Chemistry 9701/23 — May/June 2016

Background Concept

A reactive metal such as iron reacts with a dilute (non-oxidising) acid to give the corresponding metal salt and hydrogen gas. The general pattern is:

The key point here is which oxidation state of iron is produced. Dilute sulfuric acid is a non-oxidising acid: the oxidising agent is the $\text{H}^+$ ion, which is only a weak oxidiser, so iron is oxidised only as far as $\text{Fe}^{2+}$ — not to $\text{Fe}^{3+}$. The question explicitly states that all iron atoms were converted to iron(II) ions, confirming this. The sulfate ion is a spectator; it is not reduced.

Understanding the Question

The command word is write, so this is a single-marks recall task: produce the balanced symbol equation for the reaction between iron metal and dilute sulfuric acid. The stem for part (a) tells us iron(II) ions form and that air is excluded (so no oxidation to $\text{Fe}^{3+}$ occurs).

Approach

Combine two known facts: (1) metal + acid → salt + hydrogen; (2) iron is oxidised to the +2 state here. So the salt is $\text{FeSO}_4$ and the gas is $\text{H}_2$. Then balance the equation so that atoms and charges match on each side.

Step-by-Step Reasoning

Write the unbalanced equation with the correct products:

Count atoms on each side. Iron: 1 on each side. Sulfur: 1 on each side ($\text{SO}_4$). Hydrogen: 2 on the left ($\text{H}_2\text{SO}_4$) and 2 on the right ($\text{H}_2$). Oxygen: 4 on each side. Everything balances as written, so no coefficients are needed. Oxidation states confirm the redox: $\text{Fe}$ goes from 0 to $+2$ (oxidation) and $\text{H}^+$ goes from $+1$ to 0 in $\text{H}_2$ (reduction).

Key Takeaways

• Metal + dilute acid → metal salt + hydrogen.
• Dilute $\text{H}_2\text{SO}_4$ is non-oxidising, so iron gives $\text{Fe}^{2+}$, not $\text{Fe}^{3+}$.
• Always check that the equation balances both atoms and charge.

Common Mistakes

• Writing $\text{Fe}_2(\text{SO}_4)_3$ or producing $\text{Fe}^{3+}$: this is only correct with an oxidising acid such as hot concentrated sulfuric acid or nitric acid, not dilute sulfuric acid.
• Writing $\text{H}_2\text{O}$ or $\text{SO}_2$ as a product instead of $\text{H}_2$.
• Omitting balancing: the equation must balance; an unbalanced equation scores zero.

Things to Be Careful About

• State symbols are good practice and are accepted; make sure the salt is aqueous and hydrogen is a gas.
• Do not confuse the observation of gas bubbles (hydrogen) with an incorrect product such as $\text{SO}_2$.

Background Concept

A redox reaction can be constructed by combining two balanced half-equations. The rule is simple: the number of electrons lost by the reducing agent must equal the number of electrons gained by the oxidising agent. Here dichromate(VI) gains 6 electrons per ion, and each iron(II) ion loses 1 electron. So the ratio of dichromate to iron(II) must be 1:6.

Understanding the Question

The command word is effectively write (using the given half-equations). You are handed the two half-equations and must combine them into one overall equation with no electrons left over — electrons must cancel completely.

Dichromate half-equation (reduction, gain of 6 e⁻):

Iron half-equation (oxidation, loss of 1 e⁻):

Approach

Scale one or both half-equations by integers so that the number of electrons matches, then add them and cancel the electrons (and anything that appears on both sides).

Step-by-Step Reasoning

The dichromate half-equation consumes 6 electrons. The iron half-equation releases 1 electron. To balance, multiply the iron half-equation by 6:

Now add the two half-equations and cancel the 6 electrons appearing on both sides:

Check atoms and charge: Cr 2/2, O 7/7, H 14/14, Fe 6/6, and total charge is $(-2+14+12)=+24$ on the left and $(6+18)=+24$ on the right. Balanced.

Key Takeaways

• Always balance the electrons first when combining half-equations.
• The dichromate:iron(II) ratio of 1:6 is the key stoichiometric relationship used in the titration calculation later.
• Charge must balance as well as atoms.

