Redox reaction and Method for balancing Redox reaction Questions in English

(D) From the balanced chemical equation:
$MnO_4^- + 5Fe^{2+} + 8H^{+} \to Mn^{2+} + 5Fe^{3+} + 4H_2O$
$1 \ \text{mole of } KMnO_4 \text{ reacts with } 5 \ \text{moles of } FeSO_4$.
Given:
$\text{Moles of } KMnO_4 = \text{Molarity} \times \text{Volume (in L)} = 0.1 \ M \times 0.010 \ L = 0.001 \ \text{moles}$.
Since $1 \ \text{mole of } KMnO_4 \text{ reacts with } 5 \ \text{moles of } FeSO_4$,
$\text{Moles of } FeSO_4 \text{ required} = 5 \times 0.001 \ \text{moles} = 0.005 \ \text{moles}$.
Now,$\text{Volume of } FeSO_4 = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.005 \ \text{moles}}{0.1 \ M} = 0.050 \ L = 50 \ mL$.
Therefore,$10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to $50 \ mL$ of $0.1 \ M$ $FeSO_4$.

(A) The balanced redox reaction in an acidic medium is:
$3MnO_4^- + 5FeC_2O_4 + 24H^+ \to 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$
From the stoichiometry of the balanced equation,$5 \ mol$ of $FeC_2O_4$ reacts with $3 \ mol$ of $KMnO_4$.
Therefore,$1 \ mol$ of $FeC_2O_4$ will react with $3/5 \ mol$ of $KMnO_4$.

(B) The balanced redox reaction is:
$2MnO_4^- + 16H^{+} + 5C_2O_4^{2-} \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Comparing this with the given equation $yMnO_4^- + xH^{+} + C_2O_4^{2-} \to yMn^{2+} + 2CO_2 + \frac{x}{2}H_2O$,we divide the balanced equation by $5$ to match the coefficient of $C_2O_4^{2-}$ as $1$:
$\frac{2}{5}MnO_4^- + \frac{16}{5}H^{+} + C_2O_4^{2-} \to \frac{2}{5}Mn^{2+} + 2CO_2 + \frac{8}{5}H_2O$
However,the question implies integer coefficients for the balanced equation.
For $2MnO_4^- + 16H^{+} + 5C_2O_4^{2-} \to 2Mn^{2+} + 10CO_2 + 8H_2O$,we have $y = 2$ and $x = 16$.

(D) The balanced chemical equation is: $Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$.
From the balanced equation,we can observe:
$1$. The number of nitrogen atoms on the product side is $4$ (two in $Cu(NO_3)_2$ and two in $2NO_2$).
$2$. The number of water molecules $(H_2O)$ is $2$.
$3$. The net charge on the product side is $0$ as all species are neutral molecules.

(B) When $H_2S$ gas is passed through an acidic solution of $KMnO_4$,the $KMnO_4$ acts as a strong oxidizing agent.
It oxidizes $H_2S$ to elemental sulfur $(S)$.
The balanced chemical equation is:
$2KMnO_4 + 3H_2SO_4 + 5H_2S \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5S$.
Thus,sulfur $(S)$ is the product formed.

(C) In an acidic medium,$K_2Cr_2O_7$ acts as a strong oxidizing agent and is reduced to $Cr^{3+}$.
The half-reaction is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$.
The change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,the $n$-factor for $K_2Cr_2O_7$ is $6$.
The equivalent mass is calculated as: $\text{Equivalent mass} = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{294}{6} = 49$.

(B) The reduction half-reaction for $IO_4^-$ to $I_2$ in an acidic medium is:
$2IO_4^- + 16H^+ + 14e^- \rightarrow I_2 + 8H_2O$.
Here,the total number of electrons involved per mole of $IO_4^-$ is $n = \frac{14}{2} = 7$.
Therefore,the equivalent weight is given by $\frac{M}{n} = \frac{M}{7}$.

(B) In an alkaline medium,the reaction between $KMnO_4$ and $KI$ is represented as:
$2KMnO_4 + H_2O + KI \to 2MnO_2 + 2KOH + KIO_3$
From the balanced chemical equation,$1$ mole of $KI$ reacts with $2$ moles of $KMnO_4$.
Therefore,$1$ mole of $KI$ reduces $2$ moles of $KMnO_4$.

