Mix Examples-Redox Reactions Questions in English

(D) To determine if a reaction involves oxidation or reduction,we check the oxidation states of the elements involved:
$I.$ $Cs$ changes from $0$ to $+1$ (oxidation),and $H$ in $H_2O$ changes from $+1$ to $0$ (reduction). This is a redox reaction.
$II.$ $Cu$ changes from $+2$ to $+1$ (reduction),and $I$ changes from $-1$ to $0$ (oxidation). This is a redox reaction.
$III.$ $NH_4Br + KOH \to KBr + NH_3 + H_2O$. The oxidation states are: $N(-3), H(+1), Br(-1), K(+1), O(-2)$. There is no change in oxidation states for any element. This is an acid-base neutralization reaction,not a redox reaction.
$IV.$ $4KCN + Fe(CN)_2 \to K_4[Fe(CN)_6]$. This is a complex formation reaction. The oxidation state of $Fe$ remains $+2$ and all other elements remain in their respective oxidation states. This is not a redox reaction.
Therefore,reactions $III$ and $IV$ do not involve oxidation or reduction.

(D) redox reaction involves a change in the oxidation state of atoms.
$I.$ In $Cr_2O_7^{2-} + 2OH^- \to 2CrO_4^{2-} + H_2O$,the oxidation state of $Cr$ is $+6$ in both $Cr_2O_7^{2-}$ and $CrO_4^{2-}$. No change in oxidation state occurs,so it is not a redox reaction.
$II.$ In $Zn + CuSO_4 \to ZnSO_4 + Cu$,$Zn$ is oxidized $(0 \to +2)$ and $Cu$ is reduced $(+2 \to 0)$. This is a redox reaction.
$III.$ In $2MnO_4^- + 3Mn^{2+} + 4OH^- \to 5MnO_2 + 2H_2O$,$Mn$ in $MnO_4^-$ is $+7$ and in $Mn^{2+}$ is $+2$. Both are reduced to $Mn$ in $MnO_2$ $(+4)$. This is a redox reaction (disproportionation/comproportionation).
$IV.$ In $2Cu^+ \to Cu + Cu^{2+}$,$Cu^+$ is both oxidized $(+1 \to +2)$ and reduced $(+1 \to 0)$. This is a disproportionation redox reaction.
Therefore,reactions $II, III,$ and $IV$ are redox reactions.

(A) An intermolecular redox reaction is one in which one element is oxidized and another element is reduced,typically involving different molecules or atoms.
$A)$ $MgCO_3 \to MgO + CO_2$: This is a thermal decomposition reaction. The oxidation states of $Mg$ $(+2)$,$C$ $(+4)$,and $O$ $(-2)$ remain unchanged. It is not a redox reaction.
$B)$ $2H_2 + O_2 \to 2H_2O$: $H$ is oxidized $(0 \to +1)$ and $O$ is reduced $(0 \to -2)$. This is an intermolecular redox reaction.
$C)$ $K + H_2O \to KOH + 1/2H_2$: $K$ is oxidized $(0 \to +1)$ and $H$ is reduced $(+1 \to 0)$. This is an intermolecular redox reaction.
$D)$ $MnBr_3 \to MnBr_2 + 1/2Br_2$: $Mn$ is reduced $(+3 \to +2)$ and $Br$ is oxidized $(-1 \to 0)$. This is an intramolecular redox reaction (where the same molecule undergoes both oxidation and reduction).
Therefore,option $A$ is not a redox reaction at all,and option $D$ is an intramolecular redox reaction. Given the context of the question asking for 'not an intermolecular redox reaction',option $A$ is the most appropriate answer as it is not a redox reaction.

(A) The balanced redox reaction is: $3Sn^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 3Sn^{4+} + 2Cr^{3+} + 7H_2O$.
From the stoichiometry of the balanced equation,$3$ moles of $Sn^{2+}$ react with $1$ mole of $Cr_2O_7^{2-}$ (which comes from $1$ mole of $K_2Cr_2O_7$).
Therefore,$1$ mole of $Sn^{2+}$ will react with $1/3$ mole of $K_2Cr_2O_7$.

(B) In a basic medium,the reaction between $KI$ and $KMnO_4$ is given by:
$I^- + 2MnO_4^- + H_2O \rightarrow IO_3^- + 2MnO_2 + 2OH^-$
In this reaction:
- The oxidation state of $I$ changes from $-1$ to $+5$,which is an increase of $6$ units.
- The oxidation state of $Mn$ changes from $+7$ to $+4$,which is a decrease of $3$ units per $Mn$ atom.
To balance the electron transfer:
$1 \ mol$ of $I^-$ loses $6 \ mol$ of electrons.
$1 \ mol$ of $MnO_4^-$ gains $3 \ mol$ of electrons.
Therefore,$1 \ mol$ of $I^-$ will reduce $6/3 = 2 \ mol$ of $MnO_4^-$.
Thus,$1 \ mol$ of $KI$ reduces $2 \ mol$ of $KMnO_4$.

