Masterclass: 25 Solved JEE Advanced Numericals on Redox Reactions & Volumetric Analysis
Conquer electron transfer! This exhaustive guide features highly complex multi-step problems on n-factor calculations, Iodometry, Permanganometry, and Disproportionation. Click "View Solution" to reveal the step-by-step breakdown.
Redox Volumetric Analysis demands a flawless understanding of the n-factor (valency factor) and the Law of Chemical Equivalence. Before attempting these rigorous problems, ensure you can accurately track oxidation state changes across complex molecules and handle multi-component back titrations.
View Solution
Step 1: Identify the oxidation changes
In ${FeC_2O_4}$, iron is in the $+2$ state (${Fe^{2+}}$) and the oxalate ion is ${C_2O_4^{2-}}$ (where Carbon is in the $+3$ state).
When oxidized by strong acidic ${KMnO_4}$:
1. ${Fe^{2+}} \rightarrow {Fe^{3+}} + 1e^-$ (Oxidation state of Fe changes from $+2$ to $+3$).
2. ${C_2O_4^{2-}} \rightarrow 2{CO_2} + 2e^-$ (Oxidation state of C changes from $+3$ to $+4$. Since there are 2 carbon atoms, total electron loss $= 2$).
Step 2: Calculate total n-factor
In this specific compound, both the cation and the anion undergo oxidation simultaneously.
Total moles of electrons lost per mole of ${FeC_2O_4}$ = $1 \text{ (from Fe)} + 2 \text{ (from C)} = 3$.
Therefore, the n-factor = $3$.
Step 3: Calculate Equivalent Weight
Equivalent Weight ($E$) = $\frac{\text{Molar Mass}}{\text{n-factor}}$
$E = \frac{144}{3} = 48 \text{ g eq}^{-1}$.
View Solution
Step 1: Write the skeletal reaction and oxidation states
$P_4 \rightarrow PH_3 + {H_2PO_2^-}$
Oxidation state of P in $P_4 = 0$.
Oxidation state of P in $PH_3 = -3$ (Reduction).
Oxidation state of P in ${H_2PO_2^-} = +1$ (Oxidation: let ox state be $x$; $2(+1) + x + 2(-2) = -1 \implies x - 2 = -1 \implies x = +1$).
Step 2: Calculate individual n-factors for oxidation and reduction
For reduction ($P_4 \rightarrow 4PH_3$): Change per atom = $|0 - (-3)| = 3$. For 4 atoms, $n_{red} = 4 \times 3 = 12$.
For oxidation ($P_4 \rightarrow 4{H_2PO_2^-}$): Change per atom = $|0 - 1| = 1$. For 4 atoms, $n_{ox} = 4 \times 1 = 4$.
Step 3: Apply the Disproportionation Formula
When the exact same species undergoes both oxidation and reduction, the overall n-factor ($n_{total}$) is calculated using the formula for parallel resistors:
$n_{total} = \frac{n_{ox} \times n_{red}}{n_{ox} + n_{red}}$
$n_{total} = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3$.
Step 4: Calculate Equivalent Weight
$E = \frac{M}{n_{total}} = \frac{124}{3} = 41.33 \text{ g eq}^{-1}$.
View Solution
Step 1: Understand Iodometric Reactions
Reaction 1 (Liberation of Iodine):
$2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 \downarrow + I_2$
Reaction 2 (Titration of Iodine with Hypo):
$I_2 + 2{S_2O_3^{2-}} \rightarrow 2I^- + {S_4O_6^{2-}}$
Step 2: Apply the Law of Equivalence
According to the law of equivalence across sequential reactions:
Equivalents of $Cu^{2+}$ = Equivalents of $I_2$ liberated = Equivalents of ${Na_2S_2O_3}$ used.
Step 3: Calculate n-factors
For $Cu^{2+} \rightarrow Cu^+$ (in $Cu_2I_2$), the change in oxidation state is 1. So, n-factor of $CuSO_4$ = 1.
For ${S_2O_3^{2-}} \rightarrow {S_4O_6^{2-}}$, oxidation state of S changes from $+2$ to $+2.5$. Change = $0.5$ per S atom. For 2 S atoms in the formula, n-factor = $2 \times 0.5 = 1$.
Step 4: Solve the Volumetric Equation
$N_1 V_1 (CuSO_4) = N_2 V_2 ({Na_2S_2O_3})$
Since n-factors are 1, Normality = Molarity for both.
