Redox reaction and Method for balancing Redox reaction Questions in English

(D) The balanced chemical equation is:
$Cr_2O_7^{2-} + 14H^{+} + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
Comparing this with the given equation:
$X = 14$
$Y = 2$
$Z = 7$
Therefore,the sum $(X + Y + Z) = 14 + 2 + 7 = 23$.

(A) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $Mn^{2+}$.
The reaction is: $2KMnO_4 + 10KI + 8H_2SO_4 \rightarrow 6K_2SO_4 + 2MnSO_4 + 5I_2 + 8H_2O$.
The oxidation state of $Mn$ in $KMnO_4$ is $+7$.
The oxidation state of $Mn$ in $MnSO_4$ is $+2$.
The change in oxidation state is $|(+2) - (+7)| = 5$.

(A) Using the ion-electron method:
Reduction half-reaction: $Cr_2O_7^{2-} + 14H^{+} + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Oxidation half-reaction: $(SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^{+} + 2e^-) \times 3$
Adding both half-reactions:
$Cr_2O_7^{2-} + 14H^{+} + 6e^- + 3SO_3^{2-} + 3H_2O \rightarrow 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 6H^{+} + 6e^-$
Simplifying the equation:
$Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$
Comparing this with the given equation $aCr_2O_7^{2-} + bSO_3^{2-} + cH^{+} \rightarrow 2aCr^{3+} + bSO_4^{2-} + \frac{c}{2}H_2O$:
$a = 1, b = 3, c = 8$.

(C) The balanced chemical equation for the oxidation of $PbS$ by $O_3$ is:
$PbS + 4 O_3 \rightarrow PbSO_4 + 4 O_2$
From the balanced equation,we can see that $1$ mole of $PbS$ reacts with $4$ moles of $O_3$ to produce $4$ moles of $O_2$.
Therefore,$X = 4$ and $Y = 4$.
The value of $X + Y = 4 + 4 = 8$.

(A) The given reaction is a disproportionation reaction of chlorine in an alkaline medium:
$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$
Step $1$: Assign oxidation states.
$Cl_2$ $(0)$ $\rightarrow Cl^-$ $(-1)$ and $ClO^-$ $(+1)$.
Step $2$: Write half-reactions.
Reduction: $Cl_2 + 2e^- \rightarrow 2Cl^-$
Oxidation: $Cl_2 + 4OH^- \rightarrow 2ClO^- + 2H_2O + 2e^-$
Step $3$: Add the half-reactions.
$2Cl_2 + 4OH^- \rightarrow 2Cl^- + 2ClO^- + 2H_2O$
Step $4$: Simplify by dividing by $2$.
$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$
Comparing with $a Cl_2 + b OH^- \rightarrow c ClO^- + d Cl^- + e H_2O$,we get $a=1, b=2, c=1, d=1$.

(D) In an alkaline medium,the permanganate ion $(MnO_4^{-})$ acts as an oxidizing agent and oxidizes the iodide ion $(I^{-})$ to the iodate ion $(IO_3^{-})$.
The balanced chemical equation for this reaction is:
$2 MnO_4^{-} + H_2O + I^{-} \xrightarrow{\text{alkaline medium}} 2 MnO_2 + 2 OH^{-} + IO_3^{-}$

(A) The given redox reaction is: $2 MnO_4^{-} + bI^{-} + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^{-}$
Step $1$: Write the half-reactions.
Reduction half: $MnO_4^{-} \rightarrow MnO_2$
Balancing $O$ and $H$ in basic medium: $MnO_4^{-} + 2 H_2O + 3 e^{-} \rightarrow MnO_2 + 4 OH^{-}$
Multiply by $2$: $2 MnO_4^{-} + 4 H_2O + 6 e^{-} \rightarrow 2 MnO_2 + 8 OH^{-}$
Step $2$: Oxidation half.
Oxidation half: $I^{-} \rightarrow I_2$
Balancing atoms and electrons: $2 I^{-} \rightarrow I_2 + 2 e^{-}$
Multiply by $3$: $6 I^{-} \rightarrow 3 I_2 + 6 e^{-}$
Step $3$: Add the two half-reactions.
$2 MnO_4^{-} + 6 I^{-} + 4 H_2O \rightarrow 2 MnO_2 + 3 I_2 + 8 OH^{-}$
Comparing this with the given equation,the coefficient $z$ corresponds to the number of $OH^{-}$ ions,which is $8$.