Common Mistakes

• Forgetting to multiply the iron half-equation by 6, leaving electrons in the final equation.
• Miscounting the electrons in the dichromate half-equation (it is 6, because Cr goes from +6 to +3, two chromium atoms = 6 electrons).
• Dropping the $14\text{H}^+$ or the $7\text{H}_2\text{O}$, which unbalances H and O.

Things to Be Careful About

• Ions must keep the correct state/charge notation: $\text{Cr}_2\text{O}_7^{2-}$, $\text{Cr}^{3+}$, $\text{Fe}^{2+}$, $\text{Fe}^{3+}$.
• The equation must have no leftover electrons.

Background Concept

The amount in moles of a dissolved substance is given by $n = cV$, where $c$ is the concentration in $\text{mol dm}^{-3}$ and $V$ is the volume in $\text{dm}^3$. Because the titration volume is given in $\text{cm}^3$, it must be converted to $\text{dm}^3$ by dividing by 1000 (since $1\ \text{dm}^3 = 1000\ \text{cm}^3$).

Understanding the Question

The command word is calculate. You are given the concentration of potassium dichromate(VI) ($0.0250\ \text{mol dm}^{-3}$) and the titre ($32.0\ \text{cm}^3$) and must find the moles of dichromate ions used.

Approach

Use $n = cV$ directly after converting the volume to $\text{dm}^3$. This is the standard first step of any titration calculation.

Step-by-Step Reasoning

Convert the volume:

Apply the relationship:

Note that potassium dichromate(VI) provides two dichromate ions per formula unit only in terms of stoichiometry of the salt; however, $\text{K}_2\text{Cr}_2\text{O}_7$ contains one $\text{Cr}_2\text{O}_7^{2-}$ ion per formula unit (the two chromium atoms are within the single dichromate ion). So the moles of $\text{Cr}_2\text{O}_7^{2-}$ equal the moles of $\text{K}_2\text{Cr}_2\text{O}_7$, giving $8.0 \times 10^{-4}$ mol.

Key Takeaways

• $n = cV$ with $V$ in $\text{dm}^3$.
• The dichromate ion is $\text{Cr}_2\text{O}_7^{2-}$ (one per $\text{K}_2\text{Cr}_2\text{O}_7$ unit).
• Keep the answer to a sensible number of significant figures (here 2 s.f. matching the data).

Common Mistakes

• Forgetting to divide $\text{cm}^3$ by 1000, giving an answer 1000 times too large.
• Doubling the value by wrongly assuming two dichromate ions per formula unit.

Things to Be Careful About

• Significant figures: the titre (32.0) and concentration (0.0250) both support 3 s.f., but the mark scheme accepts $8 \times 10^{-4}$.
• Units must be moles (mol).

Background Concept

The mole ratio from the balanced equation is the bridge between two different species in a reaction. From the equation written in part (ii), 1 mole of dichromate(VI) reacts with 6 moles of iron(II) ions. This 1:6 ratio is the heart of the titration.

Understanding the Question

The command word is calculate. The volume of solution titrated is $25.0\ \text{cm}^3$, and your value from (iii) is the moles of dichromate in the $32.0\ \text{cm}^3$ titre. You must use the 1:6 ratio to find how many moles of $\text{Fe}^{2+}$ that dichromate consumed.

Approach

Multiply the moles of dichromate by 6, the stoichiometric coefficient of $\text{Fe}^{2+}$ in the balanced equation.

Step-by-Step Reasoning

From the balanced equation:

Substitute the value from (iii):

This is the amount of iron(II) present in the $25.0\ \text{cm}^3$ aliquot that was titrated.

Key Takeaways

• Always read the mole ratio from the balanced equation, not from the formula of the compound.
• The 1:6 dichromate-to-iron(II) ratio is the pivot of this calculation.

Common Mistakes

• Using the wrong ratio (e.g. 1:1 or 1:2), which throws off every subsequent step.
• Dividing by 6 instead of multiplying.

Things to Be Careful About

• Error carried forward: if the value from (iii) is wrong but used correctly here, this mark can still be earned.
• Quoting the answer with correct units (mol).

