Oxidizing and Reducing agent Questions in English

(B) The given reaction is: $14H^+ + Cr_2O_7^{2-} + 3Ni \to 2Cr^{3+} + 7H_2O + 3Ni^{2+}$
In this reaction,the oxidation state of $Ni$ changes from $0$ to $+2$ $(3Ni \to 3Ni^{2+} + 6e^-)$.
Since $Ni$ undergoes oxidation,it acts as a reducing agent.

(C) In $HNO_2$,the oxidation state of nitrogen is $+3$.
Since the oxidation state of nitrogen can increase (up to $+5$) and decrease (down to $-3$),it can act as both an oxidizing agent and a reducing agent.

(A) The reaction is: $\mathop{P}\limits^{+3}\mathop{Cl}\limits^{-1}_3 + \mathop{Cl}\limits^{0}_2 \to \mathop{P}\limits^{+5}\mathop{Cl}\limits^{-1}_5$.
In this reaction,the oxidation state of phosphorus increases from $+3$ to $+5$,which means $PCl_3$ undergoes oxidation.
Therefore,$PCl_3$ acts as a reducing agent.

(C) In $CO_2$,carbon is in its maximum oxidation state of $+4$.
Since it cannot be oxidized further,it cannot act as a reducing agent.
In contrast,$SO_2$ ($S$ is $+4$),$H_2O_2$ ($O$ is $-1$),and $NO_2$ ($N$ is $+4$) can all be oxidized to higher oxidation states.

(B) The reaction involves the reduction of dichromate $(Cr_2O_7^{2-})$ to $Cr^{3+}$,which means $X$ must act as a reducing agent and undergo oxidation.
In $SO_4^{2-}$,the oxidation state of sulfur is $+6$,which is its maximum oxidation state.
Therefore,$SO_4^{2-}$ cannot be oxidized further and cannot act as a reducing agent in this reaction.
Thus,$SO_4^{2-}$ cannot be $X$.

(A) In the reaction: $2H^{-1}I + H_2S^{+6}O_4 \to I_2^0 + S^{+4}O_2 + 2H_2O$,the oxidation state of sulfur in $H_2SO_4$ decreases from $+6$ to $+4$ in $SO_2$. Since $H_2SO_4$ undergoes reduction,it acts as an oxidizing agent.

(C) $SO_3$ acts as a strong oxidizing agent.
For example:
$2SO_3 + S \xrightarrow{100^{\circ}C} 3SO_2$
$5SO_3 + 2P \to 5SO_2 + P_2O_5$
$SO_3 + 2HBr \to SO_2 + Br_2 + H_2O$

(A) An oxidizing agent is a substance that gains electrons and undergoes reduction,causing the oxidation of another substance.
In the reaction $2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O$,the oxidation state of $I$ in $HI$ increases from $-1$ to $0$ (oxidation).
The oxidation state of $S$ in $H_2SO_4$ decreases from $+6$ to $+4$ (reduction).
Since $H_2SO_4$ undergoes reduction,it acts as an oxidizing agent.
In the other reactions,$H_2SO_4$ acts as an acid (neutralization or displacement reaction) where no change in oxidation states occurs.

(B) Potassium permanganate $(KMnO_4)$ is a strong oxidizing agent.
It is commonly used as an antiseptic in dilute solutions for treating skin infections,wounds,and fungal infections of the feet.

(C) reducing agent is a substance that can be oxidized to a higher oxidation state.
In $SO_2$,the oxidation state of $S$ is $+4$,which can be oxidized to $+6$.
In $NO_2$,the oxidation state of $N$ is $+4$,which can be oxidized to $+5$.
In $ClO_2$,the oxidation state of $Cl$ is $+4$,which can be oxidized to $+7$.
In $CO_2$,the oxidation state of $C$ is $+4$,which is its maximum oxidation state (group valence).
Therefore,$CO_2$ cannot be further oxidized and cannot act as a reducing agent.

(C) In the reaction $2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O$,the oxidation state of sulfur $(S)$ decreases from $+6$ in $H_2SO_4$ to $+4$ in $SO_2$,indicating reduction.
Simultaneously,the oxidation state of iodine $(I)$ increases from $-1$ in $HI$ to $0$ in $I_2$,indicating oxidation.
Since $H_2SO_4$ causes the oxidation of $HI$ while being reduced itself,it acts as an oxidizing agent.

