Oxidation, Reduction Questions in English

(D) Combustion is a chemical process in which a substance reacts rapidly with an oxidant (usually oxygen) and gives off heat.
$(a)$ Oxidation of coal in air is a combustion reaction.
$(b)$ Burning of magnesium in nitrogen is a combustion reaction (forming $Mg_3N_2$).
$(c)$ Reaction of antimony in chlorine is a combustion reaction (forming $SbCl_3$).
$(d)$ Lighting of an electric lamp is a physical process involving the heating of a filament,not a chemical combustion reaction.

(A) In the given reaction: $H_2S + NO_2 \to H_2O + NO + S$
$1$. The oxidation state of sulfur $(S)$ in $H_2S$ is $-2$.
$2$. The oxidation state of elemental sulfur $(S)$ on the product side is $0$.
$3$. Since the oxidation state of sulfur increases from $-2$ to $0$,$H_2S$ undergoes oxidation.
$4$. Therefore,$H_2S$ is oxidised.

(B) In the compound $PbO_2$,the oxidation state of $Pb$ is $+4$.
In the compound $Pb(NO_3)_2$,the oxidation state of $Pb$ is $+2$.
Since the oxidation state of $Pb$ decreases from $+4$ to $+2$,the process is a reduction.

(C) The reaction between $H_2S$ and halogens ($X_2$,where $X = Cl, Br, I$) is given by the equation: $H_2S + X_2 \to 2HX + S$.
In this reaction,the oxidation state of the halogen decreases from $0$ in $X_2$ to $-1$ in $HX$.
Therefore,the halogens are reduced.

(B) In the given reaction,the oxidation state of $Mg$ changes from $0$ to $+2$,which represents an increase in oxidation state,meaning $Mg$ is oxidized.
Conversely,the oxidation state of $N$ changes from $0$ to $-3$,which represents a decrease in oxidation state,meaning $N$ is reduced.
Therefore,Magnesium is oxidized and Nitrogen is reduced.

(A) The chemical reaction is: $SnCl_2 + 2HgCl_2 \rightarrow SnCl_4 + Hg_2Cl_2(s)$.
In this reaction,the oxidation state of $Hg$ changes from $+2$ in $HgCl_2$ to $+1$ in $Hg_2Cl_2$.
Since the oxidation state of $Hg$ decreases,$HgCl_2$ undergoes reduction.
Therefore,the correct option is $(A)$.

(A) . Oxidation is defined as the process in which an atom,molecule,or ion loses one or more electrons. This is also known as de-electronation.

(C) The reaction between copper and silver nitrate is a displacement reaction: $Cu(s) + 2AgNO_3(aq) \to Cu(NO_3)_2(aq) + 2Ag(s)$.
In this reaction,$Cu$ is oxidized to $Cu^{2+}$ ions,which impart a blue colour to the solution.
$Cu$ is placed above $Ag$ in the electrochemical series,making it a stronger reducing agent.

(A) The reaction is represented as: $Sn^{2+} \to Sn^{4+} + 2e^-$.
Since the oxidation state of $Sn$ increases from $+2$ to $+4$,it indicates the loss of two electrons.
This process is known as an oxidation reaction.

(B) In the given reaction,the oxidation state of $Zn$ decreases from $+2$ to $0$.
Since there is a gain of electrons,the process is defined as reduction.

(B) Reduction is defined as the process in which an atom,ion,or molecule gains electrons. This process is also known as electronation.

(C) The chemical equation for the reaction is: $Zn(s) + I_2(s) \rightarrow ZnI_2(s)$.
In this reaction,the oxidation state of $Zn$ increases from $0$ to $+2$,which means the $Zn$ atom is oxidised.
The oxidation state of $I$ decreases from $0$ to $-1$,which means $I_2$ is reduced.
Therefore,the $Zn$ atom is being oxidised.

(A) To determine which element is reduced,we calculate the oxidation numbers of the elements in the reaction.
In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$.
In $2Cr^{3+}$,the oxidation state of $Cr$ is $+3$.
Since the oxidation state of $Cr$ decreases from $+6$ to $+3$,it undergoes reduction.
Therefore,$Cr$ is the element that is reduced.

