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The Ultimate Masterclass on Redox Titrations for JEE Advanced
A definitive, highly descriptive study guide tailored for IIT-JEE aspirants. This guide covers every nuance of oxidation-reduction volumetric analysis, n-factor derivations, complex iodometric back-titrations, and features interactive, high-level numerical problems.
1. Introduction to Redox Titrations
Volumetric analysis, widely known as titration, is an indispensable pillar of quantitative analytical chemistry. While acid-base titrations revolve around the transfer of protons ($H^+$), redox titrations are governed by the macroscopic and microscopic transfer of electrons between reacting species. The theoretical foundation of these titrations lies in the simultaneous occurrence of oxidation (loss of electrons) and reduction (gain of electrons).
In the context of the Joint Entrance Examination (JEE) Advanced, redox titrations hold a disproportionately high weightage. The examination rarely tests simple straightforward formula applications. Instead, it demands a rigorous understanding of the Law of Chemical Equivalence, sequential reactions, disproportionation reactions, and complex iodometric estimations (like the determination of available chlorine in bleaching powder or the estimation of copper in an ore).
Unlike acid-base titrations where the $pH$ changes abruptly near the equivalence point, in redox titrations, the electrochemical potential ($E$) of the solution undergoes a sharp, vertical change. This potential can be calculated at any stage of the titration using the Nernst Equation, making redox titrations a perfect intersection of stoichiometry and electrochemistry.
2. Fundamental Redox Concepts: A Quick Refresher
Before diving into the complexities of titration, a robust grasp of the foundational definitions is strictly mandatory. Let us redefine these terms with absolute precision:
- Oxidation: Traditionally defined as the addition of oxygen or removal of hydrogen. In modern electrochemical terms, it is the loss of electrons or the algebraic increase in the oxidation state of an element. For example, $Fe^{2+} \rightarrow Fe^{3+} + e^-$.
- Reduction: The gain of electrons or the algebraic decrease in the oxidation state. For example, $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
- Oxidizing Agent (Oxidant): The chemical species that facilitates oxidation by accepting electrons. Consequently, the oxidizing agent itself gets reduced. Classic examples include $KMnO_4$, $K_2Cr_2O_7$, $HNO_3$, and $I_2$.
- Reducing Agent (Reductant): The chemical species that facilitates reduction by donating electrons. It undergoes oxidation itself. Examples include Mohr's salt ($FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$), Oxalic acid ($H_2C_2O_4$), and Sodium thiosulfate ($Na_2S_2O_3$).
2.1 The Concept of Oxidation State (Number)
The oxidation state is a theoretical, apparent charge assigned to an atom in a molecule or ion, assuming all bonds to be purely ionic. Calculating accurate oxidation states is the first hurdle in solving any JEE redox problem.
Crucial Exceptions and Special Cases for JEE: Students often stumble on molecules with coordinate bonds or peroxy linkages. Let's look at some critical examples:
Also known as the "butterfly structure". It possesses two peroxy linkages (O-O) and one double bond with oxygen.
Structure: $Cr(=O)(-O-O-)_2$.
Four oxygen atoms are in peroxy linkages (Oxidation state = -1), and one is a normal oxide (Oxidation state = -2).
Let $x$ be the O.S. of Cr: $x + 4(-1) + 1(-2) = 0 \implies x = +6$.
2. Marshall's Acid ($H_2S_2O_8$):
Peroxydisulfuric acid contains one peroxy linkage between the two sulfur atoms.
Let $x$ be the O.S. of S: $2(+1) + 2x + 6(-2) + 2(-1) = 0 \implies 2x - 12 = 0 \implies x = +6$.
3. Carbon Suboxide ($C_3O_2$):
Structure is $O=C=C=C=O$. The central carbon is bonded to two other identical carbons, thus its oxidation state is $0$. The terminal carbons are bonded to oxygen, so they have an oxidation state of $+2$. Average oxidation state = $+4/3$.
3. Equivalent Weight & The Infamous "n-Factor" Concept
The backbone of solving complex titration problems without balancing tedious full chemical equations is the Law of Chemical Equivalence. It states that in any chemical reaction, the number of equivalents (or milli-equivalents) of all reactants consumed are equal, and they produce an equal number of equivalents of the products.
