Oxidizing and Reducing agent Questions in English

(A) To determine the strongest oxidizing agent,we look at the oxidation state of the central atom and its ability to be reduced.
In $SO_3$,the oxidation state of $S$ is $+6$,which is its maximum oxidation state. It cannot be oxidized further,but it can easily be reduced to a lower oxidation state,making it a strong oxidizing agent.
In $SO_2$,$S$ is in $+4$ state.
In $NO_2$,$N$ is in $+4$ state.
In $NO_2^-$,$N$ is in $+3$ state.
Comparing these,$SO_3$ contains sulfur in its highest possible oxidation state $(+6)$,which makes it the most effective at accepting electrons and thus the strongest oxidizing agent among the given options.

(C) Starch iodide paper contains potassium iodide $(KI)$ and starch.
When an oxidizing agent comes in contact with the paper,it oxidizes the iodide ions $(I^-)$ to iodine $(I_2)$.
The liberated iodine $(I_2)$ then reacts with starch to form a deep blue-colored complex.
Therefore,starch iodide paper is used to detect the presence of an oxidizing agent.

(C) Copper sulphate solution is not stored in a zinc vessel because $Zn$ is more reactive than $Cu$ and has a higher tendency to undergo oxidation,thereby displacing $Cu$ from its aqueous solution.
The chemical reaction is: $Zn_{(s)} + CuSO_{4(aq)} \to ZnSO_{4(aq)} + Cu_{(s)}$.
$Zn$ does not form a complex with $CuSO_4$. The reason provided is scientifically incorrect. Therefore,the Assertion is true,but the Reason is false.

$(a)$ In sulphur dioxide $(SO_{2})$,the oxidation number $(O.N.)$ of $S$ is $+4$ and the range of the $O.N.$ that $S$ can have is from $+6$ to $-2$. Therefore,$SO_{2}$ can act as an oxidising as well as a reducing agent.
$(b)$ In hydrogen peroxide $(H_{2}O_{2})$,the $O.N.$ of $O$ is $-1$ and the range of the $O.N.$ that $O$ can have is from $0$ to $-2$. Hence,$H_{2}O_{2}$ can act as an oxidising as well as a reducing agent.
$(c)$ In ozone $(O_{3})$,the $O.N.$ of $O$ is $0$ and the range of the $O.N.$ that $O$ can have is from $0$ to $-2$. Therefore,the $O.N.$ of $O$ can only decrease. Hence,$O_{3}$ acts only as an oxidant.
$(d)$ In nitric acid $(HNO_{3})$,the $O.N.$ of $N$ is $+5$ and the range of the $O.N.$ that $N$ can have is from $+5$ to $-3$. Therefore,the $O.N.$ of $N$ can only decrease. Hence,$HNO_{3}$ acts only as an oxidant.

(N/A) The given reaction occurs because $XeO_6^{4-}$ acts as an oxidizing agent and $F^{-}$ acts as a reducing agent.
In the reaction: $\mathop {Xe}\limits^{+8} O_6^{4-}{(aq)} + 2\mathop {F^{-}}\limits^{-1}{(aq)} + 6H^{+}{(aq)} \to \mathop {Xe}\limits^{+6}O_{3_{(g)}} + \mathop {F_2}\limits^{0}_{(g)} + 3H_2O_{(l)}$
The oxidation number $(O.N.)$ of $Xe$ decreases from $+8$ in $XeO_6^{4-}$ to $+6$ in $XeO_3$ (reduction).
The $O.N.$ of $F$ increases from $-1$ in $F^{-}$ to $0$ in $F_2$ (oxidation).
Since $XeO_6^{4-}$ oxidizes $F^{-}$ to $F_2$,we can conclude that $Na_4XeO_6$ is a very strong oxidizing agent.

(N/A) $Ag^{+}$ and $Cu^{2+}$ act as oxidizing agents in reactions $(a)$ and $(b)$ respectively.
In reaction $(c)$,$Ag^{+}$ oxidizes $C_6H_5CHO$ to $C_6H_5COO^{-}$,but in reaction $(d)$,$Cu^{2+}$ cannot oxidize $C_6H_5CHO$.
Hence,we can conclude that $Ag^{+}$ is a stronger oxidizing agent than $Cu^{2+}$.

