Redox reaction and Method for balancing Redox reaction Questions in English

(D) The reaction is: $MnSO_4(aq) + Sr(NO_3)_2(aq) \longrightarrow SrSO_4(s) \downarrow + Mn(NO_3)_2(aq)$.
$SrSO_4$ (Strontium sulfate) is a white crystalline solid that precipitates out of the solution.
Therefore,the correct assignment for the formation of $SrSO_4$ precipitate is for white ppt.

(A) . $2Cu(NO_3)_2 + 4KCN \to 4KNO_3 + 2CuCN + (CN)_2 \uparrow$ (Redox reaction,cyanogen gas is released).
$B$. $AgNO_3 + KCN \to KNO_3 + AgCN \downarrow$ (Precipitate formation).
$C$. $Cd(NO_3)_2 + 2KCN \to 2KNO_3 + Cd(CN)_2 \downarrow$ (Precipitate formation).
$D$. $Pb(NO_3)_2 + 2KCN \to 2KNO_3 + Pb(CN)_2 \downarrow$ (Precipitate formation).

(A) The balanced redox reaction is obtained by balancing the half-reactions:
Reduction half: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Oxidation half: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
To balance the electrons,multiply the reduction half by $2$ and the oxidation half by $5$:
$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$
$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$
Adding these gives the balanced equation:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
The coefficients for the reactants $MnO_4^-$,$C_2O_4^{2-}$,and $H^+$ are $2$,$5$,and $16$ respectively.

(A) To balance the redox reaction $aIO_3^- + bI^- + 6H^+ \to cI_2 + 3H_2O$:
$1$. Identify the oxidation states: $I$ in $IO_3^-$ is $+5$,$I$ in $I^-$ is $-1$,and $I$ in $I_2$ is $0$.
$2$. Oxidation half-reaction: $2I^- \to I_2 + 2e^-$.
$3$. Reduction half-reaction: $IO_3^- + 6H^+ + 5e^- \to \frac{1}{2}I_2 + 3H_2O$.
$4$. To balance electrons,multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$:
$10I^- \to 5I_2 + 10e^-$
$2IO_3^- + 12H^+ + 10e^- \to I_2 + 6H_2O$
$5$. Adding these gives: $2IO_3^- + 10I^- + 12H^+ \to 6I_2 + 6H_2O$.
$6$. Dividing by $2$ to simplify: $1IO_3^- + 5I^- + 6H^+ \to 3I_2 + 3H_2O$.
Thus,$a = 1$,$b = 5$,and $c = 3$.

(B) The reduction half-reaction for the conversion of nitrate ion $(NO_{3}^{-})$ to hydrazine $(N_{2}H_{4})$ is determined by balancing the oxidation states.
In $NO_{3}^{-}$,the oxidation state of $N$ is $+5$.
In $N_{2}H_{4}$,the oxidation state of $N$ is $-2$.
The change in oxidation state per nitrogen atom is $5 - (-2) = 7$.
Therefore,for one mole of $NO_{3}^{-}$ to be reduced to half a mole of $N_{2}H_{4}$ (which contains one mole of $N$ atoms),$7$ moles of electrons are required.
$NO_{3}^{-} + 7e^{-} + 6H^{+} \longrightarrow \frac{1}{2} N_{2}H_{4} + 3H_{2}O$.

(B) The balanced chemical equation for the reaction between $Na_2SO_3$ and $KMnO_4$ in an acidic medium is:
$5 Na_2SO_3 + 3 H_2SO_4 + 2 KMnO_4 \rightarrow 5 Na_2SO_4 + 2 MnSO_4 + K_2SO_4 + 3 H_2O$
From the stoichiometry of the balanced equation,$2 \text{ moles of } KMnO_4$ react with $5 \text{ moles of } Na_2SO_3$.
Therefore,$1 \text{ mole of } KMnO_4$ will react with $\frac{5}{2} = 2.5 \text{ moles of } Na_2SO_3$.

(D) Thermal decomposition of most metal carbonates (like $BeCO_3$,$Na_2CO_3$,$ZnCO_3$) results in the formation of metal oxide and $CO_2$,where the oxidation state of the metal remains unchanged.
However,for silver carbonate $(Ag_2CO_3)$,the decomposition reaction is:
$2Ag_2CO_3(s) \rightarrow 4Ag(s) + 2CO_2(g) + O_2(g)$
In this reaction,the oxidation state of $Ag$ changes from $+1$ in $Ag_2CO_3$ to $0$ in metallic $Ag$.

(B) The chemical reaction for the conversion of $I_2$ to iodate ion $(IO_3^-)$ in an alkaline medium is:
$I_2 + 12OH^- \to 2IO_3^- + 6H_2O + 10e^-$
In this reaction,the oxidation state of iodine changes from $0$ in $I_2$ to $+5$ in $IO_3^-$.
Since there are two iodine atoms in $I_2$,the total change in oxidation number is $2 \times (5 - 0) = 10$.
Thus,the $n$-factor is $10$.
The equivalent weight $(E)$ is calculated as:
$E = \frac{\text{Molar mass of } I_2}{n\text{-factor}} = \frac{254}{10} = 25.4$.