To calculate equivalents, we must define the Equivalent Weight (E):
The number of equivalents can be calculated in three ways depending on the data provided in the numerical:
- Equivalents = $\frac{\text{Weight of substance in grams}}{\text{Equivalent Weight (E)}}$
- Equivalents = $\text{Number of Moles} \times \text{n-factor}$
- Equivalents = $\text{Normality (N)} \times \text{Volume in Liters (V)}$
3.1 Demystifying the n-Factor in Redox Reactions
For oxidizing and reducing agents, the n-factor is defined as the total number of electrons transferred per molecule of the reactant during the chemical change, or the total change in oxidation state per molecule.
Formula: n-factor = | Total change in oxidation state | × Number of atoms of the element in one molecule undergoing change.
Case 1: Standard Redox Reactions
Let's evaluate the n-factor for standard laboratory reagents across different pH mediums. This is highly tested.
| Reagent | Medium | Reaction (Conversion) | Change in O.S. | n-factor |
|---|---|---|---|---|
| Potassium Permanganate ($KMnO_4$) | Acidic ($H^+$) | $Mn^{7+} \rightarrow Mn^{2+}$ | |2 - 7| = 5 | 5 |
| Neutral/Faintly Basic | $Mn^{7+} \rightarrow MnO_2 (Mn^{4+})$ | |4 - 7| = 3 | 3 | |
| Strongly Basic ($OH^-$) | $Mn^{7+} \rightarrow MnO_4^{2-} (Mn^{6+})$ | |6 - 7| = 1 | 1 | |
| Potassium Dichromate ($K_2Cr_2O_7$) | Acidic ($H^+$) | $Cr_2^{6+} \rightarrow 2Cr^{3+}$ | 2 × |3 - 6| = 6 | 6 |
| Oxalic Acid / Oxalate ($C_2O_4^{2-}$) | Acidic | $C_2^{3+} \rightarrow 2C^{4+}O_2$ | 2 × |4 - 3| = 2 | 2 |
| Mohr's Salt ($Fe^{2+}$) | Acidic | $Fe^{2+} \rightarrow Fe^{3+}$ | |3 - 2| = 1 | 1 |
| Sodium Thiosulfate ($Na_2S_2O_3$) | Neutral/Acidic (with $I_2$) | $2S_2^{2+}O_3^{2-} \rightarrow S_4^{2.5+}O_6^{2-}$ | 2 × |2.5 - 2| = 1 | 1 |
Case 2: Salts that Oxidize and Reduce (Special Cases)
Sometimes, more than one element in a compound undergoes a change in oxidation state. If both elements get oxidized (or both get reduced), the n-factor of the compound is the sum of individual n-factors.
Example: Ferrous Oxalate ($FeC_2O_4$) oxidized by $KMnO_4$.
Here, $Fe^{2+}$ oxidizes to $Fe^{3+}$ (n-factor contribution = 1).
The oxalate ion $C_2O_4^{2-}$ oxidizes to $CO_2$ (n-factor contribution = 2).
Since both processes release electrons, total n-factor of $FeC_2O_4 = 1 + 2 = 3$.
Thus, 1 mole of $FeC_2O_4$ requires $\frac{3}{5}$ moles of $KMnO_4$ in acidic medium.
Case 3: Disproportionation Reactions
In a disproportionation reaction, the same element in a single oxidation state is simultaneously oxidized and reduced. The n-factor for the reactant is calculated using a parallel resistance type formula.
Let $n_1$ be the number of electrons lost per molecule of A to form B.
Let $n_2$ be the number of electrons gained per molecule of A to form C.
$$ n_{factor} = \frac{n_1 \times n_2}{n_1 + n_2} $$
Example: $3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3^- + 3H_2O$
Oxidation: $Br_2^0 \rightarrow 2Br^{+5}$ ($n_1 = 2 \times 5 = 10$ electrons lost per $Br_2$ molecule)
Reduction: $Br_2^0 \rightarrow 2Br^{-1}$ ($n_2 = 2 \times 1 = 2$ electrons gained per $Br_2$ molecule)
n-factor of $Br_2 = \frac{10 \times 2}{10 + 2} = \frac{20}{12} = \frac{5}{3}$.
Equivalent weight of $Br_2$ in this reaction = $M / (5/3) = \frac{3M}{5}$.
4. Detailed Classification of Redox Titrations
Redox titrations are broadly classified based on the titrant (standard solution) used in the burette. We will explore the three most prevalent types in rigorous detail.