(A) The balanced chemical equation for the reaction between permanganate ion and oxalate ion in acidic medium is:
$2 MnO_4^{-} + 5 C_2O_4^{2-} + 16 H^{+} \longrightarrow 2 Mn^{2+} + 10 CO_2 + 8 H_2O$
From the stoichiometry of the balanced equation,$2 \ moles$ of $MnO_4^{-}$ require $16 \ moles$ of $H^{+}$ ions.
Therefore,$1 \ mole$ of $MnO_4^{-}$ requires $\frac{16}{2} = 8 \ moles$ of $H^{+}$ ions.

(D) The balanced half-reaction for the reduction of dichromate in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Comparing this with the given equation $Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$,we get:
$X = 14$
$Y = 6$
$Z = 7$
$A = Cr^{3+}$
Therefore,the correct values are $14, 6, 7$ and $Cr^{3+}$.

(NONE OF THE ABOVE) The balanced chemical equation for the reaction between $KMnO_4$ and oxalic acid in acidic medium is:
$2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$.
According to the law of equivalence,the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid:
$n_{eq}(KMnO_4) = n_{eq}(H_2C_2O_4)$.
Equivalents are calculated as $Molarity \times Volume \times n-factor$.
For $KMnO_4$ in acidic medium,the $n-factor = 5$.
For oxalic acid $(H_2C_2O_4)$,the $n-factor = 2$.
Let the molarity of $KMnO_4$ be $M$.
$M \times 2 \ mL \times 5 = 2 \ M \times 20 \ mL \times 2$.
$10M = 80$.
$M = 8 \ M$.

(D) The balanced redox reaction in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \longrightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
From the stoichiometry of the balanced equation,$1 \text{ mole}$ of $Cr_2O_7^{2-}$ reacts with $6 \text{ moles}$ of $Fe^{2+}$ (Mohr's salt).
Alternatively,using the $n$-factor method:
$n$-factor of $Cr_2O_7^{2-} = 6$
$n$-factor of $Fe^{2+} = 1$
By the law of equivalence,$\text{moles of } Cr_2O_7^{2-} \times (n\text{-factor}) = \text{moles of } Fe^{2+} \times (n\text{-factor})$
$1 \times 6 = \text{moles of } Fe^{2+} \times 1$
Therefore,$6 \text{ moles}$ of Mohr's salt are required.
Hence,$(D)$ is the correct option.

(B) The reaction of aqueous sodium thiosulfate $(Na_2S_2O_3)$ with chlorine $(Cl_2)$ is an oxidation reaction.
$Na_2S_2O_3 + 4Cl_2 + 5H_2O \longrightarrow 2NaHSO_4 + 8HCl$
In this reaction,the sulfur in thiosulfate is oxidized to sulfate $(HSO_4^-)$,and chlorine is reduced to chloride $(HCl)$.

(B) The balanced redox reaction in neutral or faintly alkaline medium is:
$8 MnO_4^{-} + 3 S_2O_3^{2-} + H_2O \rightarrow 8 MnO_2 + 6 SO_4^{2-} + 2 OH^{-}$
From the stoichiometry of the balanced equation,$8$ moles of $MnO_4^{-}$ produce $6$ moles of $SO_4^{2-}$.
Therefore,the value of $X$ is $6$.

(B) The balanced chemical equation for the reaction of bromine with sodium carbonate in aqueous solution is:
$3 Na_2CO_3 + 3 Br_2 \rightarrow 5 NaBr + NaBrO_3 + 3 CO_2$
From the balanced equation,the coefficient of sodium bromide $(NaBr)$ is $5$.

