Redox reaction and Method for balancing Redox reaction Questions in English

(C) To balance the reaction $ClO_3^- + N_2H_4 \to NO_3^- + Cl^-$,we use the oxidation number method:
$1$. Oxidation half-reaction: $N_2H_4 \to NO_3^- + 12e^- + 16H^+$
$2$. Reduction half-reaction: $ClO_3^- + 6H^+ + 6e^- \to Cl^- + 3H_2O$
$3$. Multiplying the reduction half-reaction by $2$ to balance electrons: $2ClO_3^- + 12H^+ + 12e^- \to 2Cl^- + 6H_2O$
$4$. Adding both half-reactions: $2ClO_3^- + N_2H_4 \to 2Cl^- + 2NO_3^- + 4H^+ + 2H_2O$
(Note: The provided solution in the prompt was incorrect. The balanced equation is $2ClO_3^- + N_2H_4 \to 2NO_3^- + 2Cl^- + 4H^+ + 2H_2O$.)
$5$. The stoichiometric coefficient of $N_2H_4$ is $1$ and for $Cl^-$ is $2$.
$6$. The ratio is $1/2$. Since $1/2$ is not in the options,let us re-examine the reaction stoichiometry. The balanced equation is $2ClO_3^- + N_2H_4 \to 2NO_3^- + 2Cl^- + 4H^+ + 2H_2O$. The ratio of $N_2H_4$ to $Cl^-$ is $1:2$.

(D) The balanced chemical equation for the reaction is:
$2KMnO_4 + 16HCl \rightarrow 2MnCl_2 + 5Cl_2 + 2KCl + 8H_2O$
From the balanced equation,we can see that $5 \text{ moles of } Cl_2$ are produced along with $8 \text{ moles of } H_2O$.
Therefore,for $1 \text{ mole of } Cl_2$ liberated,the number of moles of $H_2O$ formed is:
$\frac{8}{5} \text{ moles}$.

(C) In $Fe_{0.92}O$,let the oxidation state of $Fe$ be $x$. Since the oxidation state of $O$ is $-2$,we have $0.92x + (-2) = 0$,so $x = \frac{2}{0.92}$.
When $1$ $mole$ of $Fe_{0.92}O$ is oxidized to $Fe^{3+}$,the change in oxidation state of $Fe$ is $(3 - \frac{2}{0.92}) \times 0.92 = 3 \times 0.92 - 2 = 2.76 - 2 = 0.76$.
$K_2Cr_2O_7$ acts as an oxidizing agent where $Cr$ changes from $+6$ to $+3$,a change of $3$ per $Cr$ atom,or $6$ per $K_2Cr_2O_7$ molecule.
Equating the total change in oxidation states: $n \times 6 = 0.76$.
Therefore,$n = \frac{0.76}{6}$ $moles$.

(C) Iodine reacts with sodium thiosulfate (hypo solution) to form sodium tetrathionate and sodium iodide.
The balanced chemical equation is:
$I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI$
Thus,the product formed is sodium tetrathionate $(Na_2S_4O_6)$.

(A) The reaction involves the reduction of $BrO_3^-$ to $Br^-$. The change in oxidation state of $Br$ is from $+5$ to $-1$,so the $n$-factor for $KBrO_3$ is $6$.
Equivalents of $KBrO_3 = \frac{\text{mass}}{\text{Equivalent mass}} = \frac{0.167}{167/6} = \frac{0.167 \times 6}{167} = 0.001 \times 6 = 0.006 \ eq$.
Since equivalents of $KBrO_3$ = equivalents of $Na_2S_2O_3$,we have:
$N_1 V_1 = N_2 V_2$
$0.006 = N \times 45 \times 10^{-3} \ L$
$N = \frac{0.006}{0.045} = \frac{6}{45} = \frac{2}{15} \ N$.

(A) The balanced chemical reaction is:
$2 MnO_{4}^{-} + 5 SO_{3}^{2-} + 6 H^{+} \rightarrow 2 Mn^{2+} + 5 SO_{4}^{2-} + 3 H_{2}O$
From the balanced chemical equation,we can observe that:
$5$ moles of $SO_{3}^{2-}$ react with $2$ moles of $KMnO_4$.
Therefore,$1$ mole of $SO_{3}^{2-}$ reacts with $2/5$ moles of $KMnO_4$.
Hence,the correct option is $A$.

(A) The oxidation half-reaction is: $Fe^{2+} \rightarrow Fe^{3+} + e^-$.
The reduction half-reaction for $KMnO_4$ in acidic medium is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
To balance the electrons,multiply the oxidation half-reaction by $5$: $5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$.
Adding the two half-reactions gives the balanced redox equation: $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$.
From the stoichiometry,$5$ moles of $Fe^{2+}$ require $1$ mole of $KMnO_4$.
Therefore,$1$ mole of $Fe^{2+}$ requires $1/5$ moles of $KMnO_4$.