(B) The given reaction is a displacement reaction where $Br_2$ acts as an oxidizing agent.
For the reaction to be spontaneous,the standard reduction potential of the halogen being displaced $(X_2/X^-)$ must be lower than that of $Br_2/Br^-$.
The standard reduction potentials $(E^{\circ})$ are:
$F_2/F^- = +2.87 \ V$
$Cl_2/Cl^- = +1.36 \ V$
$Br_2/Br^- = +1.09 \ V$
$I_2/I^- = +0.54 \ V$
Since $Br_2$ can only oxidize species with a lower reduction potential than itself,it can only oxidize $I^-$ to $I_2$ $(E^{\circ} = 0.54 \ V < 1.09 \ V)$.
Therefore,the reaction is correct when $X^-$ is $I^-$.

(D) Acidified $KMnO_4$ is a strong oxidizing agent.
It can oxidize $C_2O_4^{2-}$ to $CO_2$,$NO_2^{-}$ to $NO_3^{-}$,and $S^{2-}$ to $S$.
However,$F^{-}$ is the conjugate base of the weakest acid $HF$ and is the weakest reducing agent among the halides.
Fluorine $(F_2)$ is the strongest oxidizing agent,meaning $F^{-}$ cannot be oxidized by $KMnO_4$ because its oxidation potential is very low and it is already in its most stable oxidation state $(-1)$.

(C) substance can act as both an oxidising agent and a reducing agent if the oxidation state of its central atom lies between its minimum and maximum possible oxidation states.
$1$. In $HClO_2$,the oxidation state of $Cl$ is $+3$. It can be oxidised to $+7$ and reduced to $-1$.
$2$. In $SO_2$,the oxidation state of $S$ is $+4$. It can be oxidised to $+6$ and reduced to $-2$.
$3$. In $H_2O_2$,the oxidation state of $O$ is $-1$. It can be oxidised to $0$ $(O_2)$ and reduced to $-2$ $(H_2O)$.

(D) In the reaction $2HI + H_2SO_4 \to I_2 + SO_2 + 2H_2O$,the oxidation state of iodine $(I)$ increases from $-1$ to $0$,which indicates an oxidation process.
Simultaneously,the oxidation state of sulfur $(S)$ in $H_2SO_4$ decreases from $+6$ to $+4$ in $SO_2$,which indicates a reduction process.
Since $H_2SO_4$ undergoes reduction,it acts as an oxidizing agent.

(B) In the given reaction: $I_2 + 2S_2O_3^{2-} \to 2I^{-} + S_4O_6^{2-}$
$1$. The oxidation state of $I$ in $I_2$ changes from $0$ to $-1$ in $I^{-}$. Since the oxidation state decreases,$I_2$ undergoes reduction and acts as an oxidising agent.
$2$. The oxidation state of $S$ in $S_2O_3^{2-}$ changes from $+2$ to $+2.5$ in $S_4O_6^{2-}$. Since the oxidation state increases,$S_2O_3^{2-}$ undergoes oxidation and acts as a reducing agent.

(B) The acidified permanganate ion $(MnO_4^-)$ acts as a strong oxidizing agent in acidic medium.
It can oxidize $C_2O_4^{2-}$ to $CO_2$,$S^{2-}$ to $S$,and $Cl^-$ to $Cl_2$ (under specific conditions).
However,in the nitrate ion $(NO_3^-)$,nitrogen is already in its maximum oxidation state of $+5$.
Since nitrogen cannot be oxidized further,$MnO_4^-$ cannot oxidize $NO_3^-$.

(B) reducing agent is a substance that can donate electrons and undergo oxidation.
In $KI$,the oxidation state of $I$ is $-1$,which can be oxidized to $I_2$ $(0)$.
In $HNO_2$,the oxidation state of $N$ is $+3$,which can be oxidized to $+5$.
In $H_2S$,the oxidation state of $S$ is $-2$,which can be oxidized to $S$ $(0)$.
In $HNO_3$,the oxidation state of $N$ is $+5$,which is its maximum possible oxidation state.
Since $N$ cannot be further oxidized,$HNO_3$ can only act as an oxidizing agent.

(A) reductant (or reducing agent) is a substance that undergoes oxidation by losing one or more electrons during a chemical reaction.

(D) reducing agent is a substance that can be oxidized by losing electrons or increasing its oxidation state.
In $SO_2$,the oxidation state of $S$ is $+4$,which can be oxidized to $+6$.
In $Cl_2O_3$,the oxidation state of $Cl$ is $+3$,which can be oxidized to higher states.
In $NO_2$,the oxidation state of $N$ is $+4$,which can be oxidized to $+5$.
In $CO_2$,the oxidation state of $C$ is $+4$,which is the maximum oxidation state for carbon. Therefore,$CO_2$ cannot be further oxidized and cannot act as a reducing agent.