(B) The balanced chemical equation for the reaction of $KMnO_4$ with $KI$ in a weakly basic or neutral medium is:
$2KMnO_4 + KI + H_2O \longrightarrow 2MnO_2 + KIO_3 + 2KOH$ (Note: $I^-$ is oxidized to $IO_3^-$ in neutral/basic medium).
However,if the reaction is considered as $I^-$ to $I_2$ (as implied by the question context):
$2KMnO_4 + 6KI + 4H_2O \longrightarrow 2MnO_2 + 3I_2 + 8KOH$.
From the stoichiometry of the balanced equation,$2$ moles of $KMnO_4$ produce $3$ moles of $I_2$.
Therefore,for $4$ moles of $KMnO_4$,the moles of $I_2$ produced = $(4 \times 3) / 2 = 6$ moles.

(B) The balanced chemical equation is:
$6I^{-} + ClO_{3}^{-} + 6H_{2}SO_{4} \longrightarrow Cl^{-} + 6HSO_{4}^{-} + 3I_{2} + 3H_{2}O$
$1$. The stoichiometric coefficient of $HSO_{4}^{-}$ is $6$. (Statement $A$ is correct)
$2$. The oxidation state of $I$ changes from $-1$ to $0$,so $I^{-}$ is oxidized. (Statement $B$ is correct)
$3$. The oxidation state of $S$ in $H_{2}SO_{4}$ is $+6$ and in $HSO_{4}^{-}$ is $+6$. Thus,sulphur is not reduced. (Statement $C$ is incorrect)
$4$. $H_{2}O$ is formed as a product. (Statement $D$ is correct)
Therefore,the correct statements are $A, B, D$.

(A) The $n$-factor is the change in oxidation state per molecule.
$A$. $C_2O_4^{2-} \rightarrow 2CO_2$: Oxidation state of $C$ changes from $+3$ to $+4$. Total change $= 2 \times (4-3) = 2$. So,$A-P$.
$B$. $Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$: Oxidation state of $Cr$ changes from $+6$ to $+3$. Total change $= 2 \times (6-3) = 6$. So,$B-Q$.
$C$. $2S_2O_3^{2-} \rightarrow S_4O_6^{2-} + 2e^-$: Oxidation state of $S$ changes from $+2$ to $+2.5$. Total change for $2$ moles of $S_2O_3^{2-}$ is $2 \times [2 \times (2.5-2)] = 2$. However,per mole of $Na_2S_2O_3$,$n$-factor is $1$. So,$C-S$.
$D$. $MnO_4^{-} \rightarrow MnO_2$: Oxidation state of $Mn$ changes from $+7$ to $+4$. Total change $= 7-4 = 3$. So,$D-R$.
Thus,the correct match is $A-P, B-Q, C-S, D-R$.

(C) The balanced redox reaction is determined by balancing the oxidation and reduction half-reactions.
$BrO_3^{-} \rightarrow Br_2$ (Reduction: $Br^{+5}$ to $Br^0$,change of $5$ per $Br$,so $10$ per $Br_2$)
$Cr^{+3} \rightarrow HCrO_4^{-}$ (Oxidation: $Cr^{+3}$ to $Cr^{+6}$,change of $3$ per $Cr$)
To balance the electrons,multiply the reduction half by $3$ and the oxidation half by $10$ (or simplify to $3$ and $5$ for $Br_2$ and $Cr$ respectively).
$6BrO_3^{-} + 10Cr^{+3} + 22H_2O \rightarrow 3Br_2 + 10HCrO_4^{-} + 34H^{+}$
Comparing coefficients,we get $X = 6, Y = 10, Z = 22$.

(C) To find the equivalent weight,we need to determine the change in oxidation state of the species in $Cu_2S$.
The oxidation states are: $Cu$ in $Cu_2S$ is $+1$,$S$ in $Cu_2S$ is $-2$.
In the products: $Cu$ is $+2$ and $S$ in $SO_2$ is $+4$.
The change in oxidation state for $Cu_2S$ is calculated as:
$2 \times (Cu^{+1} \rightarrow Cu^{+2}) = 2 \times 1 = 2$
$1 \times (S^{-2} \rightarrow S^{+4}) = 1 \times 6 = 6$
Total change in oxidation state = $2 + 6 = 8$.
Since the equivalent weight is $M / n$-factor,and the $n$-factor is $8$,the value of $x$ is $8$.

