Oxidation and Reduction Questions in English

(A) $(i)$ In the reaction $4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$,Sodium $(Na)$ is oxidised because it gains oxygen to form $Na_2O$,and Oxygen $(O_2)$ is reduced.
$(ii)$ In the reaction $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$,Copper oxide $(CuO)$ is reduced to Copper $(Cu)$ because it loses oxygen,while Hydrogen $(H_2)$ is oxidised to Water $(H_2O)$ because it gains oxygen.

(A) In the given reaction: $2 PbO_{(s)} + C_{(s)} \longrightarrow 2 Pb_{(s)} + CO_{2(g)}$
$1$. Lead oxide $(PbO)$ loses oxygen to form lead $(Pb)$,so $PbO$ is being reduced.
$2$. Carbon $(C)$ gains oxygen to form carbon dioxide $(CO_2)$,so $C$ is being oxidised.
Evaluating the statements:
$(a)$ Lead is getting reduced: Incorrect. Lead oxide is being reduced,not lead.
$(b)$ Carbon dioxide is getting oxidised: Incorrect. Carbon is being oxidised to form carbon dioxide.
$(c)$ Carbon is getting oxidised: Correct.
$(d)$ Lead oxide is getting reduced: Correct.
Therefore,statements $(a)$ and $(b)$ are incorrect.

(N/A) $X$ is copper $(Cu)$ and the black-coloured compound formed is copper oxide $(CuO)$.
The chemical reaction that occurs when copper is heated in air is as follows:
$2Cu (s) + O_2 (g) \xrightarrow{\text{Heat}} 2CuO (s)$
In this reaction,copper reacts with oxygen to form copper$(II)$ oxide,which is black in colour.

(N/A) Reduction is defined as the loss of oxygen from a substance during a chemical reaction.
For example:
$(i)$ $\mathop {CO_2 + H_2 \to CO}\limits_{\text{Removal of oxygen - reduction}} + H_2O$
$(ii)$ $\mathop {CuO + H_2 \xrightarrow{\Delta } Cu}\limits_{\text{Loss of oxygen - reduction}} + H_2O$
In equation $(i)$, $CO_2$ is reduced to $CO$ due to the loss of oxygen, and in equation $(ii)$, $CuO$ is reduced to $Cu$ due to the loss of oxygen.

(D) Oxidising agents are substances that have the ability to add oxygen to other substances or remove hydrogen from them during a chemical reaction.
Examples include alkaline potassium permanganate $(KMnO_4)$ and acidified potassium dichromate $(K_2Cr_2O_7)$.
These substances facilitate the oxidation process by providing oxygen to the reactant.

(C) In the given reaction: $3 Fe (s) + 4 H_2 O (g) \rightarrow Fe_3 O_4 (s) + 4 H_2 (g)$
$(i)$ Iron $(Fe)$ gains oxygen to form $Fe_3 O_4$,so it is getting oxidised.
$(ii)$ Water $(H_2 O)$ loses oxygen to form $H_2$,so it is getting reduced.
$(iii)$ Since water provides oxygen to iron,it acts as an oxidising agent.
$(iv)$ Therefore,statement $(iv)$ is correct while statement $(iii)$ is incorrect.
Thus,statements $(i), (ii)$ and $(iv)$ are correct.

(B) The reaction between acidified potassium permanganate $(KMnO_4)$ and ferrous sulphate $(FeSO_4)$ is a classic redox reaction.
In this reaction,$KMnO_4$ acts as a strong oxidising agent.
It oxidises ferrous ions $(Fe^{2+})$ to ferric ions $(Fe^{3+})$.
During this process,the purple-coloured permanganate ions $(MnO_4^-)$ are reduced to colourless manganese ions $(Mn^{2+})$.
Therefore,the purple colour of the solution fades and eventually disappears as the reaction proceeds.