(A) The reaction involves the reduction of dichromate $(Cr_2O_7^{2-})$ in acidic medium to chromium $(III)$ ions $(Cr^{3+})$,which are green in color.
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$
$SO_2$ and $H_2S$ are strong reducing agents that can reduce $Cr_2O_7^{2-}$ to $Cr^{3+}$,thereby changing the color of the solution from orange to green.
$CO_2$ is an acidic oxide and does not act as a reducing agent in this context,so it cannot reduce $Cr_2O_7^{2-}$ to $Cr^{3+}$.
Therefore,$CO_2$ is not used to change the color.

(C) substance can act as both an oxidising agent and a reducing agent if the central atom is in an intermediate oxidation state.
$1$. In $KMnO_4$,$Mn$ is in $+7$ (maximum) oxidation state,so it acts only as an oxidising agent. $O_3$ acts as an oxidising agent. In $SO_3$,$S$ is in $+6$ (maximum) oxidation state,so it acts only as an oxidising agent.
$2$. In $HClO_4$,$Cl$ is in $+7$ (maximum) oxidation state,so it acts only as an oxidising agent.
$3$. In $HNO_2$,$N$ is in $+3$ oxidation state (intermediate between $-3$ and $+5$). In $SO_2$,$S$ is in $+4$ oxidation state (intermediate between $-2$ and $+6$). In $H_2O_2$,$O$ is in $-1$ oxidation state (intermediate between $-2$ and $0$). Thus,all three can act as both oxidising and reducing agents.
$4$. In $HNO_3$,$N$ is in $+5$ (maximum) oxidation state. In $H_2SO_4$,$S$ is in $+6$ (maximum) oxidation state. Thus,they cannot act as reducing agents.
Therefore,the correct option is $(C)$.

(D) $SbF_5$ is a strong Lewis acid. In this reaction,it accepts a fluoride ion $(F^-)$ from $BrCH_2CH_2F$ to form the $SbF_6^-$ anion.
Since $SbF_5$ accepts an electron pair (from the $F^-$ ion),it acts as a Lewis acid,which is a type of electrophile.

(C) $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent that is commonly used in organic chemistry to reduce carbonyl compounds like aldehydes,ketones,and carboxylic acids to alcohols. In contrast,$CrO_3 / H^{+}$ and $KMnO_4$ are strong oxidizing agents,and $O_3$ is also an oxidizing agent.

(C) In the given reaction: $\overset{+4}{H_2SO_3}_{(aq)} + Sn^{4+}_{(aq)} + H_2O_{(l)} \to Sn^{2+}_{(aq)} + \overset{+6}{HSO_4^-}_{(aq)} + 3H^{+}_{(aq)}$.
$1$. The oxidation state of $S$ in $H_2SO_3$ increases from $+4$ to $+6$,which means $H_2SO_3$ undergoes oxidation.
$2$. The oxidation state of $Sn$ decreases from $+4$ to $+2$,which means $Sn^{4+}$ undergoes reduction.
$3$. Since $H_2SO_3$ undergoes oxidation,it acts as the reducing agent.
Therefore,the correct statement is that $H_2SO_3$ is the reducing agent because it undergoes oxidation.

(A) An oxidising agent is a substance that gains electrons and undergoes reduction,meaning its oxidation state must be able to decrease.
In $I^{-}$,the iodine atom is in its lowest possible oxidation state of $-1$. Therefore,it cannot lose electrons to be reduced further; it can only act as a reducing agent by losing electrons to become $I_2$.
Conversely,$S_{(s)}$,$NO_3^{-}$,and $Cr_2O_7^{2-}$ contain elements in oxidation states that can be reduced further.

(B) Concentrated $H_2SO_4$ acts as an oxidizing agent when it undergoes reduction itself.
In reaction $(B)$,$H_2SO_4$ oxidizes $HI$ to $I_2$ and is itself reduced to $SO_2$.
The oxidation state of sulfur in $H_2SO_4$ decreases from $+6$ to $+4$ in $SO_2$,confirming its role as an oxidizing agent.
In the other reactions ($A$,$C$,and $D$),$H_2SO_4$ acts only as an acid,as there is no change in the oxidation states of the sulfur atom.