(C) Reaction $(I): \ SeO_3^{2-} + BrO_3^{-} + H^{+} \to SeO_4^{2-} + Br_2 + H_2O$ (n-factor for $SeO_3^{2-} = 2$,for $BrO_3^{-} = 5$)
Reaction $(II): \ BrO_3^{-} + AsO_2^{-} + H_2O \to Br^{-} + AsO_4^{3-} + H^{+}$ (n-factor for $BrO_3^{-} = 6$,for $AsO_2^{-} = 2$)
Step $1$: Calculate milliequivalents of $AsO_2^{-}$ used in back titration:
$meq \ of \ AsO_2^{-} = 12.5 \ mL \times (\frac{1}{25} \ M) \times 2 = 1 \ meq$
Step $2$: Calculate millimoles of excess $BrO_3^{-}$:
$meq \ of \ BrO_3^{-} = meq \ of \ AsO_2^{-} = 1 \ meq$
$mmol \ of \ BrO_3^{-} = \frac{1}{6} \ mmol$
Step $3$: Calculate millimoles of $BrO_3^{-}$ consumed in reaction $(I)$:
Total $mmol \ of \ BrO_3^{-} = 70 \ mL \times \frac{1}{60} \ M = \frac{7}{6} \ mmol$
$mmol \ of \ BrO_3^{-} \ consumed = \frac{7}{6} - \frac{1}{6} = 1 \ mmol$
Step $4$: Calculate millimoles of $SeO_3^{2-}$:
$meq \ of \ SeO_3^{2-} = meq \ of \ BrO_3^{-} \ consumed$
$mmol \ of \ SeO_3^{2-} \times 2 = 1 \ mmol \times 5$
$mmol \ of \ SeO_3^{2-} = 2.5 \ mmol = 2.5 \times 10^{-3} \ mol$

(C) From the balanced chemical equation,$2 \ \text{moles}$ of $KMnO_4$ react with $5 \ \text{moles}$ of $H_2C_2O_4$.
Using the concept of milliequivalents (Meq):
$\text{Meq of } KMnO_4 = \text{Molarity} \times n\text{-factor} \times \text{Volume (mL)}$
For $KMnO_4$,the $n$-factor is $5$ $(Mn^{+7} \to Mn^{+2})$.
$\text{Meq of } KMnO_4 = 0.1 \times 5 \times 20 = 10 \ \text{Meq}$.
For $H_2C_2O_4$,the $n$-factor is $2$ $(C_2^{+3} \to 2C^{+4} + 2e^-)$.
We need $10 \ \text{Meq}$ of $H_2C_2O_4$.
$\text{Meq} = \text{Molarity} \times 2 \times \text{Volume (mL)} = 10$.
Checking option $(C)$: $50 \ mL$ of $0.1 \ M \ H_2C_2O_4$:
$\text{Meq} = 0.1 \times 2 \times 50 = 10 \ \text{Meq}$.
Thus,$20 \ mL$ of $0.1 \ M \ KMnO_4$ is equivalent to $50 \ mL$ of $0.1 \ M \ H_2C_2O_4$.

(A) $N_2 + O_2 \xrightarrow{2000 \; K} 2 NO$ (Redox reaction)
During the reaction,the oxidation state of nitrogen changes from $0$ to $+2$ (oxidation) and the oxidation state of oxygen changes from $0$ to $-2$ (reduction). Thus,it is a redox reaction.
$3 O_2 \xrightarrow{hv} 2 O_3$ (Non-redox reaction)
$H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O$ (Neutralization reaction,non-redox)
$[Co(H_2O)_6]Cl_3 + 3 AgNO_3 \rightarrow 3 AgCl \downarrow + [Co(H_2O)_6](NO_3)_3$ (Precipitation reaction,non-redox)

(N/A) In reaction $(a)$,the compound nitric oxide is formed by the combination of elemental substances,nitrogen and oxygen; therefore,this is a combination redox reaction.
In reaction $(b)$,lead nitrate breaks down into three components; therefore,this is a decomposition redox reaction.
In reaction $(c)$,the hydrogen of water is displaced by the hydride ion to form dihydrogen gas; therefore,this is a displacement redox reaction.
In reaction $(d)$,$NO_2$ ($+4$ oxidation state) disproportionates into $NO_2^-$ ($+3$ oxidation state) and $NO_3^-$ ($+5$ oxidation state); therefore,this is a disproportionation redox reaction.

(N/A) $Pb_{3}O_{4}$ is a stoichiometric mixture of $2 \ mol$ of $PbO$ and $1 \ mol$ of $PbO_{2}$. In $PbO_{2}$,lead is in the $+4$ oxidation state,while in $PbO$,it is in the $+2$ oxidation state.
$PbO_{2}$ acts as an oxidizing agent and can oxidize the $Cl^{-}$ ion of $HCl$ to $Cl_{2}$. The reaction with $HCl$ is a combination of an acid-base reaction $(2PbO + 4HCl \rightarrow 2PbCl_{2} + 2H_{2}O)$ and a redox reaction $(PbO_{2} + 4HCl \rightarrow PbCl_{2} + Cl_{2} + 2H_{2}O)$.
In contrast,$HNO_{3}$ is itself an oxidizing agent,so it does not react with $PbO_{2}$. The reaction with $HNO_{3}$ is limited to the acid-base reaction between $PbO$ and $HNO_{3}$: $2PbO + 4HNO_{3} \rightarrow 2Pb(NO_{3})_{2} + 2H_{2}O$. The $PbO_{2}$ remains as a residue,making the overall reaction different.

(N/A) Photosynthesis involves two steps. Step $1$: $H_{2}O$ decomposes to give $H_{2}$ and $O_{2}$ $(2 H_{2}O_{(l)} \rightarrow 2 H_{2(g)} + O_{2(g)})$. Step $2$: The $H_{2}$ produced reduces $CO_{2}$ to glucose $(6 CO_{2(g)} + 12 H_{2(g)} \rightarrow C_{6}H_{12}O_{6(s)} + 6 H_{2}O_{(l)})$. The net reaction is $6 CO_{2(g)} + 12 H_{2}O_{(l)} \rightarrow C_{6}H_{12}O_{6(s)} + 6 H_{2}O_{(l)} + 6 O_{2(g)}$. This is more appropriate because water is both a reactant and a product. The path can be investigated using radioactive isotope labeling,such as $H_{2}O^{18}$.
$(b)$ $O_{2}$ is produced from both $O_{3}$ and $H_{2}O_{2}$. The reaction occurs in two steps: $O_{3(g)} \rightarrow O_{2(g)} + O_{(g)}$ and $H_{2}O_{2(l)} + O_{(g)} \rightarrow H_{2}O_{(l)} + O_{2(g)}$. The net reaction $O_{3(g)} + H_{2}O_{2(l)} \rightarrow H_{2}O_{(l)} + O_{2(g)} + O_{2(g)}$ shows the origin of the oxygen molecules. The path can be investigated using $H_{2}O_{2}^{18}$ or $O_{3}^{18}$.