Background Concept

The titration only analysed a small aliquot (portion) of the solution. The $3.35\ \text{g}$ of wire was dissolved and made up to a total volume of $250\ \text{cm}^3$. The aliquot titrated was $25.0\ \text{cm}^3$. Because concentration is uniform throughout a solution, moles are directly proportional to volume, so the whole flask contains $\frac{250}{25.0}$ times the amount found in the aliquot.

Understanding the Question

The command word is calculate. You must find the total moles of iron in the original wire, not just in the sample titrated. This means multiplying the sample moles by the dilution factor.

Approach

Use the ratio of total volume to sample volume:

Then scale the moles found in the sample up to the whole solution.

Step-by-Step Reasoning

From (iv), the $25.0\ \text{cm}^3$ aliquot contains $4.8 \times 10^{-3}$ mol $\text{Fe}$.

Because all the iron in the wire was converted to $\text{Fe}^{2+}$ and dissolved, this is the moles of iron metal in the $3.35\ \text{g}$ piece.

Key Takeaways

• Always check whether the titration used the whole solution or an aliquot; if an aliquot, apply the volume ratio.
• Moles are proportional to volume at constant concentration.

Common Mistakes

• Forgetting to scale up and reporting the aliquot value ($4.8 \times 10^{-3}$) as the total.
• Using the ratio upside down ($\frac{25}{250}$), which makes the answer ten times too small.

Things to Be Careful About

• The factor is 10 here; small slips in the ratio are the most common error.
• Keep the answer in moles for the next part.

Background Concept

Mass, moles and relative atomic mass are linked by $m = nM_r$ (or $m = nA_r$ for an element). The relative atomic mass of iron is $55.8$, which is the mass in grams of one mole of iron atoms.

Understanding the Question

The command word is calculate. You have the moles of iron from part (v) and must convert to the mass of iron in the wire sample.

Approach

Multiply the moles of iron by its relative atomic mass. The value $55.8$ is the standard $A_r$ of iron given in the periodic table.

Step-by-Step Reasoning

Rounding to three significant figures gives $2.68\ \text{g}$. The mark scheme accepts $2.68$ or $2.678$.

Key Takeaways

• $m = nM$ converts moles to grams.
• Use the relative atomic mass of the element when it is a pure element.

Common Mistakes

• Using the molar mass of a compound such as $\text{FeSO}_4$ instead of the element.
• Multiplying or dividing incorrectly, or dropping a power of ten.

Things to Be Careful About

• Error carried forward applies: a wrong moles value used correctly here still earns this mark.
• Give units (g) and a sensible number of significant figures.

Background Concept

Percentage by mass expresses how much of a sample is a particular component:

Understanding the Question

The command word is calculate. You have the mass of iron found in the wire ($2.68\ \text{g}$) and the total mass of the wire piece ($3.35\ \text{g}$), so you can find the iron content as a percentage.

Approach

Divide the mass of iron by the original mass of the wire and multiply by 100.

Step-by-Step Reasoning

The wire is therefore $80\%$ iron by mass (the remaining $20\%$ would be impurities or alloying elements).

Key Takeaways

• Percentage by mass is component mass over total mass, times 100.
• This final value summarises the whole calculation chain from titre to iron content.

Common Mistakes

• Dividing the wrong way round (total over component).
• Forgetting to multiply by 100, leaving an answer like 0.80.

Things to Be Careful About

• Use the original sample mass $3.35\ \text{g}$, not the mass of any intermediate.
• Error carried forward applies from part (vi).

Background Concept

The character of a bond between two atoms depends on the difference in electronegativity. A large difference (roughly more than 1.7 on the Pauling scale) gives predominantly ionic bonding, while a small difference gives predominantly covalent bonding. Electronegativity is the power of an atom to attract the electron pair in a covalent bond.

Importantly, for transition-metal chlorides like $\text{FeCl}_3$, the bonding is covalent even though a naive comparison with NaCl is misleading. The key comparison the question wants is with $\text{AlCl}_3$: aluminium chloride is itself covalent (dimeric, $\text{Al}_2\text{Cl}_6$).