(A) The balanced redox reaction in acidic medium is:
$K_2Cr_2O_7 + 3 SnCl_2 + 14 HCl \rightarrow 2 CrCl_3 + 3 SnCl_4 + 2 KCl + 7 H_2O$
From the stoichiometry of the balanced equation,$1$ mole of $K_2Cr_2O_7$ reacts with $3$ moles of $SnCl_2$.

(B) In the reaction,$A$ in $A_2O_x$ is oxidized to $AO_3^-$. Let the oxidation state of $A$ in $A_2O_x$ be $n$. Then $2n + x(-2) = 0$,so $n = x$. In $AO_3^-$,the oxidation state of $A$ is $y$. Then $y + 3(-2) = -1$,so $y = +5$.
Change in oxidation state per atom of $A = 5 - x$. Since there are $2$ atoms of $A$ in $A_2O_x$,total change $= 2(5 - x) = 10 - 2x$.
For $MnO_4^-$ in acidic medium,$Mn^{+7}$ is reduced to $Mn^{+2}$,so the change in oxidation state is $5$.
Equating equivalents: $n_{A_2O_x} \times (10 - 2x) = n_{KMnO_4} \times 5$.
$1.5 \times 10^{-3} \times (10 - 2x) = (0.03 \times 0.040) \times 5$.
$1.5 \times 10^{-3} \times (10 - 2x) = 0.006$.
$10 - 2x = 4 \implies 2x = 6 \implies x = 3$.
Thus,the value of $x$ is $3$ and the oxide is $A_2O_3$.

(A) The redox reaction is: $x \ MnO + y \ PbO_2 + z \ HNO_3 \to a \ HMnO_4 + b \ Pb(NO_3)_2 + c \ H_2O$.
Step $1$: Assign oxidation states. $Mn$ in $MnO$ is $+2$,$Mn$ in $HMnO_4$ is $+7$ (change of $+5$). $Pb$ in $PbO_2$ is $+4$,$Pb$ in $Pb(NO_3)_2$ is $+2$ (change of $-2$).
Step $2$: Balance the change in oxidation states. Multiply $MnO$ by $2$ and $PbO_2$ by $5$ to balance the electrons ($2 \times 5 = 10$ and $5 \times 2 = 10$).
This gives: $2 \ MnO + 5 \ PbO_2 + z \ HNO_3 \to 2 \ HMnO_4 + 5 \ Pb(NO_3)_2 + c \ H_2O$.
Step $3$: Balance Nitrogen atoms. There are $5 \times 2 = 10$ nitrogen atoms on the right,so $z = 10$.
Step $4$: Balance Hydrogen atoms. There are $10$ hydrogen atoms on the left,so $c = 4$ to give $4 \ H_2O$.
The balanced equation is: $2 \ MnO + 5 \ PbO_2 + 10 \ HNO_3 \to 2 \ HMnO_4 + 5 \ Pb(NO_3)_2 + 4 \ H_2O$.

(C) The oxidation half-reaction is: $Zn \to Zn^{2+} + 2e^-$.
The reduction half-reaction is: $NO_3^- + 10H^+ + 8e^- \to NH_4^+ + 3H_2O$.
To balance electrons,multiply the oxidation half-reaction by $4$: $4Zn \to 4Zn^{2+} + 8e^-$.
Adding the two half-reactions: $4Zn + NO_3^- + 10H^+ \to 4Zn^{2+} + NH_4^+ + 3H_2O$.
Adding nitrate ions to both sides to balance the equation: $4Zn + 10HNO_3 \to 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O$.
Comparing this with the given equation $aZn + b \, HNO_3 \to c \, Zn(NO_3)_2 + d \, NH_4NO_3 + 3H_2O$,we get $b = 10$.

(B) The oxidation states change as follows: $\mathop{CrI_3}\limits^{(+3)(3 \times -1)} \to \mathop{Cr_2O_7^{2-}}\limits^{(2 \times +6)} + \mathop{3IO_4^-}\limits^{(3 \times +7)}$.
For $Cr$: The change in oxidation state is $(+6 - (+3)) = +3$ per $Cr$ atom.
For $I$: The change in oxidation state is $(+7 - (-1)) = +8$ per $I$ atom. Since there are $3$ iodine atoms,the total change is $3 \times 8 = 24$.
Total $n$-factor $= 3 + 24 = 27$.
Equivalent mass $E = \frac{M}{n\text{-factor}} = \frac{M}{27}$.

(B) The reduction half-reaction is: $BrO_{3}^{-} \to Br^{-}$. The change in oxidation state of $Br$ is from $+5$ to $-1$,so the $n$-factor is $6$.
For $\frac{2}{3}$ mole of $BrO_{3}^{-}$,the total electrons gained $= \frac{2}{3} \times 6 = 4 \, e^{-}$.
The oxidation half-reaction is: $N_{2}H_{4} \to N_{2}$. The change in oxidation state of $N$ is from $-2$ to $0$. Since there are two $N$ atoms,the $n$-factor is $2 \times 2 = 4$.
Let $n$ be the moles of $N_{2}H_{4}$. Total electrons lost $= n \times 4$.
Equating total electrons lost and gained: $n \times 4 = 4$.
Therefore,$n = 1 \, \text{mole}$.