(A) The half-reactions are:
Oxidation: $Zn \rightarrow Zn^{+2} + 2e^{-}$ (multiplied by $4$ to balance electrons: $4Zn \rightarrow 4Zn^{+2} + 8e^{-}$)
Reduction: $NO_3^{-} + 10H^{+} + 8e^{-} \rightarrow NH_4^{+} + 3H_2O$
Adding both half-reactions gives the balanced equation:
$4Zn + NO_3^{-} + 10H^{+} \rightarrow 4Zn^{+2} + NH_4^{+} + 3H_2O$
The coefficients of $NO_3^{-}$,$Zn$,and $H^{+}$ are $1$,$4$,and $10$ respectively.

(B) reaction intermediate is a species that is produced in one step of a reaction mechanism and consumed in a subsequent step.
In step $(i)$,$ClO_2^{-}$ is formed.
In step $(ii)$,$ClO_2^{-}$ is consumed.
Therefore,$ClO_2^{-}$ is the reaction intermediate.

(D) In $KMnO_4$,the oxidation state of $Mn$ is $+7$.
In $MnSO_4$,the oxidation state of $Mn$ is $+2$.
The reduction half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
This shows that $1 \ mol$ of $KMnO_4$ requires $5 \ mol$ of electrons ($5 \ F$ of charge) to be reduced to $Mn^{2+}$.
Therefore,for $2 \ mol$ of $KMnO_4$,the total charge required is $2 \times 5 \ F = 10 \ F$.

(C) First,assign oxidation numbers to the elements in the reaction: $I_2^0 + K^{+1}Cl^{+5}O_3^{-2} \rightarrow I^{+1}Cl^{-1} + K^{+1}I^{+5}O_3^{-2}$.
In this reaction,$I_2$ (oxidation number $0$) is oxidized to $I^{+1}$ (in $ICl$) and $I^{+5}$ (in $KIO_3$).
Since $I_2$ is being oxidized,it acts as a reducing agent.
$Cl$ in $KClO_3$ changes from $+5$ to $-1$ (in $ICl$).
The change in oxidation number of $Cl$ is $|-1 - 5| = 6$. Thus,the oxidation number of $Cl$ decreases by $6$.
Therefore,the statement 'Oxidation number of $Cl$ decreases by $6$' is correct.

(B) In the reaction $3 H_3AsO_3 + BrO_3^- \rightarrow Br^- + 3 H_3AsO_4$,we assign oxidation numbers:
For $BrO_3^-$,the oxidation state of $Br$ is $+5$. In $Br^-$,it is $-1$. Since the oxidation number decreases from $+5$ to $-1$,$Br$ undergoes reduction.
For $H_3AsO_3$,the oxidation state of $As$ is $+3$. In $H_3AsO_4$,it is $+5$. Since the oxidation number increases from $+3$ to $+5$,$As$ undergoes oxidation.
Therefore,$Br$ undergoes reduction and $As$ undergoes oxidation.

(D) In the given reaction: $Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_2(aq) + H_2(g)$
$1$. The oxidation state of $Zn$ changes from $0$ to $+2$ (loss of $2$ electrons,oxidation).
$2$. The oxidation state of $H$ in $HCl$ changes from $+1$ to $0$ in $H_2$ (gain of electrons,reduction).
$3$. Since $H^+$ is being reduced,$HCl$ acts as the oxidizing agent.
$4$. The reduction half-reaction is: $2H^+ + 2e^- \longrightarrow H_2$.
$5$. This shows that $2$ moles of $H^+$ (which corresponds to $2$ moles of $HCl$) gain $2$ moles of electrons.
$6$. Therefore,$1$ mole of the oxidizing agent $(HCl)$ gains $1$ mole of electrons.

