Oxidizing and Reducing agent Questions in English

(C) Bromine water acts as an oxidising agent with a standard reduction potential of $E^{\circ} = +1.09 \ V$.
For a species to be oxidised by bromine water,the standard oxidation potential of the species must be lower than the reduction potential of bromine,or the standard reduction potential of the species must be less than $+1.09 \ V$.
$1$. $Fe^{2+} \rightarrow Fe^{3+} + e^-$,$E^{\circ}_{ox} = -0.77 \ V$ (Oxidised by $Br_2$)
$2$. $Cu^{+} \rightarrow Cu^{2+} + e^-$,$E^{\circ}_{ox} = -0.15 \ V$ (Oxidised by $Br_2$)
$3$. $Sn^{2+} \rightarrow Sn^{4+} + 2e^-$,$E^{\circ}_{ox} = -0.15 \ V$ (Oxidised by $Br_2$)
$4$. $Mn^{2+} \rightarrow MnO_4^-$,$E^{\circ}_{ox} = -1.51 \ V$. Since the oxidation potential is much more negative than the reduction potential of bromine,$Mn^{2+}$ cannot be oxidised to $MnO_4^-$ by bromine water.

(B) Any substance which is capable of oxidizing other substances and is capable of accepting or gaining electrons during a redox reaction is called an oxidizing agent or oxidant.

(A) The balanced chemical equation is $5C_2O_4^{2-} + 2MnO_4^- + 16H^{+} \to 2Mn^{2+} + 10CO_2 + 8H_2O$.
In this reaction,the oxidation state of carbon in $C_2O_4^{2-}$ increases from $+3$ to $+4$ (oxidation).
The oxidation state of manganese in $MnO_4^-$ decreases from $+7$ to $+2$ (reduction).
Since $C_2O_4^{2-}$ undergoes oxidation,it acts as the reducing agent (reductant).

(B) substance that is capable of reducing other substances and donating electrons during a chemical reaction is called a reducing agent or reductant.

(A) In the reaction $H_2S + H_2O_2 \to 2H_2O + S$,the oxidation state of sulfur in $H_2S$ changes from $-2$ to $0$ (oxidation).
The oxidation state of oxygen in $H_2O_2$ changes from $-1$ to $-2$ (reduction).
Since $H_2O_2$ undergoes reduction,it acts as an oxidizing agent.
Therefore,the reaction shows the oxidizing action of $H_2O_2$.

(B) In $NaNO_2$,the oxidation state of $N$ is $+3$. Since $N$ can be oxidized to $+5$ or reduced to lower oxidation states,it acts as both a reducing and oxidizing agent.
In $HI$,$I^-$ can be oxidized to $I_2$,so it acts as a reducing agent.
In $SnCl_2$,$Sn^{2+}$ can be oxidized to $Sn^{4+}$,so it acts as a reducing agent.
In $NaNO_3$,the oxidation state of $N$ is $+5$,which is the maximum possible oxidation state for nitrogen. Therefore,it can only be reduced and acts exclusively as an oxidizing agent.

(D) An oxidising agent is a substance that gains electrons and gets reduced in a chemical reaction.
$O_2$ is a strong oxidising agent as it readily gains electrons.
$KMnO_4$ is a well-known strong oxidising agent in various media.
$I_2$ can act as both an oxidising and a reducing agent,but it is primarily known for its ability to act as a mild oxidising agent (e.g.,in the reaction with thiosulfate).
However,in the context of standard chemistry problems,if we must choose the best fit,$I_2$ is often considered a weaker oxidising agent compared to the others,but technically all listed substances can act as oxidising agents under appropriate conditions.
Given the options,if the question implies which is the least likely or cannot act as one,the provided answer is incorrect as $I_2$ is an oxidising agent. Therefore,the correct choice is $D$ (None of these).

(A) In the reaction $C + H_2O \to CO + H_2$,the oxidation state of $C$ increases from $0$ to $+2$ (oxidation),while the oxidation state of $H$ in $H_2O$ decreases from $+1$ to $0$ (reduction).
Since $H_2O$ undergoes reduction,it acts as an oxidising agent.

