(CrO_5)$,which is deep blue in colour.
$Cr_2O_7^{2-} + 2H^+ + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$
$CrO_5$ is unstable and decomposes to form chromic sulphate,releasing oxygen gas in the process.
$4CrO_5 + 6H_2SO_4 \rightarrow 2Cr_2(SO_4)_3 + 6H_2O + 7O_2$
Therefore,both the formation of the blue colour and the evolution of oxygen are observed.

(C) The reaction between acidified $KMnO_4$ and $HCl$ is a redox reaction where $HCl$ is oxidized to chlorine gas $(Cl_2)$.
The balanced chemical equation is: $2 KMnO_4 + 16 HCl \rightarrow 2 KCl + 2 MnCl_2 + 8 H_2O + 5 Cl_2$.
Here,$X$ is chlorine gas $(Cl_2)$,which is a greenish yellow gas.

(D) Acidified $KMnO_4$ is a strong oxidizing agent.
It oxidizes $SO_2$ to $SO_4^{2-}$,$H_2O_2$ to $O_2$,and $Fe^{2+}$ to $Fe^{3+}$.
Since all these species can be oxidized by $KMnO_4$,they all cause the decolourisation of the purple $KMnO_4$ solution.
Therefore,the correct answer is $D$.

(D) $KI$ can be oxidised to $I_2$ by various oxidising agents.
$1$. $KMnO_4$ in alkaline medium: $2KMnO_4 + KI + H_2O \to 2MnO_2 + 2KOH + KIO_3$ (Note: In acidic medium it gives $I_2$,but alkaline $KMnO_4$ also acts as an oxidant).
$2$. Ozone $(O_3)$: $2KI + H_2O + O_3 \to 2KOH + I_2 + O_2$.
$3$. $CuSO_4$ solution: $2CuSO_4 + 4KI \to 2K_2SO_4 + Cu_2I_2 + I_2$.
Since all the given reagents can oxidise $KI$ to $I_2$,the correct option is $D$.

(D) redox reaction involves a change in the oxidation state of at least one element.
In the reaction $XeF_6 + H_2O \to XeO_2F_2 + HF$,the oxidation state of $Xe$ remains $+6$,$F$ remains $-1$,$H$ remains $+1$,and $O$ remains $-2$.
Since there is no change in the oxidation states of any of the elements involved,this is a non-redox reaction (hydrolysis).
In contrast,the other reactions involve the reduction of $Xe$ and the oxidation of $O$ or $F$.

(D) $(I)$ Aluminum reacts with aqueous $NaOH$ to form sodium aluminate $(NaAlO_2)$ and releases hydrogen gas $(H_2)$. This is a correct reaction.
$(II)$ White phosphorus $(P_4)$ reacts with aqueous $NaOH$ to undergo disproportionation,forming sodium hypophosphite $(NaH_2PO_2)$ and phosphine $(PH_3)$. This is a correct reaction.
$(III)$ Sulfur $(S)$ reacts with aqueous $NaOH$ to undergo disproportionation,forming sodium thiosulfate $(Na_2S_2O_3)$,sodium sulfide $(Na_2S)$,and water $(H_2O)$. This is a correct reaction.
Therefore,all three reactions are correct.

(A) The chemical formula for ferric oxalate is $Fe_2(C_2O_4)_3$.
In the oxidation process,$Fe^{2+}$ is not present (it is $Fe^{3+}$),so only the oxalate ion $(C_2O_4^{2-})$ is oxidised to $CO_2$.
The half-reaction for the oxidation of oxalate is: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$.
Since there are $3$ moles of $C_2O_4^{2-}$ in $1$ mole of $Fe_2(C_2O_4)_3$,the total electrons released per mole of ferric oxalate is $3 \times 2 = 6 \, e^-$.
For $10 \, \text{moles}$ of ferric oxalate,total electrons released $= 10 \times 6 = 60 \, e^-$.
The reduction half-reaction for $MnO_4^-$ in acidic medium is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
To balance the electrons,the number of moles of $MnO_4^-$ required is: $x = \frac{60}{5} = 12$.

(B) In basic medium,$MnO_4^-$ is reduced to $MnO_4^{2-}$ ($n$-factor $= 1$). The oxidation of $C_2O_4^{2-}$ to $CO_3^{2-}$ involves an $n$-factor of $2$. For $1 \text{ mole of } H_2C_2O_4$,$1 \times 2 = x \times 1$,so $x = 2$.
In acidic medium,$MnO_4^-$ is reduced to $Mn^{2+}$ ($n$-factor $= 5$). The oxidation of $C_2O_4^{2-}$ to $CO_2$ involves an $n$-factor of $2$. For $1 \text{ mole of } NaHC_2O_4$,$1 \times 2 = y \times 5$,so $y = 2/5 = 0.4$.
The ratio $x/y = 2 / (2/5) = 10/2 = 5/1$.