(B) The given reaction is a comproportionation reaction where $X$ is oxidized from $-1$ to $0$ and reduced from $+5$ to $0$.
Step $1$: Write the half-reactions:
Oxidation: $X^- \longrightarrow X_2 + e^-$
Reduction: $XO_3^- + 6H^+ + 5e^- \longrightarrow X_2 + 3H_2O$
Step $2$: Balance the electrons by multiplying the oxidation half-reaction by $5$:
$5X^- \longrightarrow 2.5X_2 + 5e^-$
Step $3$: Add the two half-reactions:
$5X^- + XO_3^- + 6H^+ \longrightarrow 3X_2 + 3H_2O$
From the balanced equation,the molar ratio of $X^-$ to $XO_3^-$ is $5 : 1$.

(A) In the reaction,$K_2Cr_2O_7$ acts as an oxidizing agent and $X^{n+}$ acts as a reducing agent.
The change in oxidation state of $Cr$ in $K_2Cr_2O_7$ is from $+6$ to $+3$. The change per $Cr$ atom is $3$,and since there are $2$ $Cr$ atoms,the total change is $2 \times 3 = 6$.
Total electrons gained by $K_2Cr_2O_7 = 6 \times 10^{-3} \times 6 = 36 \times 10^{-3} \text{ mol}$.
In $XO_3^-$,the oxidation state of $X$ is $+5$. The change in oxidation state of $X$ is $(5 - n)$.
Total electrons lost by $X^{n+} = 9 \times 10^{-3} \times (5 - n)$.
Equating electrons gained and lost: $36 \times 10^{-3} = 9 \times 10^{-3} \times (5 - n)$.
$4 = 5 - n$,which gives $n = 1$.

(A) The unbalanced equation is: $H^{+} + Cr_2O_7^{2-} + SO_3^{2-} \to Cr^{3+} + SO_4^{2-} + H_2O$.
Step $1$: Identify oxidation and reduction half-reactions.
Reduction: $Cr_2O_7^{2-} \to Cr^{3+}$.
Oxidation: $SO_3^{2-} \to SO_4^{2-}$.
Step $2$: Balance atoms and charges.
Reduction half: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \to 2Cr^{3+} + 7H_2O$.
Oxidation half: $SO_3^{2-} + H_2O \to SO_4^{2-} + 2H^{+} + 2e^{-}$.
Step $3$: Multiply oxidation half by $3$ to balance electrons: $3SO_3^{2-} + 3H_2O \to 3SO_4^{2-} + 6H^{+} + 6e^{-}$.
Step $4$: Add the two half-reactions: $Cr_2O_7^{2-} + 14H^{+} + 3SO_3^{2-} + 3H_2O \to 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 6H^{+}$.
Step $5$: Simplify: $Cr_2O_7^{2-} + 8H^{+} + 3SO_3^{2-} \to 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$.
The coefficient of $H^{+}$ is $8$ and the coefficient of $SO_4^{2-}$ is $3$.

(D) When $SO_2$ gas is passed through an acidic solution of potassium dichromate $(K_2Cr_2O_7)$,the orange-colored dichromate ion $(Cr_2O_7^{2-})$ is reduced to green-colored chromium$(III)$ sulfate $(Cr_2(SO_4)_3)$.
The chemical reaction is:
$K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
In this reaction,$Cr^{6+}$ is reduced to $Cr^{3+}$,which imparts a green color to the solution.

(B) $SO_2$ acts as a reducing agent and reduces bromine water ($Br_2$ in $H_2O$) to hydrobromic acid $(HBr)$,while $SO_2$ itself gets oxidized to sulfuric acid $(H_2SO_4)$.
The balanced chemical equation is:
$SO_2 + Br_2 + 2H_2O \rightarrow H_2SO_4 + 2HBr$

(D) In the reaction $2KClO_3 \to 2KCl + 3O_2$:
$1$. The oxidation state of $Cl$ in $KClO_3$ is $+5$ and in $KCl$ is $-1$. Since the oxidation state decreases,$Cl$ undergoes reduction. Thus,statement $(2)$ is True.
$2$. The oxidation state of $O$ in $KClO_3$ is $-2$ and in $O_2$ is $0$. Since the oxidation state increases,$O$ undergoes oxidation. Thus,statement $(3)$ is False.
$3$. Since both oxidation and reduction occur in the same reaction,it is a redox reaction. Thus,statement $(4)$ is False.
$4$. Since $KClO_3$ contains both the element being oxidized $(O)$ and the element being reduced $(Cl)$,it undergoes both oxidation and reduction. Thus,statement $(1)$ is True.
Therefore,the sequence of truth values is $(1)=T, (2)=T, (3)=F, (4)=F$,which corresponds to $TTFF$.

(D) redox reaction is one in which both oxidation and reduction occur simultaneously,involving a change in oxidation states of the elements.
In option $D$,the reaction is $H_2S + 3H_2SO_4 \to 4SO_2 + 4H_2O$.
Here,the oxidation state of $S$ in $H_2S$ changes from $-2$ to $+4$ in $SO_2$ (oxidation).
The oxidation state of $S$ in $H_2SO_4$ changes from $+6$ to $+4$ in $SO_2$ (reduction).
Since both oxidation and reduction occur,this is a redox reaction.
Options $A$,$B$,and $C$ are acid-base or precipitation reactions where no change in oxidation state occurs.