Understanding the Question

The command word is suggest with explain. You are given electronegativity values (Al = 1.5, Cl = 3.0, Fe = 1.8) and asked to state the bond type in $\text{FeCl}_3$ and justify it using the data. This is a 2-mark question: one mark for the bond type and one for the explanation.

Approach

Compare the electronegativity difference for Fe–Cl with that for Al–Cl. If the Fe–Cl difference is small (comparable to or smaller than Al–Cl, which is known to be covalent), the bonding must also be covalent rather than ionic.

Step-by-Step Reasoning

Calculate the differences:

• Al–Cl: $3.0 - 1.5 = 1.5$
• Fe–Cl: $3.0 - 1.8 = 1.2$

The Fe–Cl difference ($1.2$) is smaller than the Al–Cl difference ($1.5$). Since $\text{AlCl}_3$ is a covalent chloride, a still smaller difference for Fe–Cl means the iron(III) chloride bond is even less ionic — it is covalent. Unlike NaCl (difference ~2.1), where the bonding is ionic, the small difference here means the electrons in the Fe–Cl bonds are shared rather than transferred.

Key Takeaways

• The electronegativity difference determines the ionic/covalent character of a bond.
• A small difference → covalent; a large difference → ionic.
• $\text{FeCl}_3$ and $\text{AlCl}_3$ are covalent chlorides, unlike ionic Group 1 chlorides.

Common Mistakes

• Stating the bonding is ionic because iron is a metal — transition-metal chlorides with high metal oxidation states are frequently covalent.
• Simply quoting the values without comparing them; the mark requires the comparison between the Fe–Cl and Al–Cl differences.
• Confusing electronegativity difference with electronegativity itself.

Things to Be Careful About

• The explanation must reference the small(er) difference; just saying "small difference" without the comparison to Al may lose the second mark.
• Use the actual values from the table to support the answer.

Background Concept

Covalent (non-metal-like) chlorides undergo hydrolysis with water. In hydrolysis, water attacks the electron-deficient metal centre and displaces chloride ions; the water molecules then act as ligands (Lewis bases) donating lone pairs to the metal ion, forming a complex ion. For iron(III), the product is the hexaaquairon(III) ion, $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$, plus free chloride ions.

This is why iron(III) chloride is often described as hydrolysing in water to give an acidic solution — the aqua complex can lose a proton, releasing $\text{H}^+$.

Understanding the Question

The command word is suggest. You are asked for a plausible equation for the reaction between $\text{FeCl}_3$ and water. Two answers are accepted: simple hydration to the hexaaqua complex, or hydration plus one deprotonation step giving the hydroxo complex and $\text{H}^+$.

Approach

Recognise that $\text{FeCl}_3$ is a covalent, hydrolysing chloride. Water molecules coordinate to the $\text{Fe}^{3+}$ centre (6 ligands for an octahedral complex), displacing the three chloride ions. Balance the equation so atoms and charge match.

Step-by-Step Reasoning

Write the products: the hexaaquairon(III) ion and three chloride ions.

Check balance: Fe 1/1, Cl 3/3, O 6/6, H 12/12. Charge: left 0, right $(+3) + 3(-1) = 0$. Balanced.

The accepted alternative involves the aqua complex acting as an acid:

Here one water ligand has lost a proton, giving the pentaaquahydroxoiron(III) ion and a proton. Charge: right $(+2) + (+1) + 3(-1) = 0$. Balanced.

Key Takeaways

• Covalent metal chlorides hydrolyse in water to give aqua complexes and chloride ions.
• The hexaaqua ion $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ is octahedral with 6 water ligands.
• Hydrolysis of high-charge metal ions produces acidic solutions.

Common Mistakes

• Writing $\text{Fe(OH)}_3$ as a product: simple neutralisation does not occur with a covalent chloride in this way; a complex forms instead.
• Forgetting the six water molecules on the left.
• Not showing the complex in square brackets or omitting the chloride ions.

Things to Be Careful About

• The charge on the complex and the balancing of charge must both be correct.
• Either accepted equation is fine; make sure whichever you choose balances.