$M_1 \times 50 \text{ mL} = 0.1 \text{ M} \times 20 \text{ mL}$
$M_1 \times 50 = 2.0$
$M_1 = \frac{2.0}{50} = 0.04 \text{ M}$.
View Solution
Step 1: Analyze Acid-Base Titration (NaOH)
Both $H_2C_2O_4$ and $NaHC_2O_4$ are acidic and will react with $NaOH$.
Let molarity of $H_2C_2O_4$ be $x$ and $NaHC_2O_4$ be $y$.
n-factor for neutralization: $H_2C_2O_4$ is a dibasic acid ($n=2$). $NaHC_2O_4$ is a monobasic acid salt ($n=1$).
Equivalents of acid = Equivalents of base.
$(2x + 1y) \times 10 \text{ mL} = 0.1 \text{ N} \times 3.0 \text{ mL}$
$2x + y = \frac{0.3}{10} = 0.03$ --- (Equation 1)
Step 2: Analyze Redox Titration (KMnO₄)
Both compounds contain the oxalate ion (${C_2O_4^{2-}}$) which oxidizes to $CO_2$.
n-factor for oxidation: Both $H_2C_2O_4$ and $NaHC_2O_4$ release 2 electrons per molecule ($n=2$).
Equivalents of reductant = Equivalents of oxidant.
$(2x + 2y) \times 10 \text{ mL} = 0.1 \text{ N} \times 4.0 \text{ mL}$
$2x + 2y = \frac{0.4}{10} = 0.04$ --- (Equation 2)
Step 3: Solve the Simultaneous Equations
From Eq 2: $2x + 2y = 0.04$. Subtract Eq 1 ($2x + y = 0.03$):
$(2x + 2y) - (2x + y) = 0.04 - 0.03$
$y = 0.01 \text{ M}$ (Molarity of $NaHC_2O_4$).
Substitute $y$ in Eq 1: $2x + 0.01 = 0.03 \implies 2x = 0.02 \implies x = 0.01 \text{ M}$ (Molarity of $H_2C_2O_4$).
Step 4: Calculate Masses in 1 Liter
Since volume is $1.0 \text{ L}$, moles = molarity.
Moles of $H_2C_2O_4 = 0.01 \text{ mol}$. Molar mass $= 90 \text{ g/mol}$. Mass $= 0.01 \times 90 = 0.90 \text{ g}$.
Moles of $NaHC_2O_4 = 0.01 \text{ mol}$. Molar mass $= 112 \text{ g/mol}$. Mass $= 0.01 \times 112 = 1.12 \text{ g}$.
Check total mass: $0.90 + 1.12 = 2.02 \text{ g}$ (Perfect match!).
View Solution
Step 1: Calculate Average Oxidation State algebraically
Let oxidation state of Br be $x$. Oxygen is usually $-2$.
$3x + 8(-2) = 0$
$3x - 16 = 0 \implies x = +16/3$.
Since an atom cannot literally lose a fraction of an electron, $+16/3$ is simply a mathematical average.
Step 2: Draw the Structure of $Br_3O_8$
The molecule is a linear chain of three Bromine atoms.
Structure: $O_3Br - Br(O_2) - BrO_3$
The terminal Bromine atoms are each double-bonded to three Oxygen atoms.
The central Bromine atom is double-bonded to two Oxygen atoms and single-bonded to the two terminal Bromine atoms.
Step 3: Assign Absolute Oxidation States via Electronegativity
Oxygen is more electronegative than Bromine. Every bond to an Oxygen atom counts as $+1$ for Bromine.
Terminal $Br_1$: Bonded to 3 Oxygen atoms (double bonds = 6 bonds). O.S. = $+6$.
Central $Br_2$: Bonded to 2 Oxygen atoms (double bonds = 4 bonds). Bonds to other Bromines do not contribute to charge separation. O.S. = $+4$.
Terminal $Br_3$: Bonded to 3 Oxygen atoms (double bonds = 6 bonds). O.S. = $+6$.
Step 4: Verify the Average
Average = $\frac{+6 + +4 + +6}{3} = \frac{16}{3}$.