(A) The balanced half-reaction for the reduction of dichromate is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
From the stoichiometry,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons.
Therefore,for $2.5 \ mol$ of $Cr_2O_7^{2-}$,the number of moles of electrons required is:
$2.5 \ mol \times 6 \ mol \ e^- / mol \ Cr_2O_7^{2-} = 15 \ mol \ e^-$.

(C) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in an acidic medium is:
$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$
From the stoichiometry of the reaction,$2 \ mol$ of $KMnO_4$ react with $5 \ mol$ of $H_2O_2$.
Therefore,$1 \ mol$ of $KMnO_4$ will react with $\frac{5}{2} = 2.5 \ mol$ of $H_2O_2$.

(C) The balanced chemical equation for the reaction in basic medium is:
$2ClO_2 + 5H_2O_2 + 2OH^- \rightarrow 2Cl^- + 5O_2 + 6H_2O$
From the stoichiometry of the balanced equation,$2 \ mol$ of $ClO_2$ reacts with $5 \ mol$ of $H_2O_2$.
Therefore,$1 \ mol$ of $ClO_2$ will oxidize $\frac{5}{2} = 2.5 \ mol$ of $H_2O_2$.

(C) The given reaction is $M^{+x} + MnO_4^- \to MO_3^- + Mn^{+2}$.
Given that $1.67 \ mol$ of $M^{+x}$ reacts with $1 \ mol$ of $MnO_4^-$. Since $1.67 \approx 5/3$,we can write the stoichiometry as:
$\frac{5}{3} M^{+x} + 1 MnO_4^- \to \frac{5}{3} MO_3^- + 1 Mn^{+2}$.
Multiplying by $3$ to clear the fraction:
$5 M^{+x} + 3 MnO_4^- \to 5 MO_3^- + 3 Mn^{+2}$.
Now,calculate the change in oxidation states:
For $M$: $5 \times (5 - x) = 25 - 5x$ (increase in oxidation state).
For $Mn$: $3 \times (7 - 2) = 3 \times 5 = 15$ (decrease in oxidation state).
For a balanced redox reaction,the total increase in oxidation state must equal the total decrease:
$25 - 5x = 15$.
$5x = 10$.
$x = 2$.

(D) The reaction between potassium dichromate $(K_2Cr_2O_7)$ and potassium iodide $(KI)$ in an acidic medium is given by the following equation:
$Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2$
In this reaction,the dichromate ion $(Cr_2O_7^{2-})$ acts as an oxidizing agent and is reduced to the chromium$(III)$ ion $(Cr^{3+})$.
Therefore,the oxidation state of chromium changes from $+6$ in $K_2Cr_2O_7$ to $+3$ in the final product $Cr^{3+}$.

(C) The given reaction is a disproportionation reaction: $P_4 + xOH^- + yH_2O \to PH_3 + H_2PO_2^-$.
Balancing the reaction using the oxidation number method:
Oxidation half-reaction: $P_4 \to 4H_2PO_2^-$. The oxidation state of $P$ changes from $0$ to $+1$. Total increase in oxidation number is $4 \times 1 = 4$.
Reduction half-reaction: $P_4 \to 4PH_3$. The oxidation state of $P$ changes from $0$ to $-3$. Total decrease in oxidation number is $4 \times 3 = 12$.
To balance the change in oxidation numbers,multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $1$: $3P_4 + 12OH^- + 12H_2O \to 3PH_3 + 9PH_3 + 12H_2PO_2^-$.
Simplifying the equation: $4P_4 + 12OH^- + 12H_2O \to 4PH_3 + 12H_2PO_2^-$.
Dividing by $4$: $P_4 + 3OH^- + 3H_2O \to PH_3 + 3H_2PO_2^-$.
Comparing this with the given equation,$x = 3$ and $y = 3$.
Therefore,$x = y$.

(D) To find the value of $n$,we determine the change in the oxidation state of nitrogen in the reaction.
In $NO_3^-$,the oxidation state of $N$ is $x + 3(-2) = -1$,so $x = +5$.
In $NO$,the oxidation state of $N$ is $x + (-2) = 0$,so $x = +2$.
The change in oxidation state is $|+5 - (+2)| = 3$.
Since the oxidation state decreases by $3$,$3$ electrons are required for the reduction process.
Therefore,$n = 3$.

(B) To find the value of $n$,we determine the change in the oxidation state of nitrogen in the reaction.
In $NO_2^-$,the oxidation state of $N$ is $x + 2(-2) = -1$,so $x = +3$.
In $NO_3^-$,the oxidation state of $N$ is $x + 3(-2) = -1$,so $x = +5$.
The change in oxidation state is $+5 - (+3) = 2$.
Since the oxidation state increases by $2$,$2$ electrons must be released to balance the charge.
Therefore,$n = 2$.

(D) The balanced chemical equation is $3Fe + 4H_2O \to Fe_3O_4 + 4H_2$.
In this reaction,the oxidation state of $Fe$ changes from $0$ in $Fe$ to $+8/3$ in $Fe_3O_4$.
Total change in oxidation state for $3$ atoms of $Fe$ is $3 \times (8/3 - 0) = 8$.
Therefore,$8$ electrons are lost by $3$ atoms of $Fe$ during this oxidation process.