(N/A) Whenever a reaction between an oxidising agent and a reducing agent is carried out,a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows:
$(i)$ $P_4$ and $F_2$ are reducing and oxidising agents respectively.
If an excess of $P_4$ is treated with $F_2$,then $PF_3$ is produced,wherein the oxidation number $(O.N.)$ of $P$ is $+3$: $P_4 (\text{excess}) + 6F_2 \to 4PF_3$.
However,if $P_4$ is treated with an excess of $F_2$,then $PF_5$ is produced,wherein the $O.N.$ of $P$ is $+5$: $P_4 + 10F_2 (\text{excess}) \to 4PF_5$.
$(ii)$ $K$ acts as a reducing agent,whereas $O_2$ is an oxidising agent.
If an excess of $K$ reacts with $O_2$,then $K_2O$ is formed,wherein the $O.N.$ of $O$ is $-2$: $4K (\text{excess}) + O_2 \to 2K_2O$.
However,if $K$ reacts with an excess of $O_2$,then $K_2O_2$ is formed,wherein the $O.N.$ of $O$ is $-1$: $2K + O_2 (\text{excess}) \to K_2O_2$.
$(iii)$ $C$ is a reducing agent,while $O_2$ acts as an oxidising agent.
If an excess of $C$ is burnt in the presence of an insufficient amount of $O_2$,then $CO$ is produced,wherein the $O.N.$ of $C$ is $+2$: $2C (\text{excess}) + O_2 \to 2CO$.
On the other hand,if $C$ is burnt in an excess of $O_2$,then $CO_2$ is produced,wherein the $O.N.$ of $C$ is $+4$: $C + O_2 (\text{excess}) \to CO_2$.

(N/A) Oxidised: $C_6 H_6 O_2$,Reduced: $AgBr$,Oxidising agent: $AgBr$,Reducing agent: $C_6 H_6 O_2$
$(b)$ Oxidised: $HCHO$,Reduced: $[Ag(NH_3)_2]^+$,Oxidising agent: $[Ag(NH_3)_2]^+$,Reducing agent: $HCHO$
$(c)$ Oxidised: $HCHO$,Reduced: $Cu^{2+}$,Oxidising agent: $Cu^{2+}$,Reducing agent: $HCHO$
$(d)$ Oxidised: $N_2 H_4$,Reduced: $H_2 O_2$,Oxidising agent: $H_2 O_2$,Reducing agent: $N_2 H_4$
$(e)$ Oxidised: $Pb$,Reduced: $PbO_2$,Oxidising agent: $PbO_2$,Reducing agent: $Pb$

(N/A) The average oxidation number $(O.N.)$ of $S$ in $S_2O_3^{2-}$ is $+2$.
$Br_2$ is a stronger oxidizing agent than $I_2$.
$Br_2$ oxidizes the sulfur in thiosulfate to the $+6$ oxidation state (in $SO_4^{2-}$).
$I_2$ is a weaker oxidizing agent and can only oxidize the sulfur to an average oxidation state of $+2.5$ (in $S_4O_6^{2-}$).
Therefore,the extent of oxidation depends on the strength of the oxidizing agent,causing the thiosulfate to react differently.

(A) For $(a)$: $P_{4(s)} + 3OH^{-}_{(aq)} + 3H_{2}O_{(l)} \rightarrow PH_{3(g)} + 3H_{2}P{O_{2}}^{-}_{(aq)}$.
Oxidation half: $P_{4(s)} + 8OH^{-}_{(aq)} \rightarrow 4H_{2}P{O_{2}}^{-}_{(aq)} + 4e^{-}$.
Reduction half: $P_{4(s)} + 12H_{2}O_{(l)} + 12e^{-} \rightarrow 4PH_{3(g)} + 12OH^{-}_{(aq)}$.
Multiplying oxidation half by $3$ and adding to reduction half gives the balanced equation.
$P_{4}$ acts as both the oxidising and reducing agent (disproportionation reaction).

(N/A) redox reaction is a chemical reaction in which the oxidation numbers of the reactants change.
It is a combination of oxidation and reduction processes.
Uses:
$i$. These reactions are extensively used in pharmaceutical,biological,industrial,metallurgical,and agricultural sectors.
$ii$. They are used for professional purposes and in various types of fuels.
$iii$. They are used in the industrial preparation of $NaOH$.
$iv$. They are essential for the operation of dry and wet batteries.
$v$. They are used in electrochemical processes for the extraction of highly reactive metals and non-metals.
$vi$. They are used in the production of hydrogen.
$vii$. They are involved in the chemical processes leading to the development of the ozone hole in the ozone layer.