(B) To balance the reaction $5H_2O_2 + XClO_2 + 2OH^- \to XCl^- + YO_2 + 6H_2O$:
$1$. Balance Chlorine $(Cl)$: There are $X$ atoms of $Cl$ on both sides.
$2$. Balance Oxygen $(O)$: On the left side,total $O = 5(2) + 2(X) + 2 = 12 + 2X$. On the right side,total $O = 2Y + 6(1) = 2Y + 6$.
$3$. Balance Hydrogen $(H)$: On the left side,total $H = 5(2) + 2 = 12$. On the right side,total $H = 6(2) = 12$. (Already balanced).
$4$. Balance Charge: Left side charge = $-2$. Right side charge = $-X$. Thus,$X = 2$.
$5$. Substitute $X = 2$ into the Oxygen balance equation: $12 + 2(2) = 2Y + 6 \implies 16 = 2Y + 6 \implies 2Y = 10 \implies Y = 5$.
Therefore,$X = 2$ and $Y = 5$.

(A) Let us balance the reaction using the oxidation number method.
$(i) IO_3^- + I^- + H^+ \to H_2O + I_2$
$(ii)$ Oxidation number change:
$\mathop{I}\limits^{+5}O_3^- + \mathop{I}\limits^{-1} \to \mathop{I}\limits^{0}_2$
Decrease in oxidation number = $5$ units (per $I$ atom in $IO_3^-$)
Increase in oxidation number = $1$ unit (per $I$ atom in $I^-$)
$(iii)$ To balance the change,multiply $I^-$ by $5$:
$IO_3^- + 5I^- + H^+ \to H_2O + I_2$
$(iv)$ Total $I$ atoms on the left is $6$,so $I_2$ on the right must be $3$:
$IO_3^- + 5I^- + H^+ \to H_2O + 3I_2$
$(v)$ Balance $O$ atoms by adding $3H_2O$:
$IO_3^- + 5I^- + H^+ \to 3H_2O + 3I_2$
$(vi)$ Balance $H$ atoms by adding $6H^+$:
$IO_3^- + 5I^- + 6H^+ \to 3H_2O + 3I_2$
Comparing with $IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$,we get $a=5, b=6, c=3, d=3$.

(A) To balance the reaction,we analyze the charge and atom conservation.
Substituting $x = 2$,$y = 5$,and $z = 6$ into the equation:
$2\,MnO_4^- + 5\,H_2O \to 2Mn^{2+} + 5H_2O + 9O_2 + 6e^-$
Checking the charge balance:
Left side: $2 \times (-1) = -2$
Right side: $2 \times (+2) + 6 \times (-1) = 4 - 6 = -2$
Since the charges and atoms are balanced,the values are $x = 2$,$y = 5$,and $z = 6$.

(A) The balanced half-reaction for the reduction of $MnO_4^-$ in an acidic medium is:
$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$
This shows that $1 \ mol$ of $KMnO_4$ accepts $5 \ mol$ of electrons.
The oxidation of $Fe(C_2O_4)$ involves:
$Fe(C_2O_4) \to Fe^{3+} + 2CO_2 + 3e^-$
This shows that $1 \ mol$ of $Fe(C_2O_4)$ loses $3 \ mol$ of electrons.
To balance the electrons,we equate the moles of electrons gained and lost:
$5 \times (\text{moles of } KMnO_4) = 3 \times (\text{moles of } Fe(C_2O_4))$
For $1 \ mol$ of $Fe(C_2O_4)$,the moles of $KMnO_4$ required is:
$\text{Moles of } KMnO_4 = \frac{3}{5} = 0.6 \ mol$.

(A) The balanced chemical equation for the reaction between $Cr_2O_7^{2-}$ and $Sn^{2+}$ in an acidic medium is:
$Cr_2O_7^{2-} + 3Sn^{2+} + 14H^+ \to 2Cr^{3+} + 3Sn^{4+} + 7H_2O$
From the stoichiometry of the reaction,$3$ moles of $Sn^{2+}$ ions reduce $1$ mole of $Cr_2O_7^{2-}$.
Therefore,$1$ mole of $Sn^{2+}$ ions will reduce $\frac{1}{3}$ mole of $K_2Cr_2O_7$.