(N/A) An oxidising agent is a substance that gains electrons and undergoes reduction in a chemical reaction. Conversely,a reducing agent is a substance that loses electrons and undergoes oxidation.
Consider the following redox reactions:
$2 Na_{(s)} + Cl_{2_{(g)}} \rightarrow 2 Na^{+} Cl_{(s)}^{-}$
$2 Na_{(s)} + S_{(s)} \rightarrow (Na^{+})_{2} S_{(s)}^{-2}$
$2 Na_{(s)} + \frac{1}{2} O_{2_{(g)}} \rightarrow (Na^{+})_{2} O_{(s)}^{-2}$
In these reactions,$Na$ loses electrons and is oxidized,acting as a reducing agent. $Cl_{2}$,$S$,and $O_{2}$ gain electrons and are reduced,acting as oxidising agents.
For example,the reaction between $Na$ and $Cl_{2}$ can be split into two half-reactions:
Oxidation half-reaction: $2 Na_{(s)} \rightarrow 2 Na^{+}_{(g)} + 2 e^{-}$
Reduction half-reaction: $Cl_{2_{(g)}} + 2 e^{-} \rightarrow 2 Cl^{-}_{(g)}$
Overall reaction: $2 Na_{(s)} + Cl_{2_{(g)}} \rightarrow 2 NaCl_{(s)}$
$A$ redox reaction is a process involving the transfer of electrons from one reactant to another.

(B) $SO_{2}$ acts as a reducing agent when passed through an aqueous solution containing $Fe(III)$ salt.
It reduces $Fe(III)$ to $Fe(II)$,i.e.,ferric ions to ferrous ions.
The balanced chemical equation is:
$2Fe^{3+}(aq) + SO_{2}(g) + 2H_{2}O(l) \longrightarrow 2Fe^{2+}(aq) + SO_{4}^{2-}(aq) + 4H^{+}(aq)$

(N/A) $SO_{2}$: In $SO_{2}$,the oxidation number of $S$ is $+4$. The oxidation number of $S$ ranges from $-2$ to $+6$. Therefore,the oxidation number of $S$ in $SO_{2}$ can both increase and decrease,allowing it to act as either an oxidizing or a reducing agent.
$(b)$ $H_{2}O_{2}$: In $H_{2}O_{2}$,the oxidation number of $O$ is $-1$. The oxidation number of oxygen can increase to $0$ (in $O_{2}$) or decrease to $-2$ (in $H_{2}O$ or $OH^-$). Thus,$H_{2}O_{2}$ can act as both an oxidizing and a reducing agent.
$(c)$ $O_{3}$: The oxidation number of oxygen in $O_{3}$ is $0$. It can only be reduced to $-1$ or $-2$. Therefore,it acts only as an oxidizing agent.
$(d)$ $HNO_{3}$: The oxidation number of nitrogen in $HNO_{3}$ is $+5$,which is its maximum oxidation state. Therefore,it can only be reduced,meaning $HNO_{3}$ acts only as an oxidizing agent.

(N/A) In reactions $(a)$ and $(b)$,$H_3PO_2$ acts as a reducing agent,reducing $Ag^+$ and $Cu^{2+}$ to their metallic states,while $Ag^+$ and $Cu^{2+}$ act as oxidizing agents.
In reaction $(c)$,the Tollens' reagent ($Ag^+$ complex) successfully oxidizes benzaldehyde $(C_6H_5CHO)$ to benzoate ion $(C_6H_5COO^-)$.
In reaction $(d)$,$Cu^{2+}$ fails to oxidize benzaldehyde under the given conditions.
Therefore,we can infer that $Ag^+$ is a stronger oxidizing agent than $Cu^{2+}$.

(C) An oxidising agent (or oxidant) is a substance that has the ability to oxidise other substances.
In terms of chemical composition,it is a substance that provides oxygen or removes hydrogen during a chemical reaction.
In terms of electronic concept,an oxidising agent is a substance that gains electrons during a redox reaction.