(D) In the reaction $2Na_2S_2O_3 + I_2 \to Na_2S_4O_6 + 2NaI$,the oxidation state of sulfur in $Na_2S_2O_3$ is $+2$ and in $Na_2S_4O_6$ is $+2.5$.
The change in oxidation state per sulfur atom is $|2.5 - 2| = 0.5$.
Since there are $4$ sulfur atoms in $Na_2S_4O_6$,the total $n$-factor is $0.5 \times 4 = 2$.
The equivalent weight $E$ is given by the formula $E = \frac{M}{n\text{-factor}}$.
Therefore,$E = \frac{M}{2}$.

(A) The given reaction is: $xMnO + yPbO_2 + zHNO_3 \to HMnO_4 + Pb(NO_3)_2 + H_2O$.
Assign oxidation states: $Mn$ in $MnO$ is $+2$,in $HMnO_4$ is $+7$. Change in $Mn = +5$.
$Pb$ in $PbO_2$ is $+4$,in $Pb(NO_3)_2$ is $+2$. Change in $Pb = -2$.
To balance the electron transfer,multiply $Mn$ by $2$ and $Pb$ by $5$: $2MnO + 5PbO_2 + zHNO_3 \to 2HMnO_4 + 5Pb(NO_3)_2 + H_2O$.
Now balance $N$ atoms: There are $5 \times 2 = 10$ nitrogen atoms on the right,so $z = 10$.
Finally,balance $H$ and $O$: $2MnO + 5PbO_2 + 10HNO_3 \to 2HMnO_4 + 5Pb(NO_3)_2 + 4H_2O$.
Thus,$x = 2, y = 5, z = 10$.

(C) The reaction involves the reduction of the metal ion and the oxidation of the sulphite ion.
For the metal ion: $M^{3+} + (3-n)e^- \to M^{n+}$ (Reduction).
The $n$-factor for the metal is $(3-n)$.
For the sulphite ion: $SO_3^{2-} \to SO_4^{2-} + 2e^-$.
The $n$-factor for the sulphite ion is $2$.
Using the principle of equivalence: $M_1 V_1 n_1 = M_2 V_2 n_2$.
$(0.1 \ M) \times (50 \ mL) \times (3-n) = (0.1 \ M) \times (25 \ mL) \times (2)$.
$5 \times (3-n) = 5$.
$3-n = 1$.
$n = 2$.
Thus,the new oxidation number of the metal is $2$.

(D) In a weak alkaline medium,$KMnO_4$ acts as an oxidizing agent and oxidizes iodide $(I^{-})$ to iodate $(IO_3^{-})$.
The balanced chemical equation is:
$2 KMnO_4 + H_2O + KI \rightarrow 2 MnO_2 + 2 KOH + KIO_3$
In this reaction,the oxidation state of iodine changes from $-1$ in $KI$ to $+5$ in $KIO_3$.

(C) The given unbalanced equation is: $IO_3^- + HSO_3^- \to I^- + SO_4^{2-} + H^+$.
Step $1$: Assign oxidation states. $I$ in $IO_3^-$ is $+5$,$I$ in $I^-$ is $-1$. $S$ in $HSO_3^-$ is $+4$,$S$ in $SO_4^{2-}$ is $+6$.
Step $2$: Write half-reactions.
Reduction: $IO_3^- + 6H^+ + 6e^- \to I^- + 3H_2O$.
Oxidation: $HSO_3^- + H_2O \to SO_4^{2-} + 3H^+ + 2e^-$.
Step $3$: Multiply oxidation half-reaction by $3$ to balance electrons: $3HSO_3^- + 3H_2O \to 3SO_4^{2-} + 9H^+ + 6e^-$.
Step $4$: Add the two half-reactions: $IO_3^- + 3HSO_3^- + 6H^+ + 3H_2O \to I^- + 3SO_4^{2-} + 9H^+ + 3H_2O$.
Step $5$: Simplify: $IO_3^- + 3HSO_3^- \to I^- + 3SO_4^{2-} + 3H^+$.
Comparing with $xIO_3^- + yHSO_3^- \to zI^- + pH^+ + qSO_4^{2-}$,we get $x=1, y=3, z=1$.

(A) The given reaction is a redox reaction: $IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$.
Step $1$: Assign oxidation states.
In $IO_3^-$,$I$ is $+5$. In $I^-$,$I$ is $-1$. In $I_2$,$I$ is $0$.
Step $2$: Write half-reactions.
Reduction: $IO_3^- + 6H^+ + 5e^- \to \frac{1}{2}I_2 + 3H_2O$
Oxidation: $I^- \to \frac{1}{2}I_2 + e^-$
Step $3$: Balance electrons by multiplying the oxidation half-reaction by $5$:
$IO_3^- + 6H^+ + 5e^- \to \frac{1}{2}I_2 + 3H_2O$
$5I^- \to \frac{5}{2}I_2 + 5e^-$
Step $4$: Add the half-reactions:
$IO_3^- + 5I^- + 6H^+ \to 3I_2 + 3H_2O$
Comparing this with the given equation $IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$,we get $a = 5, b = 6, c = 3, d = 3$.