(A) The chemical reaction is: $SO_2 + 2H_2S \rightarrow 3S + 2H_2O$.
In this reaction,the oxidation state of sulphur in $SO_2$ is $+4$ and it is reduced to $0$ in $S$.
Since $SO_2$ undergoes reduction,it acts as an oxidising agent.
$H_2S$ undergoes oxidation (sulphur goes from $-2$ to $0$),so it acts as a reducing agent.

(B) reducing agent is a substance that donates electrons or increases the oxidation state of another species while being oxidized itself.
$HI$ (Hydrogen Iodide) is a strong reducing agent because the bond dissociation energy of the $H-I$ bond is low,and the iodide ion $(I^-)$ is easily oxidized to iodine $(I_2)$.
In contrast,$NaOCl$ and $FeCl_3$ typically act as oxidizing agents,and $KBr$ is a relatively weak reducing agent compared to $HI$.

(B) The correct answer is $(B)$.
$H_2S$ acts as a strong reducing agent because the sulfur atom in $H_2S$ is in its lowest oxidation state $(-2)$.
It can easily lose electrons to get oxidized to elemental sulfur $(S^0)$ or higher oxidation states,thereby reducing other substances.
In contrast,while $SnCl_2$ is also a reducing agent,$H_2S$ is generally considered a stronger reducing agent in this context due to the ease of oxidation of the sulfide ion.

(D) An oxidising agent is a substance that gains electrons or increases the oxidation state of another element in a chemical reaction.
$HNO_3$ (nitric acid) is a strong oxidising agent because nitrogen is in its highest oxidation state $(+5)$ and can easily be reduced.
Concentrated $H_2SO_4$ (sulfuric acid) also acts as an oxidising agent because sulfur is in its highest oxidation state $(+6)$ and can be reduced to $SO_2$.
$FeSO_4$ typically acts as a reducing agent because $Fe^{2+}$ can be oxidised to $Fe^{3+}$.
Therefore,both $(b)$ and $(c)$ can act as oxidising agents.

(B) In the given reaction: $14H^{+} + Cr_2O_7^{2-} + 3Ni \to 2Cr^{3+} + 7H_2O + 3Ni^{2+}$
$1$. The oxidation state of $Ni$ increases from $0$ (in elemental form) to $+2$ (in $Ni^{2+}$ ion).
$2$. Since $Ni$ undergoes oxidation (loss of electrons),it acts as the reducing agent.
$3$. The oxidation state of $Cr$ in $Cr_2O_7^{2-}$ decreases from $+6$ to $+3$,meaning $Cr_2O_7^{2-}$ acts as the oxidizing agent.
Therefore,the correct option is $(B)$.

(D) An oxidizing agent is a substance that accepts electrons or increases the oxidation state of another substance.
In $CO_2$,the oxidation state of carbon is $+4$,which is its maximum oxidation state.
Therefore,it can only be reduced and acts as an oxidizing agent.
$C_2H_2O_2$,$CO$,and $H_2S$ generally act as reducing agents in various chemical reactions.

(C) Chlorine water is formed by the reaction of chlorine with water: $Cl_2 + H_2O \rightarrow HCl + HOCl$.
$HOCl$ (hypochlorous acid) is unstable and decomposes to release nascent oxygen: $HOCl \rightarrow HCl + [O]$.
Since $HOCl$ provides nascent oxygen,it acts as the oxidising agent in chlorine water.

(D) In the given reaction: $Ag_2O + H_2O_2 \to 2Ag + H_2O + O_2$
$1$. The oxidation state of $Ag$ in $Ag_2O$ is $+1$,and it changes to $0$ in $Ag$. This is a reduction process.
$2$. The oxidation state of $O$ in $H_2O_2$ is $-1$,and it changes to $0$ in $O_2$. This is an oxidation process.
$3$. Since $H_2O_2$ undergoes oxidation,it acts as a reducing agent.
$4$. Additionally,$H_2O_2$ can also act as an oxidising agent in certain reactions,but in this specific reaction,it is clearly acting as a reducing agent.
$5$. However,looking at the options,option $A$ ($D$ and $B$ both) suggests it acts as both an oxidising and reducing agent. While $H_2O_2$ is known for its dual nature,in this specific reaction,it acts as a reducing agent. Given the provided options,$D$ is the most accurate description of its role here.