4.1 Permanganometry (Using $KMnO_4$)
Permanganate titrations rely on the highly intense purple/violet color of the $MnO_4^-$ ion. It acts as a powerful oxidizing agent and, crucially, as a self-indicator. When $KMnO_4$ is added to a reducing agent from a burette, the solution remains colorless (as $Mn^{2+}$ is very faint pink, practically colorless in dilute solutions). The moment the reducing agent is completely consumed, the next drop of $KMnO_4$ remains unreacted, imparting a distinct, permanent light pink color to the entire solution. This visually marks the end point.
For acidic medium titrations, only dilute $H_2SO_4$ is used.
- Why not $HCl$? $KMnO_4$ is a strong enough oxidant to oxidize $Cl^-$ to $Cl_2$ gas ($2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2 \uparrow$). This means $KMnO_4$ gets consumed by the acid itself, leading to falsely high volume readings for the actual analyte.
- Why not $HNO_3$? Nitric acid is itself a strong oxidizing agent. It will compete with $KMnO_4$ to oxidize the reducing agent, requiring less $KMnO_4$ and leading to falsely low volume readings.
Standardization of $KMnO_4$:
$KMnO_4$ is not a primary standard. It contains traces of $MnO_2$ and reacts with organic matter in distilled water. It must be standardized against a primary standard like Oxalic acid ($H_2C_2O_4 \cdot 2H_2O$) or Sodium oxalate ($Na_2C_2O_4$).
Reaction kinetics in Oxalate Titration:
The reaction between $MnO_4^-$ and $C_2O_4^{2-}$ is exceptionally slow at room temperature. Therefore, the conical flask containing oxalate solution must be heated to 60-70°C before commencing the titration. Interestingly, as the titration proceeds, the reaction speeds up even if heating is stopped. Why? Because the product, $Mn^{2+}$ ions, act as an auto-catalyst for the reaction.
$$ 2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 \uparrow + 8H_2O $$
4.2 Dichromatometry (Using $K_2Cr_2O_7$)
Potassium dichromate is an excellent primary standard. It can be weighed directly to prepare standard solutions, unlike $KMnO_4$. It is exclusively used in acidic mediums ($H_2SO_4$ or $HCl$).
Wait, didn't we just say $HCl$ cannot be used with strong oxidants? Why is it allowed here?
The standard reduction potential of $Cr_2O_7^{2-}/Cr^{3+}$ ($E^\circ = 1.33 \text{ V}$) is slightly lower than that of $Cl_2/Cl^-$ ($E^\circ = 1.36 \text{ V}$). Therefore, under normal titration conditions, dichromate does not oxidize chloride ions. This makes $K_2Cr_2O_7$ the preferred titrant for estimating iron in iron ores dissolved in $HCl$.
Indicators for Dichromate:
Dichromate is orange ($Cr^{6+}$) and reduces to green ($Cr^{3+}$). The color change from orange to green makes it impossible to visually pinpoint an endpoint accurately. Hence, an internal redox indicator is mandatory. The classic indicator used is Diphenylamine or N-Phenylanthranilic acid, which produces an intense violet-blue color at the endpoint when oxidized by the first microscopic excess drop of dichromate.
4.3 Iodometry vs. Iodimetry (The Heavyweights of JEE)
These two terms sound identical but describe completely distinct analytical procedures involving iodine. Understanding the difference is non-negotiable for advanced problems.
A. Iodimetry (Direct Titration with $I_2$)
In iodimetry, a standard solution of Iodine ($I_2$) is directly used as a titrant in the burette. Iodine is a mild oxidizing agent ($E^\circ_{I_2/I^-} = 0.54 \text{ V}$). It is used to titrate strong reducing agents like Thiosulfate ($S_2O_3^{2-}$), Arsenite ($AsO_3^{3-}$), Antimonite, and Sulfite ($SO_3^{2-}$).
Because iodine is barely soluble in water, the standard solution is prepared by dissolving $I_2$ in a concentrated solution of Potassium Iodide ($KI$), forming the highly soluble triiodide complex ($I_3^-$). $$ I_2 + I^- \rightleftharpoons I_3^- $$
B. Iodometry (Indirect Titration involving liberation of $I_2$)
This is a two-step back-titration process used to estimate strong oxidizing agents (like $Cu^{2+}$, $KMnO_4$, $K_2Cr_2O_7$, $H_2O_2$, Bleaching powder, $O_3$).