(A) The unbalanced equation is: $Cr(OH)_3 + ClO^{-} + OH^{-} \to CrO_4^{2-} + Cl^{-} + H_2O$
Step $1$: Assign oxidation states. $Cr$ changes from $+3$ to $+6$ (oxidation,loss of $3$ electrons). $Cl$ changes from $+1$ to $-1$ (reduction,gain of $2$ electrons).
Step $2$: Balance the electrons. Multiply the $Cr$ half-reaction by $2$ and the $Cl$ half-reaction by $3$: $2Cr(OH)_3 + 3ClO^{-} \to 2CrO_4^{2-} + 3Cl^{-}$.
Step $3$: Balance the charge using $OH^{-}$. The left side has a charge of $-3$ and the right side has a charge of $-4 - 3 = -7$. Adding $4OH^{-}$ to the left side balances the charge: $2Cr(OH)_3 + 3ClO^{-} + 4OH^{-} \to 2CrO_4^{2-} + 3Cl^{-} + 5H_2O$.
The coefficients are $2, 3, 2, 3$.

(D) The balanced redox reaction in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$
$Sn^{2+} \to Sn^{4+} + 2e^-$
To balance the electrons,multiply the second equation by $3$:
$3Sn^{2+} \to 3Sn^{4+} + 6e^-$
Adding the two half-reactions:
$Cr_2O_7^{2-} + 3Sn^{2+} + 14H^+ \to 2Cr^{3+} + 3Sn^{4+} + 7H_2O$
From the stoichiometry,$1 \ mol$ of $Cr_2O_7^{2-}$ reacts with $3 \ mol$ of $Sn^{2+}$.
Therefore,the number of moles of $Sn^{2+}$ oxidized is $3$.

(C) The reaction between potassium permanganate $(KMnO_4)$ and hydrochloric acid $(HCl)$ is a redox reaction where $KMnO_4$ acts as an oxidizing agent.
The balanced chemical equation is: $2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$.
In this reaction,$Cl^-$ ions are oxidized to chlorine gas $(Cl_2)$,which is a greenish-yellow gas.

(A) redox reaction is one in which oxidation and reduction occur simultaneously,involving a change in the oxidation states of the elements.
In the reaction $2Na[Ag(CN)_2] + Zn \to Na_2[Zn(CN)_4] + 2Ag$:
$1$. The oxidation state of $Ag$ changes from $+1$ to $0$ (reduction).
$2$. The oxidation state of $Zn$ changes from $0$ to $+2$ (oxidation).
Since both oxidation and reduction occur,this is a redox reaction.
Other options represent double displacement or acid-base type reactions without changes in oxidation states.

(A) In an acidic medium,the permanganate ion $(MnO_4^-)$ acts as a strong oxidizing agent.
It oxidizes the iodide ion $(I^-)$ to iodine $(I_2)$.
The balanced chemical equation for this reaction is:
$10I^- + 2MnO_4^- + 16H^+ \rightarrow 5I_2 + 2Mn^{2+} + 8H_2O$.
Therefore,the product formed from $I^-$ is $I_2$.

(C) The oxidation half-reaction is: $Br^- + 6OH^- \to BrO_3^- + 3H_2O + 6e^-$
The reduction half-reaction is: $MnO_4^- + 2H_2O + 3e^- \to MnO_2 + 4OH^-$
To balance the electrons,multiply the reduction half-reaction by $2$:
$2MnO_4^- + 4H_2O + 6e^- \to 2MnO_2 + 8OH^-$
Adding the two half-reactions:
$2MnO_4^- + Br^- + 4H_2O \to 2MnO_2 + BrO_3^- + 3H_2O + 2OH^-$
Simplifying the water molecules:
$2MnO_4^- + Br^- + H_2O \to 2MnO_2 + BrO_3^- + 2OH^-$
The coefficients of $MnO_4^-$,$BrO_3^-$,and $OH^-$ are $2, 1, 2$ respectively.

(C) The given redox reaction is: $xP_4 + yHNO_3 \to H_3PO_4 + NO_2 + H_2O$.
First,identify the oxidation and reduction half-reactions:
Oxidation: $P_4^0 \to 4P^{+5} + 20e^-$.
Reduction: $N^{+5} + 1e^- \to N^{+4}$.
To balance the electrons,multiply the reduction half-reaction by $20$:
$P_4 + 20HNO_3 \to 4H_3PO_4 + 20NO_2 + 4H_2O$.
Comparing this with the given equation $xP_4 + yHNO_3 \to H_3PO_4 + NO_2 + H_2O$,we get $x = 1$ and $y = 20$.