(A) The reduction half-reaction involves the change of chlorine from $ClO_3^-$ to $Cl^-$.
In $ClO_3^-$,the oxidation state of $Cl$ is $+5$.
In $Cl^-$,the oxidation state of $Cl$ is $-1$.
The change in oxidation state is $(+5) - (-1) = 6$.
Therefore,$6$ electrons are added to the reduction half-reaction: $ClO_3^- + 6e^- \to Cl^-$.
The correct option is $A$.

(B) The oxidation state of nitrogen in $NO_3^-$ is $+5$.
When $4$ moles of electrons are added,the total change in oxidation state is $-4$.
The final oxidation state of nitrogen will be $+5 - 4 = +1$.
Among the options,$N_2O$ has nitrogen in the $+1$ oxidation state.
The balanced half-reaction is: $NO_3^- + 4e^- + 2H^+ \longrightarrow \frac{1}{2} N_2O + \frac{5}{2} H_2O$.
Thus,the product is $1/2 \ mol \ N_2O$.

(B) To balance the redox reaction,we determine the change in oxidation states:
$1$. Oxidation: $O$ in $H_2O_2$ (oxidation state $-1$) is oxidized to $O_2$ (oxidation state $0$). The increase in oxidation state per oxygen atom is $1$. Since there are two oxygen atoms in $H_2O_2$,the total increase is $2$.
$2$. Reduction: $Cl$ in $ClO_2$ (oxidation state $+4$) is reduced to $Cl^-$ (oxidation state $-1$). The decrease in oxidation state is $5$.
$3$. To balance the electrons,we multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$.
$4$. This gives $x = 2$ and $y = 5$ in the balanced equation: $5H_2O_2 + 2ClO_2 + 2OH^- \to 2Cl^- + 5O_2 + 6H_2O$.

(C) The given reaction is a disproportionation reaction where $Cl_2$ is both oxidized and reduced.
$Cl_2^0 \to ClO_3^- (+5)$ (Oxidation,loss of $5$ electrons per $Cl$ atom)
$Cl_2^0 \to Cl^- (-1)$ (Reduction,gain of $1$ electron per $Cl$ atom)
To balance the electrons,we multiply the reduction half-reaction by $5$ and the oxidation half-reaction by $1$:
$1Cl_2 + 5Cl_2 + 12OH^- \to 2ClO_3^- + 10Cl^- + 6H_2O$
Simplifying the equation by dividing by $2$:
$3Cl_2 + 6OH^- \to ClO_3^- + 5Cl^- + 3H_2O$
Comparing this with the given equation $xCl_2 + 6OH^- \to ClO_3^- + yCl^- + 3H_2O$,we get $x = 3$ and $y = 5$.

(A) The given unbalanced equation is: $IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$
$1$. Assign oxidation states:
In $IO_3^-$,$I$ is $+5$.
In $I^-$,$I$ is $-1$.
In $I_2$,$I$ is $0$.
$2$. Identify oxidation and reduction half-reactions:
Oxidation: $I^- \to I_2$ (Change in oxidation state per $I$ atom is $1$,for $2$ atoms it is $2$)
Reduction: $IO_3^- \to I_2$ (Change in oxidation state per $I$ atom is $5$,for $2$ atoms it is $10$)
$3$. Balance the electrons:
Multiply the oxidation half-reaction by $5$ to balance the electron change $(5 \times 2 = 10)$:
$5I^- \to 2.5I_2$ (or $10I^- \to 5I_2$)
$4$. Combine and balance atoms:
$IO_3^- + 5I^- + 6H^+ \to 3I_2 + 3H_2O$
Comparing this with the given equation $IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$,we get:
$a = 5$
$b = 6$
$c = 3$
$d = 3$
Thus,the correct option is $A$.

(A) The given half-reaction is the reduction of the permanganate ion in an acidic medium.
Step $1$: Balance the $Mn$ atoms.
$2MnO_4^- + yH^+ + ze^- \to 2Mn^{2+} + 4H_2O$
Step $2$: Balance the oxygen atoms by adding $H_2O$.
Since there are $8$ oxygen atoms on the left $(2 \times 4)$,we need $8$ oxygen atoms on the right. The equation already has $4H_2O$,so we need $8H_2O$ on the right.
$2MnO_4^- + yH^+ + ze^- \to 2Mn^{2+} + 8H_2O$
Step $3$: Balance the hydrogen atoms.
There are $16$ hydrogen atoms on the right $(8 \times 2)$,so $y = 16$.
$2MnO_4^- + 16H^+ + ze^- \to 2Mn^{2+} + 8H_2O$
Step $4$: Balance the charge.
Left side charge: $2(-1) + 16(+1) + z(-1) = -2 + 16 - z = 14 - z$
Right side charge: $2(+2) + 0 = 4$
$14 - z = 4 \implies z = 10$
Thus,$x = 2, y = 16, z = 10$. Since the provided options do not match this result exactly,we identify the closest logical structure. Given the standard reduction half-reaction $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$,multiplying by $2$ gives $2MnO_4^- + 16H^+ + 10e^- \to 2Mn^{2+} + 8H_2O$.