(N/A) To determine if a reaction involves oxidation or reduction,we look at the change in oxidation states:
$(i)$ $FeSO_4 + Mg \to MgSO_4 + Fe$: Here,$Mg$ (oxidation state $0$) is oxidized to $Mg^{2+}$ (in $MgSO_4$),and $Fe^{2+}$ (in $FeSO_4$) is reduced to $Fe$ $(0)$. This is a redox reaction.
$(ii)$ $Cu + 4HNO_3 \to Cu(NO_3)_2 + 2NO_2 + 2H_2O$: Here,$Cu$ $(0)$ is oxidized to $Cu^{2+}$ $(+2)$,and $N$ ($+5$ in $HNO_3$) is reduced to $N$ ($+4$ in $NO_2$). This is a redox reaction.
$(iii)$ $H_2S + Cl_2 \to S + 2HCl$: Here,$S$ ($-2$ in $H_2S$) is oxidized to $S$ $(0)$,and $Cl$ ($0$ in $Cl_2$) is reduced to $Cl$ ($-1$ in $HCl$). This is a redox reaction.
All three reactions are redox reactions involving both oxidation and reduction.

(A) In reaction $(i)$,the oxidation state of $S$ increases from $0$ to $+6$ in $H_2SO_4$,which is an oxidation process.
In reaction $(ii)$,the oxidation state of $K$ increases from $0$ to $+1$ (oxidation) and $Br$ decreases from $0$ to $-1$ (reduction). However,considering the overall reaction as a redox process,the formation of $KBr$ involves the oxidation of $K$ and the reduction of $Br_2$. Given the context of identifying the nature of the reaction,$(i)$ is an oxidation of $S$ and $(ii)$ is a redox reaction where $K$ is oxidized and $Br_2$ is reduced.

(N/A) Carbon $(C)$ is a reducing agent and $O_2$ is an oxidizing agent. When $C$ is in excess,$CO$ is formed (oxidation state of $C = +2$). When $O_2$ is in excess,$CO_2$ is formed (oxidation state of $C = +4$).
$2C(s) + O_2(g) \rightarrow 2CO(g)$ (Excess $C$)
$C(s) + O_2(g) \rightarrow CO_2(g)$ (Excess $O_2$)
$(b)$ $P_4$ is a reducing agent and $Cl_2$ is an oxidizing agent. When $P_4$ is in excess,$PCl_3$ is formed (oxidation state of $P = +3$). When $Cl_2$ is in excess,$PCl_5$ is formed (oxidation state of $P = +5$).
$P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(l)$ (Excess $P_4$)
$P_4(s) + 10Cl_2(g) \rightarrow 4PCl_5(s)$ (Excess $Cl_2$)
$(c)$ $Na$ is a reducing agent and $O_2$ is an oxidizing agent. When $Na$ is in excess,$Na_2O$ is formed (oxidation state of $O = -2$). When $O_2$ is in excess,$Na_2O_2$ is formed (oxidation state of $O = -1$).
$4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$ (Excess $Na$)
$2Na(s) + O_2(g) \rightarrow Na_2O_2(s)$ (Excess $O_2$)

(N/A) Toluene is a non-polar organic compound,while $KMnO_4$ is an ionic,polar inorganic compound. They are immiscible in water. Alcohol acts as a mutual solvent for both,facilitating the reaction.
The balanced redox equation for the oxidation of toluene to benzoate in an alkaline medium is:
$C_6H_5CH_3 + 2MnO_4^- + OH^- \rightarrow C_6H_5COO^- + 2MnO_2 + 2H_2O$
$(b)$ Concentrated $H_2SO_4$ acts as both an acid and an oxidising agent. With chloride salts,it produces $HCl$ gas,which is not further oxidised by $H_2SO_4$. However,with bromide salts,the $HBr$ produced is a stronger reducing agent and is further oxidised by concentrated $H_2SO_4$ to $Br_2$ (red vapours):
$2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O$

(N/A)

Reaction	Oxidised Species	Reduced Species	Oxidising Agent	Reducing Agent
$(a)$	$C_6H_6O_{2(aq)}$	$AgBr_{(s)}$	$AgBr_{(s)}$	$C_6H_6O_{2(aq)}$
$(b)$	$HCHO_{(l)}$	$[Ag(NH_3)_2]^+_{(aq)}$	$[Ag(NH_3)_2]^+_{(aq)}$	$HCHO_{(l)}$
$(c)$	$HCHO_{(l)}$	$Cu^{2+}_{(aq)}$	$Cu^{2+}_{(aq)}$	$HCHO_{(l)}$
$(d)$	$N_2H_{4(l)}$	$H_2O_{2(l)}$	$H_2O_{2(l)}$	$N_2H_{4(l)}$
$(e)$	$Pb_{(s)}$	$PbO_{2(s)}$	$PbO_{2(s)}$	$Pb_{(s)}$

(N/A) In $S_2O_3^{2-}$,the average oxidation number of $S$ is $+2$. In $S_4O_6^{2-}$,the average oxidation number of $S$ is $+2.5$. In $SO_4^{2-}$,the oxidation number of $S$ is $+6$.
$Br_2$ is a stronger oxidizing agent than $I_2$. Therefore,$Br_2$ is capable of oxidizing $S_2O_3^{2-}$ (where $S$ is $+2$) to $SO_4^{2-}$ (where $S$ is $+6$).
$I_2$ is a weaker oxidizing agent. Therefore,it only oxidizes $S_2O_3^{2-}$ $(S=+2)$ to $S_4O_6^{2-}$ $(S=+2.5)$.
Thus,the difference in the oxidizing strength of the halogens leads to different products.