View Solution
Step 1: Decode "Volume Strength"
A label of "22.4 V" means that $1.0 \text{ Liter}$ of this $H_2O_2$ solution will decompose to yield exactly $22.4 \text{ Liters}$ of Oxygen gas ($O_2$) at STP.
Step 2: Reaction Stoichiometry
$2H_2O_2 \rightarrow 2H_2O + O_2$
According to the balanced equation, $2 \text{ moles}$ of $H_2O_2$ produce $1 \text{ mole}$ of $O_2$.
At STP, $1 \text{ mole}$ of $O_2$ occupies $22.4 \text{ Liters}$.
Since $1 \text{ L}$ of our solution produced $22.4 \text{ L}$ of $O_2$, it means $1 \text{ mole}$ of $O_2$ was produced.
Therefore, moles of $H_2O_2$ present in $1 \text{ L}$ of solution = $2 \text{ moles}$.
Molarity ($M$) = $2.0 \text{ M}$.
Step 3: Calculate Normality
The n-factor for $H_2O_2$ in redox reactions (whether acting as an oxidant or reductant) is $2$.
Normality ($N$) = Molarity $\times$ n-factor = $2.0 \times 2 = 4.0 \text{ N}$.
Step 4: Calculate Percentage Strength ($\text{w/v}$)
Molar mass of $H_2O_2 = 34 \text{ g/mol}$.
Mass of $H_2O_2$ in $1 \text{ Liter}$ ($1000 \text{ mL}$) = Moles $\times$ Molar mass = $2.0 \text{ mol} \times 34 \text{ g/mol} = 68 \text{ g}$.
Percentage strength is mass per $100 \text{ mL}$.
$\% \text{ w/v} = \frac{68 \text{ g}}{1000 \text{ mL}} \times 100 = 6.8\%$.
View Solution
Step 1: Separate into Oxidation and Reduction Half-Reactions
Reduction half: $Cl_2 \rightarrow Cl^-$
Oxidation half: $Cl_2 \rightarrow ClO_3^-$
Step 2: Balance Reduction Half
Balance atoms: $Cl_2 \rightarrow 2Cl^-$
Balance charge by adding electrons: $Cl_2 + 2e^- \rightarrow 2Cl^-$ --- (Eq 1)
Step 3: Balance Oxidation Half (Basic Medium)
Balance Cl atoms: $Cl_2 \rightarrow 2ClO_3^-$
Balance O atoms: Add $6H_2O$ to the left side.
$Cl_2 + 6H_2O \rightarrow 2ClO_3^-$
Balance H atoms: Add $12H^+$ to the right side.
$Cl_2 + 6H_2O \rightarrow 2ClO_3^- + 12H^+$
Since it's a basic medium, neutralize $H^+$ by adding $12OH^-$ to BOTH sides.
$Cl_2 + 6H_2O + 12OH^- \rightarrow 2ClO_3^- + 12H_2O$
Simplify water: $Cl_2 + 12OH^- \rightarrow 2ClO_3^- + 6H_2O$
Balance charge: Left side is $-12$, right side is $-2$. Add $10e^-$ to the right.
$Cl_2 + 12OH^- \rightarrow 2ClO_3^- + 6H_2O + 10e^-$ --- (Eq 2)
Step 4: Equate Electrons and Add
Multiply Eq 1 by $5$ to get $10e^-$:
$5Cl_2 + 10e^- \rightarrow 10Cl^-$
Add to Eq 2:
$6Cl_2 + 12OH^- \rightarrow 10Cl^- + 2ClO_3^- + 6H_2O$
Divide by 2 to get simplest integer form:
$3Cl_2 + 6OH^- \rightarrow 5Cl^- + ClO_3^- + 3H_2O$
Step 5: Sum of Coefficients
Sum = $3 (Cl_2) + 6 (OH^-) + 5 (Cl^-) + 1 (ClO_3^-) + 3 (H_2O) = 18$.
View Solution
Step 1: Establish Equivalence Chain
By the Law of Equivalence, all reactants and products in a sequential series react in equal equivalents.
Equivalents of $H_2O_2$ = Equivalents of $I_2$ liberated = Equivalents of $Na_2S_2O_3$ used.
Step 2: Calculate Equivalents of Thiosulfate
Milliequivalents (meq) of $Na_2S_2O_3$ = Normality $\times$ Volume (in mL)
meq = $0.2 \text{ N} \times 30 \text{ mL} = 6.0 \text{ meq}$.