(A) The half-reactions are:
Reduction: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$
Oxidation: $C_2O_4^{2-} \to 2CO_2 + 2e^-$
To balance the electrons,multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:
$2MnO_4^- + 16H^+ + 10e^- \to 2Mn^{2+} + 8H_2O$
$5C_2O_4^{2-} \to 10CO_2 + 10e^-$
Adding these gives the balanced equation:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Comparing this with the given equation,we get $x = 2, y = 5, z = 16$.

(C) redox reaction involves a change in oxidation number $(O.N.)$ of the reacting species.
In the reaction $N_2 + O_2 \to 2NO$,the $O.N.$ of $N$ changes from $0$ to $+2$ (oxidation) and the $O.N.$ of $O$ changes from $0$ to $-2$ (reduction).
Therefore,this is a redox reaction.

(D) The given skeletal equation is $2MnO_4^- (aq) + Br^- (aq) \to 2MnO_2 (s) + BrO_3^- (aq)$.
Step $1$: Balance the charge. The total charge on the left is $-3$ and on the right is $-1$. To balance the charge in a basic medium,we add $2 \ OH^-$ ions to the right side.
$2MnO_4^- (aq) + Br^- (aq) \to 2MnO_2 (s) + BrO_3^- (aq) + 2 \ OH^- (aq)$
Step $2$: Balance the hydrogen atoms. There are $2 \ H$ atoms on the right,so we add one $H_2O$ molecule to the left side.
$2MnO_4^- (aq) + Br^- (aq) + H_2O (l) \to 2MnO_2 (s) + BrO_3^- (aq) + 2 \ OH^- (aq)$
Thus,both $(a)$ and $(b)$ are correct steps to balance the reaction.

(A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
The reaction proceeds as follows:
$M + HNO_3 \to MNO_3 + [H]$ (Nascent hydrogen)
$2HNO_3 + 2[H] \to 2NO_2 + 2H_2O$
Overall,the nascent hydrogen produced by the metal-acid reaction reduces the nitric acid to nitrogen dioxide $(NO_2)$.

(C) The Assertion is correct because the breathalyzer test uses the oxidation of ethanol by acidic potassium dichromate $(K_2Cr_2O_7)$,which changes color from orange to green as $Cr^{6+}$ is reduced to $Cr^{3+}$.
The Reason is incorrect because the color change is due to a redox reaction (oxidation of alcohol),not complexation.
The chemical reaction is:
$2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_5OH \longrightarrow 2K_2SO_4 + 2Cr_2(SO_4)_3 + 3CH_3COOH + 11H_2O$
Therefore,the correct option is $C$.

(C) In neutral or faintly alkaline medium,$KMnO_4$ acts as an oxidizing agent and reduces to $MnO_2$.
The balanced chemical equation for the reaction between $KMnO_4$ and potassium iodide $(KI)$ is:
$2KMnO_4 + KI + H_2O \longrightarrow 2MnO_2 + KIO_3 + 2KOH$.
Here,the iodide ion $(I^-)$ is oxidized to the iodate ion $(IO_3^-)$.
Therefore,'$X$' is $IO_3^-$.

(B) The reduction half-reaction is: $MnO_{4}^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_{2}O$ ... $(I)$
The oxidation half-reaction is: $C_{2}O_{4}^{2-} \longrightarrow 2CO_{2} + 2e^{-}$ ... $(II)$
To balance the electrons,multiply $(I)$ by $2$ and $(II)$ by $5$:
$2MnO_{4}^{-} + 16H^{+} + 10e^{-} \longrightarrow 2Mn^{2+} + 8H_{2}O$
$5C_{2}O_{4}^{2-} \longrightarrow 10CO_{2} + 10e^{-}$
Adding these two equations gives the balanced redox reaction:
$2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+} \longrightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$
Thus,the coefficients for $MnO_{4}^{-}$,$C_{2}O_{4}^{2-}$,and $H^{+}$ are $2$,$5$,and $16$ respectively.

(N/A) The reaction $2 Na ( s ) + H_2 ( g ) \rightarrow 2 NaH ( s )$ involves the formation of an ionic compound,which can be represented as $Na^{+} H^{-} ( s )$.
This process can be split into two half-reactions:
$1. \text{Oxidation half-reaction: } 2 Na ( s ) \rightarrow 2 Na^{+} ( s ) + 2 e^-$
$2. \text{Reduction half-reaction: } H_2 ( g ) + 2 e^- \rightarrow 2 H^{-} ( s )$
In this reaction,$Na$ loses electrons (is oxidized) and $H_2$ gains electrons (is reduced).
Since both oxidation and reduction occur simultaneously,the reaction is a redox change.

(N/A) Assigning oxidation numbers to each species:
$2 \overset{+1}{Cu}_{2} \overset{-2}{O}_{(s)} + \overset{+1}{Cu}_{2} \overset{-2}{S}_{(s)}$ $\rightarrow 6 \overset{0}{Cu}_{(s)} + \overset{+4}{S} \overset{-2}{O}_{2(g)}$
In this reaction,the oxidation state of $Cu$ decreases from $+1$ to $0$ (reduction),and the oxidation state of $S$ increases from $-2$ to $+4$ (oxidation).
Since both oxidation and reduction occur simultaneously,it is a redox reaction.
Species reduced: $Cu$ (in $Cu_{2}O$ and $Cu_{2}S$).
Species oxidised: $S$ (in $Cu_{2}S$).
Oxidant (Oxidizing agent): $Cu_{2}O$ (as it provides $Cu$ to be reduced).
Reductant (Reducing agent): $Cu_{2}S$ (as it provides $S$ to be oxidised).