(D) An oxidizing agent is a substance that undergoes reduction (decrease in oxidation state) during a redox reaction.
In the given reaction: $\mathop {HAsO_2}\limits^{+3} + \mathop {Sn}\limits^{+2} \to \mathop {As}\limits^{0} + \mathop {Sn}\limits^{+4} + H_2O$
$1$. The oxidation state of $As$ changes from $+3$ in $HAsO_2$ to $0$ in $As$. Since the oxidation state decreases,$HAsO_2$ is reduced.
$2$. The oxidation state of $Sn$ changes from $+2$ in $Sn^{2+}$ to $+4$ in $Sn^{4+}$. Since the oxidation state increases,$Sn^{2+}$ is oxidized.
Therefore,$HAsO_2$ acts as the oxidizing agent.

(B) The correct option is $B$.
In $HNO_2$,the oxidation number of $N$ is $+3$. Since the maximum oxidation state of $N$ is $+5$,it can be oxidized to $+5$ (acting as a reductant) or reduced to a lower state (acting as an oxidant).
In $HNO_3$,the oxidation number of $N$ is $+5$,which is the maximum possible oxidation state for nitrogen. Therefore,it can only be reduced,acting exclusively as an oxidant.

(D) When an element is in its lowest oxidation state,it can easily lose electrons to reach a higher oxidation state.
Therefore,it acts as a strong reducing agent,exhibiting the highest reducing property.

(B) To prevent rancidification of food material,we add anti-oxidants,which are substances that act as oxidation inhibitors.

(C) An oxidising agent is a substance that can be reduced (its oxidation state decreases),and a reducing agent is a substance that can be oxidised (its oxidation state increases).
In $Al_2O_3$,the oxidation state of $Al$ is $+3$,which is its maximum possible oxidation state (group $13$).
Therefore,$Al$ cannot be further oxidised,meaning $Al_2O_3$ cannot act as a reducing agent.
In the other compounds,the central atom is in an intermediate oxidation state,allowing them to act as both oxidising and reducing agents.

(D) The correct answer is $(D)$.
Reducing agents are species that donate electrons easily.
Among the halide ions $(Cl^{-}, Br^{-}, I^{-})$,the reducing power increases down the group because the atomic size increases and the valence electrons are held less tightly by the nucleus.
Since $I^{-}$ has the largest size among the given halide ions,it loses its valence electron most easily,making it the strongest reducing agent.

(B) An oxidizing agent is a substance that can accept electrons (its oxidation state decreases),while a reducing agent is a substance that can donate electrons (its oxidation state increases).
For a compound to act as both,its central atom must be in an intermediate oxidation state.
$1$. In $H_2O_2$,oxygen is in $-1$ state. It can go to $0$ (oxidizing) or $-2$ (reducing).
$2$. In $SO_2$,sulfur is in $+4$ state. It can go to $+6$ (reducing) or $0$ (oxidizing).
$3$. In $Cl_2$,chlorine is in $0$ state. It can go to $-1$ (oxidizing) or $+1$ (reducing).
$4$. In $H_2$,hydrogen is in $0$ state. It can only be oxidized to $+1$ state. Therefore,$H_2$ acts only as a reducing agent.

(B) The oxidation state of nitrogen in $HNO_2$ is $+3$.
Since the maximum oxidation state of nitrogen is $+5$ and the minimum is $-3$,nitrogen in $HNO_2$ can be oxidized to $+5$ (acting as a reducing agent) or reduced to lower oxidation states like $+2, +1, 0, -3$ (acting as an oxidizing agent).
Therefore,$HNO_2$ can act as both an oxidizing and a reducing agent.
Option $B$ is the correct statement.

(C) The chemical reaction is: $3CuO + 2NH_3 \to 3Cu + N_2 + 3H_2O$.
In this reaction,$CuO$ is reduced to $Cu$ (loss of oxygen) and $NH_3$ is oxidized to $N_2$ (loss of hydrogen).
Since ammonia reduces copper oxide to copper,it acts as a reducing agent.