- Step 1 (Liberation): The unknown oxidizing agent is reacted with an excess of unmeasured Potassium Iodide ($KI$) in acidic medium. The oxidant reduces itself and oxidizes $I^-$ to free $I_2$. The quantity of $I_2$ liberated is stoichiometrically equivalent to the amount of the original oxidizing agent.
- Step 2 (Titration): The liberated $I_2$ is immediately titrated against a standard solution of Sodium Thiosulfate ($Na_2S_2O_3$, commonly known as hypo) using starch as an indicator near the endpoint.
Equivalents of Oxidizing Agent = Equivalents of Liberated $I_2$ = Equivalents of $Na_2S_2O_3$ consumed.
$$ \text{Eq}_{\text{Oxidant}} = \text{Eq}_{I_2} = N_{\text{hypo}} \times V_{\text{hypo}} $$
The Starch Indicator Magic:
Aqueous iodine is brown/yellow depending on concentration. Starch forms a deep, intense blue-black adsorption complex with molecular $I_2$ (specifically with the amylose helical structure trapping $I_5^-$ chains).
If starch is added at the beginning of the titration when iodine concentration is very high, it forms an extremely stable, irreversible coagulated complex. The iodine gets "trapped" so tightly that sodium thiosulfate cannot react with it, leading to a sluggish and erroneous endpoint. Therefore, we titrate the brown iodine solution until it becomes pale yellow (straw color), indicating most iodine is consumed. Then we add starch, which instantly turns the solution deep blue. We continue titrating dropwise until the blue color sharply disappears, leaving a colorless solution of $I^-$ and $S_4O_6^{2-}$.
Classic Iodometric Estimation: Copper in an Ore
When $Cu^{2+}$ reacts with $I^-$, a fascinating reaction occurs. Cupric iodide ($CuI_2$) is highly unstable and instantly decomposes into insoluble white cuprous iodide ($Cu_2I_2$ or $CuI$) and free iodine.
An interesting nuance for advanced problems: The precipitate of $Cu_2I_2$ tends to adsorb some $I_2$ on its surface, which escapes the hypo titration, making the endpoint fuzzy. To release this trapped iodine, Potassium Thiocyanate (KSCN) is added near the endpoint. The $SCN^-$ replaces the adsorbed $I_2$ due to the lower solubility product of $CuSCN$, sharpening the endpoint beautifully.
5. The Theory of Redox Indicators
How does a redox indicator know when to change color? Unlike acid-base indicators which respond to $[H^+]$ concentration (pH), redox indicators respond to the electrochemical potential of the system.
A true redox indicator is an organic compound that exhibits different colors in its oxidized and reduced forms. Let the indicator half-reaction be: $$ In_{ox} + n e^- \rightleftharpoons In_{red} $$
Using the Nernst equation for the indicator system: $$ E_{system} = E^\circ_{In} - \frac{0.0591}{n} \log \frac{[In_{red}]}{[In_{ox}]} \quad (\text{at 298 K}) $$
Similar to the human eye's limitation with acid-base indicators, we perceive the color of the oxidized form if $[In_{ox}] / [In_{red}] \ge 10$, and the reduced form if $[In_{red}] / [In_{ox}] \ge 10$. Substituting these ratios into the Nernst equation gives the transition potential range of the indicator:
To select a suitable indicator for a given titration, the $E^\circ_{In}$ (standard potential of the indicator) must fall precisely within the steep vertical region of the titration curve near the equivalence point. For example, for the titration of $Fe^{2+}$ with $Ce^{4+}$ ($E_{eq} \approx 1.06\text{ V}$), Ferroin (1,10-phenanthroline iron(II) complex, $E^\circ = 1.06\text{ V}$) is the perfect indicator, changing from pale blue (oxidized) to blood red (reduced).
6. Redox Titration Curves & Equivalence Point Potential
Let's analyze a general redox titration where a reducing agent $Red_1$ is titrated with an oxidizing agent $Ox_2$. $$ a \, Red_1 + b \, Ox_2 \rightleftharpoons a \, Ox_1 + b \, Red_2 $$ Where 'a' electrons are involved in the half-reaction of substance 2, and 'b' electrons in substance 1.
At any point during the titration, after thermodynamic equilibrium is rapidly established, the potential of the system ($E_{system}$) is identical for both half-cell couples. We can calculate $E_{system}$ using the Nernst equation for either couple.