(N/A) Step $1:$ The skeletal ionic equation is: $Cr_{2}{O_{7}}^{2-}_{(aq)} + S{O_{3}}^{2-}_{(aq)} \rightarrow Cr^{3+}_{(aq)} + S{O_{4}}^{2-}_{(aq)}$
Step $2:$ Assign oxidation numbers for $Cr$ and $S$: $\mathop {Cr_{2}O_{7}^{2-}}\limits^{+6, -2}_{(aq)} + \mathop {SO_{3}^{2-}}\limits^{+4, -2}_{(aq)} \to \mathop {Cr^{3+}}\limits^{+3}_{(aq)} + \mathop {SO_{4}^{2-}}\limits^{+6, -2}_{(aq)}$. This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.
Step $3:$ Calculate the increase and decrease of oxidation number and make them equal. The oxidation state of chromium changes from $+6$ to $+3$ (decrease of $3$ per $Cr$ atom). The oxidation state of sulphur changes from $+4$ to $+6$ (increase of $2$ per $S$ atom). To balance,multiply $Cr$ by $2$ and $S$ by $3$: $Cr_{2}{O_{7}}^{2-}_{(aq)} + 3S{O_{3}}^{2-}_{(aq)} \to 2Cr^{3+}_{(aq)} + 3S{O_{4}}^{2-}_{(aq)}$.
Step $4:$ As the reaction occurs in an acidic medium,balance the charge by adding $8H^{+}$ on the left: $Cr_{2}{O_{7}}^{2-}_{(aq)} + 3S{O_{3}}^{2-}_{(aq)} + 8H^{+}_{(aq)} \to 2Cr^{3+}_{(aq)} + 3S{O_{4}}^{2-}_{(aq)}$.
Step $5:$ Balance the hydrogen atoms by adding $4H_{2}O$ on the right: $Cr_{2}{O_{7}}^{2-}_{(aq)} + 3S{O_{3}}^{2-}_{(aq)} + 8H^{+}_{(aq)} \to 2Cr^{3+}_{(aq)} + 3S{O_{4}}^{2-}_{(aq)} + 4H_{2}O_{(l)}$.

(N/A) Step $1$: The skeletal ionic equation is: $MnO_4^{-}{(aq)} + Br^{-}{(aq)} \rightarrow MnO_{2(s)} + BrO_3^{-}{(aq)}$
Step $2$: Assign oxidation numbers for $Mn$ and $Br$: $\mathop{Mn}\limits^{+7}O_4^{-}{(aq)} + \mathop{Br^{-}}\limits^{-1}{(aq)}$ $\rightarrow \mathop{Mn}\limits^{+4}O_{2(s)} + \mathop{Br}\limits^{+5}O_3^{-}{(aq)}$. This indicates that permanganate ion is the oxidant and bromide ion is the reductant.
Step $3$: Calculate the increase and decrease of oxidation number,and make the increase equal to the decrease: $2MnO_4^{-}{(aq)} + Br^{-}{(aq)} \rightarrow 2MnO_{2(s)} + BrO_3^{-}{(aq)}$.
Step $4$: As the reaction occurs in the basic medium,and the ionic charges are not equal on both sides,add $2OH^{-}$ ions on the right to make ionic charges equal: $2MnO_4^{-}{(aq)} + Br^{-}{(aq)} \rightarrow 2MnO_{2(s)} + BrO_3^{-}{(aq)} + 2OH^{-}{(aq)}$.
Step $5$: Finally,count the hydrogen atoms and add appropriate number of water molecules (i.e.,one $H_2O$ molecule) on the left side to achieve balanced redox change: $2MnO_4^{-}{(aq)} + Br^{-}{(aq)} + H_2O_{(l)} \rightarrow 2MnO_{2(s)} + BrO_3^{-}{(aq)} + 2OH^{-}{(aq)}$.

(N/A) Step $1:$ Write the skeletal ionic equation:
$MnO_4^-(aq) + I^-(aq) \rightarrow MnO_2(s) + I_2(s)$
Step $2:$ Identify the two half-reactions:
Oxidation half: $I^-(aq) \rightarrow I_2(s)$
Reduction half: $MnO_4^-(aq) \rightarrow MnO_2(s)$
Step $3:$ Balance $I$ atoms in the oxidation half-reaction:
$2I^-(aq) \rightarrow I_2(s) + 2e^-$
Step $4:$ Balance $O$ and $H$ atoms in the reduction half-reaction in basic medium:
$MnO_4^-(aq) + 2H_2O(l) + 3e^- \rightarrow MnO_2(s) + 4OH^-(aq)$
Step $5:$ Equalize the number of electrons by multiplying the oxidation half-reaction by $3$ and the reduction half-reaction by $2$:
$6I^-(aq) \rightarrow 3I_2(s) + 6e^-$
$2MnO_4^-(aq) + 4H_2O(l) + 6e^- \rightarrow 2MnO_2(s) + 8OH^-(aq)$
Step $6:$ Add the two half-reactions to obtain the balanced net ionic equation:
$2MnO_4^-(aq) + 6I^-(aq) + 4H_2O(l) \rightarrow 2MnO_2(s) + 3I_2(s) + 8OH^-(aq)$

