Redox Reactions

To determine the oxidation state of carbon in both isomers:
1. In $\text{HCN}$ ($\text{H}-\text{C}\equiv\text{N}$):
Electronegativity order is $\text{N} > \text{C} > \text{H}$.
Hydrogen is assigned $+1$.
Nitrogen is assigned $-3$.
Let $x$ be the oxidation number of carbon:
$(+1) + x + (-3) = 0$
$x - 2 = 0$
$x = +2$
2. In $\text{HNC}$ ($\text{H}-\text{N}\equiv\text{C}$):
Hydrogen is assigned $+1$.
Nitrogen is assigned $-3$.
Let $y$ be the oxidation number of carbon:
$(+1) + (-3) + y = 0$
$y - 2 = 0$
$y = +2$
Therefore, the oxidation numbers of $\text{C}$ in $\text{HCN}$ and $\text{HNC}$ are $+2$ and $+2$, respectively.

Potassium ($\text{K}$) belongs to Group 1 of the periodic table and consistently shows an oxidation number of $+1$ in its compounds.
In $\text{KO}_2$:
Let $x$ be the oxidation state of potassium and $y$ be the oxidation state of oxygen.
Since $\text{KO}_2$ is a superoxide containing the superoxide ion ($\text{O}_2^-$), oxygen is in the $-\frac{1}{2}$ oxidation state.
Setting up the neutral charge equation:
$x + 2 \times (-\frac{1}{2}) = 0$
$x - 1 = 0$
$x = +1$
Thus, the oxidation number of $\text{K}$ in $\text{KO}_2$ is $+1$.

To calculate the oxidation state of $\text{Xe}$ in $\text{Ba}_2\text{XeO}_6$:
1. Assign the known oxidation states:
Oxidation state of $\text{Ba} = +2$
Oxidation state of $\text{O} = -2$
2. Let $x$ be the oxidation state of $\text{Xe}$.
3. Set up the equation for a neutral compound:
$2(+2) + x + 6(-2) = 0$
$+4 + x - 12 = 0$
$x - 8 = 0$
$x = +8$
Therefore, the oxidation state of $\text{Xe}$ in $\text{Ba}_2\text{XeO}_6$ is $+8$.

To find the change in oxidation state of $\text{Cr}$:
1. Oxidation state of $\text{Cr}$ in $\text{K}_2\text{Cr}_2\text{O}_7$:
Let $x$ be the oxidation number of $\text{Cr}$.
$2(+1) + 2x + 7(-2) = 0$
$+2 + 2x - 14 = 0$
$2x = 12$
$x = +6$
2. Oxidation state of $\text{Cr}$ in $\text{K}_2\text{CrO}_4$:
Let $y$ be the oxidation number of $\text{Cr}$.
$2(+1) + y + 4(-2) = 0$
$+2 + y - 8 = 0$
$y = +6$
3. Change in oxidation number:
$\text{Change} = \vert{}+6 - (+6)\vert{} = 0$
Thus, the change in oxidation number of $\text{Cr}$ is 0.

Evaluating the oxidation numbers for the elements involved in the reaction:
1. Bromine ($\text{Br}$):
* In $\text{Br}_2$: Oxidation state is $0$.
* In $\text{Br}^-$: Oxidation state is $-1$ (Reduction, gain of electrons).
* In $\text{BrO}_3^-$: Let $x$ be the oxidation state of $\text{Br}$.
$x + 3(-2) = -1 \implies x = +5$ (Oxidation, loss of electrons).
2. Carbon ($\text{C}$):
* In $\text{CO}_3^{2-}$: Oxidation state is $+4$.
* In $\text{HCO}_3^-$: Oxidation state is $+4$.
There is no change in the oxidation state of carbon.
3. Hydrogen ($\text{H}$) and Oxygen ($\text{O}$):
* Oxidation states remain $+1$ for hydrogen and $-2$ for oxygen throughout the reaction.
Therefore, bromine is both oxidized and reduced.

Analyzing the oxidation states of the key elements in each reaction:
1. Reaction A:
$\text{Br}_2$ (oxidation state $0$) is reduced to $\text{Br}^-$ (oxidation state $-1$) and oxidized to $\text{BrO}_3^-$ (oxidation state $+5$). This is a disproportionation reaction.
2. Reaction B:
$\text{P}_4$ (oxidation state $0$) is reduced to $\text{PH}_3$ (oxidation state $-3$) and oxidized to $\text{H}_2\text{PO}_2^-$ (oxidation state $+1$). This is a disproportionation reaction.
3. Reaction C:
Sulfur in $\text{H}_2\text{S}$ has an oxidation state of $-2$, and sulfur in $\text{SO}_2$ has an oxidation state of $+4$. Both react to form elemental sulfur $\text{S}$ with an oxidation state of $0$. Since two different oxidation states converge to one intermediate oxidation state, this is a comproportionation reaction, not a disproportionation reaction.
4. Reaction D:
Oxygen in $\text{H}_2\text{O}_2$ (oxidation state $-1$) is reduced to $\text{H}_2\text{O}$ (oxidation state $-2$) and oxidized to $\text{O}_2$ (oxidation state $0$). This is a disproportionation reaction.
Therefore, the reaction in option C is not a disproportionation reaction.