(B) reducing agent is a substance that undergoes oxidation by losing electrons during a chemical reaction.
Alternatively,it can be defined as a substance that donates hydrogen or removes oxygen from another substance.

(A) In the reaction $CH_3CHO + Ag_2O \to CH_3COOH + 2Ag$,the oxidation state of carbon in the aldehyde group $(-CHO)$ increases from $+1$ to $+3$ in the carboxylic acid group $(-COOH)$.
This indicates that $CH_3CHO$ is oxidized.
The oxidation state of silver in $Ag_2O$ decreases from $+1$ to $0$ in $Ag$.
This indicates that $Ag_2O$ is reduced.
$A$ substance that undergoes oxidation acts as a reducing agent.
Therefore,$CH_3CHO$ is the reducing agent.

(B) In the given reaction,the oxidation state of $Na$ increases from $0$ to $+1$ (oxidation).
The oxidation state of $H$ in $H_2O$ decreases from $+1$ to $0$ in $H_2$ (reduction).
Since $H_2O$ undergoes reduction,it acts as an oxidising agent.

(C) $I$. $NaH$ (sodium hydride) contains hydrogen in the $-1$ oxidation state,making it a strong reducing agent,not an oxidising agent. Thus,statement $I$ is false.
$II$. In pyridine,the nitrogen atom has a lone pair of electrons that is not involved in the aromatic sextet. This lone pair is available for donation,which makes pyridine basic in nature. Thus,statement $II$ is true.

(C) In equation $(A)$: $HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$. The oxidation state of $Cl$ changes from $+1$ in $HOCl$ to $-1$ in $Cl^{-}$. Thus,$HOCl$ is reduced and $H_2O_2$ acts as a reducing agent.
In equation $(B)$: $I_2 + H_2O_2 + 2 OH^{-} \rightarrow 2 I^{-} + 2 H_2O + O_2$. The oxidation state of $I$ changes from $0$ in $I_2$ to $-1$ in $I^{-}$. Thus,$I_2$ is reduced and $H_2O_2$ acts as a reducing agent.
Therefore,in both equations $(A)$ and $(B)$,$H_2O_2$ acts as a reducing agent.

(C) The correct option is $C$.
When copper powder is heated in air,it reacts with oxygen to form copper$(II)$ oxide,which is black in color: $2Cu(s) + O_{2}(g) \stackrel{\Delta}{\longrightarrow} 2CuO(s)$ (Black).
When this black copper$(II)$ oxide is heated with hydrogen gas $(H_{2})$,it undergoes a reduction reaction to form copper metal,which is brown in color: $CuO(s) + H_{2}(g) \stackrel{\Delta}{\longrightarrow} Cu(s) + H_{2}O(g)$.
Thus,$H_{2}$ acts as a reducing agent to convert the black oxide back into brown copper.

(C) .
When a filter paper soaked in salt $X$ is exposed to $HNO_{3}$ vapor,it turns brown due to the liberation of iodine gas $(I_{2})$,which is brown in color.
This reaction occurs because $KI$ acts as a strong reducing agent and reduces $HNO_{3}$ to $NO_{2}$ (a brown gas) while $I^{-}$ is oxidized to $I_{2}$.
The chemical reaction is: $2KI + 4HNO_{3} \longrightarrow I_{2} + 2NO_{2} + 2KNO_{3} + 2H_{2}O$.

(B) In acidic solution,potassium dichromate $(K_2Cr_2O_7)$ acts as an oxidizing agent. The reduction half-reaction is given by:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is calculated as:
$2x + 7(-2) = -2$ $\Rightarrow 2x - 14 = -2$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$
In $Cr^{3+}$,the oxidation state of $Cr$ is $+3$.
Thus,the oxidation state of chromium changes from $+6$ to $+3$.

(A) An oxidizing agent is a substance that gains electrons and undergoes reduction,meaning its oxidation state must decrease.
In the $N^{3-}$ ion,nitrogen is in its lowest possible oxidation state $(-3)$.
Since it cannot lose more electrons to be reduced further,it cannot function as an oxidizing agent.
In contrast,$S$ in $SO_4^{2-}$,$Br$ in $BrO_3^{-}$,and $Mn$ in $MnO_4^{-}$ are in high oxidation states and can be reduced.