(N/A) To justify that the reaction is a redox reaction,we assign oxidation numbers to each atom:
$\overset{+1}{H_2} \overset{-2}{O} + \overset{0}{F_2}$ $\rightarrow \overset{+1}{H} \overset{-1}{F} + \overset{+1}{H} \overset{-2}{O} \overset{+1}{F}$
$1$. In this reaction,the oxidation number of $F$ in $F_2$ is $0$.
$2$. In $HF$,the oxidation number of $F$ is $-1$. Since the oxidation number decreases from $0$ to $-1$,$F_2$ is reduced.
$3$. In $HOF$,the oxidation number of $F$ is $+1$. Since the oxidation number increases from $0$ to $+1$,$F_2$ is oxidized.
$4$. Since both oxidation and reduction occur simultaneously,this is a redox reaction.

$(a)$ Oxidation half: $2I^{-} \rightarrow I_2 + 2e^{-}$. Reduction half: $MnO_4^{-} + 2H_2O + 3e^{-} \rightarrow MnO_2 + 4OH^{-}$. Multiplying by $3$ and $2$ respectively: $6I^{-} + 2MnO_4^{-} + 4H_2O \rightarrow 3I_2 + 2MnO_2 + 8OH^{-}$.
$(b)$ Oxidation half: $SO_2 + 2H_2O \rightarrow HSO_4^{-} + 3H^{+} + 2e^{-}$. Reduction half: $MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O$. Multiplying by $5$ and $2$ respectively: $2MnO_4^{-} + 5SO_2 + 2H_2O + H^{+} \rightarrow 2Mn^{2+} + 5HSO_4^{-}$.
$(c)$ Oxidation half: $Fe^{2+} \rightarrow Fe^{3+} + e^{-}$. Reduction half: $H_2O_2 + 2H^{+} + 2e^{-} \rightarrow 2H_2O$. Multiplying oxidation by $2$: $2Fe^{2+} + H_2O_2 + 2H^{+} \rightarrow 2Fe^{3+} + 2H_2O$.
$(d)$ Oxidation half: $SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^{+} + 2e^{-}$. Reduction half: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O$. Multiplying oxidation by $3$: $Cr_2O_7^{2-} + 3SO_2 + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O$.

(N/A) The given reaction can be represented as:
$Mn_{(aq)}^{3+} \to Mn_{(aq)}^{2+} + MnO_{2_{(s)}} + H_{(aq)}^{+}$
The oxidation half-reaction is:
$Mn_{(aq)}^{3+} \to MnO_{2_{(s)}}$
Balancing $O$ atoms by adding $2H_2O$ and $H$ atoms by adding $4H^{+}$:
$Mn_{(aq)}^{3+} + 2H_2O_{(l)} \to MnO_{2_{(s)}} + 4H_{(aq)}^{+}$
Balancing charge by adding $e^{-}$:
$Mn_{(aq)}^{3+} + 2H_2O_{(l)} \to MnO_{2_{(s)}} + 4H_{(aq)}^{+} + e^{-} \dots (i)$
The reduction half-reaction is:
$Mn_{(aq)}^{3+} + e^{-} \to Mn_{(aq)}^{2+} \dots (ii)$
Adding equations $(i)$ and $(ii)$ gives the balanced ionic equation:
$2Mn_{(aq)}^{3+} + 2H_2O_{(l)} \to Mn_{(aq)}^{2+} + MnO_{2_{(s)}} + 4H_{(aq)}^{+}$

(N/A) The given redox reaction can be represented as:
$Cl_{2(aq)} + SO_{2(aq)} + 2H_2O_{(l)} \to 2Cl^{-}_{(aq)} + SO_{4(aq)}^{2-} + 4H^{+}_{(aq)}$
The oxidation half-reaction is:
$\overset{+4}{S}O_{2(aq)} \to \overset{+6}{S}O_{4(aq)}^{2-}$
Balancing the oxidation number by adding two electrons:
$SO_{2(aq)} \to SO_{4(aq)}^{2-} + 2e^-$
Balancing the charge by adding $4H^{+}$ ions:
$SO_{2(aq)} \to SO_{4(aq)}^{2-} + 4H^{+}_{(aq)} + 2e^-$
Balancing $O$ atoms and $H^{+}$ ions by adding $2H_2O$ molecules:
$SO_{2(aq)} + 2H_2O_{(l)} \to SO_{4(aq)}^{2-} + 4H^{+}_{(aq)} + 2e^- \quad \dots(i)$
The reduction half-reaction is:
$Cl_{2(aq)} \to Cl^{-}_{(aq)}$
Balancing chlorine atoms:
$\overset{0}{Cl}_{2(aq)} \to 2\overset{-1}{Cl}^{-}_{(aq)}$
Balancing the oxidation number by adding electrons:
$Cl_{2(aq)} + 2e^- \to 2Cl^{-}_{(aq)} \quad \dots(ii)$
Adding equation $(i)$ and $(ii)$ gives the balanced chemical equation:
$Cl_{2(aq)} + SO_{2(aq)} + 2H_2O_{(l)} \to 2Cl^{-}_{(aq)} + SO_{4(aq)}^{2-} + 4H^{+}_{(aq)}$