(D) When $SO_{2}$ is passed through an acidified $K_{2}Cr_{2}O_{7}$ solution,the orange dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to the green chromium$(III)$ ion $(Cr^{3+})$.
The balanced chemical equation is:
$K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \rightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$
As a result,the solution turns green due to the formation of $Cr_{2}(SO_{4})_{3}$.

(C) In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $1 + x + 4(-2) = 0$,so $x = +7$.
In $Mn_2O_3$,the oxidation state of $Mn$ is calculated as: $2x + 3(-2) = 0$,so $2x = +6$,which means $x = +3$.
The change in oxidation state per $Mn$ atom is $|7 - 3| = 4$.
Since there are $2$ $Mn$ atoms in $Mn_2O_3$,the total number of electrons transferred is $2 \times 4 = 8$.
However,the question asks for the number of electrons transferred per $Mn$ atom involved in the reduction process from $KMnO_4$ to $Mn_2O_3$.
Considering the stoichiometry $2KMnO_4 \rightarrow Mn_2O_3$,the total change is $2 \times (7 - 3) = 8$ electrons.
Given the options provided,the change per $Mn$ atom is $4$.

(A) To find the value of $x$,we need to balance the charge on both sides of the equation.
In $BiO_3^{-}$,the oxidation state of $Bi$ is $x + 3(-2) = -1$,so $x = +5$.
In $Bi^{3+}$,the oxidation state of $Bi$ is $+3$.
The change in oxidation state is $+5 - (+3) = 2$.
However,we must also balance the total charge.
Total charge on the left side: $(-1) + 6(+1) + x(-1) = 5 - x$.
Total charge on the right side: $(+3) + 3(0) = +3$.
Equating the charges: $5 - x = 3$,which gives $x = 2$.

(C) The given reaction is: $x CuO + y NH_3 \rightarrow x Cu + N_2 + x H_2 O$.
Step $1$: Balance the atoms other than $O$ and $H$. Since $x$ is the coefficient for $CuO$ and $Cu$,the $Cu$ atoms are balanced.
Step $2$: Balance $N$ atoms. There are $2$ nitrogen atoms in $N_2$,so we need $2$ $NH_3$ molecules to provide $2$ $N$ atoms. Thus,$y = 2$.
Step $3$: Balance $O$ atoms. There are $x$ oxygen atoms in $x CuO$,so we need $x$ $H_2 O$ molecules. This matches the equation.
Step $4$: Balance $H$ atoms. There are $3y$ hydrogen atoms in $y NH_3$ and $2x$ hydrogen atoms in $x H_2 O$. Therefore,$3y = 2x$.
Step $5$: Substitute $y = 2$ into the equation $3y = 2x$: $3(2) = 2x$ $\Rightarrow 6 = 2x$ $\Rightarrow x = 3$.
Thus,$x = 3$ and $y = 2$. The balanced equation is $3 CuO + 2 NH_3 \rightarrow 3 Cu + N_2 + 3 H_2 O$.

(B) To balance the redox reaction,we look at the change in oxidation states:
$Mn^{2+} \rightarrow MnO_2$: $Mn$ changes from $+2$ to $+4$ (oxidation,loss of $2e^-$).
$ClO_3^- \rightarrow ClO_2$: $Cl$ changes from $+5$ to $+4$ (reduction,gain of $1e^-$).
To balance the electrons,we multiply the reduction half-reaction by $2$:
$Mn^{2+} + 2 ClO_3^- \rightarrow MnO_2 + 2 ClO_2$.
Comparing this to the given equation $Mn^{2+} + x ClO_3^- \rightarrow MnO_2 + x ClO_2$,we find $x=2$.