(B) In the reaction $SO_2 + 2H_2S \to 3S + 2H_2O$,the oxidation state of $S$ in $SO_2$ changes from $+4$ to $0$ in $S$.
Thus,$SO_2$ acts as the oxidizing agent.
The change in oxidation number per molecule of $SO_2$ is $|4 - 0| = 4$.
Therefore,the $n$-factor for $SO_2$ is $4$.
The molar mass $(M_W)$ of $SO_2$ is $32 + (2 \times 16) = 64 \ g/mol$.
The equivalent mass $(E)$ is given by $E = \frac{M_W}{n\text{-factor}} = \frac{64}{4} = 16$.

(A) In $Fe_3O_4$,the average oxidation state of $Fe$ is $+\frac{8}{3}$.
In $Fe_2O_3$,the oxidation state of $Fe$ is $+3$.
The change in oxidation state per $Fe$ atom is $3 - \frac{8}{3} = \frac{1}{3}$.
Since there are $3$ atoms of $Fe$ in one molecule of $Fe_3O_4$,the total change in oxidation state ($n$-factor) is $3 \times \frac{1}{3} = 1$.
The equivalent weight $E_w = \frac{\text{Molar mass } (M)}{n\text{-factor}} = \frac{M}{1} = M$.

(D) To balance the half-reaction $Cr_2O_7^{2-} + ZH^{+} + e^- \to Cr^{3+} + H_2O$:
$1$. Balance the $Cr$ atoms: $Cr_2O_7^{2-} + ZH^{+} + e^- \to 2Cr^{3+} + H_2O$.
$2$. Balance the $O$ atoms by adding $H_2O$: $Cr_2O_7^{2-} + ZH^{+} + e^- \to 2Cr^{3+} + 7H_2O$.
$3$. Balance the $H$ atoms: There are $14$ hydrogen atoms on the right side,so $Z = 14$.
$4$. Balance the charge: The left side has $-2 + 14 = +12$ charge (assuming $6$ electrons are added). The right side has $2 \times (+3) = +6$. To balance,we need $6$ electrons: $Cr_2O_7^{2-} + 14H^{+} + 6e^- \to 2Cr^{3+} + 7H_2O$. Thus,$Z = 14$.

(C) The given unbalanced reaction is $Cr_2O_7^{2-} + Fe^{2+} + C_2O_4^{2-} \to Cr^{3+} + Fe^{3+} + CO_2$.
Step $1$: Oxidation half-reactions:
$Fe^{2+} \to Fe^{3+} + e^-$ $(i)$
$C_2O_4^{2-} \to 2CO_2 + 2e^-$ $(ii)$
Adding $(i)$ and $(ii)$: $Fe^{2+} + C_2O_4^{2-} \to Fe^{3+} + 2CO_2 + 3e^-$.
Step $2$: Reduction half-reaction:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$ $(iii)$.
Step $3$: To balance the electrons,multiply the oxidation half-reaction by $2$ and the reduction half-reaction by $1$:
$2(Fe^{2+} + C_2O_4^{2-} \to Fe^{3+} + 2CO_2 + 3e^-) \implies 2Fe^{2+} + 2C_2O_4^{2-} \to 2Fe^{3+} + 4CO_2 + 6e^-$.
Step $4$: Adding the balanced half-reactions:
$Cr_2O_7^{2-} + 2Fe^{2+} + 2C_2O_4^{2-} + 14H^+ \to 2Cr^{3+} + 2Fe^{3+} + 4CO_2 + 7H_2O$.
The total number of electrons involved in the balanced redox reaction is $6$.

(C) The unbalanced equation is: $Na_2HAsO_3 + NaBrO_3 + HCl \longrightarrow NaBr + H_3AsO_4 + NaCl$
$1$. Identify oxidation states:
$As$ in $Na_2HAsO_3$ is $+3$,and in $H_3AsO_4$ is $+5$ (oxidation,change of $+2$).
$Br$ in $NaBrO_3$ is $+5$,and in $NaBr$ is $-1$ (reduction,change of $-6$).
$2$. Balance the change in oxidation states:
To balance the electron transfer,multiply the $As$ half-reaction by $3$ and the $Br$ half-reaction by $1$.
$3 Na_2HAsO_3 + 1 NaBrO_3 \longrightarrow 3 H_3AsO_4 + 1 NaBr$
$3$. Balance the remaining atoms:
Total $Na$ on the left is $3 \times 2 + 1 = 7$. On the right,we have $1$ from $NaBr$ and $6$ from $NaCl$ to make $7$.
Total $Cl$ on the right is $6$,so $Z = 6$.
The balanced equation is: $3 Na_2HAsO_3 + 1 NaBrO_3 + 6 HCl \longrightarrow 1 NaBr + 3 H_3AsO_4 + 6 NaCl$
Thus,$X = 3$,$Y = 1$,and $Z = 6$.