The most critical calculation is the Potential at the Equivalence Point ($E_{eq}$). At equivalence, the amounts of titrant and analyte added are stoichiometrically exact. By setting up the simultaneous Nernst equations for both couples and adding them (weighted by their electron transfer numbers), we arrive at the golden formula:
$E_1^\circ$ = Standard reduction potential of Analyte couple.
$E_2^\circ$ = Standard reduction potential of Titrant couple.
$n_1$ = Number of electrons in Analyte half-reaction.
$n_2$ = Number of electrons in Titrant half-reaction.
*Note: This simplified formula assumes no hydrogen or hydroxide ions are involved in the overall stoichiometry, or that standard states ($[H^+]=1M$) are maintained. If $H^+$ is involved, the $pH$ term must be incorporated into the Nernst expressions before equating.
7. Solved JEE Advanced Level Problems
The theory means nothing without rigorous application. Here are 5 meticulously crafted, highly complex numerical problems that simulate the exact rigor of the JEE Advanced examination. Click on "Show Detailed Solution" to reveal the step-by-step logic. Attempt to solve them on your own first!
Problem 1: The Classic Double Titration (Mixture Analysis)
A $5.0 \text{ g}$ mixture of oxalic acid ($H_2C_2O_4 \cdot 2H_2O$) and sodium oxalate ($Na_2C_2O_4$) was dissolved in water and the solution made up to $500 \text{ mL}$.
$50 \text{ mL}$ of this solution required $30 \text{ mL}$ of $0.1 \text{ M}$ $NaOH$ for complete neutralization in presence of phenolphthalein.
Another $50 \text{ mL}$ of the same original solution required $V \text{ mL}$ of $0.02 \text{ M}$ $KMnO_4$ for complete oxidation in acidic medium.
Calculate the mass percentage of sodium oxalate in the mixture and the value of volume $V$. (Molar mass of $H_2C_2O_4 \cdot 2H_2O = 126 \text{ g/mol}$, $Na_2C_2O_4 = 134 \text{ g/mol}$)
Show Detailed Solution
Step 1: Understand the chemistry of the two titrations.
- Titration 1 (Acid-Base with NaOH): Only the acid ($H_2C_2O_4$) reacts with NaOH. Sodium oxalate is a neutral salt of a strong base and does not react with NaOH.
$H_2C_2O_4$ is dibasic, so its n-factor for acid-base = $2$. - Titration 2 (Redox with KMnO4): Both oxalic acid and sodium oxalate contain the oxalate ion ($C_2O_4^{2-}$). Therefore, both will get oxidized by $KMnO_4$.
n-factor of both $H_2C_2O_4$ and $Na_2C_2O_4$ for redox = $2$ ($C_2^{3+} \rightarrow 2C^{4+}$).
n-factor of $KMnO_4$ in acidic medium = $5$.
Step 2: Solve Titration 1 to find moles of Oxalic Acid.
Let moles of $H_2C_2O_4 \cdot 2H_2O$ in $500 \text{ mL}$ be $x$ and $Na_2C_2O_4$ be $y$.
In $50 \text{ mL}$ aliquot, moles of acid = $x/10$.
Equivalents of Acid = Equivalents of Base
$(x/10) \times 2 = M_{\text{NaOH}} \times V_{\text{NaOH}} \text{ (in Liters)}$
$2x/10 = 0.1 \times 30 \times 10^{-3}$
$2x = 0.1 \times 30 \times 10^{-2} = 3 \times 10^{-2}$
$x = 1.5 \times 10^{-2} \text{ moles}$ (Total moles of acid in 500 mL)
Step 3: Calculate mass percentage.
Mass of Oxalic acid = $x \times 126 = 1.5 \times 10^{-2} \times 126 = 1.89 \text{ g}$.
Since total mass = $5.0 \text{ g}$, Mass of Sodium oxalate = $5.0 - 1.89 = 3.11 \text{ g}$.
Moles of Sodium oxalate ($y$) = $3.11 / 134 \approx 2.32 \times 10^{-2} \text{ moles}$.
Mass % of $Na_2C_2O_4$ = $(3.11 / 5.0) \times 100 = 62.2\%$
Step 4: Solve Titration 2 to find Volume V of KMnO4.
In $50 \text{ mL}$ aliquot, we have $(x/10)$ moles of acid and $(y/10)$ moles of salt.