(D) $KI$ can be oxidized to $I_2$ by various oxidizing agents:
$1$. $KMnO_4$ in neutral or slightly alkaline medium: $2 KMnO_4 + KI + H_2O \rightarrow 2 MnO_2 + 2 KOH + KIO_3$ (Note: $KI$ is oxidized to $I_2$ or $IO_3^-$ depending on conditions).
$2$. Ozone $(O_3)$ in alkaline solution: $2 KI + O_3 + H_2O \rightarrow I_2 + 2 KOH + O_2$.
$3$. $CuSO_4$ solution: $2 CuSO_4 + 4 KI \rightarrow Cu_2I_2 + 2 K_2SO_4 + I_2$.
Since all the given reagents can oxidize $KI$ to $I_2$,the correct option is $D$.

(A) The balanced redox reaction in acidic medium is: $5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \rightarrow 10CO_2 + 2Mn^{2+} + 8H_2O$.
Number of millimoles of $KMnO_4 = \text{Molarity} \times \text{Volume (in mL)} = 1 \, M \times 20 \, mL = 20 \, mmol$.
From the stoichiometry,$2 \, mol$ of $MnO_4^-$ reacts with $5 \, mol$ of $C_2O_4^{2-}$.
Therefore,$mmol$ of $C_2O_4^{2-} = \frac{5}{2} \times 20 = 50 \, mmol$.
Since $1 \, mol$ of $CaC_2O_4$ contains $1 \, mol$ of $Ca^{2+}$,$mmol$ of $Ca^{2+} = 50 \, mmol$.
Mass of $Ca^{2+} = \frac{50}{1000} \times 40 \, g = 2 \, g$.

(A) Mercuric chloride $(HgCl_2)$ reacts with stannous chloride $(SnCl_2)$ in two steps.
First,$HgCl_2$ is reduced to mercurous chloride $(Hg_2Cl_2)$:
$2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$
When $SnCl_2$ is in excess,the mercurous chloride $(Hg_2Cl_2)$ is further reduced to metallic mercury $(Hg)$:
$Hg_2Cl_2 + SnCl_2 \rightarrow 2Hg + SnCl_4$
Thus,the final products are liquid $Hg$ and $SnCl_4$.

(C) The reaction between $I_2$ and sodium thiosulphate (commonly known as Hypo,$Na_2S_2O_3$) is a standard redox titration reaction.
The reaction is: $2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2NaI$.
In this reaction,$I_2$ acts as an oxidizing agent and is reduced to $I^-$,while the thiosulphate ion $(S_2O_3^{2-})$ acts as a reducing agent and is oxidized to tetrathionate $(S_4O_6^{2-})$.
Therefore,Hypo can reduce $I_2$.

(B) In the reaction,$A$ in $A_2O_x$ is oxidized to $AO_3^-$. Let the oxidation state of $A$ in $A_2O_x$ be $n$. Then $2n + x(-2) = 0$,so $n = x$. In $AO_3^-$,the oxidation state of $A$ is $y$. Then $y + 3(-2) = -1$,so $y = +5$. The change in oxidation state per atom of $A$ is $(5 - x)$. Since there are $2$ atoms of $A$ in $A_2O_x$,the total change is $2(5 - x) = 10 - 2x$.
For $MnO_4^-$ in acidic medium,$Mn^{+7} + 5e^- \rightarrow Mn^{+2}$,so the change in oxidation state is $5$.
Equating the equivalents: $n_f \times \text{moles of } A_2O_x = n_f \times \text{moles of } KMnO_4$.
$(10 - 2x) \times (1.5 \times 10^{-3}) = 5 \times (0.03 \times 0.040)$.
$(10 - 2x) \times 1.5 \times 10^{-3} = 5 \times 1.2 \times 10^{-3}$.
$10 - 2x = \frac{6}{1.5} = 4$.
$2x = 6$,so $x = 3$.
Thus,the value of $x$ is $3$ and the oxide is $A_2O_3$.

(D) In $HNO_2$,the oxidation state of $N$ is $+3$. It can be oxidized to $+5$ or reduced to lower oxidation states.
In $H_2O_2$,the oxidation state of $O$ is $-1$. It can be oxidized to $0$ $(O_2)$ or reduced to $-2$ $(H_2O)$.
In $SO_2$,the oxidation state of $S$ is $+4$. It can be oxidized to $+6$ or reduced to lower oxidation states.
Therefore,all of these compounds can act as both oxidizing and reducing agents.

(A) The yellow color of $FeCl_3$ solution is due to the presence of $Fe^{3+}$ ions.
When zinc is added to the acidic $FeCl_3$ solution,it reacts with the $HCl$ produced by the hydrolysis of $FeCl_3$ to generate nascent hydrogen $([H])$.
The reaction is: $Zn + 2HCl \rightarrow ZnCl_2 + 2[H]$.
This nascent hydrogen reduces the $Fe^{3+}$ ions to $Fe^{2+}$ ions: $2FeCl_3 + 2[H] \rightarrow 2FeCl_2 + 2HCl$.
The $Fe^{2+}$ ions are light green in color,which causes the solution to turn from yellow to light green.