Step-by-step balancing of the redox reaction:
1. Oxidation half-reaction:
$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$
2. Reduction half-reaction:
$\text{NO}_3^- \rightarrow \text{NH}_4^+$
Oxidation state of nitrogen changes from $+5$ in $\text{NO}_3^-$ to $-3$ in $\text{NH}_4^+$ (gain of $8e^-$):
$\text{NO}_3^- + 8e^- \rightarrow \text{NH}_4^+$
Balance oxygen atoms by adding $3\text{H}_2\text{O}$ to the right side:
$\text{NO}_3^- + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O}$
Balance hydrogen atoms by adding $10\text{H}^+$ to the left side:
$\text{NO}_3^- + 10\text{H}^+ + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O}$
Convert to basic medium by adding $10\text{OH}^-$ to both sides:
$\text{NO}_3^- + 10\text{H}_2\text{O} + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O} + 10\text{OH}^-$
Simplify water molecules:
$\text{NO}_3^- + 7\text{H}_2\text{O} + 8e^- \rightarrow \text{NH}_4^+ + 10\text{OH}^-$
3. Equalize electrons by multiplying the oxidation half-reaction by 4:
$4\text{Zn} \rightarrow 4\text{Zn}^{2+} + 8e^-$
4. Combine both half-reactions:
$4\text{Zn} + \text{NO}_3^- + 7\text{H}_2\text{O} \rightarrow 4\text{Zn}^{2+} + \text{NH}_4^+ + 10\text{OH}^-$
The coefficients of $\text{Zn}$, $\text{NO}_3^-$, and $\text{OH}^-$ in the balanced equation are 4, 1, and 10, respectively.

To find the n-factor for the reaction $\text{KMnO}_4 \xrightarrow{\text{H}^+} \text{Mn}^{2+}$, we first determine the oxidation numbers of the manganese atom in both the reactant and the product. In potassium permanganate ($\text{KMnO}_4$), oxygen has an oxidation state of $-2$ and potassium is $+1$, making manganese $+7$. In the product $\text{Mn}^{2+}$, the oxidation state is clearly $+2$. The magnitude of the change is $7 - 2 = 5$. Because there is one manganese atom per molecule, the n-factor is 5.

To find the n-factor for the reaction $\text{KMnO}_4 \xrightarrow{\text{H}_2\text{O}} \text{Mn}^{4+}$, we identify the oxidation states of Manganese ($\text{Mn}$). In the permanganate ion ($\text{KMnO}_4$), Manganese is in the $+7$ oxidation state. In the product ion $\text{Mn}^{4+}$, it is in the $+4$ oxidation state. The difference in oxidation state is $|7 - 4| = 3$. Since there is one $\text{Mn}$ atom per molecule of $\text{KMnO}_4$, the n-factor is $3 \times 1 = 3$.

To find the n-factor for the reaction $\text{KMnO}_4 \xrightarrow{\text{OH}^- \text{ (Conc. basic medium)}} \text{Mn}^{6+}$, we look at the change in the oxidation number of Manganese ($\text{Mn}$). In potassium permanganate ($\text{KMnO}_4$), Manganese holds an oxidation state of $+7$. In the product $\text{Mn}^{6+}$, it is in the $+6$ state. The magnitude of the change is $|7 - 6| = 1$. Multiplying by the number of $\text{Mn}$ atoms (which is 1), we get an n-factor of 1.

To find the n-factor for the reaction $\text{K}_2\text{Cr}_2\text{O}_7 \xrightarrow{\text{H}^+} \text{Cr}^{3+}$, we must determine the oxidation number of Chromium in both the reactant and the product. In dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$), Chromium is in the $+6$ oxidation state. In the product $\text{Cr}^{3+}$, the state is $+3$. The difference in oxidation state is $|6 - 3| = 3$. Because there are two Chromium atoms per formula unit of dichromate, the total n-factor is $3 \times 2 = 6$.

To find the n-factor for the conversion of $\text{C}_2\text{O}_4^{2-}$ to $\text{CO}_2$, we calculate the change in the oxidation state of Carbon. In the oxalate ion ($\text{C}_2\text{O}_4^{2-}$), the oxidation state of Carbon is $+3$. In $\text{CO}_2$, the oxidation state of Carbon is $+4$. The magnitude of the change is $|4 - 3| = 1$. Since there are two Carbon atoms in the oxalate ion, the total n-factor is $1 \times 2 = 2$.