Total equivalents of oxalate in $50 \text{ mL}$ = $(\frac{x}{10} \times 2) + (\frac{y}{10} \times 2) = \frac{2}{10}(x+y)$
Equivalents of $KMnO_4$ = $M \times V \times n\text{-factor} = 0.02 \times (V \times 10^{-3}) \times 5$
Equating them:
$0.2 \times (1.5 \times 10^{-2} + 2.32 \times 10^{-2}) = 0.1 \times V \times 10^{-3}$
$0.2 \times 3.82 \times 10^{-2} = 10^{-4} \times V$
$7.64 \times 10^{-3} = 10^{-4} \times V$
$V = 76.4 \text{ mL}$
Answers: Mass % = 62.2%, V = 76.4 mL
Problem 2: The Iodometric Back Titration
$2.0 \text{ g}$ of a sample containing $MnO_2$ (and inert impurities) is treated with excess of concentrated $HCl$ and heated. The chlorine gas evolved is passed into a solution of excess Potassium Iodide ($KI$). The liberated iodine requires $60.0 \text{ mL}$ of $0.2 \text{ M}$ sodium thiosulfate ($Na_2S_2O_3$) solution for complete reaction using starch as an indicator.
Calculate the percentage purity of $MnO_2$ in the sample. (Atomic weights: Mn=55, O=16)
Show Detailed Solution
Step 1: Write down the sequence of reactions.
Reaction 1: Generation of Chlorine
$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 \uparrow + 2H_2O$
Reaction 2: Liberation of Iodine
$Cl_2 + 2KI \rightarrow 2KCl + I_2$
Reaction 3: Titration with Hypo
$I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$
Step 2: Apply Law of Equivalence bridging all steps.
Because these are sequential quantitative reactions:
Equivalents of $MnO_2$ reacted = Equivalents of $Cl_2$ produced.
Equivalents of $Cl_2$ reacted = Equivalents of $I_2$ liberated.
Equivalents of $I_2$ reacted = Equivalents of Hypo consumed.
Therefore: Equivalents of $MnO_2$ = Equivalents of $Na_2S_2O_3$
Step 3: Calculate n-factors.
n-factor of $MnO_2$: $Mn^{4+} \rightarrow Mn^{2+}$ (Change is 2. So, $n=2$).
n-factor of $Na_2S_2O_3$: $2S_2^{2+} \rightarrow S_4^{2.5+}$ (Change is $1 \text{ } e^-$ per molecule of hypo. So, $n=1$).
Step 4: Execute calculations.
Let the mass of pure $MnO_2$ in the sample be '$w$' grams.
Molar mass of $MnO_2 = 55 + 2(16) = 87 \text{ g/mol}$.
Equivalents of $MnO_2$ = $\frac{w}{\text{Equivalent Weight}} = \frac{w}{M/n} = \frac{w}{87/2} = \frac{2w}{87}$
Equivalents of Hypo = $M \times V \times n = 0.2 \times (60 \times 10^{-3}) \times 1 = 0.012 \text{ equivalents}$.
Equating them:
$\frac{2w}{87} = 0.012$
$2w = 1.044$
$w = 0.522 \text{ g}$ of pure $MnO_2$.
Step 5: Calculate Percentage Purity.
$\% \text{ Purity} = \frac{\text{Mass of pure substance}}{\text{Total mass of sample}} \times 100$
$\% \text{ Purity} = \frac{0.522}{2.0} \times 100 = 26.1\%$
Answer: Percentage Purity = 26.1%
Problem 3: The Fe2+/Fe3+ Mixture Mystery
A $2.0 \text{ g}$ sample containing a mixture of $Fe_2O_3$ and $FeO$ (and some inert silica) was dissolved in dilute $H_2SO_4$. The solution required $20 \text{ mL}$ of $0.05 \text{ M}$ $KMnO_4$ for complete titration.
Subsequently, the resulting solution from the flask was treated with excess of Zinc dust and dilute $H_2SO_4$ to completely reduce all iron back to the $+2$ state. The unreacted zinc was filtered off. The new filtrate required $50 \text{ mL}$ of the same $0.05 \text{ M}$ $KMnO_4$ solution to reach the endpoint.
Calculate the mass percentages of $FeO$ and $Fe_2O_3$ in the original sample. (Fe=56, O=16)
Show Detailed Solution
Step 1: Analyze the first titration.
The original mixture has $Fe^{2+}$ (from $FeO$) and $Fe^{3+}$ (from $Fe_2O_3$).