(D) In $SO_2$,the oxidation state of $S$ is $+4$. Since $S$ can be oxidized to $+6$,$SO_2$ acts as a strong reducing agent.
$SO_2$ reduces $Cl_2$ water to $HCl$ $(SO_2 + Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl)$.
$SO_2$ reduces acidified $Cr_2O_7^{2-}$ to $Cr^{3+}$ $(Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O)$.
$SO_2$ reduces acidified $MnO_4^-$ to $Mn^{2+}$ $(2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+)$.
Therefore,$SO_2$ can reduce all of these.

(D) $Na_2S_2O_3$ (sodium thiosulfate) exhibits the following properties:
$1$. It reduces $Cu^{2+}$ to $Cu^{+}$: $2Cu^{2+} + 2S_2O_3^{2-} \rightarrow 2Cu^{+} + S_4O_6^{2-}$.
$2$. It reduces $I_2$ to $I^{-}$: $I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}$.
$3$. It dissolves $AgBr$ by forming a soluble complex: $AgBr + 2S_2O_3^{2-} \rightarrow [Ag(S_2O_3)_2]^{3-} + Br^{-}$.
Since all statements are correct,the answer is $D$.

(C) $CO_2$ and $SO_2$ both turn lime water milky,so the lime water test cannot distinguish them.
$SO_2$ is a reducing agent and can decolorize acidified $KMnO_4$ solution ($purple$ to $colorless$),whereas $CO_2$ does not react with $KMnO_4/H^+$.
Therefore,$KMnO_4/H^+$ is the correct reagent to distinguish between these two gases.

(D) Oxidation is defined as the loss of electrons or an increase in the oxidation state of an element.
In option $(A)$,$Na$ (oxidation state $0$) is converted to $NaOH$ where $Na$ has an oxidation state of $+1$. Thus,$Na$ is oxidised.
In option $(B)$,$Cu$ (oxidation state $0$) loses $2$ electrons to form $Cu^{2+}$ (oxidation state $+2$). Thus,$Cu$ is oxidised.
In option $(C)$,$Cu^{2+}$ gains electrons to form $Cu$,which is a reduction process.
Therefore,both $(A)$ and $(B)$ represent the oxidation of a metal.

(D) substance can act as both an oxidizing agent $(O.A.)$ and a reducing agent $(R.A.)$ if the central atom is in an intermediate oxidation state,meaning it is between its maximum and minimum possible oxidation states.
In $H_2O_2$,the oxidation state of oxygen is $-1$. Oxygen can be oxidized to $0$ (in $O_2$) or reduced to $-2$ (in $H_2O$).
Similarly,$H_2$ (oxidation state $0$) can be oxidized to $+1$ or reduced to $-1$ (in metal hydrides).
$I_2$ (oxidation state $0$) can be oxidized to higher states (e.g.,$+1, +5, +7$) or reduced to $-1$ (in $I^-$).
Therefore,all the given substances can act as both oxidizing and reducing agents.

(C) Iodine $(I_2)$ is an oxidizing agent. It reacts with reducing agents to form iodide ions $(I^-)$,which are colorless.
$Na_2SO_3$ (sodium sulfite) reduces $I_2$ to $I^-$.
$Na_2S_2O_3$ (sodium thiosulfate) reduces $I_2$ to $I^-$ (forming tetrathionate).
$NaOH$ reacts with $I_2$ in a disproportionation reaction to form $I^-$ and $IO_3^-$.
$NaCl$ (sodium chloride) does not react with $I_2$ because chloride ions $(Cl^-)$ cannot be oxidized by iodine under standard conditions. Thus,$NaCl$ does not decolourise iodine.

(C) In a weak alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $MnO_2$.
The balanced chemical equation for the reaction between $KMnO_4$ and $KI$ in a weak alkaline medium is:
$2 KMnO_4 + H_2O + KI \rightarrow 2 MnO_2 + 2 KOH + KIO_3$
In this reaction,the iodide ion $(I^-)$ is oxidized to the iodate ion $(IO_3^-)$.

(C) In a redox reaction,the oxidation state of at least one element must change.
In the reaction $CrO_4^{2-} \to Cr_2O_7^{2-}$,the oxidation state of $Cr$ in $CrO_4^{2-}$ is $+6$ $(x + 4(-2) = -2 \implies x = +6)$.
In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is also $+6$ $(2x + 7(-2) = -2 \implies 2x = 12 \implies x = +6)$.
Since there is no change in the oxidation state of any element,this reaction is not a redox reaction.

(D) In $CrO_5$ (butterfly structure),the oxidation state of $Cr$ is $+6$.
In $Fe_3O_4$ (mixed oxide of $FeO$ and $Fe_2O_3$),the average oxidation state of $Fe$ is $+\frac{8}{3}$.
In $Na-Hg$ (sodium amalgam),$Na$ is in its metallic state,so its oxidation number is $0$.
Therefore,all the given options are incorrectly matched.