To find the n-factor for the reaction $\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3$, we identify the oxidation numbers of Iron ($\text{Fe}$). In iron(II) sulfate ($\text{FeSO}_4$), Iron is in the $+2$ oxidation state. In iron(III) oxide ($\text{Fe}_2\text{O}_3$), Iron is in the $+3$ oxidation state. The difference in oxidation state is $|3 - 2| = 1$. Because the balanced chemical equation shows that $2$ atoms of $\text{Fe}$ are involved in the transition to $\text{Fe}_2\text{O}_3$, the total n-factor is $1 \times 2 = 2$.

To find the n-factor for the reaction $\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3$, we identify the oxidation numbers of Iron ($\text{Fe}$). In iron(II) sulfate ($\text{FeSO}_4$), Iron is in the $+2$ oxidation state. In iron(III) oxide ($\text{Fe}_2\text{O}_3$), Iron is in the $+3$ oxidation state. The difference in oxidation state is $|3 - 2| = 1$. Because the balanced chemical equation shows that $2$ atoms of $\text{Fe}$ are involved in the transition to $\text{Fe}_2\text{O}_3$, the total n-factor is $1 \times 2 = 2$.

Given oxidation states:
Oxidation state of $\text{A} = +6$
Oxidation state of $\text{B} = -1$
Oxidation state of $\text{C} = -2$
To find a valid neutral molecular formula, test the total sum of oxidation states for the given options:
1. For $\text{AB}_2\text{C}_2$:
$\text{Sum} = 1(+6) + 2(-1) + 2(-2)$
$\text{Sum} = +6 - 2 - 4 = 0$
Since the total charge equals zero, $\text{AB}_2\text{C}_2$ represents a neutral and valid molecular formula for the compound (another valid formula would be $\text{AB}_4\text{C}$).

1. Identify the given values:
Mass of solute ($W$) = $15.8\text{ g}$
Volume of solution ($V$) = $50\text{ mL}$
Molar mass of $\text{KMnO}_4 = 158\text{ g/mol}$
2. Calculate the equivalent weight ($E$):
In acidic medium, valence factor = $5$.
$E = \frac{158}{5} = 31.6\text{ g/eq}$
3. Calculate the normality ($N$):
$N = \frac{W \times 1000}{E \times V(\text{mL})}$
$N = \frac{15.8 \times 1000}{31.6 \times 50}$
$N = \frac{15800}{1580}$
$N = 10\text{ N}$
Therefore, the normality of the solution is $10\text{ N}$.

1. Identify the molarity and valence factor:
Molarity ($M$) = $5\text{ M}$
Valence factor for $\text{K}_2\text{Cr}_2\text{O}_7$ in acidic medium = $6$
2. Calculate the normality using the relationship formula:
$\text{Normality} = \text{Molarity} \times \text{Valence factor}$
$\text{Normality} = 5 \times 6 = 30\text{ N}$
Therefore, the normality of the $\text{K}_2\text{Cr}_2\text{O}_7$ solution in acidic medium is $30\text{ N}$.

1. Determine the valence factor ($\text{v.f.}$) for each reactant:
* For $\text{KMnO}_4$ in acidic medium:
$\text{v.f.} = \vert{}+7 - (+2)\vert{} = 5$
* For $\text{Cu}_2\text{S}$:
$\text{Cu}_2\text{S} \rightarrow 2\text{Cu}^{2+} + \text{SO}_2$
Change for $2\text{Cu}^+$: $2 \times (+2 - 1) = 2$
Change for $\text{S}^{2-}$: $+4 - (-2) = 6$
Total $\text{v.f.} = 2 + 6 = 8$
2. Apply the law of equivalence:
$\text{Equivalents of } \text{KMnO}_4 = \text{Equivalents of } \text{Cu}_2\text{S}$
$\text{Moles of } \text{KMnO}_4 \times 5 = \text{Moles of } \text{Cu}_2\text{S} \times 8$
3. Given $1$ mole of $\text{Cu}_2\text{S}$:
$n_2 \times 5 = 1 \times 8$
$n_2 = \frac{8}{5} = 1.6$
Therefore, $1.6$ moles of $\text{KMnO}_4$ are required.

1. Calculate the mass of $\text{NaOH}$ present in $1000\text{ mL}$ of $0.15\text{ M}$ solution:
$\text{Molar mass of NaOH} = 23 + 16 + 1 = 40\text{ g/mol}$
$\text{Amount of NaOH in } 1000\text{ mL} = 0.15 \times 40 = 6\text{ g}$
2. Calculate the mass of $\text{NaOH}$ contained per $\text{mL}$ of this solution:
$1\text{ mL of solution contains } \frac{6}{1000}\text{ g} = 6 \times 10^{-3}\text{ g of NaOH}$
3. Calculate the volume required for $1.184\text{ g}$ of $\text{NaOH}$:
$\text{Volume} = \frac{1.184}{6 \times 10^{-3}} = \frac{1184}{6} = 197.33\text{ mL}$
Therefore, $197.33\text{ mL}$ of $0.15\text{ M}$ $\text{NaOH}$ solution is required.