When titrated with $KMnO_4$, only the $Fe^{2+}$ will be oxidized to $Fe^{3+}$. The $Fe^{3+}$ already present does not react with $KMnO_4$.
Let moles of $FeO$ be '$x$'. (So moles of $Fe^{2+}$ = $x$).
n-factor of $Fe^{2+} \rightarrow Fe^{3+}$ is $1$.
Equivalents of $FeO$ = Equivalents of $KMnO_4$ used initially.
$x \times 1 = (0.05 \text{ M}) \times (20 \times 10^{-3} \text{ L}) \times 5 \text{ (n-factor of KMnO4)}$
$x = 0.05 \times 0.02 \times 5 = 0.005 \text{ moles}$ of $FeO$.
Step 2: Calculate mass of FeO.
Molar mass of $FeO = 56 + 16 = 72 \text{ g/mol}$.
Mass of $FeO = 0.005 \times 72 = 0.36 \text{ g}$.
Step 3: Analyze the Zinc reduction and second titration.
After the first titration, ALL iron in the flask is now in the $+3$ state (the original $Fe^{3+}$ plus the newly oxidized $Fe^{2+}$).
Adding Zinc (a strong reducing agent) reduces ALL $Fe^{3+}$ down to $Fe^{2+}$.
So, the total moles of $Fe^{2+}$ present before the second titration = Total moles of Iron atoms in the original sample.
Let moles of $Fe_2O_3$ in original sample be '$y$'.
Moles of Fe atoms from $FeO$ = $x$
Moles of Fe atoms from $Fe_2O_3$ = $2y$ (since each molecule has 2 Fe atoms).
Total moles of $Fe^{2+}$ titrated = $x + 2y$.
Equivalents of total $Fe^{2+}$ = Equivalents of $KMnO_4$ used in second titration.
$(x + 2y) \times 1 = (0.05 \text{ M}) \times (50 \times 10^{-3} \text{ L}) \times 5$
$x + 2y = 0.0125 \text{ moles}$
Step 4: Solve for y.
We already know $x = 0.005$.
$0.005 + 2y = 0.0125$
$2y = 0.0075 \implies y = 0.00375 \text{ moles}$ of $Fe_2O_3$.
Step 5: Calculate mass and percentages.
Molar mass of $Fe_2O_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \text{ g/mol}$.
Mass of $Fe_2O_3 = 0.00375 \times 160 = 0.60 \text{ g}$.
Total sample mass = $2.0 \text{ g}$.
$\% \text{ of } FeO = (0.36 / 2.0) \times 100 = 18\%$
$\% \text{ of } Fe_2O_3 = (0.60 / 2.0) \times 100 = 30\%$
Answers: FeO = 18%, Fe2O3 = 30%
Problem 4: Advanced Disproportionation & Equivalents
$H_2O_2$ can act as both an oxidizing and reducing agent. In an experiment, $20 \text{ mL}$ of an $H_2O_2$ solution reacts completely with $30 \text{ mL}$ of $0.1 \text{ N}$ $KMnO_4$ in acidic medium.
The same volume ($20 \text{ mL}$) of the same $H_2O_2$ solution is allowed to undergo complete thermal disproportionation into $H_2O$ and $O_2$ gas. Calculate the volume of $O_2$ gas produced at STP (Standard Temperature and Pressure: $1 \text{ atm}$, $273 \text{ K}$).
Show Detailed Solution
Step 1: Determine the Normality of $H_2O_2$ from the titration.
In reaction with $KMnO_4$ (an oxidant), $H_2O_2$ acts as a reductant and oxidizes to $O_2$.
$H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$ (n-factor of $H_2O_2 = 2$).
From equivalence: $N_{H_2O_2} \times V_{H_2O_2} = N_{KMnO_4} \times V_{KMnO_4}$
$N_{H_2O_2} \times 20 = 0.1 \times 30$
$N_{H_2O_2} = 3 / 20 = 0.15 \text{ N}$.
Since $N = M \times \text{n-factor}$, Molarity ($M$) of $H_2O_2 = 0.15 / 2 = 0.075 \text{ M}$.
Step 2: Calculate moles of $H_2O_2$ in $20 \text{ mL}$.
Moles of $H_2O_2$ = $M \times V(\text{Liters}) = 0.075 \times 0.02 = 1.5 \times 10^{-3} \text{ moles}$.
Step 3: Analyze the disproportionation reaction.