(D) The given reaction is: $x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$
Step $1$: Write the half-reactions.
Reduction half-reaction: $ClO_4^{-} \rightarrow ClO_2^{-}$
Balancing oxygen and charge: $ClO_4^{-} + 4 H^{+} + 4 e^{-} \rightarrow ClO_2^{-} + 2 H_2O$
Step $2$: Oxidation half-reaction.
$H_2O_2 \rightarrow O_2 + 2 H^{+} + 2 e^{-}$
Step $3$: Equalize the number of electrons.
Multiply the oxidation half-reaction by $2$ to match the $4 e^{-}$ in the reduction half-reaction:
$2 H_2O_2 \rightarrow 2 O_2 + 4 H^{+} + 4 e^{-}$
Step $4$: Add the two half-reactions.
$(ClO_4^{-} + 4 H^{+} + 4 e^{-}) + (2 H_2O_2) \rightarrow (ClO_2^{-} + 2 H_2O) + (2 O_2 + 4 H^{+} + 4 e^{-})$
Simplifying,we get:
$2 H_2O_2 + ClO_4^{-} \rightarrow 2 O_2 + ClO_2^{-} + 2 H_2O$
Comparing this with the given equation $x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$,we find $x = 2$.

(D) In acidic medium,$KMnO_4$ acts as an oxidising agent and gets reduced to $Mn^{2+}$. The half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
Sulphide ion $(S^{2-})$ is oxidised to elemental sulphur $(S)$: $S^{2-} \rightarrow S + 2e^-$.
To balance the electrons,we multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:
$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$
$5S^{2-} \rightarrow 5S + 10e^-$
Adding these,the balanced equation is: $2MnO_4^- + 5S^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 5S + 8H_2O$.
From the stoichiometry,$5$ moles of $S^{2-}$ react with $2$ moles of $KMnO_4$.
Therefore,$1$ mole of $S^{2-}$ reacts with $2/5$ moles of $KMnO_4$.

(B) In a faintly alkaline or neutral medium,the permanganate ion $(MnO_4^-)$ acts as an oxidizing agent and oxidizes the iodide ion $(I^-)$ to the iodate ion $(IO_3^-)$.
The balanced chemical equation for this reaction is:
$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$
Therefore,the product obtained is $IO_3^-$.

(A) The balanced chemical equation for the reaction is:
$Cr_{2}O_{7}^{2-} + 6I^{-} + 14H^{+} \longrightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O$
From the stoichiometry of the reaction,$1$ mole of $K_{2}Cr_{2}O_{7}$ produces $3$ moles of $I_{2}$.
Therefore,to liberate $6$ moles of $I_{2}$,the required moles of $K_{2}Cr_{2}O_{7}$ is:
$\frac{1 \text{ mol } K_{2}Cr_{2}O_{7}}{3 \text{ mol } I_{2}} \times 6 \text{ mol } I_{2} = 2 \text{ moles of } K_{2}Cr_{2}O_{7}$.

(C) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $MnO_2$.
The reaction with iodide ion $(I^-)$ is as follows:
$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-$
Thus,the iodide ion is oxidized to iodate ion $(IO_3^-)$.

(C) Acidified $K_{2}Cr_{2}O_{7}$ acts as a strong oxidizing agent and oxidizes $SO_{2}$ to $SO_{4}^{2-}$ ions (present in $Cr_{2}(SO_{4})_{3}$).
The balanced chemical equation is:
$3SO_{2} + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} \longrightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$
In $SO_{2}$,the oxidation state of sulphur is $+4$.
In $Cr_{2}(SO_{4})_{3}$,the sulphate ion $(SO_{4}^{2-})$ contains sulphur in the $+6$ oxidation state.
Thus,the oxidation state of sulphur changes from $+4$ to $+6$.

(D) The balanced redox reaction is: $5H_2S + 2MnO_4^{-} + 6H^{+} \rightarrow 5S + 2Mn^{2+} + 8H_2O$.
In this reaction,the oxidation state of $S$ in $H_2S$ increases from $-2$ to $0$,so $H_2S$ is oxidised to $S$.
The oxidation state of $Mn$ in $MnO_4^{-}$ decreases from $+7$ to $+2$,so $MnO_4^{-}$ is reduced to $Mn^{2+}$.