Let us evaluate each statement individually:
1. Statement (a): Gram molecular weight ($\text{GMW}$) is defined as the molecular weight expressed in grams, which corresponds to the mass of $1\text{ mole}$ ($6.022 \times 10^{23}$) of molecules. Thus, statement (a) is correct.
2. Statement (b): One mole of any substance contains Avogadro's number ($N_A$) of fundamental particles, which equals $6.022 \times 10^{23}$ molecules. Thus, statement (b) is correct.
3. Statement (c): The term "gram-molecule" is synonymous with one mole of molecules. Therefore, $1\text{ mole} = 1\text{ gram-molecule}$, making statement (c) correct.
Since statements (a), (b), and (c) are all correct, the correct choice is (d).

A substance shows oxidising behaviour when it undergoes reduction (decrease in oxidation state). In the reaction $\text{2HI} + \text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + \text{SO}_2 + \text{2H}_2\text{O}$, the oxidation state of Sulfur ($\text{S}$) in $\text{H}_2\text{SO}_4$ is $+6$ and it decreases to $+4$ in $\text{SO}_2$. This reduction of $\text{H}_2\text{SO}_4$ confirms its role as an oxidising agent. In the other three reactions, the oxidation state of sulfur remains unchanged at $+6$, meaning no redox reaction involving $\text{H}_2\text{SO}_4$ as an oxidising agent occurs.

To identify the reaction where $\text{H}_2\text{O}_2$ acts as a reducing agent, we look for the process where it undergoes oxidation (increase in oxidation state). In reaction (b), $\text{Cl}_2 + \text{H}_2\text{O}_2 \rightarrow \text{2HCl} + \text{O}_2$, the oxygen in $\text{H}_2\text{O}_2$ has an oxidation state of $-1$. It changes to $\text{O}_2$, where the oxidation state is $0$. Since the oxidation state increases, $\text{H}_2\text{O}_2$ is being oxidized and is therefore acting as the reducing agent. In the other three reactions, $\text{H}_2\text{O}_2$ is reduced to water, making it an oxidising agent in those cases.

Analyzing the changes in oxidation states for each reaction:
1. Reaction A ($\text{CrO}_4^{2-} \rightarrow \text{Cr}_2\text{O}_7^{2-}$):
* In chromate ion ($\text{CrO}_4^{2-}$):
$x + 4(-2) = -2$
$x = +6$
* In dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$):
$2x + 7(-2) = -2$
$2x = 12$
$x = +6$
Since the oxidation state of chromium remains $+6$ on both sides, this reaction involves neither oxidation nor reduction.
2. Reaction B ($\text{Cr} \rightarrow \text{CrCl}_3$):
Chromium changes its oxidation state from $0$ to $+3$, which is an oxidation process.
3. Reaction C ($\text{Na} \rightarrow \text{Na}^+$):
Sodium changes its oxidation state from $0$ to $+1$, which is an oxidation process.
4. Reaction D ($2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-}$):
Sulfur changes its average oxidation state from $+2$ in $\text{S}_2\text{O}_3^{2-}$ to $+2.5$ in $\text{S}_4\text{O}_6^{2-}$, which is an oxidation process.

Step-by-step balancing of the redox reaction in acidic medium:
1. Oxidation half-reaction:
$\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^-$
2. Reduction half-reaction:
$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
3. Equalize the electrons transferred by multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2:
$5\text{H}_2\text{O}_2 \rightarrow 5\text{O}_2 + 10\text{H}^+ + 10e^-$
$2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$
4. Combine both half-reactions and simplify $\text{H}^+$ ions:
$6\text{H}^+ + 5\text{H}_2\text{O}_2 + 2\text{MnO}_4^- \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}$
The stoichiometric coefficients of $\text{H}^+$, $\text{H}_2\text{O}_2$, and $\text{MnO}_4^-$ are 6, 5, and 2, respectively.

Step-by-step balancing of the reaction:
1. Oxidation half-reaction:
$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$
2. Reduction half-reaction:
Nitrogen in $\text{NO}_3^-$ ($+5$) is reduced to $\text{NH}_4^+$ ($-3$), taking $8$ electrons:
$\text{NO}_3^- + 10\text{H}^+ + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O}$
3. Multiply the oxidation half-reaction by 4 to equalize electrons transferred ($8e^-$):
$4\text{Zn} \rightarrow 4\text{Zn}^{2+} + 8e^-$
4. Combine both ionic half-reactions:
$4\text{Zn} + \text{NO}_3^- + 10\text{H}^+ \rightarrow 4\text{Zn}^{2+} + \text{NH}_4^+ + 3\text{H}_2\text{O}$
5. Convert to molecular equation by adding $9\text{NO}_3^-$ ions to both sides:
$4\text{Zn} + 10\text{HNO}_3\text{(dil)} \rightarrow 4\text{Zn(NO}_3\text{)}_2 + 3\text{H}_2\text{O} + \text{NH}_4\text{NO}_3$
Therefore, the balanced stoichiometric coefficients for $\text{Zn}$, $\text{HNO}_3$, and $\text{Zn(NO}_3\text{)}_2$ are 4, 10, and 4, respectively.