The thermal decomposition equation is:
$2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$
Stoichiometry tells us that 2 moles of $H_2O_2$ yield 1 mole of $O_2$ gas.
Therefore, moles of $O_2$ produced = $(1/2) \times (\text{moles of } H_2O_2)$
Moles of $O_2$ = $0.5 \times 1.5 \times 10^{-3} = 0.75 \times 10^{-3} \text{ moles}$.
Step 4: Calculate Volume of $O_2$ at STP.
At STP, 1 mole of an ideal gas occupies $22.4 \text{ Liters}$ (or $22400 \text{ mL}$).
Volume of $O_2$ = $0.75 \times 10^{-3} \text{ moles} \times 22400 \text{ mL/mole}$
Volume of $O_2$ = $16.8 \text{ mL}$.
Answer: Volume of O2 at STP = 16.8 mL
Problem 5: Complex Iodometry involving a precipitate
A $5.0 \text{ g}$ sample of bleaching powder ($Ca(OCl)Cl$) was suspended in water and made up to $500 \text{ mL}$. $25 \text{ mL}$ of this suspension was acidified with acetic acid and an excess of $KI$ was added. The iodine liberated required $20 \text{ mL}$ of $0.1 \text{ M}$ Sodium Thiosulfate to reach the starch endpoint.
Calculate the percentage of "Available Chlorine" in the bleaching powder.
Show Detailed Solution
Definition Note: "Available Chlorine" is defined as the mass of chlorine gas liberated when $100 \text{ g}$ of bleaching powder reacts with dilute acid. Essentially, it's the percentage by mass of $Cl_2$ that can be generated.
Reaction: $CaOCl_2 + 2CH_3COOH \rightarrow Ca(CH_3COO)_2 + H_2O + Cl_2 \uparrow$
Step 1: The Iodometric Sequence.
The generated $Cl_2$ reacts with $KI$:
$Cl_2 + 2KI \rightarrow 2KCl + I_2$
Titration of $I_2$:
$I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$
From equivalence: Equivalents of $Cl_2$ = Equivalents of $I_2$ = Equivalents of Hypo.
Let's calculate for the $25 \text{ mL}$ aliquot.
Equivalents of Hypo used = $M \times V \times n\text{-factor}$
Eq = $0.1 \times (20 \times 10^{-3}) \times 1 = 2 \times 10^{-3} \text{ equivalents}$.
Step 2: Scale up to the total volume.
The $25 \text{ mL}$ aliquot contains $2 \times 10^{-3}$ equivalents of $Cl_2$.
The total $500 \text{ mL}$ suspension will contain = $(500 / 25) \times 2 \times 10^{-3} = 20 \times 2 \times 10^{-3} = 0.04$ equivalents of $Cl_2$.
Step 3: Convert equivalents to mass of Chlorine.
Equivalent weight of $Cl_2$ (n-factor = 2 for $Cl_2 \rightarrow 2Cl^-$) is $M/2 = 71/2 = 35.5 \text{ g/eq}$.
Mass of $Cl_2$ available in the entire sample = Equivalents $\times$ Eq. Weight
Mass of $Cl_2$ = $0.04 \times 35.5 = 1.42 \text{ g}$.
Step 4: Calculate % Available Chlorine.
$\% \text{ Available } Cl_2 = \frac{\text{Mass of available } Cl_2}{\text{Mass of sample}} \times 100$
$\% \text{ Available } Cl_2 = \frac{1.42 \text{ g}}{5.0 \text{ g}} \times 100 = 28.4\%$
Answer: % Available Chlorine = 28.4%
8. Frequently Asked Questions (FAQ)
Why can't we use Nitric acid ($HNO_3$) to provide the acidic medium for $KMnO_4$ titration?
What is the role of Zimmermann-Reinhardt (ZR) reagent?
1. $MnSO_4$ lowers the reduction potential of the $MnO_4^- / Mn^{2+}$ couple, making it too weak to oxidize $Cl^-$.
2. $H_3PO_4$ complexes with the formed $Fe^{3+}$ to create a colorless complex $[Fe(PO_4)_2]^{3-}$, preventing the yellow color of aqueous $Fe^{3+}$ from interfering with observing the pale pink endpoint.
Why do we heat the oxalic acid solution before titrating with $KMnO_4$?
How is equivalent weight defined for a salt that doesn't undergo redox?
Is there any difference between Molarity and Normality for Sodium Thiosulfate in Iodine titrations?
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