(B) In the brown ring test,$FeSO_4$ acts as a reducing agent.
It reduces the nitrate ion $(NO_3^-)$ to nitric oxide $(NO)$.
The $NO$ then reacts with the remaining $Fe^{2+}$ ions to form the brown-colored complex $[Fe(H_2O)_5(NO)]SO_4$.
Therefore,the reducing power of $FeSO_4$ is the fundamental reason for this test.

(D) Acidified potassium permanganate $(KMnO_4)$ is a strong oxidizing agent.
It can oxidize $C_2O_4^{2-}$ to $CO_2$,$S^{2-}$ to $S$,and many other species.
However,the fluoride ion $(F^-)$ is the conjugate base of a strong acid $(HF)$ and has an extremely low tendency to lose electrons.
Therefore,$F^-$ cannot be oxidized by $KMnO_4$ under standard conditions.

(C) The chemical reaction is: $2NH_3 + 3CuO \xrightarrow{\Delta} N_2 + 3Cu + 3H_2O$.
In this reaction,the oxidation state of nitrogen increases from $-3$ to $0$ (oxidation),and the oxidation state of copper decreases from $+2$ to $0$ (reduction).
Since ammonia reduces copper oxide to metallic copper,it acts as a reducing agent.

(D) In $SO_2$,the oxidation state of sulfur is $+4$. Sulfur can be oxidized to $+6$ or reduced to $0$ or $-2$. However,$SO_2$ acts as an oxidizing agent when it reacts with stronger reducing agents,such as $H_2S$,where sulfur is reduced from $+4$ to $0$. Among the given options,$SO_2$ is the most common oxidizing agent.

(B) In $SO_2$,the oxidation state of sulfur is $+4$. Since the stable oxidation states of sulfur range from $-2$ to $+6$,sulfur in $SO_2$ can be oxidized to $+6$ (acting as a reducing agent) or reduced to lower oxidation states like $0$ or $-2$ (acting as an oxidizing agent). Therefore,$SO_2$ acts as both an oxidizing and a reducing agent.

(C) In $CO_2$,the oxidation state of carbon is $+4$,which is its maximum oxidation state.
Since carbon cannot be further oxidized,$CO_2$ cannot act as a reducing agent.
In contrast,$SO_2$,$H_2O_2$,and $NO_2$ contain elements in intermediate oxidation states that can be further oxidized.

(C) Ozone $(O_3)$ acts as a strong oxidizing agent.
In $KMnO_4$,the oxidation state of Manganese $(Mn)$ is $+7$,which is its maximum possible oxidation state.
Therefore,$Mn$ cannot be oxidized further.
In contrast,$KI$ ($I^-$ to $I_2$),$FeSO_4$ ($Fe^{2+}$ to $Fe^{3+}$),and $K_2MnO_4$ ($Mn^{6+}$ to $Mn^{7+}$) can all be oxidized by $O_3$.

(C) Fluorine $(F)$ never acts as a reducing agent because it has the highest electronegativity and the highest standard reduction potential in the periodic table. \\ It only undergoes reduction and cannot be oxidized.

(D) An oxidizing agent is a substance that gains electrons in a chemical reaction.
$I^-$ has a complete octet and a negative charge,meaning it has a low tendency to gain more electrons.
In fact,$I^-$ acts as a reducing agent because it can easily lose electrons to form $I_2$.
Therefore,$I^-$ will not act as an oxidizing agent.

(B) An oxidising agent is a substance that oxidises another substance and gets reduced itself.
Reduction is defined as the gain of electrons.
Therefore,an oxidising agent accepts electrons.

(B) An oxidant (oxidizing agent) is a substance that can accept electrons,meaning the central atom must be in its highest possible oxidation state.
In $Fe^{+3}$,the iron is in its maximum oxidation state of $+3$.
In $NO_3^-$,the nitrogen is in its maximum oxidation state of $+5$.
In $SO_3$,the sulfur is in its maximum oxidation state of $+6$.
Since all these species have their central atoms in their maximum oxidation states,they can only act as oxidizing agents.

(A) substance can act as both an oxidizing and a reducing agent if the central atom is in an intermediate oxidation state.
In $SO_2$,the oxidation state of sulfur is $+4$.
The maximum oxidation state of sulfur is $+6$ and the minimum is $-2$.
Since $+4$ lies between $-2$ and $+6$,$SO_2$ can be oxidized to $+6$ (acting as a reducing agent) or reduced to lower oxidation states (acting as an oxidizing agent).
Therefore,$SO_2$ acts as both an oxidizing and a reducing agent.