(A) $(a) \ CuO_{(s)} + H_{2(g)} \to Cu_{(s)} + H_2O_{(g)}$
Oxidation number of $Cu$ decreases from $+2$ to $0$ (reduction) and $H$ increases from $0$ to $+1$ (oxidation). Thus,it is a redox reaction.
$(b) \ Fe_2O_{3(s)} + 3CO_{(g)} \to 2Fe_{(s)} + 3CO_{2(g)}$
Oxidation number of $Fe$ decreases from $+3$ to $0$ (reduction) and $C$ increases from $+2$ to $+4$ (oxidation). Thus,it is a redox reaction.
$(c) \ 4BCl_{3(g)} + 3LiAlH_{4(s)} \to 2B_2H_{6(g)} + 3LiCl_{(s)} + 3AlCl_{3(s)}$
Oxidation number of $B$ decreases from $+3$ to $-3$ (reduction) and $H$ increases from $-1$ to $+1$ (oxidation). Thus,it is a redox reaction.
$(d) \ 2K_{(s)} + F_{2(g)} \to 2K^+F^-_{(s)}$
Oxidation number of $K$ increases from $0$ to $+1$ (oxidation) and $F$ decreases from $0$ to $-1$ (reduction). Thus,it is a redox reaction.
$(e) \ 4NH_{3(g)} + 5O_{2(g)} \to 4NO_{(g)} + 6H_2O_{(g)}$
Oxidation number of $N$ increases from $-3$ to $+2$ (oxidation) and $O$ decreases from $0$ to $-2$ (reduction). Thus,it is a redox reaction.

(N/A) $(i)$ In one situation,the reagent itself is intensely coloured,e.g.,permanganate ion,$MnO_{4}^{-}.$ Here $MnO_{4}^{-}$ acts as the self-indicator. The visible end point in this case is achieved after the last of the reductant (e.g.,$Fe^{2+}$ or $C_{2}O_{4}^{2-}$) is oxidised and the first lasting tinge of pink colour appears at $MnO_{4}^{-}$ concentration as low as $10^{-6} \ mol \ L^{-1}$. This ensures a minimal "overshoot" in colour beyond the equivalence point,the point where the reductant and the oxidant are equal in terms of their mole stoichiometry.
$(ii)$ If there is no dramatic auto-colour change (as with $MnO_{4}^{-}$ titration),there are indicators which are oxidised immediately after the last bit of the reactant is consumed,producing a dramatic colour change. The best example is afforded by $Cr_{2}O_{7}^{2-}$,which is not a self-indicator,but oxidises the indicator substance diphenylamine just after the equivalence point to produce an intense blue colour,thus signalling the end point.
$(iii)$ There is yet another method which is interesting and quite common. Its use is restricted to those reagents which are able to oxidise $I^{-}$ ions,for example: $2Cu_{(aq)}^{2+} + 4I_{(aq)}^{-} \rightarrow Cu_{2}I_{2(s)} + I_{2(aq)}$.
This method relies on the fact that iodine itself gives an intense blue colour with starch and has a very specific reaction with thiosulphate ions $(S_{2}O_{3(aq)}^{2-})$,which is also a redox reaction: $I_{2(aq)} + 2S_{2}O_{3(aq)}^{2-} \rightarrow 2I_{(aq)}^{-} + S_{4}O_{6(aq)}^{2-}$.
$I_{2}$,though insoluble in water,remains in solution containing $KI$ as $KI_{3}$.
On addition of starch after the liberation of iodine from the reaction of $Cu^{2+}$ ions on iodide ions,an intense blue colour appears. This colour disappears as soon as the iodine is consumed by the thiosulphate ions. Thus,the end-point can easily be tracked and the rest is the stoichiometric calculation only.

(N/A) $(1) \ 4Zn + 2NO_3^- + 10H^+ \to 4Zn^{2+} + N_2O + 5H_2O$
$(2) \ 4I^- + O_2 + 2H_2O \to 2I_2 + 4OH^-$
$(3) \ 4MnO_4^- + 5C_2H_5OH + 12H^+ \to 4Mn^{2+} + 5CH_3COOH + 11H_2O$

The reduction of $KMnO_4$ depends on the $pH$ of the medium:
$1$. In acidic medium $(pH < 7)$: $MnO_4^-$ is reduced to $Mn^{2+}$ (colourless solution). The reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
$2$. In neutral or faintly alkaline medium $(pH \approx 7)$: $MnO_4^-$ is reduced to $MnO_2$ (brown precipitate). The reaction is: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$.
$3$. In strongly alkaline medium $(pH > 7)$: $MnO_4^-$ is reduced to $MnO_4^{2-}$ (green solution). The reaction is: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$.

(D) Both dilute and concentrated $HCl$ react with $Zn$ to produce $ZnCl_2$ and $H_2$ gas.
$Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$.
Since both reactions occur,the statement that $(ii)$ is not possible is false.

(N/A) The oxidising behaviour of $KMnO_{4}$ depends on the $pH$ of the solution.
$1$. In acidic medium $(pH < 7)$:
$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$
Here,$Mn$ is reduced from $+7$ to $+2$ oxidation state,resulting in a colourless solution.
$2$. In alkaline medium $(pH > 7)$:
$MnO_{4}^{-} + e^{-} \rightarrow MnO_{4}^{2-}$ (Green solution)
Here,$Mn$ is reduced from $+7$ to $+6$ oxidation state.
$3$. In neutral or weakly alkaline medium $(pH \approx 7)$:
$MnO_{4}^{-} + 2H_{2}O + 3e^{-} \rightarrow MnO_{2} + 4OH^{-}$ (Brown precipitate)
Here,$Mn$ is reduced from $+7$ to $+4$ oxidation state.

(N/A) In equation $(i)$,the oxidation number of the elements does not change,so this reaction is not a redox reaction. It is an acid-base reaction.
In equation $(ii)$,the oxidation number of $Pb$ decreases from $+4$ to $+2$ (reduction) and the oxidation number of $Cl$ increases from $-1$ to $0$ (oxidation). Thus,$PbO_2$ acts as an oxidising agent and $Cl^-$ is oxidised to $Cl_2$.