(C) To act as a reducing agent,a species must be able to increase its oxidation state (i.e.,undergo oxidation).
In $SO_2$,the oxidation state of $S$ is $+4$ (can increase to $+6$).
In $SO_3^{2-}$,the oxidation state of $S$ is $+4$ (can increase to $+6$).
In $S^{2-}$,the oxidation state of $S$ is $-2$ (can increase to $0$ or higher).
In $H_2SO_4$,the oxidation state of $S$ is $+6$,which is the maximum possible oxidation state for sulphur.
Therefore,$H_2SO_4$ cannot be further oxidized and cannot act as a reducing agent.

(C) Acidified potassium dichromate $(K_2Cr_2O_7)$ acts as a strong oxidizing agent in an acidic medium.
It oxidizes stannous ions $(Sn^{2+})$ to stannic ions $(Sn^{4+})$ while being reduced to chromic ions $(Cr^{3+})$.
The balanced chemical equation is:
$Cr_2O_7^{2-} + 14H^+ + 3Sn^{2+} \rightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O$
Thus,$Sn^{2+}$ is oxidized to $Sn^{4+}$.

(D) reducing agent is a substance that donates electrons and gets oxidized in a chemical reaction.
In $(COOH)_2$ (oxalic acid),the carbon atom is in the $+3$ oxidation state.
It can be oxidized to $CO_2$,where carbon is in the $+4$ oxidation state.
Since it can increase its oxidation state,$(COOH)_2$ acts as a reducing agent.
Conversely,$HNO_3$,$KMnO_4$,and $H_2SO_4$ are strong oxidizing agents because the central atoms ($N$,$Mn$,$S$) are in their highest possible oxidation states and tend to get reduced.

(C) An oxidising agent is a species that gains electrons and gets reduced. The strength of an oxidising agent is determined by its standard reduction potential $(E^{\circ}_{red})$. Higher positive values of $E^{\circ}_{red}$ indicate a stronger tendency to gain electrons,making the species a stronger oxidising agent.
Among the given options:
$1$. $Li$ is a strong reducing agent (it loses electrons easily).
$2$. $Li^{+}$ has a very negative reduction potential and is a very weak oxidising agent.
$3$. $F_2$ has the highest standard reduction potential $(+2.87 \ V)$,making it the strongest oxidising agent among the options.
$4$. $F^{-}$ is the reduced form of fluorine and cannot be further reduced,so it acts as a reducing agent.
Therefore,$F_2$ is the strongest oxidising agent.

(B) reducing agent is a species that donates electrons and gets oxidized itself.
For a species to act as a reducing agent,it must be able to lose electrons.
$Li$ is a strong reducing agent because it easily loses an electron to form $Li^+$.
$F_2$ is a strong oxidizing agent as it easily gains electrons.
$F^-$ has a complete octet and a very high electronegativity,making it extremely difficult to remove an electron from it.
$Li^+$ has a stable noble gas configuration $(1s^2)$ and cannot lose further electrons easily to act as a reducing agent.
However,comparing $Li^+$ and $F^-$,$Li^+$ is a cation and cannot be oxidized further,making it the weakest reducing agent among the given options.

(B) In the given reaction: $Fe_{(s)} + Cu_{(aq)}^{2+} \rightarrow Fe_{(aq)}^{2+} + Cu_{(s)}$
$1$. The oxidation state of $Fe$ increases from $0$ to $+2$,so $Fe$ is oxidized.
$2$. The oxidation state of $Cu$ decreases from $+2$ to $0$,so $Cu^{2+}$ is reduced.
$3$. An oxidizing agent is a substance that gets reduced in a chemical reaction.
$4$. Since $Cu_{(aq)}^{2+}$ undergoes reduction,it acts as the oxidizing agent.