Step-by-step balancing using the change in oxidation number method:
1. Oxidation process ($\text{CrI}_3 \rightarrow \text{CrO}_4^{2-} + 3\text{IO}_4^-$):
* For $\text{Cr}$: $+3 \rightarrow +6$ (increase of $3$)
* For $3\text{I}$: $3 \times (-1) \rightarrow 3 \times (+7)$ (increase of $24$)
Total increase in oxidation number per $\text{CrI}_3 = 3 + 24 = 27$
2. Reduction process ($\text{Cl}_2 \rightarrow 2\text{Cl}^-$):
* For $\text{Cl}_2$: $2 \times 0 \rightarrow 2 \times (-1)$ (decrease of $2$)
3. Equalize increase and decrease in oxidation numbers:
Multiply $\text{CrI}_3$ by $2$ and $\text{Cl}_2$ by $27$:
$2\text{CrI}_3 + 27\text{Cl}_2 \rightarrow 2\text{K}_2\text{CrO}_4 + 6\text{KIO}_4 + 54\text{KCl}$
4. Balance potassium ($\text{K}$) atoms by adding $\text{KOH}$:
Right side total $\text{K} = 2(2) + 6(1) + 54(1) = 4 + 6 + 54 = 64\text{ K}$
Add $64\text{ KOH}$ to the left side:
$2\text{CrI}_3 + 64\text{KOH} + 27\text{Cl}_2 \rightarrow 2\text{K}_2\text{CrO}_4 + 6\text{KIO}_4 + 54\text{KCl} + 32\text{H}_2\text{O}$
The stoichiometric coefficients of $\text{CrI}_3$, $\text{KOH}$, and $\text{Cl}_2$ are 2, 64, and 27, respectively.

Step-by-step balancing of the disproportionation reaction:
1. Analyze changes in oxidation states of phosphorus:
* Oxidation state of $\text{P}$ in $\text{P}_2\text{H}_4$: $2x + 4(+1) = 0 \Rightarrow x = -1$
* Oxidation state of $\text{P}$ in $\text{PH}_3$: $x + 3(+1) = 0 \Rightarrow x = -3$
* Oxidation state of $\text{P}$ in $\text{P}_4$: $0$
2. Calculate electron changes per unit:
* Reduction ($\text{P}^{-1} \rightarrow \text{P}^{-3}$): gain of $2e^-$ per $\text{P}$ atom.
* Oxidation ($4\text{P}^{-1} \rightarrow \text{P}_4^0$): loss of $4 \times 1 = 4e^-$ per $\text{P}_4$ molecule formed.
3. Balance electrons by adjusting the ratio of reduced to oxidized products:
To balance $4e^-$ lost in forming $1$ molecule of $\text{P}_4$, $2$ molecules of $\text{PH}_3$ must be formed (since forming $1$ $\text{PH}_3$ gains $2e^-$).
Thus, $\text{PH}_3 : \text{P}_4 = 8 : 1$ to satisfy hydrogen and total atom balances:
$6\text{P}_2\text{H}_4 \rightarrow 8\text{PH}_3 + \text{P}_4$
4. Verify atom counts:
* Left side: $12\text{ P}$, $24\text{ H}$
* Right side: $8 + 4 = 12\text{ P}$, $8 \times 3 = 24\text{ H}$
The balanced equation is $6\text{P}_2\text{H}_4 \rightarrow 8\text{PH}_3 + \text{P}_4$. The coefficients are 6, 8, and 1, respectively.