Therefore, meq of $H_2O_2$ originally present in the $20 \text{ mL}$ sample is $6.0 \text{ meq}$.
Step 3: Calculate Normality of H₂O₂
$N_1 \times V_1 = \text{meq}$
$N_{H_2O_2} \times 20 \text{ mL} = 6.0 \text{ meq}$
$N_{H_2O_2} = \frac{6.0}{20} = 0.3 \text{ N}$.
Step 4: Calculate Mass per Liter (Strength)
n-factor of $H_2O_2$ is 2. Equivalent weight = Molar Mass / 2 = $34 / 2 = 17 \text{ g/eq}$.
Strength ($\text{g/L}$) = Normality $\times$ Equivalent Weight
Strength = $0.3 \text{ eq/L} \times 17 \text{ g/eq} = 5.1 \text{ g/L}$.
View Solution
Step 1: Analyze Acidic Medium
In an acidic medium, $MnO_4^-$ reduces to $Mn^{2+}$. Change in O.S. is from $+7$ to $+2$. n-factor = 5.
For $FeSO_4$, $Fe^{2+}$ oxidizes to $Fe^{3+}$. n-factor = 1.
Equivalents of $KMnO_4$ = Equivalents of $FeSO_4$
$(M_1 \times n_1) \times V_1 = (M_2 \times n_2) \times V_2$
$(0.1 \times 5) \times V_1 = (0.1 \times 1) \times 50$
$0.5 \times V_1 = 5.0 \implies V_1 = 10 \text{ mL}$.
Step 2: Analyze Faintly Alkaline/Neutral Medium
In a neutral/faintly alkaline medium, $MnO_4^-$ reduces to solid $MnO_2$. Change in O.S. is from $+7$ to $+4$. n-factor = 3.
The n-factor for $FeSO_4$ remains 1.
$(M_1 \times n_1) \times V_1 = (M_2 \times n_2) \times V_2$
$(0.1 \times 3) \times V_1 = (0.1 \times 1) \times 50$
$0.3 \times V_1 = 5.0 \implies V_1 = \frac{5.0}{0.3} = 16.67 \text{ mL}$.
View Solution
Step 1: Identify the Redox Active Component
In $KNO_3$, Nitrogen is already in its highest oxidation state ($+5$). It cannot be oxidized further. Therefore, $KMnO_4$ only reacts with $KNO_2$ (where N is $+3$).
Step 2: Determine n-factors
$KMnO_4$ in acidic medium: n-factor = 5.
$KNO_2$ oxidizes to $KNO_3$ ($N^{3+} \rightarrow N^{5+}$): n-factor = 2.
Step 3: Apply Law of Equivalence
Equivalents of $KNO_2$ = Equivalents of $KMnO_4$
$\frac{\text{Mass of } KNO_2}{\text{Equivalent Weight}} = N \times V(\text{in L})$
Equivalent weight of $KNO_2 = \frac{M}{n} = \frac{85}{2} = 42.5 \text{ g/eq}$.
Normality of $KMnO_4 = M \times n = 0.1 \times 5 = 0.5 \text{ N}$.
$\frac{\text{Mass}}{42.5} = 0.5 \text{ eq/L} \times 0.020 \text{ L}$
$\frac{\text{Mass}}{42.5} = 0.010 \implies \text{Mass of } KNO_2 = 0.010 \times 42.5 = 0.425 \text{ g}$.
Step 4: Calculate Percentage
$\% \text{ of } KNO_2 = \left( \frac{0.425}{1.52} \right) \times 100 \approx 27.96\%$.
View Solution
Step 1: The Algebraic Trap
Assuming all Oxygen atoms have an oxidation state of $-2$:
$x + 5(-2) = 0 \implies x = +10$.
This is physically impossible because Chromium (Group 6, $[Ar] 4s^1 3d^5$) only has a maximum of 6 valence electrons to lose. The highest possible oxidation state for Cr is $+6$.
Step 2: Structural Analysis (The Butterfly)
$CrO_5$ contains peroxide linkages. Its structure features one Chromium atom double-bonded to one Oxygen atom, and single-bonded to four Oxygen atoms arranged in two peroxide rings ($-O-O-$). It looks like a butterfly.
Step 3: Assigning Correct Oxidation States
The one double-bonded Oxygen (oxide) is $-2$.