(C) The oxidation states of the elements in the reaction are: $\stackrel{+3}{C_2}O_4^{2-} + \stackrel{+7}{Mn}O_4^{-} + H^{+}$ $\longrightarrow \stackrel{+2}{Mn^{2+}} + \stackrel{+4}{C}O_2 + H_2O$.
In this reaction,the oxidation number of carbon $(C)$ increases from $+3$ in $C_2O_4^{2-}$ to $+4$ in $CO_2$.
Since the oxidation number increases,$C_2O_4^{2-}$ undergoes oxidation and acts as the reducing agent (reductant).

(A) To identify the reductant,we determine the oxidation states of the elements in the reaction:
$H_2S^{-2} + NO_2^{+4} \rightarrow H_2O + NO^{+2} + S^0$
In $H_2S$,the oxidation state of $S$ increases from $-2$ to $0$. This is an oxidation process.
Since $H_2S$ undergoes oxidation,it acts as the reducing agent (reductant).
In $NO_2$,the oxidation state of $N$ decreases from $+4$ to $+2$. This is a reduction process,so $NO_2$ acts as the oxidizing agent.
Therefore,$H_2S$ is the reductant.

(B) To identify the reducing agent,we analyze the change in oxidation states:
$1$. In $H_2O_{2(aq)}$,the oxidation state of oxygen is $-1$. In $O_{2(g)}$,the oxidation state of oxygen is $0$.
$2$. Since the oxidation state of oxygen increases from $-1$ to $0$,$H_2O_{2(aq)}$ undergoes oxidation.
$3$. $A$ substance that undergoes oxidation acts as a reducing agent.
$4$. Therefore,$H_2O_{2(aq)}$ is the reducing agent.

(D) In the given reaction: $CH_{4(g)} + 2O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$
$1$. The oxidation state of carbon in $CH_4$ increases from $-4$ to $+4$ (oxidation).
$2$. The oxidation state of oxygen in $O_2$ decreases from $0$ to $-2$ (reduction).
$3$. $A$ substance that undergoes oxidation acts as a reducing agent.
$4$. Since $CH_4$ is oxidized,it is the reducing agent.

(B) $SO_2$ can act both as an oxidising agent and a reducing agent because the oxidation state of sulphur in $SO_2$ is $+4$,which is an intermediate oxidation state (between $-2$ and $+6$).
As a reducing agent: $SO_2$ reduces $Fe^{3+}$ to $Fe^{2+}$ and decolourises acidified $KMnO_4$ solution.
$2Fe^{3+} + SO_2 + 2H_2O \rightarrow 2Fe^{2+} + SO_4^{2-} + 4H^+$
As an oxidising agent: $SO_2$ reacts with $H_2S$ to produce sulphur and water,where the oxidation state of sulphur in $SO_2$ decreases from $+4$ to $0$.
$SO_2 + 2H_2S \rightarrow 3S + 2H_2O$

(A) The oxidizing power of an ion depends on its standard reduction potential $(E^\circ)$.
$Li^+$ has the most negative standard reduction potential $(E^\circ = -3.04 \ V)$ among the given alkali metal ions,which makes it the weakest oxidizing agent.
While $Li^+$ has a small ionic size and high hydration energy,its tendency to remain in the ionic state in aqueous solution is very high,making it extremely difficult to reduce to $Li(s)$.
Therefore,$Li^+$ is the weakest oxidizing agent among the choices provided.

(D) Potassium dichromate $(K_2Cr_2O_7)$ is a strong oxidizing agent in acidic medium.
It reacts with ferrous ions $(Fe^{2+})$ to oxidize them into ferric ions $(Fe^{3+})$ according to the following reaction:
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
Therefore,the correct statement is that it acts as an oxidizing agent in acidic media.

(A) Acidified potassium dichromate $(K_2Cr_2O_7)$ is a strong oxidizing agent.
It oxidizes stannous ions $(Sn^{2+})$ to stannic ions $(Sn^{4+})$ in an acidic medium.
The balanced chemical equation is:
$Cr_2O_7^{2-} + 14H^+ + 3Sn^{2+} \rightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O$
Therefore,$Sn^{2+}$ changes to $Sn^{4+}$.

(A) In the second reaction,$Zn(s)$ is oxidized to $[Zn(CN)_{4}]^{2-}(aq)$ as its oxidation state increases from $0$ to $+2$.
Since $Zn$ undergoes oxidation,it acts as a reducing agent by reducing $Ag^{+}$ ions to $Ag(s)$.