(D) In a basic medium,the reduction of the permanganate ion $(MnO_4^-)$ to manganese dioxide $(MnO_2)$ is represented by the following half-reaction:
$MnO_4^- + 2H_2O + 3e^- \to MnO_2 + 4OH^-$
As shown in the equation,$3$ electrons are involved in this reduction process.

(C) The oxidation state of $Mn$ in $KMnO_4$ is $+7$. The number of electrons transferred is the difference in oxidation states of $Mn$ in the reactant and product.
$1$. For $MnO_4^{2-}$: $Mn$ is in $+6$ state. Electrons transferred = $7 - 6 = 1$.
$2$. For $MnO_2$: $Mn$ is in $+4$ state. Electrons transferred = $7 - 4 = 3$.
$3$. For $Mn_2O_3$: $Mn$ is in $+3$ state. Since there are two $Mn$ atoms,total change = $2 \times (7 - 3) = 8$. However,per $Mn$ atom basis or standard reaction stoichiometry for $KMnO_4$ reduction,the change is $4$ electrons per $Mn$ atom ($7$ to $3$).
$4$. For $Mn^{2+}$: $Mn$ is in $+2$ state. Electrons transferred = $7 - 2 = 5$.
Thus,the sequence is $1, 3, 4, 5$.

(B) In the presence of dilute $H_2SO_4$,$K_2Cr_2O_7$ acts as an oxidizing agent.
The balanced half-reaction is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$
Since $6$ electrons are involved in the reduction of one mole of $K_2Cr_2O_7$,the n-factor is $6$.
Therefore,the equivalent weight of $K_2Cr_2O_7 = \frac{\text{Molecular weight}}{6}$.

(D) In alkaline medium,the reduction reaction is:
$MnO_4^- + e^- \to MnO_4^{2-}$
The oxidation state of $Mn$ changes from $+7$ to $+6$.
The change in oxidation number ($n$-factor) is $|7 - 6| = 1$.
The equivalent weight of $KMnO_4$ is calculated as:
$\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}$
Molar mass of $KMnO_4 = 39 + 55 + (4 \times 16) = 158 \ g/mol$.
$\text{Equivalent weight} = \frac{158}{1} = 158$.

(A) The balanced half-reaction for the reduction of dichromate is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
In this reaction,the change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change in oxidation number is $2 \times (6 - 3) = 6$.
Therefore,the number of electrons gained per mole of $Cr_2O_7^{2-}$ is $6$.
The equivalent weight is calculated as:
$\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{M}{6}$

(A) Let us balance the reaction using the oxidation number method.
$Step (i)$: Write the unbalanced equation:
$MnO_4^- + C_2O_4^{2-} + H^+ \to Mn^{2+} + CO_2 + H_2O$
$Step (ii)$: Identify oxidation number changes:
$Mn$ changes from $+7$ to $+2$ (decrease of $5$ per atom).
$C$ changes from $+3$ to $+4$ (increase of $1$ per atom).
$Step (iii)$: Equalize the change in oxidation number:
Total increase in $C = 2 \times 1 = 2$.
Total decrease in $Mn = 1 \times 5 = 5$.
To balance,multiply $MnO_4^-$ by $2$ and $C_2O_4^{2-}$ by $5$:
$2MnO_4^- + 5C_2O_4^{2-} + H^+ \to 2Mn^{2+} + 10CO_2 + H_2O$
$Step (iv)$: Balance $H$ and $O$ atoms:
There are $16$ oxygen atoms on the left (excluding $H_2O$) and $20$ on the right. Balancing $H^+$ and $H_2O$ gives:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
The coefficients for $MnO_4^-$,$C_2O_4^{2-}$,and $H^+$ are $2, 5, 16$ respectively.

(A) The given reaction is: $IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$.
Step-$(I)$: Identify oxidation and reduction half-reactions.
Oxidation: $I^- \to I_2$
Reduction: $IO_3^- \to I_2$
Step-$(II)$: Balance the reduction half-reaction:
$2IO_3^- + 12H^+ + 10e^- \to I_2 + 6H_2O$
Step-$(III)$: Balance the oxidation half-reaction:
$2I^- \to I_2 + 2e^-$
Step-$(IV)$: Multiply the oxidation half-reaction by $5$ to balance the electrons:
$10I^- \to 5I_2 + 10e^-$
Step-$(V)$: Add both half-reactions:
$2IO_3^- + 10I^- + 12H^+ \to 6I_2 + 6H_2O$
Step-$(VI)$: Simplify by dividing by $2$:
$IO_3^- + 5I^- + 6H^+ \to 3H_2O + 3I_2$
Comparing with $IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$,we get $a = 5, b = 6, c = 3, d = 3$.