The four Oxygen atoms in peroxide linkages are each $-1$.
Let oxidation state of Cr be $x$:
$x + 1(-2) + 4(-1) = 0$
$x - 2 - 4 = 0 \implies x = +6$.
View Solution
Step 1: Identify the Oxidizable Species
Look at every component in Mohr's salt:
- ${SO_4^{2-}}$: Sulfur is in $+6$, cannot be oxidized further.
- ${NH_4^+}$: Nitrogen is in $-3$. While theoretically oxidizable, under standard acidic titration conditions with Dichromate or Permanganate, the ammonium ion is highly stable and does not undergo oxidation.
- ${H_2O}$: Oxygen is $-2$, does not oxidize under these conditions.
- ${Fe^{2+}}$: Iron is $+2$. This is the only species that oxidizes to ${Fe^{3+}}$.
Step 2: Calculate n-factor
${Fe^{2+}} \rightarrow {Fe^{3+}} + 1e^-$.
Since there is exactly 1 mole of Iron per mole of Mohr's salt, the total n-factor = $1$.
Step 3: Calculate Equivalent Weight
Molar mass of Mohr's salt = $56 + 32 + 64 + 2(14 + 4) + 32 + 64 + 6(18) = 392 \text{ g/mol}$.
Equivalent weight = Molar Mass / n-factor = $392 / 1 = 392 \text{ g/eq}$.
View Solution
Step 1: Concept of "Available Chlorine"
"Available Chlorine" refers to the mass of $Cl_2$ gas that can be liberated from the sample upon action with dilute acids, expressed as a percentage of the total mass of the sample.
Reaction: $CaOCl_2 + 2CH_3COOH \rightarrow Ca(CH_3COO)_2 + H_2O + Cl_2 \uparrow$.
Step 2: Iodometry Equivalence
The liberated $Cl_2$ reacts with $KI$ to liberate an equivalent amount of $I_2$, which is then titrated with Hypo.
Equivalents of $Cl_2$ in $25 \text{ mL}$ = Equivalents of Hypo used.
meq of $Cl_2$ (in $25 \text{ mL}$) = $0.1 \text{ N} \times 20 \text{ mL} = 2.0 \text{ meq}$.
Step 3: Scale up to Total Volume
If $25 \text{ mL}$ contains $2.0 \text{ meq}$ of $Cl_2$, then the total $500 \text{ mL}$ suspension contains:
Total meq of $Cl_2 = 2.0 \times \left(\frac{500}{25}\right) = 2.0 \times 20 = 40 \text{ meq} = 0.040 \text{ eq}$.
Step 4: Calculate Mass and Percentage
Equivalent weight of $Cl_2$ gas = Molar mass / 2 = $71 / 2 = 35.5 \text{ g/eq}$.
Mass of available $Cl_2 = 0.040 \text{ eq} \times 35.5 \text{ g/eq} = 1.42 \text{ g}$.
$\% \text{ Available Chlorine} = \left(\frac{1.42 \text{ g}}{3.55 \text{ g}}\right) \times 100 = 40\%$.
View Solution
Step 1: Write the balanced reaction
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
Step 2: Assign Oxidation States
In $NH_4^+$, Nitrogen is in $-3$ state.
In $NO_3^-$, Nitrogen is in $+5$ state.
In the product $N_2O$, both Nitrogen atoms are in an average $+1$ state.
Step 3: Analyze the Electron Transfer (Comproportionation)
This is a comproportionation reaction, the exact opposite of disproportionation. Two different atoms of the same element in the reactant meet at an intermediate oxidation state in the product.
Oxidation half: $N^{-3} \rightarrow N^{+1}$ (Loss of $4e^-$).
Reduction half: $N^{+5} \rightarrow N^{+1}$ (Gain of $4e^-$).
The number of electrons transferred internally per mole of the compound is exactly 4.
View Solution
Step 1: Understand Indicator Chemistry
Wait, standard double titration uses Phenolphthalein (Ph) first, then Methyl Orange (MeOH). Let's review the prompt. Ah, it gives two independent titrations from the start.
With Phenolphthalein: Neutralizes all strong base ($NaOH$) and exactly *half* of the carbonate ($Na_2CO_3 \rightarrow NaHCO_3$). Let equivalents of $NaOH$ be $a$ and equivalents of full $Na_2CO_3$ be $b$.