(B) To balance the redox reaction in basic medium:
$1$. Oxidation half-reaction: $S_2 O_3^{2-} \longrightarrow 2 SO_4^{2-} + 8 e^-$
$2$. Reduction half-reaction: $MnO_4^{-} + 2 H_2 O + 3 e^- \longrightarrow MnO_2 + 4 OH^-$
$3$. To balance electrons,multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $8$:
$3 S_2 O_3^{2-} \longrightarrow 6 SO_4^{2-} + 24 e^-$
$8 MnO_4^{-} + 16 H_2 O + 24 e^- \longrightarrow 8 MnO_2 + 32 OH^-$
$4$. Adding both equations:
$8 MnO_4^{-} + 3 S_2 O_3^{2-} + 16 H_2 O \longrightarrow 8 MnO_2 + 6 SO_4^{2-} + 32 OH^-$
Simplifying $H_2 O$ and $OH^-$ on both sides:
$8 MnO_4^{-} + 3 S_2 O_3^{2-} + H_2 O \longrightarrow 8 MnO_2 + 6 SO_4^{2-} + 2 OH^-$
Comparing with the given equation,$a = 8$ and $y = 6$.

(B) The balanced chemical equation for the reaction is:
$ 2MnO_{4}^{-} + 5H_{2}C_{2}O_{4} + 6H^{+} \rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O $
Comparing this with the given equation $ xMnO_{4}^{-} + yH_{2}C_{2}O_{4} + zH^{+} \rightarrow mMn^{2+} + nCO_{2} + pH_{2}O $,we get:
$ x = 2 $
$ y = 5 $
$ m = 2 $
$ n = 10 $
Thus,the correct values are $ 2, 5, 2, 10 $.

(B) To find the value of $x$,we must balance the charge on both sides of the equation.
In the reactant side,the total charge is: $(-1) + 6(+1) + x(-1) = 5 - x$.
In the product side,the total charge is: $(+3) + 3(0) = +3$.
Equating the charges: $5 - x = 3$.
Solving for $x$: $x = 5 - 3 = 2$.
Thus,the balanced equation is $BiO_{3}^{-} + 6H^{+} + 2e^{-} \longrightarrow Bi^{3+} + 3H_{2}O$.

(C) The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O$.
In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$.
The change in oxidation state for $2$ atoms of $Cr$ is $2 \times (6 - 3) = 6$.
Thus,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons.
For $0.2 \ mol$ of $Cr_2O_7^{2-}$,the number of moles of electrons required is $0.2 \times 6 = 1.2 \ mol$.

(A) The given reaction is: $SnCl_{2} + Cl_{2} \longrightarrow SnCl_{4}$.
In this reaction,the oxidation state of $Sn$ changes from $+2$ in $SnCl_{2}$ to $+4$ in $SnCl_{4}$.
The $n$-factor is the change in oxidation state per molecule,which is $|4 - 2| = 2$.
The molar mass of $SnCl_{2} = 118.7 + 2 \times 35.5 = 118.7 + 71 = 189.7 \approx 190 \ g/mol$.
The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{190}{2} = 95$.

(B) In neutral or faintly alkaline medium,the iodide ion $(I^{-})$ is oxidised to the iodate ion $(IO_3^{-})$.
The balanced chemical equation is:
$2 MnO_4^{-} + H_2O + I^{-} \rightarrow 2 MnO_2 + 2 OH^{-} + IO_3^{-}$
Thus,$X$ is the iodate ion.

(B) The balanced stoichiometric equation is:
$P_4 + 3 OH^{-} + 3 H_2 O \rightarrow PH_3 + 3 H_2 PO_2^{-}$
Comparing this to the given equation,$X^{-}$ is $H_2 PO_2^{-}$.
The conjugate acid is formed by adding a proton $(H^{+})$ to the base.
Conjugate acid of $H_2 PO_2^{-}$ is $H_3 PO_2$.
$H_3 PO_2$ is known as hypophosphorous acid.