(A) The chemical reaction for the combustion of sugar is: $C_{12}H_{22}O_{11} + 12O_2 \to 12CO_2 + 11H_2O$.
In this reaction,the oxidation state of carbon in $C_{12}H_{22}O_{11}$ is $0$ (average),while in $CO_2$,it is $+4$.
Since the oxidation state of carbon increases from $0$ to $+4$,the process is an oxidation reaction.

(A) The change from $Fe^{2+}$ to $Fe^{3+}$ involves the loss of one electron.
This process is known as oxidation.
The half-reaction is: $Fe^{2+} \to Fe^{3+} + e^-$.
Therefore,the correct option is $A$.

(D) The oxidation state of $Mn$ in $MnO_4^-$ is $+7$ and in $Mn^{2+}$ is $+2$.
The change in oxidation state is $7 - 2 = 5$.
Since the oxidation state decreases,it is a reduction process involving $5$ electrons.
The balanced half-reaction is: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$.

(B) The chemical reaction is: $Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$.
In this reaction,the oxidation state of $Cu$ changes from $+2$ in $CuSO_4$ to $0$ in $Cu$.
Since the oxidation state decreases,$Cu^{2+}$ undergoes reduction.

(A) The increase in the oxidation state of an atom during a chemical reaction is known as oxidation.
For example,in the reaction $Cu \rightarrow Cu^{2+} + 2e^-$,the oxidation number of copper increases from $0$ to $+2$.

(B) In an oxidation process,the oxidation number of an element always increases due to the loss of electrons.

(A) The formation of a sulphide ion $(S^{2-})$ from a sulphur atom or molecule involves the gain of electrons to complete its octet. The process is represented as: $S + 2e^- \to S^{2-}$. Therefore,the correct option is $A$.

(C) The ease of oxidation of halide ions to their corresponding dihalogens depends on their standard oxidation potentials.
The standard oxidation potential values are:
$2I^{-} \rightarrow I_2 + 2e^{-}$ $(E^o_{ox} = -0.54 \ V)$
$2Br^{-} \rightarrow Br_2 + 2e^{-}$ $(E^o_{ox} = -1.07 \ V)$
$2Cl^{-} \rightarrow Cl_2 + 2e^{-}$ $(E^o_{ox} = -1.36 \ V)$
$A$ higher (less negative) oxidation potential indicates a greater tendency to undergo oxidation.
Therefore,the order of ease of oxidation is $I^{-} > Br^{-} > Cl^{-}$.

(A) The oxidation state of carbon increases as we move from hydrocarbons to alcohols,aldehydes,carboxylic acids,and finally to carbon dioxide.
$CO_2$ represents the state where carbon is in its highest oxidation state $(+4)$ resulting from the complete combustion of hydrocarbons.
Therefore,$CO_2$ is the most oxidized form.
$CH_3-CH_3 + 3.5 O_2 \to 2 CO_2 + 3 H_2O$

(B) When $Cu$ is added to a solution of $AgNO_3$,a displacement reaction occurs: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.
$Cu$ undergoes oxidation to form $Cu^{2+}$ ions.
The presence of $Cu^{2+}$ ions in the aqueous solution imparts a blue color to it.

(B) In the given reaction,the oxidation state of $Mn$ in $KMnO_4$ is $+7$.
After the reaction,$Mn$ is present in $MnCl_2$ with an oxidation state of $+2$.
Since the oxidation state of $Mn$ decreases from $+7$ to $+2$,$KMnO_4$ undergoes reduction.
Therefore,$MnCl_2$ is the reduction product.

(A) $Cu$ is placed above $Ag$ in the electrochemical series,hence it can displace $Ag$ from its salt solution.
The reaction is: $Cu(s) + 2AgNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2Ag(s)$.
In this reaction,$Cu$ loses electrons to form $Cu^{2+}$ ions,which is an oxidation process $(Cu^0 \rightarrow Cu^{2+} + 2e^-)$.