(B) The balanced chemical equation for the reaction of $KMnO_4$ with $Br^-$ in a basic medium is:
$2MnO_4^- + Br^- + H_2O \to 2MnO_2 + BrO_3^- + 2OH^-$
In $MnO_4^-$,the oxidation state of $Mn$ is $+7$.
In $MnO_2$,the oxidation state of $Mn$ is $+4$.
Therefore,the oxidation state of $Mn$ changes from $7$ to $4$.

(C) The reduction half-reaction for $ClO_2$ in alkaline medium is:
$ClO_2 + 2H_2O + 5e^- \to Cl^- + 4OH^-$
The oxidation half-reaction for $H_2O_2$ in alkaline medium is:
$H_2O_2 + 2OH^- \to O_2 + 2H_2O + 2e^-$
To balance the electrons,multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:
$2ClO_2 + 4H_2O + 10e^- \to 2Cl^- + 8OH^-$
$5H_2O_2 + 10OH^- \to 5O_2 + 10H_2O + 10e^-$
Adding these reactions gives the balanced equation:
$2ClO_2 + 5H_2O_2 + 2OH^- \to 2Cl^- + 5O_2 + 6H_2O$
From the stoichiometry,$2 \ mol$ of $ClO_2$ react with $5 \ mol$ of $H_2O_2$.
Therefore,$1 \ mol$ of $ClO_2$ reacts with $2.5 \ mol$ of $H_2O_2$.

(D) To balance the charge in the given half-reaction: $A^{-n_2} + x e^- \to A^{-n_1}$.
Let the charge on the left side be $C_L = -n_2 + x(-1) = -n_2 - x$.
Let the charge on the right side be $C_R = -n_1$.
Equating the charges: $-n_2 - x = -n_1$.
Solving for $x$: $x = n_1 - n_2$.

(A) $1$. Identify the oxidation and reduction half-reactions.
$2$. In the given reaction,$H_2O_2$ is oxidized to $O_2$ $(H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-)$.
$3$. Since $H_2O_2$ loses electrons,it acts as the reductant.
$4$. The change in oxidation state of oxygen in $H_2O_2$ $(-1)$ to $O_2$ $(0)$ involves a change of $1$ unit per oxygen atom. For two oxygen atoms in $H_2O_2$,the total change in oxidation number is $2$.
$5$. The equivalent weight of a substance is calculated as $\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}$.
$6$. Molar mass of $H_2O_2 = (2 \times 1) + (2 \times 16) = 34 \ g/mol$.
$7$. The n-factor for $H_2O_2$ in this reaction is $2$.
$8$. Therefore,$\text{Equivalent weight} = \frac{34}{2} = 17$.

(B) In the given redox reaction:
$2Fe^{3+} + Sn^{2+} \to 2Fe^{2+} + A$
$1$. Analyze the oxidation states:
- $Fe^{3+}$ is reduced to $Fe^{2+}$ (gain of electrons).
- $Sn^{2+}$ is oxidized to $Sn^{4+}$ (loss of electrons).
$2$. Balance the charge:
- Left side total charge: $2(+3) + (+2) = +8$.
- Right side total charge: $2(+2) + x = +4 + x$.
- For the reaction to be balanced,$+8 = +4 + x$,so $x = +4$.
Therefore,'$A$' is $Sn^{4+}_{(aq)}$.