(B) In acidic medium,$KMnO_4$ acts as a strong oxidizing agent.
When $H_2S$ reacts with $KMnO_4$ in an acidic medium,$H_2S$ is oxidized to elemental sulphur,which precipitates out.
The reaction is:
$5H_2S + 2MnO_4^- + 6H^+ \longrightarrow 2Mn^{2+} + 8H_2O + 5S$
Thus,the precipitation of sulphur from $H_2S$ is a characteristic reaction of $KMnO_4$ in acidic medium.

(C) The reaction of iodine with chlorine in the presence of water is an oxidation reaction where iodine is oxidized to iodic acid $(HIO_3)$ and chlorine is reduced to hydrochloric acid $(HCl)$.
The balanced chemical equation is:
$I_2 + 5Cl_2 + 6H_2O \longrightarrow 2HIO_3 + 10HCl$
Comparing this with the given reaction $I_2 + 5Cl_2 + 6H_2O \longrightarrow X + Y$,we identify $X$ as $HIO_3$ and $Y$ as $HCl$.

(C) The balanced chemical equation for the reaction of $KMnO_4$ with $H_2S$ in an acidic medium is:
$2KMnO_4 + 3H_2SO_4 + 5H_2S \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5S$
From the stoichiometry of the balanced equation,$2$ moles of $KMnO_4$ react with $5$ moles of $H_2S$.
Therefore,the number of moles of $KMnO_4$ reacting with $1$ mole of $H_2S$ is $\frac{2}{5} = 0.4$ moles.

(D) The balanced equation is $P_4 + 3OH^{-} + 3H_2O \longrightarrow PH_3 + 3H_2PO_2^{-}$.
Comparing coefficients: $a=1, b=3, c=3, d=1, e=3$.
Statement $I$: $a+b+c = 1+3+3 = 7$. Thus,$I$ is incorrect.
Statement $II$: $b+c-e = 3+3-3 = 3$. Thus,$II$ is correct.
Statement $III$: In $H_2PO_2^{-}$,let oxidation state of $P$ be $x$. $2(+1) + x + 2(-2) = -1 \implies 2 + x - 4 = -1 \implies x = +1$. Thus,$III$ is correct.
Therefore,statements $II$ and $III$ are correct.

(A) The given reaction is a disproportionation reaction of sulfur $(S_8)$.
To balance the reaction $xS_8 + yOH^- \rightarrow zS^{2-} + 2S_2O_3^{2-} + 6H_2O$:
Step $1$: Balance sulfur atoms. There are $8$ sulfur atoms on the left. On the right,there are $z$ atoms in $S^{2-}$ and $2 \times 2 = 4$ atoms in $2S_2O_3^{2-}$. Thus,$z + 4 = 8$,which gives $z = 4$.
Step $2$: Balance oxygen atoms. There are $y$ oxygen atoms on the left. On the right,there are $2 \times 3 = 6$ atoms in $2S_2O_3^{2-}$ and $6$ atoms in $6H_2O$. Thus,$y = 6 + 6 = 12$.
Step $3$: Balance hydrogen atoms. There are $y$ hydrogen atoms on the left and $6 \times 2 = 12$ on the right. This confirms $y = 12$.
Step $4$: The coefficient $x$ for $S_8$ is $1$ to satisfy the sulfur balance.
Therefore,$x = 1, y = 12, z = 4$.

(A) To balance the redox reaction,we use the ion-electron method.
Oxidation half-reaction: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
Reduction half-reaction: $Cr_2 O_7^{2-} + 14 H^{+} + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2 O$
Multiplying the oxidation half-reaction by $6$ to balance the electrons:
$6 Fe^{2+} \rightarrow 6 Fe^{3+} + 6 e^-$
Adding both half-reactions:
$6 Fe^{2+} + Cr_2 O_7^{2-} + 14 H^{+} \rightarrow 6 Fe^{3+} + 2 Cr^{3+} + 7 H_2 O$
Comparing this with the given equation $x Fe^{2+} + y H^{+} + z Cr_2 O_7^{2-} \rightarrow \frac{y}{2} H_2 O + x Fe^{3+} + 2 z Cr^{3+}$,we get $x = 6, y = 14, z = 1$.