(A) $(a) 6 Fe^{+2} + 14 H^{+} + Cr_2O_7^{-2} \to 2 Cr^{+3} + 6 Fe^{+3} + 7 H_2O$
$(b) I_2 + 10 NO_3^- + 8 H^{+} \to 10 NO_2 + 2 IO_3^- + 4 H_2O$
$(c) I_2 + 2 S_2O_3^{-2} \to 2 I^{-} + S_4O_6^{-2}$
$(d) 2 MnO_4^- + 16 H^{+} + 5 C_2O_4^{-2} \to 2 Mn^{+2} + 10 CO_2 + 8 H_2O$

(A, C) To identify a redox reaction,we check for changes in the oxidation numbers of the elements involved.
$(a) 3HCl + HNO_3 \to Cl_2 + NOCl + 2H_2O$
In this reaction,the oxidation number of $Cl$ increases from $-1$ to $0$ (oxidation),and the oxidation number of $N$ decreases from $+5$ to $+3$ (reduction). Thus,it is a redox reaction. $HCl$ is the reducing agent and $HNO_3$ is the oxidizing agent.
$(b) HgCl_2 + 2KI \to HgI_2 + 2KCl$
In this reaction,the oxidation numbers of all elements $(Hg: +2, Cl: -1, K: +1, I: -1)$ remain unchanged. Thus,it is not a redox reaction.
$(c) Fe_2O_3 + 3CO \to 2Fe + 3CO_2$
In this reaction,the oxidation number of $Fe$ decreases from $+3$ to $0$ (reduction),and the oxidation number of $C$ increases from $+2$ to $+4$ (oxidation). Thus,it is a redox reaction. $CO$ is the reducing agent and $Fe_2O_3$ is the oxidizing agent.

(E) For reaction $(d)$: $PCl_{3(l)} + 3H_2O_{(l)} \to 3HCl_{(aq)} + H_3PO_{3(aq)}$
Calculating oxidation states: $P$ in $PCl_3$ is $+3$,$P$ in $H_3PO_3$ is $+3$. $Cl$ is $-1$ in both $PCl_3$ and $HCl$. $H$ is $+1$ and $O$ is $-2$ in both reactants and products.
Since there is no change in oxidation states,this is not a redox reaction.
For reaction $(e)$: $4NH_{3(aq)} + 3O_{2(g)} \to 2N_{2(g)} + 6H_2O_{(g)}$
Calculating oxidation states: $N$ in $NH_3$ is $-3$,$N$ in $N_2$ is $0$. $O$ in $O_2$ is $0$,$O$ in $H_2O$ is $-2$.
Here,$N$ is oxidized ($-3$ to $0$) and $O$ is reduced ($0$ to $-2$).
Therefore,this is a redox reaction.
$O_2$ acts as the oxidising agent and $NH_3$ acts as the reducing agent.

(D) The balanced redox reaction involves the oxidation of $Fe^{2+}$ to $Fe^{3+}$ and $C_2O_4^{2-}$ to $CO_2$ by $Cr_2O_7^{2-}$ in acidic medium.
The $n$-factor for $K_2Cr_2O_7$ is $6$ $(Cr^{+6} \rightarrow Cr^{+3})$.
The $n$-factor for ferrous oxalate $(FeC_2O_4)$ is $3$ ($Fe^{+2} \rightarrow Fe^{+3}$ and $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$).
Molar mass of $FeC_2O_4 = 56 + 2 \times 12 + 4 \times 16 = 144 \ g \ mol^{-1}$.
Equating the equivalents: $M_1 \times V_1 \times n_1 = \frac{\text{mass}}{M_w} \times n_2$.
$0.02 \times V \times 6 = \frac{0.288}{144} \times 3$.
$0.12 \times V = 0.002 \times 3 = 0.006$.
$V = \frac{0.006}{0.12} \times 1000 \ mL = 50 \ mL$.

(A) $K_{2}Cr_{2}O_{7}$ acts as a strong oxidizing agent in acidic medium.
It can oxidize sulfite ions $(SO_{3}^{2-})$ to sulfate ions $(SO_{4}^{2-})$.
The balanced ionic reaction is:
$Cr_{2}O_{7}^{2-} + 3SO_{3}^{2-} + 8H^{+} \rightarrow 2Cr^{3+} + 3SO_{4}^{2-} + 4H_{2}O$
Other ions like $CO_{3}^{2-}$,$SO_{4}^{2-}$,and $NO_{3}^{-}$ are either already in their highest oxidation state or do not undergo redox reaction with dichromate under standard conditions.

(C) The balanced redox reaction in acidic medium is: $6Fe^{2+} + Cr_{2}O_{7}^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_{2}O$.
According to the law of equivalence,the number of equivalents of $Fe^{2+}$ equals the number of equivalents of $Cr_{2}O_{7}^{2-}$.
$n_{eq} (Fe^{2+}) = n_{eq} (Cr_{2}O_{7}^{2-})$.
$(Molarity \times Volume \times n-factor)_{Fe^{2+}} = (Molarity \times Volume \times n-factor)_{Cr_{2}O_{7}^{2-}}$.
For $Fe^{2+} \rightarrow Fe^{3+}$,the $n-factor = 1$.
For $Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+}$,the $n-factor = 6$.
$\left(\frac{15 \times M_{Fe^{2+}}}{1000}\right) \times 1 = \left(\frac{20 \times 0.03}{1000}\right) \times 6$.
$15 \times M_{Fe^{2+}} = 20 \times 0.03 \times 6$.
$15 \times M_{Fe^{2+}} = 3.6$.
$M_{Fe^{2+}} = \frac{3.6}{15} = 0.24 \ M$.
$0.24 \ M = 24 \times 10^{-2} \ M$.
Thus,the value is $24$.