Volume for Ph ($V_1$): $a + \frac{b}{2} \propto 30 \text{ mL}$.
Step 2: Analyze Methyl Orange
Methyl Orange (MeOH) acts at a lower pH ($\approx 4$), meaning it neutralizes everything completely: all $NaOH$ and all $Na_2CO_3 \rightarrow CO_2 + H_2O$.
Volume for MeOH ($V_2$): $a + b \propto 45 \text{ mL}$.
Step 3: Solve Algebraically
Subtract Ph from MeOH:
$(a + b) - \left(a + \frac{b}{2}\right) = 45 - 30$
$\frac{b}{2} = 15 \implies b \propto 30 \text{ mL}$ of $0.1 \text{ M } HCl$.
Step 4: Calculate Mass of Na₂CO₃
Equivalents of $Na_2CO_3 (b) = N \times V = (0.1 \times 1) \times 0.030 \text{ L} = 0.003 \text{ eq}$.
Mass = Equivalents $\times$ Equivalent Weight.
Equivalent weight of $Na_2CO_3 = 106 / 2 = 53 \text{ g/eq}$.
Mass = $0.003 \times 53 = 0.159 \text{ g}$.
View Solution
Step 1: Oxidation Half-Reaction (Hydrazine)
$N_2H_4 \rightarrow N_2$.
Oxidation state of N goes from $-2$ to $0$. Change = $2$ per N atom. For 2 N atoms, total change = $4$.
n-factor of $N_2H_4 = 4$.
Step 2: Reduction Half-Reaction (Iodate)
In $KIO_3$, Iodine is in the $+5$ state.
In the product $ICl$ (Iodine monochloride), Chlorine is more electronegative ($-1$), so Iodine is forced into a $+1$ state.
Change = $|5 - 1| = 4$.
n-factor of $KIO_3 = 4$.
Step 3: Apply Equivalence
Equivalents of $N_2H_4$ = Equivalents of $KIO_3$
Moles of $N_2H_4 \times 4$ = Moles of $KIO_3 \times 4$
$1 \times 4 = \text{Moles of } KIO_3 \times 4 \implies \text{Moles of } KIO_3 = 1$.
View Solution
Step 1: Understand the 109% Label
"$109\%$" means $100 \text{ g}$ of this oleum produces exactly $109 \text{ g}$ of pure $H_2SO_4$ when fully hydrated.
Therefore, our $10 \text{ g}$ sample, when fully diluted in water, will yield $10.9 \text{ g}$ of pure $H_2SO_4$.
Step 2: Calculate Molarity/Normality of the 1L Solution
Molar mass of $H_2SO_4 = 98 \text{ g/mol}$.
Moles of $H_2SO_4$ in 1 Liter = $10.9 / 98 = 0.1112 \text{ M}$.
Since it's dibasic, Normality ($N$) = $0.1112 \times 2 = 0.2224 \text{ N}$.
Step 3: Titration Equation
$N_{\text{acid}} \times V_{\text{acid}} = N_{\text{base}} \times V_{\text{base}}$
The $NaOH$ is $0.5 \text{ M}$, which is $0.5 \text{ N}$ (since $n=1$).
$0.2224 \times 50 \text{ mL} = 0.5 \times V_{\text{base}}$
$11.12 = 0.5 \times V_{\text{base}}$
$V_{\text{base}} = \frac{11.12}{0.5} = 22.24 \text{ mL}$.
View Solution
Step 1: Write the chemical equation
With hot, concentrated base, halogens (except F) disproportionate into halide and halate ions.
$3I_2 + 6OH^- \rightarrow 5I^- + IO_3^- + 3H_2O$
Step 2: Calculate individual n-factors for I₂
Reduction: $I_2^0 \rightarrow 2I^{-1}$. Change = $|0 - (-1)| \times 2 = 2$. ($n_{red} = 2$).
Oxidation: $I_2^0 \rightarrow 2I^{+5}$. Change = $|0 - 5| \times 2 = 10$. ($n_{ox} = 10$).
Step 3: Calculate the total n-factor
Using the disproportionation formula:
$n_{total} = \frac{n_{ox} \times n_{red}}{n_{ox} + n_{red}} = \frac{10 \times 2}{10 + 2} = \frac{20}{12} = \frac{5}{3}$.