(N/A) In $4 NH_{3} + 5 O_{2} \longrightarrow 4 NO + 6 H_{2}O$,the oxidation state of nitrogen in $NH_{3}$ increases from $-3$ to $+2$. Thus,$NH_{3}$ is the reducing agent.
$(b)$ In $H_{2}O + F_{2} \longrightarrow HF + HOF$,the oxidation state of oxygen in $H_{2}O$ increases from $-2$ to $0$ (in $HOF$). Thus,$H_{2}O$ is the reducing agent.
$(c)$ In $Fe_{2}O_{3} + 3 CO \longrightarrow 2 Fe + 3 CO_{2}$,the oxidation state of carbon in $CO$ increases from $+2$ to $+4$. Thus,$CO$ is the reducing agent.
$(d)$ In $2 H_{2} + O_{2} \longrightarrow 2 H_{2}O$,the oxidation state of hydrogen in $H_{2}$ increases from $0$ to $+1$. Thus,$H_{2}$ is the reducing agent.
$A$ reducing agent is a substance that donates electrons or increases its own oxidation state during a chemical reaction.

(A) An oxidising agent is a substance that gains electrons or provides oxygen (or removes hydrogen) in a chemical reaction.
$(a) Pb_{3}O_{4}$: Lead is reduced from $+8/3$ to $+2$,so $Pb_{3}O_{4}$ acts as the oxidising agent.
$(b) O_{2}$: Oxygen is reduced from $0$ to $-2$,so $O_{2}$ acts as the oxidising agent.
$(c) CuSO_{4}$: Copper is reduced from $+2$ to $0$,so $CuSO_{4}$ acts as the oxidising agent.
$(d) V_{2}O_{5}$: Vanadium is reduced from $+5$ to $0$,so $V_{2}O_{5}$ acts as the oxidising agent.
$(e) H_{2}O$: Hydrogen is reduced from $+1$ to $0$,so $H_{2}O$ acts as the oxidising agent.
$(f) CuO$: Copper is reduced from $+2$ to $0$,so $CuO$ acts as the oxidising agent.

(N/A) No, aluminium is not getting reduced. Since oxygen is added to aluminium, it is getting oxidised to form aluminium oxide $(Al_2O_3)$.
$(b)$ No, $MnO_2$ is not getting oxidised. Since manganese has lost oxygen, $MnO_2$ is getting reduced to manganese $(Mn)$.

(N/A) The reduction process is widely used in metallurgy for the extraction of metals from their ores. For example,metal oxides are reduced to their respective pure metals by using reducing agents like carbon or hydrogen.

(N/A) In the given reaction: $CuO + H_{2} \longrightarrow Cu + H_{2}O$
$1$. The oxidising agent is $CuO$ because it provides oxygen to $H_{2}$ and gets reduced to $Cu$.
$2$. The reducing agent is $H_{2}$ because it removes oxygen from $CuO$ and gets oxidised to $H_{2}O$.

(A) When copper metal is heated in the presence of air (oxygen), it undergoes an oxidation reaction.
The copper reacts with oxygen to form copper $(II)$ oxide, which is represented by the chemical formula $CuO$.
This copper $(II)$ oxide forms a black-coloured coating on the surface of the copper metal.
The chemical equation for this reaction is: $2Cu(s) + O_2(g) \rightarrow 2CuO(s)$.

(N/A) No,oxidation or reduction cannot take place alone. This is because electrons lost during oxidation must be gained by another substance during reduction. Therefore,these processes always occur simultaneously. Such reactions are known as $Redox$ reactions.

(N/A) When copper is heated in air,it reacts with oxygen to form copper$(II)$ oxide,which is a black-colored substance.
Equation: $2Cu(s) + O_2(g) \xrightarrow{\text{Heat}} 2CuO(s)$
$(b)$ When the product obtained (copper$(II)$ oxide) is heated in hydrogen gas,it undergoes a reduction reaction to form copper metal and water.
Equation: $CuO(s) + H_2(g) \xrightarrow{\text{Heat}} Cu(s) + H_2O(l)$
The black coating turns brown as copper metal is formed.

(B) combustion reaction is classified as an oxidation reaction because it involves the addition of oxygen to a substance or the removal of hydrogen from it.
In most combustion processes,the fuel reacts with oxygen to form oxides,which is a fundamental characteristic of oxidation.
For example,the combustion of methane is represented as:
$CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)$
In this reaction,oxygen is added to carbon and hydrogen,confirming it as an oxidation process.