(A) The oxidation state of $S$ in $SO_2$ is $+4$.
The range of oxidation states for $S$ is from $-2$ to $+6$.
Since the oxidation state of $S$ in $SO_2$ $(+4)$ lies between its minimum $(-2)$ and maximum $(+6)$ oxidation states,it can either be oxidized to $+6$ or reduced to $-2$.
Therefore,$SO_2$ can act as both an oxidizing and a reducing agent.

(A) The chemical reaction is: $2H_2S + SO_2 \to 2H_2O + 3S$.
In this reaction,the oxidation state of sulphur in $H_2S$ increases from $-2$ to $0$ (oxidation),while the oxidation state of sulphur in $SO_2$ decreases from $+4$ to $0$ (reduction).
Since $SO_2$ undergoes reduction,it acts as an oxidising agent.

(B) In the reaction $2Ag + 2H_2SO_4 \to Ag_2SO_4 + 2H_2O + SO_2$,the oxidation state of $Ag$ increases from $0$ to $+1$,so it acts as a reducing agent.
In $H_2SO_4$,the oxidation state of $S$ decreases from $+6$ to $+4$ in $SO_2$,which indicates that $H_2SO_4$ undergoes reduction.
Therefore,$H_2SO_4$ acts as an oxidising agent.

(A) The mixture of potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulfuric acid $(H_2SO_4)$ is commonly referred to as chromic acid.
It is a powerful oxidizing agent used in laboratory cleaning solutions.

(A) The correct answer is $A$.
$Cl_2 + 2KBr \to 2KCl + Br_2$
$A$ more electronegative halogen can displace a less electronegative halogen from its salt solution. Since chlorine is more electronegative than bromine,it displaces bromine from $KBr$.

(C) The reaction $2KBr + I_2 \to 2KI + Br_2$ will not occur.
This is because the oxidizing power of halogens decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.
Since $I_2$ is a weaker oxidizing agent than $Br_2$,it cannot displace $Br^-$ ions from $KBr$ solution.

(D) $CuSO_4$,$K_2Cr_2O_7$,and $HNO_3$ are oxidizing agents that oxidize $I^-$ to $I_2$.
$2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2$
$Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2$
$2HNO_3 + 2HI \rightarrow 2NO_2 + 2H_2O + I_2$
$HCl$ is a non-oxidizing acid and does not oxidize $I^-$ to $I_2$.
Therefore,$HCl$ does not liberate iodine.

(A) The chemical equation for the preparation of chlorine is: $MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$.
In this reaction,the oxidation state of $Mn$ changes from $+4$ in $MnO_2$ to $+2$ in $MnCl_2$.
Since the oxidation state of $Mn$ decreases,$MnO_2$ undergoes reduction.
$A$ substance that undergoes reduction acts as an oxidising agent.
Therefore,$MnO_2$ acts as an oxidising agent.

(A) The conversion of $Br^{-}$ to $Br_2$ is an oxidation process.
Chlorine $(Cl_2)$ is a stronger oxidizing agent than bromine $(Br_2)$.
Therefore,$Cl_2$ can displace $Br^{-}$ from its solution according to the reaction:
$Cl_2 + 2Br^{-} \rightarrow 2Cl^{-} + Br_2$.

(C) $F$ cannot act as a reducing agent because it has the highest standard reduction potential.
$F_2 + 2e^- \to 2F^-; E^o = +2.87 \ V$
Since fluorine is the most electronegative element and has the highest reduction potential,it only undergoes reduction and never oxidation.

(B) $LiAlH_4$ acts as a strong reducing agent.
It provides hydride ions $(H^-)$ to reduce various functional groups like aldehydes,ketones,and carboxylic acids to alcohols.
Example: $CH_3CHO + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2OH$

(C) The starch-iodide paper is primarily used to detect the presence of oxidizing agents. When an oxidizing agent reacts with the iodide ions $(I^-)$ present in the paper,it oxidizes them to iodine $(I_2)$. The liberated iodine then reacts with the starch to form a blue-colored complex. Therefore,it is used for the test of oxidizing agents.