(C) In an acidic medium,the reduction half-reaction for potassium dichromate $(K_2Cr_2O_7)$ is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Here,the change in the oxidation state of Chromium $(Cr)$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change in oxidation number is $2 \times (6 - 3) = 6$.
Therefore,the n-factor for $K_2Cr_2O_7$ in an acidic medium is $6$.
The equivalent weight is calculated as:
$\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{M}{6}$

(A) In the given reaction: ${2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-}$.
First,determine the change in oxidation state of sulfur.
In ${S_2O_3^{2-}}$,the average oxidation state of $S$ is $+2$.
In ${S_4O_6^{2-}}$,the average oxidation state of $S$ is $+2.5$.
The change in oxidation state per sulfur atom is ${2.5 - 2 = 0.5}$.
Since there are $2$ sulfur atoms in ${S_2O_3^{2-}}$,the total change in oxidation state per molecule of ${S_2O_3^{2-}}$ is ${2 \times 0.5 = 1}$.
Therefore,the n-factor for ${S_2O_3^{2-}}$ is $1$.
The equivalent weight is defined as $\frac{\text{Molecular weight}}{\text{n-factor}}$.
Thus,the equivalent weight is $\frac{\text{Molecular weight}}{1}$.

(D) To balance the redox reaction,we use the ion-electron method.
Step $1$: Write the half-reactions.
Reduction: $MnO_4^- \rightarrow Mn^{2+}$
Oxidation: $C_2O_4^{2-} \rightarrow CO_2$
Step $2$: Balance atoms and charges.
Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Oxidation: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
Step $3$: Equalize the number of electrons.
Multiply reduction half-reaction by $2$ and oxidation half-reaction by $5$.
$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$
$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$
Step $4$: Add the half-reactions.
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
The stoichiometric coefficient of the oxalate ion $(C_2O_4^{2-})$ is $5$.

(D) In the given reaction: $2CuSO_4 + 4KI \rightarrow Cu_2I_2 + 2K_2SO_4 + I_2$.
The oxidation state of $Cu$ in $CuSO_4$ is $+2$.
The oxidation state of $Cu$ in $Cu_2I_2$ (which is $CuI$) is $+1$.
The change in oxidation state per $Cu$ atom is $2 - 1 = 1$.
Since there is one $Cu$ atom in $CuSO_4$,the $n$-factor for $CuSO_4$ is $1$.
The equivalent weight is given by the formula: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{n\text{-factor}}$.
Therefore,$\frac{\text{Equivalent weight}}{\text{Molecular weight}} = \frac{1}{n\text{-factor}} = \frac{1}{1} = 1$.

(C) The equivalent weight of an oxidizing or reducing agent is calculated by dividing its molecular weight $(M)$ by the number of electrons gained or lost per molecule ($n$-factor).
In the given half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$.
The total number of electrons gained by one molecule of $Cr_2O_7^{2-}$ is $6$.
Therefore,the $n$-factor for $Cr_2O_7^{2-}$ is $6$.
Equivalent weight = $\frac{\text{Molecular weight}}{6}$.

(D) The balanced chemical equation for the disproportionation of potassium manganate in acidic medium is:
$3K_2MnO_4 + 2H_2SO_4 \rightarrow 2KMnO_4 + K_2SO_4 + MnO_2 + 2H_2O$
From the balanced equation,the stoichiometric coefficients for the reactants are $3$ for $K_2MnO_4$ and $2$ for $H_2SO_4$.
Therefore,the molar ratio of $K_2MnO_4$ to $H_2SO_4$ is $3 : 2$.

(A) 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:
$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$

(C) In an acidic medium,the oxidation state of $Mn$ in $MnO_4^-$ changes from $+7$ to $+2$.
The half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
Therefore,one mole of $MnO_4^-$ accepts $5$ moles of electrons.

(D) The reaction between potassium dichromate $(K_2Cr_2O_7)$ and potassium iodide $(KI)$ in an acidic medium is given by:
$K_2Cr_2O_7 + 7H_2SO_4 + 6KI \rightarrow 4K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3I_2$
In the product $Cr_2(SO_4)_3$,chromium exists as $Cr^{3+}$ ion.
Therefore,the oxidation state of chromium in the final product is $+3$.

(C) Acidic $KMnO_4$ acts as a strong oxidizing agent.
Mohr's salt,which is $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$,contains $Fe^{2+}$ ions.
When it reacts with acidic $KMnO_4$,the $Fe^{2+}$ ions are oxidized to $Fe^{3+}$ ions,and the purple $MnO_4^-$ ions are reduced to colorless $Mn^{2+}$ ions.
Therefore,the solution becomes colorless.