Step 4: Calculate Equivalent Weight
$E = \frac{M}{n_{total}} = \frac{254}{5/3} = \frac{254 \times 3}{5} = 152.4 \text{ g/eq}$.
View Solution
Step 1: The Algebraic Failure
If we use algebra: $2(+1) + 2x + 3(-2) = 0 \implies 2x - 4 = 0 \implies x = +2$.
This implies both sulfur atoms have an oxidation state of $+2$, which contradicts the chemical structure of the molecule.
Step 2: Structural Analysis
The thiosulfate ion (${S_2O_3^{2-}}$) is structurally derived from the sulfate ion (${SO_4^{2-}}$) by replacing one Oxygen atom with a Sulfur atom. It consists of a central Sulfur atom bonded to three Oxygen atoms and one terminal Sulfur atom.
Structure: $[S=S(O_2)-O]^{2-}$
Step 3: Assigning Formal Oxidation States
By IUPAC convention, bonds between identical atoms (like S=S) are assumed to have electrons shared equally, unless they are in vastly different environments. Here, the central Sulfur is bonded to highly electronegative Oxygens, pulling electron density away, making it more electronegative relative to the terminal Sulfur.
However, standard structural assignment treats the terminal Sulfur almost like an Oxide (since it replaced an Oxygen).
Terminal Sulfur: Behaves as $S^{2-}$. Oxidation state = $-2$.
Central Sulfur: Bonded to three Oxygens ($+4$ contribution) and one terminal Sulfur acting as $-2$ ($+2$ contribution). Oxidation state = $+6$.
Step 4: Verify Average
Average = $\frac{+6 + (-2)}{2} = \frac{+4}{2} = +2$. The average matches the algebraic result, but the individual states are $+6$ and $-2$.
View Solution
Step 1: Volumetric Analysis
Reaction during titration: $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$.
n-factor of Dichromate = 6 (Chromium goes from $+6$ to $+3$, for 2 atoms = $2 \times 3 = 6$).
n-factor of $Fe^{2+}$ = 1.
Equivalents of $Fe^{2+}$ = Equivalents of $K_2Cr_2O_7$
$\text{Moles of } Fe \times 1 = \text{Molarity} \times \text{n-factor} \times \text{Volume in L}$
Moles of $Fe = 0.1 \text{ M} \times 6 \times 0.040 \text{ L} = 0.024 \text{ moles}$.
Step 2: Calculate Mass of Iron
Mass of $Fe = 0.024 \text{ moles} \times 56 \text{ g/mol} = 1.344 \text{ g}$.
Step 3: Calculate Percentage
Percentage of $Fe = \left(\frac{1.344}{2.0}\right) \times 100 = 67.2\%$.
View Solution
Case A: As an Oxidizing Agent (Redox)
The active species is the Dichromate ion (${Cr_2O_7^{2-}}$).
In acidic medium, it reduces to $Cr^{3+}$.
$Cr^{+6} \rightarrow Cr^{+3}$. Change = 3 electrons per Chromium. Since there are 2 Chromium atoms, total change = $6e^-$.
n-factor = 6. Equivalent weight = $M / 6$.
Case B: As a Precipitating Salt (Non-Redox)
In a double displacement or precipitation reaction, no electrons are transferred. The n-factor is defined by the total positive (or negative) charge on the ions in one formula unit.
$BaCr_2O_7$ consists of $Ba^{2+}$ and $Cr_2O_7^{2-}$. Total positive charge = 2.
n-factor = 2. Equivalent weight = $M / 2$.
View Solution
Step 1: Calculate Total Oxidizing Equivalents
Since both are oxidants, their oxidizing powers are additive.
Eq of $KMnO_4 = M \times n \times V = 0.1 \times 5 \times 0.100 \text{ L} = 0.05 \text{ eq}$.
Eq of $K_2Cr_2O_7 = M \times n \times V = 0.05 \times 6 \times 0.100 \text{ L} = 0.03 \text{ eq}$.
Total Oxidizing Equivalents = $0.05 + 0.03 = 0.08 \text{ eq}$.
Step 2: Set up Mohr's Salt Equation
Mohr's salt contains $Fe^{2+}$, which oxidizes to $Fe^{3+}$ (n-factor = 1).
Eq of Mohr's salt = Total Oxidizing Eq
$M \times n \times V = 0.08$
$0.5 \times 1 \times V = 0.08$
$V = \frac{0.08}{0.5} = 0.16 \text{ Liters} = 160 \text{ mL}$.