(B) The balanced redox reaction is:
$8CrO_{4}^{2-} + 3S_{2}O_{3}^{2-} + 10H_{2}O \rightarrow 8Cr(OH)_{4}^{-} + 6SO_{4}^{2-} + 2OH^{-}$
From the stoichiometry,the $n$-factor for $CrO_{4}^{2-}$ ($Cr^{+6}$ to $Cr^{+3}$) is $3$,and for $S_{2}O_{3}^{2-}$ ($S^{+2}$ to $S^{+6}$),the total change in oxidation state is $2 \times (6-2) = 8$.
Using the law of equivalence: $n_{1}M_{1}V_{1} = n_{2}M_{2}V_{2}$
$3 \times 0.154 \times V = 8 \times 0.25 \times 40$
$0.462 \times V = 80$
$V = \frac{80}{0.462} \approx 173.16 \ mL$
Rounding to the nearest integer,we get $173 \ mL$.

(C) The balanced redox reaction in acidic medium is:
$MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O$
Equivalents of $KMnO_{4} = \text{Equivalents of } FeSO_{4}$
$n_{factor} \times M_{1} \times V_{1} = n_{factor} \times M_{2} \times V_{2}$
For $KMnO_{4}$,$n_{factor} = 5$ (reduction of $Mn^{+7}$ to $Mn^{+2}$).
For $FeSO_{4}$,$n_{factor} = 1$ (oxidation of $Fe^{+2}$ to $Fe^{+3}$).
Given: $V_{1} = V_{2} = 10 \, mL$,$M_{2} = 0.1 \, M$.
$5 \times M_{1} \times 10 = 1 \times 0.1 \times 10$
$M_{1} = \frac{0.1}{5} = 0.02 \, M$
Molar mass of $KMnO_{4} = 39 + 55 + (4 \times 16) = 158 \, g/mol$.
Strength in $g/L = Molarity \times Molar \, mass = 0.02 \times 158 = 3.16 \, g/L$.
$3.16 \, g/L = 316 \times 10^{-2} \, g/L$.
Thus,the value is $316$.

(C) In a basic medium,potassium permanganate $(KMnO_4)$ reacts with hydrogen peroxide $(H_2O_2)$ to produce manganese dioxide $(MnO_2)$,oxygen gas $(O_2)$,water $(H_2O)$,and potassium hydroxide $(KOH)$.
The balanced chemical equation is:
$2KMnO_4 + 3H_2O_2 \longrightarrow 2MnO_2 + 3O_2 + 2H_2O + 2KOH$
In the product $MnO_2$,let the oxidation state of $Mn$ be $x$.
$x + 2(-2) = 0$
$x - 4 = 0$
$x = +4$
Therefore,the oxidation state of manganese in the product is $4$.

(C) The balanced redox reaction in acidic medium is: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$.
According to the law of equivalence,the number of equivalents of $K_2Cr_2O_7$ equals the number of equivalents of $Fe^{2+}$.
$n.f$ of $K_2Cr_2O_7 = 6$ (as $Cr$ changes from $+6$ to $+3$,$2 \times 3 = 6$).
$n.f$ of $Fe^{2+} = 1$ (as $Fe$ changes from $+2$ to $+3$).
Equivalents of $K_2Cr_2O_7 = Molarity \times Volume \times n.f = 0.02 \times 20 \times 6 = 2.4$.
Equivalents of $Fe^{2+} = M \times 10 \times 1 = 10M$.
Equating both: $10M = 2.4$.
$M = 0.24 \, M$.
$M = 24 \times 10^{-2} \, M$.
Therefore,the value is $24$.

(D) The reactions of $C$,$S$,and $P$ with hot concentrated $H_{2}SO_{4}$ are as follows:
$C + 2H_{2}SO_{4} \longrightarrow CO_{2} + 2SO_{2} + 2H_{2}O$
$S + 2H_{2}SO_{4} \longrightarrow 3SO_{2} + 2H_{2}O$
$P_{4} + 10H_{2}SO_{4} \longrightarrow 4H_{3}PO_{4} + 10SO_{2} + 4H_{2}O$
As shown in the equations,all three elements ($C$,$S$,and $P$) produce $SO_{2}$ gas when reacted with hot concentrated $H_{2}SO_{4}$.

(B)
Oxidation-reduction (redox) reactions are those in which there is a change in the oxidation state of at least one element.
$(a)$ $H_{2} (0) + Br_{2} (0) \longrightarrow 2H(+1)Br(-1)$: Oxidation states change.
$(b)$ $Na(+1)Cl(-1) + Ag(+1)N(+5)O_{3}(-2) \longrightarrow Na(+1)N(+5)O_{3}(-2) + Ag(+1)Cl(-1)$: There is no change in the oxidation state of any element. This is a double displacement reaction.
$(c)$ $2Na_{2}S_{2}O_{3} + I_{2} \longrightarrow Na_{2}S_{4}O_{6} + 2NaI$: The oxidation state of sulfur changes.
$(d)$ $Cl_{2} (0) + H_{2}O \longrightarrow H(+1)Cl(-1) + H(+1)O(-2)Cl(+1)$: Chlorine undergoes disproportionation.
Therefore,the reaction in $(b)$ is not a redox reaction.

(D) The balanced redox reaction in acidic medium is: $2KMnO_4 + 10KI + 8H_2SO_4 \longrightarrow 2MnSO_4 + 5I_2 + 6K_2SO_4 + 8H_2O$.
In this reaction,$KMnO_4$ acts as an oxidizing agent where the oxidation state of $Mn$ changes from $+7$ to $+2$,so the $n$-factor for $KMnO_4$ is $5$.
For $KI$,the oxidation state of $I$ changes from $-1$ to $0$,so the $n$-factor for $KI$ is $1$.
According to the law of equivalence,the number of equivalents of $KMnO_4$ must be equal to the number of equivalents of $KI$.
$\text{Equivalents of } KMnO_4 = \text{Equivalents of } KI$.
$n_{KMnO_4} \times n\text{-factor}_{KMnO_4} = \text{Equivalents of } KI$.
$n_{KMnO_4} \times 5 = 1$.
$n_{KMnO_4} = 1/5$.