(N/A) $(i)$ The black coloured substance is formed because copper reacts with atmospheric oxygen upon heating to form copper $(II)$ oxide.
$(ii)$ This black substance is copper $(II)$ oxide $(CuO)$.
$(iii)$ The chemical equation for the reaction is: $2Cu(s) + O_2(g) \xrightarrow{\text{Heat}} 2CuO(s)$

(N/A) In the reaction $2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s)$:
$1$. Sodium $(Na)$ loses electrons to form sodium ions $(Na^+)$. The process of loss of electrons is called oxidation. Since sodium undergoes oxidation and facilitates the reduction of chlorine,it acts as a reducing agent.
$2$. Chlorine $(Cl_2)$ gains electrons to form chloride ions $(Cl^-)$. The process of gain of electrons is called reduction. Since chlorine undergoes reduction and facilitates the oxidation of sodium,it acts as an oxidising agent.

(A) In a redox reaction,the reducing agent is the substance that gets oxidized (loses electrons or gains oxygen/loses hydrogen).
In the given reaction: $MnO_{2}(s) + 4HCl(aq) \longrightarrow MnCl_{2}(s) + 2H_{2}O(l) + Cl_{2}(g)$
$1$. The oxidation state of $Cl$ in $HCl$ changes from $-1$ to $0$ in $Cl_{2}$. Since it loses electrons,$HCl$ is oxidized.
$2$. The oxidation state of $Mn$ in $MnO_{2}$ changes from $+4$ to $+2$ in $MnCl_{2}$. Since it gains electrons,$MnO_{2}$ is reduced.
$3$. Since $HCl$ undergoes oxidation,it acts as the reducing agent.
Therefore,the correct option is $A$.

(B) In the given reaction: $CuO(s) + H_2(g) \longrightarrow Cu(s) + H_2O(l)$
$1$. Copper oxide $(CuO)$ loses oxygen to form copper $(Cu)$. The loss of oxygen is defined as reduction.
$2$. Hydrogen $(H_2)$ gains oxygen to form water $(H_2O)$. The gain of oxygen is defined as oxidation.
$3$. An oxidising agent is a substance that provides oxygen or removes hydrogen,thereby undergoing reduction itself.
$4$. Since $CuO$ provides oxygen to $H_2$ and gets reduced to $Cu$,$CuO$ acts as the oxidising agent.
Therefore,the correct option is $B$.

(B) When copper powder is heated in a china dish, it reacts with atmospheric oxygen to form copper $(II)$ oxide $(CuO)$.
The chemical reaction is: $2Cu(s) + O_2(g) \xrightarrow{\Delta} 2CuO(s)$.
Copper $(II)$ oxide is a black-coloured compound.
Therefore, the surface of the copper powder turns black.

(C) When hydrogen gas is passed over heated copper oxide $(CuO)$, a redox reaction occurs.
In this reaction, hydrogen reduces copper oxide to copper metal.
The chemical equation for this reaction is: $CuO(s) + H_2(g) \xrightarrow{\Delta} Cu(s) + H_2O(g)$.
Copper oxide $(CuO)$ is black in color, while the copper metal $(Cu)$ formed is reddish-brown in color.
Therefore, the black coating on the surface turns brown.

(B) In the given reaction: $3 Fe (s) + 4 H_2 O (g) \longrightarrow Fe_3 O_4 (s) + 4 H_2 (g)$
$1$. Iron $(Fe)$ gains oxygen to form $Fe_3 O_4$,which means iron is getting oxidised.
$2$. Water $(H_2 O)$ loses oxygen to form $H_2$,which means water is getting reduced.
$3$. Since water provides oxygen for the oxidation of iron,it acts as an oxidising agent.
$4$. Therefore,statements $(i)$,$(ii)$,and $(iv)$ are correct.