(C) In the given reaction,the oxidation state of iron $(Fe)$ increases from $+2$ in $FeSO_{4}$ to $+3$ in $Fe_{2}(SO_{4})_{3}$.
This indicates that $FeSO_{4}$ is being oxidised.
Hydrogen peroxide $(H_{2} O_{2})$ is a powerful oxidising agent that accepts electrons and gets reduced to water $(H_{2} O)$.
Therefore,$H_{2} O_{2}$ acts as the oxidising agent in this reaction.

(D) $O_3$ acts as a strong oxidizing agent.
$KI$ is oxidized to $I_2$.
$FeSO_4$ is oxidized to $Fe_2(SO_4)_3$.
$K_2MnO_4$ is oxidized to $KMnO_4$.
In $KMnO_4$,the oxidation state of $Mn$ is $+VII$,which is the maximum possible oxidation state for $Mn$.
Therefore,$KMnO_4$ cannot be further oxidized by $O_3$.
Hence,the correct option is $D$.

(C) reducing agent is a substance that can be oxidized,meaning its central atom must be able to increase its oxidation state.
In $SO_2$,the oxidation state of $S$ is $+4$,and its maximum oxidation state is $+6$,so it can be oxidized.
In $NO_2$,the oxidation state of $N$ is $+4$,and its maximum oxidation state is $+5$,so it can be oxidized.
In $CO_2$,the oxidation state of $C$ is $+4$,which is its maximum group oxidation state (Group $14$). Therefore,it cannot be oxidized further and cannot act as a reducing agent.
In $ClO_2$,the oxidation state of $Cl$ is $+4$,and its maximum oxidation state is $+7$,so it can be oxidized.

(C) The standard reduction potential of $Cu^{2+} / Cu^{+}$ is $+0.15 \ V$,while the oxidation potential of $I^{-} / I_2$ is $-0.54 \ V$.
Because the $Cu^{2+}$ ion has a high reduction potential,it acts as a strong oxidizing agent.
It readily oxidizes the iodide ion $(I^{-})$ to iodine $(I_2)$ and itself gets reduced to $Cu^{+}$.
Therefore,$2Cu^{2+} + 4I^{-} \rightarrow 2CuI + I_2$ occurs,and $CuI_2$ cannot be formed.

(A) In the given reaction,the oxidation state of oxygen in $O_3$ changes from $0$ to $-2$ (in $OH^-$) and $0$ (in $O_2$).
Specifically,the oxygen atom in ozone gains electrons to form $OH^-$,which is a reduction process.
Since $O_3$ undergoes reduction,it causes the oxidation of $I^-$ to $I_2$ (where the oxidation state of $I$ changes from $-1$ to $0$).
Therefore,$O_3$ acts as an oxidizing agent.

(D) Iron $(Fe)$ is used to obtain metallic $Cu$ from $CuSO_{4(aq)}$ because $Fe$ is more reactive than $Cu$ (it has a more negative standard reduction potential),thus it displaces copper from $CuSO_{4}$ solution to give metallic copper $(Cu)$.
$CuSO_{4(aq)} + Fe_{(s)} \longrightarrow FeSO_{4(aq)} + Cu_{(s)}$

(B) $KMnO_4$ (Potassium permanganate) acts as a strong oxidizing agent. It is widely used as an antiseptic and disinfectant in dilute solutions for treating skin infections and wounds.

(A) $K_2Cr_2O_7$ acts as a strong oxidizing agent in an acidic medium.
It reacts with reducing agents by oxidizing them,while the dichromate ion $(Cr_2O_7^{2-})$ is reduced to the green-colored chromium$(III)$ ion $(Cr^{3+})$.
Among the given species,$Fe^{2+}$,$Sn^{2+}$,$I^-$,and $S^{2-}$ are all reducing agents.
Therefore,they will reduce $K_2Cr_2O_7$ and change its orange color to green.
Thus,the correct set is $Fe^{2+}, Sn^{2+}, I^-, S^{2-}$.