View Solution
Step 1: Write the decomposition reaction
$2H_2O_2 \rightarrow 2H_2O + O_2$
Step 2: Assign Oxidation States
Oxygen in Peroxide ($H_2O_2$) = $-1$.
Oxygen in Water ($H_2O$) = $-2$ (Reduction).
Oxygen in Oxygen gas ($O_2$) = $0$ (Oxidation).
Step 3: Calculate individual n-factors per mole of $H_2O_2$
For reduction ($H_2O_2 \rightarrow H_2O$): O.S. changes from $-1$ to $-2$. Change = 1 per Oxygen. Since $H_2O_2$ has 2 Oxygens, $n_{red} = 2$.
For oxidation ($H_2O_2 \rightarrow O_2$): O.S. changes from $-1$ to $0$. Change = 1 per Oxygen. Since $H_2O_2$ has 2 Oxygens, $n_{ox} = 2$.
Step 4: Combine using Disproportionation Formula
$n_{total} = \frac{n_{ox} \times n_{red}}{n_{ox} + n_{red}} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1$.
Step 5: Equivalent Weight
$E = \frac{M}{n} = \frac{34}{1} = 34 \text{ g/eq}$.
Note: When acting purely as an oxidant or purely as a reductant in a titration, the n-factor of $H_2O_2$ is 2. But during its own disproportionation, the effective overall n-factor evaluates to 1.
View Solution
Step 1: Equivalents of Oxidant
$K_2Cr_2O_7$ n-factor = 6.
Eq of Dichromate = $M \times n \times V = 0.1 \times 6 \times 0.020 = 0.012 \text{ eq}$.
Step 2: Equivalents of Reductant
The metal goes from $+2$ to $+4$, so its n-factor = $2$.
Let atomic mass be $X$. Equivalent weight = $X / 2$.
Equivalents of Metal = $\frac{\text{Mass}}{\text{Eq. Wt}} = \frac{0.5}{X/2} = \frac{1.0}{X}$.
Step 3: Equate and Solve
$\frac{1.0}{X} = 0.012$
$X = \frac{1.0}{0.012} = 83.33 \text{ g/mol}$.
View Solution
Step 1: First Titration (Cu²⁺ analysis)
In an acidic/neutral medium, $Cu^{2+}$ reacts with $I^-$ to liberate $I_2$, but Arsenite ($AsO_3^{3-}$) does not react with $I_2$ (the equilibrium favors the reverse reaction without a basic buffer).
So, the first titration strictly measures $Cu^{2+}$.
n-factor of $Cu^{2+} \rightarrow Cu^+$ is 1. n-factor of Hypo is 1.
moles of $Cu^{2+} \times 1 = 30 \text{ mL} \times 0.1 \text{ M} \times 1$
Millimoles of $Cu^{2+} = 3.0 \text{ mmol}$.
Step 2: Second Titration (Arsenite analysis)
When the buffer is added, the $I_2$ can now quantitatively oxidize Arsenite to Arsenate.
$AsO_3^{3-} + I_2 + H_2O \rightarrow AsO_4^{3-} + 2I^- + 2H^+$
Arsenic goes from $+3$ to $+5$, so n-factor = 2.
The titrant is $I_2$ solution. $I_2$ goes from $0$ to $-1$ (2 atoms), so n-factor = 2.
Equivalents of Arsenite = Equivalents of $I_2$
$\text{mmol of } AsO_3^{3-} \times 2 = \text{Molarity of } I_2 \times \text{n-factor} \times \text{Volume}$
$\text{mmol of } AsO_3^{3-} \times 2 = 0.05 \times 2 \times 20$
$\text{mmol of } AsO_3^{3-} \times 2 = 2.0 \implies \text{mmol of } AsO_3^{3-} = 1.0 \text{ mmol}$.
Mastering the Exchange of Electrons
Congratulations on powering through these 25 highly rigorous numericals. Redox reactions form the absolute backbone of analytical volumetric chemistry. The secret to conquering JEE Advanced questions in this domain is letting go of mole-to-mole stoichiometry and fully embracing the Law of Chemical Equivalence. By quickly and accurately identifying the n-factor—especially in tricky cases like disproportionation and double salts—you can solve complex titration problems in a fraction of the time.
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