(C) In the reaction $MnO_2 + 4HCl \to MnCl_2 + 2H_2O + Cl_2$:
$1$. The oxidation state of $Mn$ in $MnO_2$ is $+4$ and in $MnCl_2$ is $+2$. Since the oxidation state decreases,$Mn^{4+}$ is reduced.
$2$. The oxidation state of $Cl$ in $HCl$ is $-1$ and in $Cl_2$ is $0$. Since the oxidation state increases,$Cl^-$ is oxidised.
Therefore,statements $(2)$ and $(3)$ are correct.

(B) To check if the equation is balanced,we verify the number of atoms and the total charge on both sides for option $B$:
Left side: $5 \times Bi$,$15 \times O$,$14 \times H$,$2 \times Mn$. Total charge: $(-5) + (+14) + (+4) = +13$.
Right side: $5 \times Bi$,$7 \times O + 8 \times O = 15 \times O$,$14 \times H$,$2 \times Mn$. Total charge: $(+15) + (-2) = +13$.
Since both atoms and charges are balanced,option $B$ is correct.

(C) The reduction half-reaction for $K_2Cr_2O_7$ in an acidic medium is:
$Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \to 2Cr^{3+} + 7H_2O$
In this reaction,the change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change in oxidation number is $2 \times (6 - 3) = 6$.
Thus,the $n$-factor for $K_2Cr_2O_7$ is $6$.
The equivalent weight is calculated as:
$\text{Equivalent weight} = \frac{\text{Molecular weight}}{n\text{-factor}} = \frac{M}{6}$.

(C) In acidic medium,the dichromate ion $(Cr_2O_7^{2-})$ acts as an oxidizing agent and is reduced to chromium$(III)$ ion $(Cr^{3+})$.
The half-reaction is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$.
The change in oxidation state of chromium is from $+6$ to $+3$ for each $Cr$ atom. Since there are two $Cr$ atoms in $K_2Cr_2O_7$,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,the $n$-factor for $K_2Cr_2O_7$ is $6$.
The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{n\text{-factor}} = \frac{294}{6} = 49 \ g$.

(A) The chemical reaction is: $3Sn + K_2Cr_2O_7 + 14HCl \rightarrow 3SnCl_4 + 2CrCl_3 + 2KCl + 7H_2O$.
In this reaction,$Sn$ is oxidized from $Sn^0$ to $Sn^{4+}$,so the n-factor for $Sn$ is $4$.
Equivalents of $Sn = \frac{\text{mass}}{\text{equivalent weight}} = \frac{1}{118.7/4} = \frac{4}{118.7} \approx 0.0337 \ eq$.
For a decinormal solution,normality $N = 0.1 \ N$.
Equivalents of $K_2Cr_2O_7 = N \times V(L) = 0.1 \times V$.
Since equivalents of $Sn$ must equal equivalents of $K_2Cr_2O_7$:
$0.1 \times V = 0.0337$.
$V = 0.337 \ L = 337 \ mL$.
However,using the standard stoichiometry where $3 \ mol$ of $Sn$ reacts with $1 \ mol$ of $K_2Cr_2O_7$ $(6 \ eq)$:
$1 \ g$ of $Sn = \frac{1}{118.7} \ mol$.
$mol$ of $K_2Cr_2O_7 = \frac{1}{3} \times \frac{1}{118.7} \ mol$.
Equivalents of $K_2Cr_2O_7 = 6 \times \frac{1}{3 \times 118.7} = \frac{2}{118.7} \approx 0.01685 \ eq$.
$0.1 \times V(L) = 0.01685 \implies V = 0.1685 \ L = 168.5 \ mL$.

(B) The equivalent weight is given by the formula: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{n\text{-factor}}$.
For the equivalent weight to be half of the molecular weight,the $n$-factor must be $2$.
In $MnSO_4$,the oxidation state of $Mn$ is $+2$.
In $MnO_2$,the oxidation state of $Mn$ is $+4$.
The change in oxidation state is $|4 - 2| = 2$.
Thus,the reaction $Mn^{2+} \rightarrow Mn^{4+} + 2e^-$ involves the loss of $2$ electrons,making the $n$-factor $2$.