(N/A) The reaction between fluorine and water is represented as:
$2F_{2(g)} + 2H_{2}O_{(l)} \to 4H_{(aq)}^{+} + 4F_{(aq)}^{-} + O_{2(g)}$
In this reaction,the oxidation states change as follows:
$F_{2}$ (oxidation state $0$) is reduced to $F^{-}$ (oxidation state $-1$). Since the oxidation state decreases,$F_{2}$ is reduced.
$H_{2}O$ (oxygen oxidation state $-2$) is oxidized to $O_{2}$ (oxidation state $0$). Since the oxidation state increases,water is oxidized.
Thus,$F_{2}$ acts as the oxidizing agent and is reduced,while $H_{2}O$ acts as the reducing agent and is oxidized.

(N/A) Combination reaction: $A$ combination reaction may be denoted as $A + B \rightarrow C$. Either $A$ or $B$ or both must be in the elemental form for such a reaction to be a redox reaction.
$C_{(s)} + O_{2(g)} \xrightarrow{\Delta} CO_{2(g)}$
$3 Mg_{(s)} + N_{2(g)} \xrightarrow{\Delta} Mg_{3}N_{2(s)}$
$CH_{4(g)} + 2 O_{2(g)} \xrightarrow{\Delta} CO_{2(g)} + 2 H_{2}O_{(l)}$
$(b)$ Decomposition reaction: These are the opposite of combination reactions. $A$ decomposition reaction leads to the breakdown of a compound into two or more components,at least one of which must be in the elemental state.
$2 H_{2}O_{(l)} \xrightarrow{\Delta} 2 H_{2(g)} + O_{2(g)}$
$2 NaH_{(s)} \xrightarrow{\Delta} 2 Na_{(s)} + H_{2(g)}$
$2 KClO_{3(s)} \xrightarrow{\Delta} 2 KCl_{(s)} + 3 O_{2(g)}$
Note: All decomposition reactions are not redox. Example: $CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_{2(g)}$
$(c)$ Displacement reaction: An ion or atom in a compound is replaced by an ion or atom of another element,denoted as $X + YZ \rightarrow XZ + Y$. These are categorized into metal and non-metal displacement.
$(i)$ Metal displacement: $A$ metal in a compound is displaced by another metal in the uncombined state. Examples:
$CuSO_{4(aq)} + Zn_{(s)} \rightarrow Cu_{(s)} + ZnSO_{4(aq)}$
$V_{2}O_{5(s)} + 5 Ca_{(s)} \rightarrow 2 V_{(s)} + 5 CaO_{(s)}$
$TiCl_{4(l)} + 2 Mg_{(s)} \rightarrow Ti_{(s)} + 2 MgCl_{2(s)}$
$(ii)$ Non-metal displacement: Includes hydrogen displacement and rare oxygen displacement reactions.

(N/A) Photosynthesis is a complex process occurring in multiple steps. Experimental evidence using isotopic labeling shows that all the oxygen evolved in photosynthesis comes from water,not from carbon dioxide. Therefore,$12$ molecules of $H_2O$ are involved in the reaction,and $6$ molecules are released as a byproduct.
$(b)$ In the reaction between ozone $(O_3)$ and hydrogen peroxide $(H_2O_2)$,the two oxygen molecules produced originate from different reactants. Specifically,one $O_2$ molecule comes from $O_3$ and the other comes from $H_2O_2$. Writing it as $O_{2(g)} + O_{2(g)}$ highlights this distinct origin.
Technique: The path of these redox reactions can be investigated using isotopic labeling (e.g.,using $^{18}O$ isotope). For reaction $(a)$,water labeled with $^{18}O$ $(H_2^{18}O)$ is used to confirm that the evolved $O_2$ contains the $^{18}O$ isotope. For reaction $(b)$,labeling either $O_3$ or $H_2O_2$ with $^{18}O$ allows researchers to track the source of the oxygen gas produced.

(N/A) The oxidation number method can be explained by the following steps:
Step-$1$: Write the correct chemical formula for each reactant and product.
Step-$2$: Identify the atoms that undergo a change in oxidation number by assigning oxidation numbers to all elements in the reaction.
Step-$3$: Calculate the increase or decrease in the oxidation number per atom and for the entire molecule or ion. If the total increase and decrease are not equal,multiply by suitable coefficients to make them equal.
Step-$4$: Account for the involvement of ions if the reaction takes place in an aqueous medium. Add $H^{+}$ or $OH^{-}$ ions to the appropriate side so that the total ionic charges on both sides are balanced. Use $H^{+}$ for acidic solutions and $OH^{-}$ for basic solutions.
Step-$5$: Balance the number of hydrogen atoms by adding water $(H_{2}O)$ molecules to the reactants or products. Finally,verify the number of oxygen atoms. If the number of oxygen atoms is equal on both sides,the equation is balanced.