Step-by-step balancing of the redox reaction:
1. Analyze oxidation state changes:
* In $\text{Ca}_3(\text{PO}_4)_2$, oxidation state of $\text{P} = +5$.
* In $\text{P}_4$, oxidation state of $\text{P} = 0$.
* Change per $\text{P}$ atom = $5e^-$ gained.
* For $1$ molecule of $\text{P}_4$ ($4$ atoms of $\text{P}$), total electrons gained = $4 \times 5 = 20e^-$.
* In elemental $\text{C}$, oxidation state = $0$.
* In $\text{CO}$, oxidation state of $\text{C} = +2$.
* Change per $\text{C}$ atom = $2e^-$ lost.
2. Equalize electron loss and gain:
* Number of carbon atoms needed = $\frac{20e^-}{2e^-/\text{C}} = 10\text{ C}$.
* This forms $10\text{ CO}$.
3. Balance phosphorus ($\text{P}$):
* To produce $1\text{ P}_4$ ($4\text{ P}$ atoms), we need $2$ formula units of $\text{Ca}_3(\text{PO}_4)_2$.
4. Balance calcium ($\text{Ca}$) and silicon ($\text{Si}$):
* $2\text{Ca}_3(\text{PO}_4)_2$ contains $6\text{ Ca}$ atoms.
* Therefore, $6\text{ CaSiO}_3$ are formed on the right side.
* This requires $6\text{ SiO}_2$ on the left side.
5. Verify oxygen ($\text{O}$) balance:
* Left side: $2 \times 8 \text{ (from phosphate)} + 6 \times 2 \text{ (from silica)} = 16 + 12 = 28\text{ O}$.
* Right side: $6 \times 3 \text{ (from calcium silicate)} + 10 \times 1 \text{ (from CO)} = 18 + 10 = 28\text{ O}$.
The balanced equation is:
$2\text{Ca}_3(\text{PO}_4)_2 + 6\text{SiO}_2 + 10\text{C} \rightarrow 6\text{CaSiO}_3 + \text{P}_4 + 10\text{CO}$
The respective stoichiometric coefficients are 2, 6, 10, 6, 1, and 10.

1. Calculate the net charge on the left side:
$\text{Charge on } \text{NO}_3^- = -1$
$\text{Charge on } 4\text{H}^+ = +4$
$\text{Total left side charge} = -1 + 4 = +3$
2. Calculate the net charge on the right side:
$\text{Charge on } 2\text{H}_2\text{O} = 0$
$\text{Charge on } \text{NO} = 0$
$\text{Total right side charge} = 0$
3. Determine the number of electrons required to balance the charge:
$\text{Charge difference} = +3 - 0 = +3$
Adding $3$ electrons ($3e^-$) to the left side reduces its overall charge from $+3$ to $0$, matching the right side:
$\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow 2\text{H}_2\text{O} + \text{NO}$
Therefore, $3$ electrons are required on the left side.

1. Identify the reductant:
In $\text{H}_2\text{S}$, sulfur has an oxidation state of $-2$.
In elemental sulfur ($\text{S}$), the oxidation state is $0$.
Since the oxidation state of sulfur increases from $-2$ to $0$, $\text{H}_2\text{S}$ is oxidized and acts as the reductant.
2. Calculate the electron loss per mole of reductant:
$\text{S}^{2-} \rightarrow \text{S}^0 + 2e^-$
3. Determine the electrons involved:
For $1\text{ mole}$ of $\text{H}_2\text{S}$, $2\text{ moles}$ of electrons are lost during oxidation.
Therefore, $2$ moles of electrons are involved in the oxidation of $1$ mole of reductant.

1. Identify the reductant:
In $\text{H}_2\text{S}$, sulfur has an oxidation state of $-2$.
In elemental sulfur ($\text{S}$), the oxidation state is $0$.
Since the oxidation state of sulfur increases from $-2$ to $0$, $\text{H}_2\text{S}$ is oxidized and acts as the reductant.
2. Calculate the electron loss per mole of reductant:
$\text{S}^{2-} \rightarrow \text{S}^0 + 2e^-$
3. Determine the electrons involved:
For $1\text{ mole}$ of $\text{H}_2\text{S}$, $2\text{ moles}$ of electrons are lost during oxidation.
Therefore, $2$ moles of electrons are involved in the oxidation of $1$ mole of reductant.

Step-by-step balancing of the redox reaction:
1. Oxidation half-reaction:
Iron ($\text{Fe}$) is oxidized from $0$ to $+2$:
$\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$
2. Reduction half-reaction:
Nitrogen in nitrate ($\text{NO}_3^-$) is reduced from $+5$ to $-3$ in ammonium ($\text{NH}_4^+$):
$\text{NO}_3^- + 10\text{H}^+ + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O}$
3. Equalize electrons by multiplying the oxidation half-reaction by 4:
$4\text{Fe} \rightarrow 4\text{Fe}^{2+} + 8e^-$
4. Combine both half-reactions:
$4\text{Fe} + \text{NO}_3^- + 10\text{H}^+ \rightarrow 4\text{Fe}^{2+} + \text{NH}_4^+ + 3\text{H}_2\text{O}$
5. Convert to molecular form by adding $9\text{NO}_3^-$ to both sides:
$4\text{Fe} + 10\text{HNO}_3 \rightarrow 4\text{Fe(NO}_3\text{)}_2 + \text{NH}_4\text{NO}_3 + 3\text{H}_2\text{O}$
The stoichiometric coefficients of $\text{HNO}_3$, $\text{Fe(NO}_3\text{)}_2$, and $\text{NH}_4\text{NO}_3$ are 10, 4, and 1, respectively. Thus, the ratio is $10 : 4 : 1$.