(A) . Due to the reduction of copper ions.
$Zn(s) + CuSO_4(aq) \to Cu(s) + ZnSO_4(aq)$
In this displacement reaction,$Zn$ acts as a reducing agent and reduces $Cu^{2+}$ ions to metallic $Cu$.

(A) The reaction is: $Fe(s) + CuSO_4(aq) \to FeSO_4(aq) + Cu(s)$.
In this reaction,$Fe$ is oxidized to $Fe^{+2}$ and $Cu^{+2}$ is reduced to $Cu(s)$.
Therefore,the precipitation of $Cu$ occurs due to the reduction of $Cu^{+2}$ ions.

(C) In the given reaction: $\mathop {Ag}\limits^{ + 1} _2O + H_2O + 2e^- \to 2\mathop {Ag}\limits^0 + 2OH^-$
The oxidation number of $Ag$ decreases from $+1$ in $Ag_2O$ to $0$ in $Ag$.
Since the oxidation number decreases,$Ag$ undergoes reduction.

(A) In the reduction of the permanganate ion $(MnO_4^-)$ to the manganese$(II)$ ion $(Mn^{2+})$,the oxidation state of $Mn$ changes from $+7$ to $+2$.
Change in oxidation state = $|(+2) - (+7)| = 5$.
Therefore,the reaction involves the gain of $5$ electrons:
$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$.

(A) In an acidic medium,the reduction half-reaction for $KMnO_4$ is:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Here,the oxidation state of $Mn$ changes from $+7$ in $MnO_4^-$ to $+2$ in $Mn^{2+}$.
Therefore,the change in the oxidation state of $Mn$ is from $+7$ to $+2$.

(A) In the reaction $3CuO + 2\overset{-3}{N}H_3 \to \overset{0}{N_2} + 3H_2O + 3Cu$,the oxidation state of nitrogen changes from $-3$ in $NH_3$ to $0$ in $N_2$.
For each nitrogen atom,$3$ electrons are lost $(0 - (-3) = 3)$.
Since $N_2$ contains $2$ nitrogen atoms,the total number of electrons lost per mole of $N_2$ is $2 \times 3 = 6$ electrons.

(C) Reduction is defined as the gain of electrons or a decrease in the oxidation state of an atom.
Therefore,when an atom undergoes reduction,its oxidation number decreases.

(A) The reaction between sodium $(Na)$ and water $(H_2O)$ is: $2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$.
In this reaction,the oxidation state of $Na$ increases from $0$ to $+1$ (it loses electrons,so it is oxidized).
The oxidation state of $H$ in $H_2O$ decreases from $+1$ to $0$ (it gains electrons,so it is reduced).
Since $Na$ loses electrons,it acts as a reducing agent,and the water molecules are reduced to produce $H_2$ gas.

(A) The reaction is $Cu(s) + 2AgNO_3(aq) \longrightarrow Cu(NO_3)_2(aq) + 2Ag(s)$.
In this reaction,the oxidation state of $Cu$ increases from $0$ to $+2$,which means $Cu$ undergoes oxidation.
The oxidation state of $Ag$ decreases from $+1$ to $0$,which means $Ag^+$ undergoes reduction.

(C) The reaction between acidified potassium permanganate $(KMnO_4)$ and ferrous ammonium sulphate $(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O)$ involves the oxidation of ferrous ions to ferric ions.
The balanced ionic equation is: $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$.
Here,$Fe^{2+}$ is oxidised to $Fe^{3+}$ while $MnO_4^-$ is reduced to $Mn^{2+}$.
Therefore,the correct option is $C$.

(D) To determine if nitrogen undergoes oxidation,we calculate the change in the oxidation state of nitrogen in each reaction:
$A) N_2 \to HN_3$: The oxidation state of $N$ changes from $0$ to $-1/3$. This is a reduction.
$B) N_2O_4 \to 2NO_2$: The oxidation state of $N$ remains $+4$ in both $N_2O_4$ and $NO_2$. No change.
$C) NO_3^- \to N_2O_5$: The oxidation state of $N$ remains $+5$ in both $NO_3^-$ and $N_2O_5$. No change.
$D) N_2O \to NO$: The oxidation state of $N$ changes from $+1$ in $N_2O$ to $+2$ in $NO$. Since the oxidation state increases,this process represents oxidation.