(A) The given reaction is a redox reaction involving the reduction of permanganate and oxidation of oxalate.
Reduction half-reaction:
$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ ... $(i)$
Oxidation half-reaction:
$C_2O_4^{2-} \to 2CO_2 + 2e^-$ ... $(ii)$
To balance the electrons,multiply equation $(i)$ by $2$ and equation $(ii)$ by $5$:
$2MnO_4^- + 16H^+ + 10e^- \to 2Mn^{2+} + 8H_2O$
$5C_2O_4^{2-} \to 10CO_2 + 10e^-$
Adding these two equations gives:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Comparing this with the given equation $X\,MnO_4^- + Y\,C_2O_4^{2-} + Z\,H^+ \rightleftharpoons X\,Mn^{2+} + 2Y\,CO_2 + \frac{Z}{2}\,H_2O$,we get:
$X = 2$,$Y = 5$,$Z = 16$.

(A) The balanced chemical equation for the reaction is:
$5C_2O_4^{2-} + 2MnO_4^- + 16H^{+} \to 10CO_2 + 2Mn^{2+} + 8H_2O$
In this reaction,the oxalate ion $(C_2O_4^{2-})$ is oxidized to $CO_2$.
The half-reaction for the oxidation is:
$C_2O_4^{2-} \to 2CO_2 + 2e^-$
This shows that $2$ electrons are released when $2$ molecules of $CO_2$ are produced.
Therefore,for the production of one molecule of $CO_2$,the number of electrons involved is $2/2 = 1$.

(A) The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular mass}}{n\text{-factor}}$.
In an acidic medium,the reduction half-reaction for the dichromate ion is:
$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$.
Here,the total number of electrons gained per mole of $Cr_{2}O_{7}^{2-}$ is $6$.
Therefore,the $n\text{-factor} = 6$.
Thus,the equivalent weight is $\frac{\text{Molecular weight}}{6}$.

(B) In a dilute alkaline or neutral medium,the permanganate ion $(MnO_4^-)$ is reduced to manganese dioxide $(MnO_2)$.
The balanced half-reaction is:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$

(B) The balanced redox reaction involves the reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$ and the oxidation of $Sn^{2+}$ to $Sn^{4+}$.
For $K_2Cr_2O_7$: The change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change is $2 \times (6 - 3) = 6$. Thus,the $n$-factor is $6$.
For $Sn^{2+}$: The change in oxidation state of $Sn$ is from $+2$ to $+4$. The change is $4 - 2 = 2$. Thus,the $n$-factor is $2$.
Using the law of equivalence: $\text{Equivalents of } K_2Cr_2O_7 = \text{Equivalents of } Sn^{2+}$
$n_{K_2Cr_2O_7} \times 6 = n_{Sn^{2+}} \times 2$
Given $n_{Sn^{2+}} = 1 \ mol$,we have:
$n_{K_2Cr_2O_7} \times 6 = 1 \times 2$
$n_{K_2Cr_2O_7} = \frac{2}{6} = \frac{1}{3} \ mol$.

(D) In the reaction $MnO_2 + 4HCl \to MnCl_2 + 2H_2O + Cl_2$,the oxidation state of $Cl$ changes from $-1$ in $HCl$ to $0$ in $Cl_2$.
The change in oxidation number for one $Cl$ atom is $1$. Since there are $2$ $Cl$ atoms in $Cl_2$,the total change in oxidation number for $4$ moles of $HCl$ is $2$.
Therefore,the $n$-factor for $4 \ mol$ of $HCl$ is $2$.
The $n$-factor for $1 \ mol$ of $HCl$ is $\frac{2}{4} = \frac{1}{2}$.
Equivalent weight of $HCl = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{M}{1/2} = 2M$.

(A) The given unbalanced equation is: $IO_3^- + aI^- + bH^+ \longrightarrow cH_2O + dI_2$
To balance the redox reaction:
$1$. Oxidation half-reaction: $2I^- \longrightarrow I_2 + 2e^-$
$2$. Reduction half-reaction: $IO_3^- + 6H^+ + 5e^- \longrightarrow \frac{1}{2}I_2 + 3H_2O$
Multiplying the oxidation half-reaction by $5$ and the reduction half-reaction by $2$ to balance electrons:
$10I^- \longrightarrow 5I_2 + 10e^-$
$2IO_3^- + 12H^+ + 10e^- \longrightarrow I_2 + 6H_2O$
Adding them together:
$2IO_3^- + 10I^- + 12H^+ \longrightarrow 6I_2 + 6H_2O$
Dividing by $2$ to simplify:
$IO_3^- + 5I^- + 6H^+ \longrightarrow 3H_2O + 3I_2$
Comparing with the given equation,we get $a=5, b=6, c=3, d=3$.
Thus,the correct option is $A$.

(B) In acidic medium,$KMnO_4$ acts as an oxidizing agent and undergoes reduction as follows:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
The change in oxidation state of $Mn$ is from $+7$ to $+2$,which is a change of $5$ units.
Molecular mass of $KMnO_4 = 39 + 55 + (4 \times 16) = 158 \ g/mol$.
Equivalent weight = $\frac{\text{Molecular mass}}{\text{n-factor}} = \frac{158}{5} = 31.6$.