(A) In $HNO_2$,the central atom nitrogen has an oxidation state of $+3$.
Since nitrogen can exhibit oxidation states ranging from $-3$ to $+5$,it can either increase its oxidation state (acting as a reducing agent) or decrease its oxidation state (acting as an oxidising agent).
Therefore,$HNO_2$ can act as both an oxidising and a reducing agent.

(B) substance can act as both an oxidizing and a reducing agent if the central atom is in an intermediate oxidation state.
$K_2Cr_2O_7$: Chromium is in its highest oxidation state $(+6)$,so it acts only as an oxidizing agent.
$SO_2$: Sulphur is in the $+4$ oxidation state. Since the range of oxidation states for sulphur is $-2$ to $+6$,it can be oxidized to $+6$ (reducing agent) or reduced to lower states (oxidizing agent).
$H_2S$: Sulphur is in its lowest oxidation state $(-2)$,so it acts only as a reducing agent.
$NH_3$: Nitrogen is in its lowest oxidation state $(-3)$,so it acts only as a reducing agent.

(C) reducing agent is a substance that loses electrons or increases its oxidation state in a chemical reaction.
$CaH_2$ contains hydrogen in the $-1$ oxidation state,which can be oxidized to $0$,acting as a reducing agent.
$H_2S$ contains sulfur in the $-2$ oxidation state,which can be oxidized to $0$ or higher,acting as a reducing agent.
$HIO_4$ (periodic acid) contains iodine in the $+7$ oxidation state,which is the maximum possible oxidation state for iodine. Therefore,it can only be reduced and acts as a strong oxidizing agent.
$PbCl_2$ is generally stable,but $Pb$ is in the $+2$ state,which is the most stable state for lead,making it a poor reducing agent compared to the others,but $HIO_4$ is definitively an oxidizing agent.

(C) reducing agent is a substance that can be oxidized,meaning its central atom must be in an intermediate oxidation state and not in its maximum possible oxidation state.
In $SO_2$,the oxidation state of $S$ is $+4$,which can be oxidized to $+6$.
In $NO_2$,the oxidation state of $N$ is $+4$,which can be oxidized to $+5$.
In $CO_2$,the oxidation state of $C$ is $+4$,which is the maximum oxidation state for carbon (group $14$). Therefore,it cannot be oxidized further.
In $ClO_2$,the oxidation state of $Cl$ is $+4$,which can be oxidized to higher states like $+5$ or $+7$.
Thus,$CO_2$ cannot act as a reducing agent.

(D) In $N_2O_5$,the oxidation state of nitrogen is $+5$,which is the maximum possible oxidation state for nitrogen (as it belongs to group $15$ and has a valence shell configuration of $ns^2 np^3$).
Since nitrogen is already in its highest oxidation state,it cannot be further oxidized.
Therefore,$N_2O_5$ can only act as an oxidizing agent and cannot act as a reducing agent.

(B) reducing agent is a substance that loses electrons and undergoes oxidation. $F$ (Fluorine) is the most electronegative element in the periodic table. It has the highest tendency to gain electrons to achieve a stable octet configuration $(F + e^- \rightarrow F^-)$. Because it only undergoes reduction and never oxidation,it acts exclusively as an oxidizing agent and never as a reducing agent.

(D) An oxidizing agent is a substance that gains electrons and gets reduced in a chemical reaction.
For a substance to act as an oxidizing agent,the central atom must be in an oxidation state from which it can be reduced to a lower oxidation state.
In $O_2$,oxygen is in $0$ oxidation state and can be reduced to $-2$.
In $KMnO_4$,manganese is in $+7$ oxidation state,which is its maximum,so it can be reduced.
In $H_2O_2$,oxygen is in $-1$ oxidation state and can be reduced to $-2$.
In $I^-$,iodine is in $-1$ oxidation state,which is its minimum possible oxidation state for iodine. Therefore,it can only be oxidized to $I_2$ or other higher oxidation states,but it cannot be reduced further.
Thus,$I^-$ acts only as a reducing agent.

(A) In $ClO_4^{-}$,$F_2$,and $MnO_4^{-}$,the central atoms ($Cl$,$F$,and $Mn$) are in their maximum possible oxidation states. Therefore,they cannot be oxidized further and can only act as oxidizing agents.
In $O_3$,the oxygen atom has an oxidation state of $0$. Since oxygen can exist in oxidation states ranging from $-2$ to $+2$,$O_3$ can undergo both oxidation and reduction,allowing it to act as both an oxidizing and a reducing agent.

(A) In $SO_3$,the oxidation state of $S$ is $+6$,which is its maximum oxidation state.
Since it cannot be oxidized further,it cannot act as a reducing agent.