1. Calculate the equivalent weight of $\text{Na}_2\text{C}_2\text{O}_4$:
$\text{Molar mass of } \text{Na}_2\text{C}_2\text{O}_4 = 134\text{ g/mol}$
$\text{n-factor} = 2$
$\text{Equivalent weight} = \frac{134}{2} = 67\text{ g/eq}$
2. Calculate the number of gram equivalents:
$\text{Gram equivalents} = \frac{\text{Weight in g}}{\text{Equivalent weight}} = \frac{134}{67} = 0.2\text{ eq}$
3. Calculate the normality:
$\text{Normality } (N) = \frac{0.2}{\frac{100}{1000}} = \frac{0.2}{0.1} = 2\text{ N}$
Therefore, the normality of the solution is $2\text{ N}$.

1. Determine the n-factor for each species:
* For $\text{HNO}_3 \rightarrow \text{NO}$:
Change in oxidation state of nitrogen $= \vert{}+5 - (+2)\vert{} = 3$
$\text{n-factor of } \text{HNO}_3 = 3$
* For $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$:
Change in oxidation state of iron $= \vert{}+3 - (+2)\vert{} = 1$
$\text{n-factor of } \text{Fe}^{2+} = 1$
2. Calculate the equivalent weight of $\text{Fe}^{2+}$:
$\text{Atomic mass of Fe} = 56\text{ g/mol}$
$\text{Equivalent weight of Fe}^{2+} = \frac{56}{1} = 56\text{ g/eq}$
3. Apply the principle of equivalence:
$\text{Meq of } \text{HNO}_3 = \text{Meq of } \text{Fe}^{2+}$
$M \times \text{n-factor} \times V = \frac{\text{Mass of Fe}^{2+}}{\text{Equivalent weight}} \times 1000$
$6 \times 3 \times V = \frac{8}{56} \times 1000$
$18 \times V = \frac{8000}{56}$
$18 \times V = 142.857$
$V = \frac{142.857}{18} = 7.936\text{ mL}$
Therefore, $7.936\text{ mL}$ of $6\text{ M}$ $\text{HNO}_3$ solution is required.

1. Determine the n-factor for $\text{K}_2\text{Cr}_2\text{O}_7$ in acidic medium:
$\text{n-factor} = 6$
2. Calculate the normality of both solutions:
$N_1 = M_1 \times \text{n-factor} = 5 \times 6 = 30\text{ N}$
$N_2 = M_2 \times \text{n-factor} = 2 \times 6 = 12\text{ N}$
3. Calculate the resulting normality ($N_f$) of the mixture:
$N_f = \frac{N_1 V_1 + N_2 V_2}{V_1 + V_2}$
$N_f = \frac{(30 \times 50) + (12 \times 50)}{50 + 50}$
$N_f = \frac{1500 + 600}{100} = \frac{2100}{100} = 21\text{ N}$
Alternatively, calculating using molarity directly:
$M_f = \frac{(5 \times 50) + (2 \times 50)}{100} = \frac{350}{100} = 3.5\text{ M}$
$N_f = M_f \times \text{n-factor} = 3.5 \times 6 = 21\text{ N}$
Therefore, the normality of the final solution is $21\text{ N}$.

1. Define the variables:
Let the volume of $6\text{ M } \text{HCl}$ required be $x\text{ L}$.
Since the total required volume is $2\text{ L}$, the volume of $2\text{ M } \text{HCl}$ required is $(2 - x)\text{ L}$.
2. Set up the mixture equation:
$M_1 V_1 + M_2 V_2 = M_3 V_3$
$(6 \times x) + (2 \times (2 - x)) = 3 \times 2$
3. Solve for $x$:
$6x + 4 - 2x = 6$
$4x + 4 = 6$
$4x = 2$
$x = 0.5\text{ L}$
4. Calculate the required volumes:
* Volume of $6\text{ M } \text{HCl} = x = 0.5\text{ L}$
* Volume of $2\text{ M } \text{HCl} = 2 - 0.5 = 1.5\text{ L}$
Therefore, $0.5\text{ L}$ of $6\text{ M } \text{HCl}$ and $1.5\text{ L}$ of $2\text{ M } \text{HCl}$ are required.

1. Calculate milliequivalents of $\text{H}_2\text{SO}_4$:
$\text{Meq of } \text{H}_2\text{SO}_4 = 45.6 \times 0.235 = 10.716\text{ Meq}$
2. Calculate the mass of pure $\text{Na}_2\text{CO}_3$ required:
$\text{Meq of pure } \text{Na}_2\text{CO}_3 = \text{Meq of } \text{H}_2\text{SO}_4$
$\frac{W_{\text{pure}}}{53} \times 1000 = 10.716$
$W_{\text{pure}} = \frac{10.716 \times 53}{1000} = 0.5679\text{ g}$
3. Calculate the weight of the $85\%$ pure sample:
$\text{Weight of sample} = 0.5679 \times \frac{100}{85} = 0.6681\text{ g}$
Therefore, approximately $0.668\text{ g}$ of $85\%$ pure $\text{Na}_2\text{CO}_3$ is required.