(C) The given reaction is: $8 \ Al + 3 \ Fe_3O_4 \to 4 \ Al_2O_3 + 9 \ Fe$
In this reaction,$Al$ is the reducing agent (reductant) as its oxidation state changes from $0$ to $+3$.
The change in oxidation number for one atom of $Al$ is $|3 - 0| = 3$.
Therefore,the number of moles of electrons transferred by $1 \ mol$ of $Al$ is $3$.

(D) In the process $NO_2^- \to NO_3^-$,the oxidation state of nitrogen changes from $+3$ to $+5$.
Since the oxidation state increases,this process represents oxidation.
$\overset{+3}{N}O_{2}^{-} \to \overset{+5}{N}O_{3}^{-}$

(B) The chemical reaction is: $2NH_3 + 3CuO \to N_2 + 3H_2O + 3Cu$
In this reaction,the oxidation state of nitrogen changes from $-3$ in $NH_3$ to $0$ in $N_2$.
The change in oxidation state per nitrogen atom is $0 - (-3) = +3$.
Therefore,each nitrogen atom in the ammonia molecule loses $3$ electrons.

(D) To determine the change in oxidation number,we calculate the oxidation state of nitrogen in each reaction:
$A) 2NO_2 (+4) \longrightarrow N_2O_4 (+4)$. No change.
$B) NH_3 (-3) + H_2O \longrightarrow NH_4^+ (-3) + OH^-$. No change.
$C) N_2O_5 (+5) + H_2O \longrightarrow 2HNO_3 (+5)$. No change.
$D) N_2 (0) + 3H_2 \longrightarrow 2NH_3 (-3)$. The oxidation number of nitrogen changes from $0$ to $-3$.

(B) The reaction between ammonia $(NH_3)$ and cupric oxide $(CuO)$ is given by: $2NH_3 + 3CuO \rightarrow N_2 + 3Cu + 3H_2O$.
In this reaction,the oxidation state of nitrogen changes from $-3$ in $NH_3$ to $0$ in $N_2$.
For one molecule of $NH_3$,the nitrogen atom loses $3$ electrons to go from an oxidation state of $-3$ to $0$.
Thus,each molecule of ammonia loses $3$ electrons.

(A) In the given reaction $H_2S + Cl_2 \to S + 2HCl$,the oxidation state of sulfur $(S)$ in $H_2S$ changes from $-2$ to $0$ in $S$.
Since there is an increase in the oxidation number of sulfur,$H_2S$ undergoes oxidation.
Conversely,the oxidation state of chlorine $(Cl)$ changes from $0$ in $Cl_2$ to $-1$ in $HCl$,indicating that $Cl_2$ undergoes reduction.

(A) To determine if oxidation occurs,we calculate the oxidation state of nitrogen in each reaction:
$1$. In $NH_4^+ \to N_2$: The oxidation state of $N$ changes from $-3$ to $0$. Since the oxidation state increases,this is an oxidation reaction.
$2$. In $NO_3^- \to NO$: The oxidation state of $N$ changes from $+5$ to $+2$. This is a reduction.
$3$. In $NO_2 \to NO_2^-$: The oxidation state of $N$ changes from $+4$ to $+3$. This is a reduction.
$4$. In $NO_3^- \to NH_4^+$: The oxidation state of $N$ changes from $+5$ to $-3$. This is a reduction.
Therefore,the correct option is $A$.

(N/A) $(i)$ $H_{2}S$ is oxidised because a more electronegative element,chlorine,is added to hydrogen (or a more electropositive element,hydrogen,has been removed from $S$). Chlorine is reduced due to the addition of hydrogen to it.
$(ii)$ Aluminium is oxidised because oxygen is added to it. Ferrous ferric oxide $(Fe_{3}O_{4})$ is reduced because oxygen has been removed from it.
$(iii)$ With the careful application of the concept of electronegativity,we may infer that sodium is oxidised and hydrogen is reduced.