(A) The given reaction is a disproportionation reaction: $P_4 + OH^{\Theta} \to PH_3 + H_2PO_2^{\Theta}$.
Step $1$: Oxidation half-reaction: $P_4 \to H_2PO_2^{\Theta}$. Balancing $P$ atoms: $P_4 \to 4 H_2PO_2^{\Theta}$. Balancing $O$ and $H$ atoms using $OH^{\Theta}$ and $H_2O$: $P_4 + 8 OH^{\Theta} \to 4 H_2PO_2^{\Theta} + 4 e^-$.
Step $2$: Reduction half-reaction: $P_4 \to PH_3$. Balancing $P$ atoms: $P_4 \to 4 PH_3$. Balancing $H$ atoms: $P_4 + 12 H_2O + 12 e^- \to 4 PH_3 + 12 OH^{\Theta}$.
Step $3$: Multiply the oxidation half-reaction by $3$ to equalize electrons: $3 P_4 + 24 OH^{\Theta} \to 12 H_2PO_2^{\Theta} + 12 e^-$.
Step $4$: Add the two half-reactions: $(3 P_4 + 24 OH^{\Theta}) + (P_4 + 12 H_2O + 12 e^-) \to (12 H_2PO_2^{\Theta} + 12 e^-) + (4 PH_3 + 12 OH^{\Theta})$.
Simplifying: $4 P_4 + 12 OH^{\Theta} + 12 H_2O \to 4 PH_3 + 12 H_2PO_2^{\Theta}$.
Dividing by $4$: $P_4 + 3 OH^{\Theta} + 3 H_2O \to PH_3 + 3 H_2PO_2^{\Theta}$.
The coefficients for $P_4$,$OH^{\Theta}$,and $H_2PO_2^{\Theta}$ are $1, 3, 3$ respectively.

(A) To balance the redox reaction,the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction.
Reduction half-reaction: $S_2O_8^{2-} + 2e^- \longrightarrow 2SO_4^{2-}$ (Gain of $2e^-$)
Oxidation half-reaction: $Mn^{2+} + 4H_2O \longrightarrow MnO_4^- + 8H^+ + 5e^-$ (Loss of $5e^-$)
To equate the electrons,multiply the reduction half-reaction by $5$ and the oxidation half-reaction by $2$:
$5S_2O_8^{2-} + 10e^- \longrightarrow 10SO_4^{2-}$
$2Mn^{2+} + 8H_2O \longrightarrow 2MnO_4^- + 16H^+ + 10e^-$
From the balanced equation,$2 \ moles$ of $Mn^{2+}$ require $5 \ moles$ of $S_2O_8^{2-}$.
Therefore,$1 \ mole$ of $Mn^{2+}$ requires $5/2 = 2.5 \ moles$ of $S_2O_8^{2-}$.

(B) In an alkaline medium,the permanganate ion $(MnO_4^-)$ acts as an oxidizing agent and is reduced to manganese dioxide $(MnO_2)$.
The iodide ion $(I^-)$ is oxidized to the iodate ion $(IO_3^-)$ in the presence of an alkaline medium.
The balanced chemical equation for this reaction is:
$I^- + 2MnO_4^- + H_2O \rightarrow IO_3^- + 2MnO_2 + 2OH^-$

(B) The chemical reaction is: $K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \to K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
In $K_2Cr_2O_7$,the oxidation state of $Cr$ is calculated as: $2(+1) + 2(x) + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$.
In $Cr_2(SO_4)_3$,the oxidation state of $Cr$ is calculated as: $2(x) + 3(-2) = 0 \implies 2x = 6 \implies x = +3$.
Thus,the oxidation state of chromium changes from $+6$ to $+3$.

(A) In a redox reaction,the oxidation state of at least one element increases (oxidation) and another decreases (reduction).
For the reaction $H_2 + Br_2 \to 2HBr$:
- The oxidation state of $H$ changes from $0$ to $+1$ (oxidation).
- The oxidation state of $Br$ changes from $0$ to $-1$ (reduction).
Therefore,this reaction involves both oxidation and reduction.

(D) The reaction between potassium permanganate $(KMnO_4)$ and hydrochloric acid $(HCl)$ is a redox reaction.
In this reaction,$Mn$ in $KMnO_4$ is reduced from an oxidation state of $+7$ to $+2$ in $MnCl_2$.
Simultaneously,$Cl^-$ ions from $HCl$ are oxidized to chlorine gas $(Cl_2)$.
The balanced chemical equation is: $2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$.