1. Determine the oxidation state of sulfur in reactant and product:
* Oxidation state of $\text{S}$ in $\text{S}_2\text{O}_3^{2-} = +2$
* Oxidation state of $\text{S}$ in $\text{S}_4\text{O}_6^{2-} = +2.5$
2. Calculate the valence factor ($\text{v.f.}$):
Change in oxidation state per sulfur atom $= 2.5 - 2 = 0.5$
Since there are $2$ sulfur atoms in $\text{Na}_2\text{S}_2\text{O}_3$:
$\text{v.f.} = 0.5 \times 2 = 1$
3. Calculate equivalent weight:
$\text{Eq. wt.} = \frac{\text{Mol. wt.}}{1} = \text{Mol. wt.}$
Therefore, the equivalent weight of $\text{Na}_2\text{S}_2\text{O}_3$ is equal to its molecular weight ($\text{Mol. wt.}$).

1. Calculate total milliequivalents of $\text{As}_2\text{O}_3$ taken:
$\text{Equivalent weight of } \text{As}_2\text{O}_3 = \frac{\text{Molar mass}}{4} = \frac{198}{4} = 49.5\text{ g/eq}$
$\text{Meq of } \text{As}_2\text{O}_3 = \frac{0.594}{49.5} \times 1000 = 12\text{ Meq}$
2. Calculate milliequivalents of $\text{KMnO}_4$ used for back-titration:
$\text{Normality of } \text{KMnO}_4 = \text{Molarity} \times \text{n-factor} = \frac{1}{50} \times 5 = 0.1\text{ N}$
$\text{Meq of } \text{KMnO}_4 = 0.1 \times 45 = 4.5\text{ Meq}$
3. Calculate milliequivalents of $\text{MnO}_2$ reacted:
$\text{Meq of } \text{MnO}_2 = \text{Meq of } \text{As}_2\text{O}_3 \text{ total} - \text{Meq of } \text{KMnO}_4$
$\text{Meq of } \text{MnO}_2 = 12 - 4.5 = 7.5\text{ Meq}$
4. Calculate mass of pure $\text{MnO}_2$:
$\text{Equivalent weight of } \text{MnO}_2 = \frac{87}{2} = 43.5\text{ g/eq}$
$\text{Mass of pure } \text{MnO}_2 = \frac{7.5 \times 43.5}{1000} = 0.32625\text{ g}$
5. Calculate percentage of $\text{MnO}_2$ in pyrolusite:
$\% \text{ of } \text{MnO}_2 = \frac{0.32625}{0.5} \times 100 = 65.25\%$
Therefore, the percentage purity of $\text{MnO}_2$ in pyrolusite is $65.25\%$.

Let $x$ be the moles of $\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$ and $y$ be the moles of $\text{KHC}_2\text{O}_4 \cdot \text{H}_2\text{O}$ present in $1.2\text{ g}$ of the sample mixture.
1. Neutralization with $\text{NaOH}$ (using $1.2\text{ g}$ sample):
$\text{Meq of } \text{NaOH} = \text{Normality} \times \text{Volume (mL)} = 0.25 \times 40.0 = 10\text{ Meq} = 0.010\text{ eq}$
$\text{Equivalents of acid} = 2x + 1y$
$2x + y = 0.010$ --- (Equation 1)
2. Redox titration with $\text{KMnO}_4$ (using $0.4\text{ g}$ sample):
Since $0.4\text{ g}$ is $\frac{1}{3}\text{rd}$ of $1.2\text{ g}$, the moles in $0.4\text{ g}$ sample are $\frac{x}{3}$ and $\frac{y}{3}$.
$\text{Meq of } \text{KMnO}_4 = \text{Normality} \times \text{Volume (mL)} = 0.125 \times 40.0 = 5\text{ Meq} = 0.005\text{ eq}$
In redox reaction, both components have an n-factor of $2$ due to oxalate ion:
$2 \left(\frac{x}{3}\right) + 2 \left(\frac{y}{3}\right) = 0.005$
$2x + 2y = 3 \times 0.005 = 0.015$
$x + y = 0.0075$ --- (Equation 2)
3. Solve Equations 1 and 2:
Subtracting Equation 2 from Equation 1:
$(2x + y) - (x + y) = 0.010 - 0.0075$
$x = 0.0025\text{ moles}$
Substitute $x = 0.0025$ into Equation 2:
$0.0025 + y = 0.0075 \Rightarrow y = 0.0050 - 0.0025 = 0.0025\text{ moles}$
4. Calculate the molar ratio:
$\text{Molar ratio} = \frac{x}{y} = \frac{0.0025}{0.0025} = 1 : 1$
Therefore, the molar ratio of $\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$ to $\text{KHC}_2\text{O}_4 \cdot \text{H}_2\text{O}$ is $1 : 1$.