An oxidation reaction is defined as a chemical process in which oxygen or an electronegative element is added to a substance,or hydrogen or an electropositive element is removed from a substance.
Examples:
$1$. Addition of oxygen:
$2 Mg_{(s)} + O_{2(g)} \rightarrow 2 MgO_{(s)}$
$S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
In these reactions,$Mg$ and $S$ are oxidized due to the addition of oxygen.
$2$. Addition of an electronegative element:
$Mg_{(s)} + F_{2(g)} \rightarrow MgF_{2(s)}$
$Mg_{(s)} + Cl_{2(g)} \rightarrow MgCl_{2(s)}$
$3$. Removal of an electropositive element:
$2 K_{4}[Fe(CN)_{6}]_{(aq)} + H_{2}O_{2(aq)} \rightarrow 2 K_{3}[Fe(CN)_{6}]_{(aq)} + 2 KOH_{(aq)}$
In this reaction,the removal of the electropositive element $K$ from potassium ferrocyanide is an oxidation process.

Reduction is defined as the removal of oxygen or an electronegative element from a substance,or the addition of hydrogen or an electropositive element to a substance.
Examples:
$1. \text{ Removal of oxygen: } 2HgO_{(s)} \rightarrow 2Hg_{(l)} + O_{2(g)}$
$2. \text{ Removal of electronegative element: } 2FeCl_{3(aq)} + H_{2(g)} \rightarrow 2FeCl_{2(aq)} + 2HCl_{(aq)}$
$3. \text{ Addition of hydrogen: } CH_{2}=CH_{2(g)} + H_{2(g)} \rightarrow CH_{3}-CH_{3(g)}$
$4. \text{ Addition of electropositive element: } 2HgCl_{2(aq)} + SnCl_{2(aq)} \rightarrow Hg_{2}Cl_{2(s)} + SnCl_{4(aq)}$

(N/A) The reaction of iron with $HCl$ produces hydrogen gas $(H_2)$:
$Fe + 2HCl \rightarrow FeCl_2 + H_2$
The liberation of hydrogen gas creates a reducing atmosphere,which prevents the oxidation of ferrous chloride $(FeCl_2)$ to ferric chloride $(FeCl_3)$.

(A) The primary function of $O_2$ intake through respiration is to facilitate the cellular respiration process.
In this process,$O_2$ acts as an oxidizing agent to break down glucose $(C_6H_{12}O_6)$ into $CO_2$ and $H_2O$,releasing energy in the form of $ATP$.
The chemical equation is: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}$.

(N/A) redox couple is defined as having together the oxidised and reduced form of a substance taking part in an oxidation or reduction half-reaction. It is represented as $\text{Oxidised form} / \text{Reduced form}$.

(N/A) Redox reactions can be defined as reactions involving the transfer of electrons from one species to another.
Consider the following reactions:
$(1)$ $2 Na_{(s)} + Cl_{2_{(g)}} \rightarrow 2 NaCl_{(s)}$
$(2)$ $4 Na_{(s)} + O_{2_{(g)}} \rightarrow 2 Na_{2}O_{(s)}$
$(3)$ $2 Na_{(s)} + S_{(s)} \rightarrow Na_{2}S_{(s)}$
In these reactions,$Na$ loses electrons to form $Na^{+}$ ions,undergoing oxidation. The non-metals $(Cl_2, O_2, S)$ gain these electrons to form negative ions,undergoing reduction.
For example,in reaction $(1)$:
$2 Na_{(s)} \rightarrow 2 Na^{+} + 2 e^{-}$ (Oxidation)
$Cl_{2_{(g)}} + 2 e^{-} \rightarrow 2 Cl^{-}$ (Reduction)
Overall,$Na$ acts as a reducing agent (electron donor) and the non-metal acts as an oxidizing agent (electron acceptor).