(B) The balanced redox reaction involves the reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$ and the oxidation of $Sn^{2+}$ to $Sn^{4+}$.
For $K_2Cr_2O_7$: $Cr$ changes from $+6$ to $+3$. The $n$-factor $= 2 \times (6 - 3) = 6$.
For $Sn^{2+}$: $Sn$ changes from $+2$ to $+4$. The $n$-factor $= 4 - 2 = 2$.
According to the law of equivalence,the equivalents of $K_2Cr_2O_7$ must equal the equivalents of $Sn^{2+}$.
Let $x$ be the moles of $K_2Cr_2O_7$.
$x \times (n\text{-factor of } K_2Cr_2O_7) = \text{moles of } Sn^{2+} \times (n\text{-factor of } Sn^{2+})$.
$x \times 6 = 1 \times 2$.
$x = \frac{2}{6} = \frac{1}{3} \, mol$.

(B) The given redox reaction is: $xBrO_3^- + yCr^{3+} + zH_2O \to aBr_2 + bHCrO_4^- + cH^+$.
Step $1$: Identify oxidation and reduction half-reactions.
Reduction: $BrO_3^- \to Br_2$ (Bromine changes from $+5$ to $0$).
Oxidation: $Cr^{3+} \to HCrO_4^-$ (Chromium changes from $+3$ to $+6$).
Step $2$: Balance the half-reactions.
Reduction: $2BrO_3^- + 12H^+ + 10e^- \to Br_2 + 6H_2O$.
Oxidation: $Cr^{3+} + 4H_2O \to HCrO_4^- + 7H^+ + 3e^-$.
Step $3$: Multiply to equalize electrons.
Multiply reduction by $3$ and oxidation by $10$:
$6BrO_3^- + 36H^+ + 30e^- \to 3Br_2 + 18H_2O$.
$10Cr^{3+} + 40H_2O \to 10HCrO_4^- + 70H^+ + 30e^-$.
Step $4$: Add the equations.
$6BrO_3^- + 10Cr^{3+} + 22H_2O \to 3Br_2 + 10HCrO_4^- + 34H^+$.
Comparing with the given equation,$x = 6$,$y = 10$,and $z = 22$.

(A) To balance the given redox reaction:
$C_2H_5OH + xI_2 + yOH^{-} \to CHI_3 + HCO_2^- + 5I^{-} + 5H_2O$
$1$. Oxidation half-reaction:
$C_2H_5OH + 5OH^{-} \to HCO_2^- + 4H_2O + 4e^-$
$2$. Reduction half-reaction:
$I_2 + 2e^- \to 2I^-$
To balance electrons,multiply the reduction half-reaction by $2$:
$2I_2 + 4e^- \to 4I^-$
Adding the two half-reactions:
$C_2H_5OH + 2I_2 + 5OH^{-} \to CHI_3 + HCO_2^- + 4I^- + 4H_2O$
Wait,checking the stoichiometry for $I$ atoms:
Reactants: $2I_2$ ($4I$ atoms) + $5I^-$ (from the provided equation structure).
Balancing the full equation:
$C_2H_5OH + 4I_2 + 6OH^{-} \to CHI_3 + HCO_2^- + 5I^{-} + 5H_2O$
Comparing with the given equation,$x = 4$ and $y = 6$.

(C) In the reaction,the oxidation state of $N$ in $N_2H_4$ changes from $-2$ to $0$. The change per atom is $2$,and for $2$ atoms of $N$,the total change in oxidation state is $4$. Thus,the $n$-factor for $N_2H_4$ is $4$.
Equivalent mass of $N_2H_4 = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{32}{4} = 8$.
The oxidation state of $I$ in $IO_3^-$ changes from $+5$ to $+1$ in $ICl$. The change in oxidation state is $5 - 1 = 4$. Thus,the $n$-factor for $KIO_3$ is $4$.
Equivalent mass of $KIO_3 = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{214}{4} = 53.5$.

(D) The given reaction is: $x BrO_3^- + y Cr^{3+} + z H_2O \to Br_2 + CrO_4^{2-} + H^{+}$
Step $1$: Identify oxidation and reduction half-reactions.
Reduction: $BrO_3^- \to Br_2$ (Bromine oxidation state changes from $+5$ to $0$)
Oxidation: $Cr^{3+} \to CrO_4^{2-}$ (Chromium oxidation state changes from $+3$ to $+6$)
Step $2$: Balance the half-reactions.
Reduction: $2 BrO_3^- + 12 H^+ + 10 e^- \to Br_2 + 6 H_2O$
Oxidation: $Cr^{3+} + 4 H_2O \to CrO_4^{2-} + 8 H^+ + 3 e^-$
Step $3$: Equalize electrons by multiplying reduction by $3$ and oxidation by $10$.
$6 BrO_3^- + 36 H^+ + 30 e^- \to 3 Br_2 + 18 H_2O$
$10 Cr^{3+} + 40 H_2O \to 10 CrO_4^{2-} + 80 H^+ + 30 e^-$
Step $4$: Add the two half-reactions.
$6 BrO_3^- + 10 Cr^{3+} + 22 H_2O \to 3 Br_2 + 10 CrO_4^{2-} + 44 H^{+}$
Comparing with the given equation,$x = 6$,$y = 10$,and $z = 22$.