(D) .
Oxalic acid,when treated with potassium permanganate $(KMnO_4)$ in the presence of an acid $(H^{+})$,undergoes oxidation to produce carbon dioxide $(CO_2)$ gas.
The balanced chemical equation is:
$2 KMnO_4 + 5 H_2C_2O_4 + 6 H^{+} \longrightarrow 2 Mn^{2+} + 2 K^{+} + 10 CO_2 + 8 H_2O$

(D) The balanced chemical equations are:
$(i)$ $3 \,Cu + 8 \,HNO_3 \rightarrow 3 \,Cu(NO_3)_2 + 2 \,NO + 4 \,H_2O$
$(ii)$ $2 \,CuI_2 \rightarrow Cu_2I_2 + I_2$
$(iii)$ $2 \,Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2 \,NaI$
$(a)$ The coefficients are: $a=3, b=8, c=3, d=2, e=4$.
$(b)$ The coefficients are: $f=2, g=1, h=1$.
$(c)$ The coefficients are: $i=2, j=1, k=1, l=2$.
$(d)$ Moles of $I_2 = \frac{2.54 \,g}{254 \,g/mol} = 0.01 \,mol$.
From equation $(ii)$,$1 \,mol$ of $I_2$ is produced from $2 \,mol$ of $CuI_2$,which corresponds to $2 \,mol$ of $Cu$.
Therefore,moles of $Cu = 2 \times 0.01 \,mol = 0.02 \,mol$.
Mass of $Cu = 0.02 \,mol \times 63.5 \,g/mol = 1.27 \,g$.
Percentage of $Cu = \frac{1.27 \,g}{2.0 \,g} \times 100 = 63.5 \%$.

(B) The reaction between copper and concentrated sulfuric acid is a redox reaction where copper is oxidized and sulfuric acid is reduced.
The balanced chemical equation is:
$Cu_{(s)} + 2H_2SO_{4(conc.)} \rightarrow CuSO_{4(aq)} + SO_{2(g)} + 2H_2O_{(l)}$
In this reaction,$SO_2$ is the sulphur-containing compound $X$ produced.
Therefore,the correct option is $B$.

(B) The reaction of $SO_2$ with acidic $KMnO_4$ is: $2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$. Here,$X$ is $MnSO_4$,where the oxidation state of $Mn$ is $+2$.
Under neutral conditions,$MnSO_4$ reacts with $KMnO_4$ in the presence of $ZnO$ (which acts as a buffer): $3MnSO_4 + 2KMnO_4 + 2H_2O \rightarrow 5MnO_2 + K_2SO_4 + 2H_2SO_4$. Here,$Y$ is $MnO_2$,where the oxidation state of $Mn$ is $+4$.
Thus,the oxidation states of $Mn$ in $X$ and $Y$ are $+2$ and $+4$ respectively.

(D) When $K_2Cr_2O_7$ paper acidified with dilute $H_2SO_4$ is exposed to $SO_2$ gas,the orange dichromate ion $(Cr_2O_7^{2-})$ is reduced to green chromium$(III)$ ion $(Cr^{3+})$.
The balanced chemical equation is:
$Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O$
Thus,the paper turns green due to the formation of $Cr^{3+}$ ions.

(A) In acidic medium,$KMnO_4$ acts as a strong oxidizing agent and oxidizes iodide ions $(I^{-})$ to iodine $(I_2)$:
$2 MnO_4^{-} + 10 I^{-} + 16 H^{+} \rightarrow 2 Mn^{2+} + 5 I_2 + 8 H_2 O$
In neutral or faintly alkaline medium,$KMnO_4$ oxidizes iodide ions $(I^{-})$ to iodate ions $(IO_3^{-})$:
$2 MnO_4^{-} + I^{-} + H_2 O \rightarrow 2 MnO_2 + IO_3^{-} + 2 OH^{-}$
Therefore,the products are $I_2$ and $IO_3^{-}$ respectively.

(A) decomposition redox reaction involves the breakdown of a single compound into two or more products where the oxidation states of the elements change.
In the reaction $2Pb(NO_3)_{2(s)} \longrightarrow 2PbO_{(s)} + 4NO_{2(g)} + O_{2(g)}$,the oxidation state of $N$ changes from $+5$ to $+4$ and $O$ changes from $-2$ to $0$. Since a single reactant breaks down into multiple products with a change in oxidation states,it is a decomposition redox reaction.
Option $A$ is the correct answer.

(A) The given reaction is $2[Au(CN)_2]_{(aq)}^{-} + Zn_{(s)} \rightarrow 2Au_{(s)} + [Zn(CN)_4]_{(aq)}^{2-}$.
Assigning oxidation states: $2[\stackrel{+1}{Au}(CN)_2]_{(aq)}^{-} + \stackrel{0}{Zn}_{(s)}$ $\rightarrow 2\stackrel{0}{Au}_{(s)} + [\stackrel{+2}{Zn}(CN)_4]_{(aq)}^{2-}$.
Here,$Zn$ is oxidized from $0$ to $+2$ and $Au$ is reduced from $+1$ to $0$. Since both oxidation and reduction occur,it is a redox reaction.
$Zn$ displaces $Au$ from the complex,making it a displacement reaction.
Therefore,both $A$ and $B$ are correct.