(D) The balanced chemical equation for the reaction of potassium dichromate $(K_2Cr_2O_7)$ with ferrous sulphate $(FeSO_4)$ in an acidic medium is given by:
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
From the stoichiometry of the reaction,$1 \text{ mole}$ of $K_2Cr_2O_7$ reacts with $6 \text{ moles}$ of $Fe^{2+}$ ions.
Therefore,$1 \text{ mole}$ of potassium dichromate oxidises $6 \text{ moles}$ of ferrous sulphate.

(C) In acidic medium,$MnO_4^-$ oxidizes ferrous oxalate $(FeC_2O_4)$ as follows:
$3MnO_4^- + 5FeC_2O_4 + 24H^+ \rightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$
From the balanced chemical equation,$5$ moles of ferrous oxalate require $3$ moles of $MnO_4^-$.
Therefore,$1$ mole of ferrous oxalate requires $\frac{3}{5} = 0.6$ moles of $MnO_4^-$.

(C) When copper reacts with concentrated $HNO_3$,it acts as a strong oxidizing agent and produces copper$(II)$ nitrate,water,and nitrogen dioxide gas.
The balanced chemical equation is:
$Cu(s) + 4HNO_3(conc.) \longrightarrow Cu(NO_3)_2(aq) + 2H_2O(l) + 2NO_2(g)$
The reddish-brown gas evolved is $NO_2$.

(D) $KMnO_4$ $(Mn^{7+})$ is reduced to $Mn^{2+}$,involving $5$ electrons per mole of $KMnO_4$.
For the same number of moles of the compound,the one requiring the least number of electrons for oxidation will require the least amount of $KMnO_4$.
$(a)$ For $FeSO_3$:
$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$
$SO_3^{2-} \rightarrow SO_4^{2-} + 2e^-$
Total $e^- = 1 + 2 = 3$
$(b)$ For $FeC_2O_4$:
$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$
$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
Total $e^- = 1 + 2 = 3$
$(c)$ For $Fe(NO_2)_2$:
$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$
$2NO_2^- \rightarrow 2NO_3^- + 4e^-$
Total $e^- = 1 + 4 = 5$
$(d)$ For $FeSO_4$:
$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$
Total $e^- = 1$
Since $FeSO_4$ involves the minimum number of electrons $(1)$,it requires the least amount of $KMnO_4$.

(B) The reaction of aqueous $KMnO_4$ with $H_2O_2$ in acidic medium is $3H_2SO_4 + 2KMnO_4 + 5H_2O_2 \rightarrow 5O_2 + 2MnSO_4 + 8H_2O + K_2SO_4$.
In the above reaction,$KMnO_4$ oxidises $H_2O_2$ to $O_2$ and itself $[MnO_4^-]$ gets reduced to $Mn^{2+}$ ion as $MnSO_4$.
Hence,aqueous solution of $KMnO_4$ with $H_2O_2$ yields $Mn^{2+}$ and $O_2$ in acidic conditions.

(C) The half-equations of the reaction are:
$MnO_4^{-} \rightarrow Mn^{2+}$
$C_2O_4^{2-} \rightarrow CO_2$
The balanced half-equations are:
$(MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O) \times 2$
$(C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-}) \times 5$
Equating the number of electrons,we get:
$2MnO_4^{-} + 16H^{+} + 10e^{-} \rightarrow 2Mn^{2+} + 8H_2O$
$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^{-}$
On adding both the equations,we get:
$2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
Thus,the values of $x, y$ and $z$ are $2, 5$ and $16,$ respectively.

(C) The oxidation half-reaction is: $2I^{-} \to I_2 + 2e^{-}$ (multiplied by $3$ to balance electrons: $6I^{-} \to 3I_2 + 6e^{-}$).
The reduction half-reaction is: $ClO_3^{-} + 6H^{+} + 6e^{-} \to Cl^{-} + 3H_2O$.
Adding these,we get: $6I^{-} + ClO_3^{-} + 6H^{+} \to 3I_2 + Cl^{-} + 3H_2O$.
Since $H_2SO_4$ provides $H^{+}$,we add $6HSO_4^{-}$ to both sides: $6I^{-} + ClO_3^{-} + 6H_2SO_4 \to 3I_2 + Cl^{-} + 3H_2O + 6HSO_4^{-}$.
Comparing the coefficients,the stoichiometric coefficient of $HSO_4^{-}$ is $6$.