(D) The reaction between acidified potassium permanganate $(KMnO_4)$ and ferrous sulphate $(FeSO_4)$ is a classic redox reaction.
In this reaction,$KMnO_4$ acts as a strong oxidising agent.
It oxidises the ferrous ions $(Fe^{2+})$ present in the $FeSO_4$ solution to ferric ions $(Fe^{3+})$.
During this process,the purple-coloured permanganate ions $(MnO_4^-)$ are reduced to colourless manganese ions $(Mn^{2+})$.
As the concentration of $MnO_4^-$ decreases,the purple colour of the solution fades and eventually disappears.

(C) Voltaic cell (also known as a Galvanic cell) is an electrochemical device that generates electrical energy from spontaneous chemical reactions occurring within it.
In this cell,redox reactions (oxidation and reduction) take place at the electrodes.
The chemical potential energy stored in the reactants is converted into electrical energy,which can then be used to power an external circuit.

(D) reducing agent is a substance that donates electrons or increases the oxidation state of another substance while being oxidized itself.
In the context of metallurgy and chemical reactions, metals like $Al$ (Aluminum) and $Fe$ (Iron) act as strong reducing agents.
However, $Al$ is a much stronger reducing agent than $Fe$, commonly used in the thermite reaction to reduce metal oxides (e.g., $Fe_{2}O_{3} + 2Al \rightarrow 2Fe + Al_{2}O_{3} + \text{Heat}$).
Therefore, $Al$ is the most appropriate answer among the given choices as it is widely recognized as a powerful reducing agent in chemical processes.

(A) In an electrolytic cell,the anode is the positive electrode. During electrolysis,negatively charged ions (anions) migrate towards the anode. At the anode,these anions lose electrons to form neutral atoms or molecules. The loss of electrons is defined as oxidation. Therefore,oxidation always occurs at the anode.

(D) In an electrolytic cell,the cathode is the negatively charged electrode.
During electrolysis,positively charged ions (cations) move towards the cathode.
At the cathode,these cations gain electrons to become neutral atoms or molecules.
The process of gaining electrons is defined as reduction.
Therefore,reduction always occurs at the cathode.

(A) When copper oxide $(CuO)$ is heated in the presence of dihydrogen gas $(H_2)$, a redox reaction occurs.
In this reaction, $CuO$ loses oxygen to form copper metal $(Cu)$, which is a reduction process.
Simultaneously, $H_2$ gains oxygen to form water $(H_2O)$, which is an oxidation process.
The chemical equation for this reaction is: $CuO(s) + H_2(g) \xrightarrow{\Delta} Cu(s) + H_2O(g)$.
Thus, copper metal is obtained as a product.

(A) When sulfur dioxide $(SO_2)$ gas is passed through an acidic potassium dichromate $(K_2Cr_2O_7)$ solution,it acts as a reducing agent.
The orange-colored dichromate ion $(Cr_2O_7^{2-})$ is reduced to the green-colored chromium$(III)$ ion $(Cr^{3+})$.
The chemical reaction is: $K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
Due to the formation of chromium$(III)$ sulfate,the solution turns green.

(D) The chemical reaction is: $SO_{2} + 2H_{2}S \rightarrow 3S + 2H_{2}O$.
In this reaction,$SO_{2}$ oxidizes $H_{2}S$ to $S$ and itself gets reduced to $S$.
Since $SO_{2}$ causes the oxidation of $H_{2}S$,it acts as an oxidizing agent.
Option $D$ states that it acts as a weak bleaching agent. While $SO_{2}$ is a bleaching agent,this property is not relevant to the chemical reaction described in the question. Therefore,the statement in option $D$ is incorrect in the context of this specific reaction.

(B) When sulfur dioxide $(SO_2)$ reacts with an acidic solution of potassium dichromate $(K_2Cr_2O_7)$,the sulfur dioxide acts as a reducing agent.
In this reaction,the oxidation state of sulfur in $SO_2$ increases from $+4$ to $+6$ (in $SO_4^{2-}$),meaning $SO_2$ undergoes oxidation.
Simultaneously,the chromium in $K_2Cr_2O_7$ is reduced from $+6$ to $+3$ $(Cr^{3+})$.
Therefore,the overall process involves the oxidation of sulfur dioxide by potassium dichromate.