(B) Acidified $KMnO_4$ is a strong oxidizing agent.
It reacts with reducing agents to get reduced,resulting in the loss of its purple color.
Among the given anions,the sulfide ion $(S^{2-})$ acts as a reducing agent and is oxidized to elemental sulfur $(S)$ by $KMnO_4$.
The balanced chemical equation is:
$2KMnO_4 + 3H_2SO_4 + 5H_2S \to K_2SO_4 + 2MnSO_4 + 8H_2O + 5S$

(C) Chromic acid is a strong oxidizing agent prepared by mixing potassium dichromate $(K_2Cr_2O_7)$ with concentrated sulfuric acid $(H_2SO_4)$.
Therefore,the correct mixture is $(K_2Cr_2O_7 + \text{conc. } H_2SO_4)$.

(D) $KMnO_4$ is a strong oxidizing agent.
It oxidizes reducing agents like $NO_2^-$,$S^{2-}$,and $Cl^-$ (in specific conditions) to their higher oxidation states,resulting in the decolourisation of the purple $KMnO_4$ solution.
$CO_3^{2-}$ contains carbon in its maximum oxidation state of $+4$,so it cannot be further oxidized by $KMnO_4$.
Therefore,$CO_3^{2-}$ will not decolourise the $KMnO_4$ solution.

(B) The oxidizing power of a substance is determined by its ability to provide nascent oxygen or its reduction potential.
In an acidic medium,$KMnO_4$ acts as a stronger oxidizing agent compared to $K_2Cr_2O_7$.
Specifically,$KMnO_4$ provides $5$ atoms of nascent oxygen per molecule,whereas $K_2Cr_2O_7$ provides $3$ atoms of nascent oxygen per molecule.
Therefore,the statement that $KMnO_4$ is a weaker oxidizing substance than $K_2Cr_2O_7$ is incorrect.

(A) $Ag_2O$ is a mild oxidising agent.
$KMnO_4$ and $K_2Cr_2O_7$ are strong oxidising agents in acidic medium.
$Cl_2$ is also a strong oxidising agent.
Therefore,the correct option is $A$.

(C) $KMnO_4$ will not be oxidized further by ozone $(O_3)$ because manganese is already present in its highest possible oxidation state,which is $+7$.

(B) $HNO_3$ is a strong oxidizing agent.
When $Zn$ reacts with $HNO_3$,the hydrogen gas $(H_2)$ initially produced is immediately oxidized by $HNO_3$ to water $(H_2O)$.
Consequently,$H_2$ gas is not evolved in the reaction.

(C) $Mn^{2+}$ is oxidized to $Mn^{7+}$ (in the form of $MnO_4^-$) by strong oxidizing agents.
$PbO_2$ acts as a strong oxidizing agent in acidic medium.
The reaction is: $2Mn^{2+} + 5PbO_2 + 4H^+ \rightarrow 2MnO_4^- + 5Pb^{2+} + 2H_2O$.

(C) Starch-iodide paper is used to detect the presence of oxidizing agents.
When an oxidizing agent comes in contact with the paper,it oxidizes the iodide ions $(I^-)$ present in the paper to iodine $(I_2)$.
The liberated iodine then reacts with the starch present in the paper to form a blue-black complex,indicating the presence of an oxidizing agent.

(B) Concentrated $HNO_3$ acts as a strong oxidizing agent.
$CuS$ (Copper sulfide) reacts with concentrated $HNO_3$ to form soluble copper$(II)$ nitrate,sulfur,and nitrogen dioxide gas.
The reaction is: $CuS + 8HNO_3 \rightarrow Cu(NO_3)_2 + S + 8NO_2 + 4H_2O$.
$BaSO_4$ is insoluble in acids.
$PbS$ and $HgS$ are generally insoluble in concentrated $HNO_3$ under standard conditions compared to $CuS$.

(B) substance containing an element in its lowest oxidation state acts as a strong reducing agent.
In $H_2S$,the oxidation state of sulfur is $-2$,which is its lowest possible oxidation state.
Therefore,$H_2S$ acts as a strong reducing agent.

(D) In the given reaction,$H_2SO_4$ acts as an acid.
Additionally,the oxidation state of $S$ changes from $+6$ to $+4$ in the reaction:
${H_2}\mathop S\limits^{+6}{O_4} \to \mathop S\limits^{+4}{O_2}$
Since the oxidation state of sulfur decreases,it undergoes reduction,meaning it acts as an oxidizing agent.
Therefore,it acts as both an acid and an oxidizing agent.