Oxidation, Reduction Questions in English

$A$ redox couple is defined as having together the oxidized and reduced forms of a substance taking part in an oxidation or reduction half-reaction.
$(a) \ Cu^{+2}/Cu \ \text{and} \ Zn^{+2}/Zn$
$(b) \ Mg^{+2}/Mg \ \text{and} \ Fe^{+2}/Fe$
$(c) \ Br_{2}/Br^{-} \ \text{and} \ Cl_{2}/Cl^{-}$
$(d) \ Fe^{+2}/Fe \ \text{and} \ Cd^{+2}/Cd$

(D) In the given reaction: $SO_2 + 2H_2S \longrightarrow 3S + 2H_2O$
$1$. Oxidation is defined as the loss of electrons or the addition of oxygen/removal of hydrogen.
$2$. In $H_2S$,the oxidation state of sulfur is $-2$. In the product $S$,the oxidation state of sulfur is $0$.
$3$. Since the oxidation state of sulfur increases from $-2$ to $0$,$H_2S$ is losing electrons and is therefore being oxidised to $S$.
$4$. Conversely,in $SO_2$,the oxidation state of sulfur is $+4$,which decreases to $0$ in $S$,meaning $SO_2$ is reduced.
Thus,the substance that is oxidised is $H_2S$.

(A) To determine the number of electrons transferred,we calculate the change in the oxidation state of the central atom:
$1$. In $MnO_4^{-}$,the oxidation state of $Mn$ is $+7$. In $Mn^{2+}$,it is $+2$. The change is $7 - 2 = 5$ electrons.
$2$. In $CrO_4^{2-}$,the oxidation state of $Cr$ is $+6$. In $Cr^{3+}$,it is $+3$. The change is $6 - 3 = 3$ electrons.
$3$. In $NO_3^{-}$,the oxidation state of $N$ is $+5$. In $NH_4^{+}$,it is $-3$. The change is $5 - (-3) = 8$ electrons.
$4$. In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$. In $2Cr^{3+}$,the total charge change for two $Cr$ atoms is $2 \times (6 - 3) = 6$ electrons.
Thus,the change involving the transfer of $5$ electrons is $MnO_4^{-} \longrightarrow Mn^{2+}$.

(C) Metals,such as $Mg$,$Ni$,and $Cu$,generally have a low ionization energy and exhibit a tendency to lose electrons,which is known as oxidation.
Non-metals,such as $O$,have a high electronegativity and a strong tendency to gain electrons,which is known as reduction.

(C) Reduction is defined as the process that involves the gain of electrons by an atom,ion,or molecule.
It is also characterized by a decrease in the oxidation number of an element.
Conversely,the gain of oxygen is an oxidation process,and the loss of electrons is also an oxidation process.

(C) In the given reaction: $Ag_2O + H_2O + 2e^- \rightarrow 2Ag + 2OH^-$
$1$. The oxidation state of $Ag$ in $Ag_2O$ is $+1$,and it changes to $0$ in $Ag$ metal.
$2$. Since the oxidation number of $Ag$ decreases from $+1$ to $0$,it undergoes reduction.
$3$. Therefore,$Ag$ is reduced.
$4$. The oxidation states of $H$ $(+1)$ and $O$ $(-2)$ remain unchanged in $H_2O$ and $OH^-$,so neither oxidation nor reduction occurs for these elements.

(A) In the given redox reaction,the oxidation state of $Cr$ changes from $+6$ in $Cr_2O_7^{2-}$ to $+3$ in $Cr^{3+}$.
Since the oxidation state of $Cr$ decreases,$Cr$ is reduced.
The oxidation state of $I$ changes from $-1$ in $I^{-}$ to $0$ in $I_2$.
Since the oxidation state of $I$ increases,$I$ is oxidized.
Therefore,the element reduced is $Cr$.

(B) In the reaction $3 \ Mg + N_2 \longrightarrow Mg_3N_2$,the oxidation state of $Mg$ changes from $0$ (in elemental form) to $+2$ (in $Mg_3N_2$).
Since there is an increase in the oxidation state of $Mg$,it undergoes oxidation.
Similarly,the oxidation state of $N$ changes from $0$ (in $N_2$) to $-3$ (in $Mg_3N_2$),which means $N_2$ undergoes reduction.
Therefore,$Mg$ is oxidised.

(D) Oxidation is defined as the addition of oxygen,the addition of an electronegative element,or the removal of hydrogen.
Removal of an electronegative element is actually a characteristic of reduction.
Therefore,option $(D)$ is not true for oxidation.

(C) In reaction $I$,the oxidation state of $H$ in $H_2O$ changes from $+1$ to $0$ (in $H_2$),which is a decrease in oxidation number,hence water is reduced.
In reaction $II$,the oxidation state of $O$ in $H_2O$ changes from $-2$ to $0$ (in $O_2$),which is an increase in oxidation number,hence water is oxidized.
Therefore,in reaction $I$ water is reduced and in reaction $II$ water is oxidized.

(B) Assign oxidation states to each element:
In $KClO_3$,the oxidation state of $K$ is $+1$,$Cl$ is $+5$,and $O$ is $-2$.
In $KCl$,the oxidation state of $K$ is $+1$ and $Cl$ is $-1$.
In $O_2$,the oxidation state of $O$ is $0$.
Comparing the oxidation states:
$Cl$ changes from $+5$ to $-1$ (decrease in oxidation state,so $Cl$ is reduced).
$O$ changes from $-2$ to $0$ (increase in oxidation state,so $O$ is oxidized).
Therefore,$Cl$ is reduced and $O$ is oxidized.

(C) The reduction of $O_2$ by four electrons is represented by the equation: $O_2 + 4e^- \rightarrow 2O^{2-}$.
Each oxygen atom gains $2e^-$ to form an oxide ion $(O^{2-})$.

(A) In an aqueous alkaline medium,the reduction of the hydroperoxide ion $(HO_2^-)$ by two electrons produces hydroxide ions $(OH^{-})$.
The balanced half-reaction is: $HO_2^- + H_2O + 2e^- \rightarrow 3OH^{-}$.