(a) Write down what you would see, and write equations for the reactions that occur, when silicon(IV) chloride and phosphorus(V) chloride are separately mixed with water. silicon(IV) chloride phosphorus(V) chloride (b) Iron(III) chloride, $\text{FeCl}_3$, is used to dissolve unwanted copper from printed circuit boards (PCBs) by the following reaction. $$ 2\text{FeCl}_3(\text{aq}) + \text{Cu}(\text{s}) \rightarrow 2\text{FeCl}_2(\text{aq}) + \text{CuCl}_2(\text{aq}) $$ A…

(a) SiCl4: white solid/steamy fumes; SiCl4 + 2H2O -> SiO2 + 4HCl. PCl5: fizzes/steamy fumes; PCl5 + 4H2O -> H3PO4 + 5HCl · (b)(i) MnO4^- + 8H^+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+ · (b)(ii) 5 : 1 · (b)(iii) 3.00 × 10^-4 mol · (b)(iv) 1.50 × 10^-3 mol · (b)(v) 0.600 mol dm^-3 · (b)(vi) 2.86 g · (c) -252 kJ mol^-1 · (d)(i) Ca2Si + 6H2O -> 2Ca(OH)2 + SiO2 + 4H2 · (d)(ii) Oxidised: silicon; reduced: hydrogen

Chemistry 9701/42 — October/November 2012

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

120

minutes

Topics Introduction to A Level Organic Chemistry · Electrochemistry · Analytical Techniques · Reaction Kinetics · Nitrogen Compounds · Carboxylic Acids and Derivatives · +7 more

Q1Halogen CompoundsTransition ElementsChemical EnergeticsElectrochemistryFree sample

(a)

Write down what you would see, and write equations for the reactions that occur, when silicon(IV) chloride and phosphorus(V) chloride are separately mixed with water.

silicon(IV) chloride

phosphorus(V) chloride

DifficultyMedium-Easy

Worked solution

Answer

silicon(IV) chloride: white solid / white steamy fumes.

$\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$

phosphorus(V) chloride: fizzes / white steamy fumes.

$\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}$

Final answer

SiCl4: white solid/steamy fumes; SiCl4 + 2H2O -> SiO2 + 4HCl. PCl5: fizzes/steamy fumes; PCl5 + 4H2O -> H3PO4 + 5HCl

Detailed explanation

Background Concept

Silicon(IV) chloride and phosphorus(V) chloride are covalent non-metal chlorides. When they react with water they undergo hydrolysis: the chlorine atoms are removed as hydrogen chloride, HCl, and the central element forms an oxide or oxoacid. The HCl(g) appears as white/steamy fumes, and any solid oxide product appears as a white solid.

Understanding the Question

You are asked to give both the observation and a balanced equation for each separate hydrolysis. There are two marks per chloride: one for the observation and one for the equation.

Approach

For each chloride, decide the oxidation state of the central element (Si +4, P +5). The hydrolysis product keeps that oxidation state: Si forms SiO2 (silicon +4) and P forms H3PO4 (phosphorus +5). All chlorine leaves as HCl. Balance the equation by atoms.

Step-by-Step Reasoning

For SiCl4: one Si gives one SiO2; four Cl need four H, so four HCl. Add water to supply the oxygen: SiCl4 + 2H2O -> SiO2 + 4HCl. Check atoms: Si 1, Cl 4, H 8, O 2 on each side.
Observation: SiO2 is a white solid; HCl(g) gives white/steamy fumes.
For PCl5: one P gives one H3PO4; five Cl need five HCl. The water must supply the H and O: PCl5 + 4H2O -> H3PO4 + 5HCl. Check atoms: P 1, Cl 5, H 3+5=8, O 4 on left; right H 3+5=8, O 4.
Observation: the reaction is vigorous/fizzes and HCl gives white/steamy fumes.

Key Takeaways

Covalent chlorides hydrolyse to give HCl and the oxide/oxoacid of the central element in the same oxidation state. Observations to quote: white/steamy fumes for HCl, white solid for SiO2, fizzing for PCl5.

Common Mistakes

Writing Cl2 instead of HCl as the chlorine-containing product.
Forgetting to balance water or HCl.
Giving only an equation and no observation, or only an observation and no equation.
Saying "chlorine gas" rather than "hydrogen chloride fumes".

Things to Be Careful About

Use state symbols where possible: SiO2 is a white solid, HCl is a gas (steamy fumes), H3PO4 is aqueous.
Balance both atoms and charges (though these are neutral molecular equations).
The mark scheme accepts "white solid" or "white/steamy fumes" for SiCl4, and "fizzes" or "white/steamy fumes" for PCl5.

Techniques used

write balanced hydrolysis equationsidentify observations of hydrolysisbalance atoms in hydrolysis reactions

(b)

Iron(III) chloride, $\text{FeCl}_3$ , is used to dissolve unwanted copper from printed circuit boards (PCBs) by the following reaction.

2\text{FeCl}_3(\text{aq}) + \text{Cu}(\text{s}) \rightarrow 2\text{FeCl}_2(\text{aq}) + \text{CuCl}_2(\text{aq})

A solution in which $[\text{Fe}^{3+}(\text{aq})]$ was originally equal to $1.50 \text{ mol dm}^{-3}$ was re-used several times to dissolve copper from the PCBs, and was then titrated as follows.

A $2.50 \text{ cm}^3$ sample of the partially-used-up solution was acidified and titrated with $0.0200 \text{ mol dm}^{-3}$ $\text{KMnO}_4$ .
This oxidised any $\text{FeCl}_2$ in the solution back to $\text{FeCl}_3$ .
It was found that $15.0 \text{ cm}^3$ of $\text{KMnO}_4(\text{aq})$ was required to reach the end point.

(i)

Construct an ionic equation for the reaction between $\text{Fe}^{2+}$ and $\text{MnO}_4^-$ in acid solution.

DifficultyMedium-Easy

Worked solution

Answer

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$

Final answer

MnO4^- + 8H^+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+

Detailed explanation

Background Concept

In acidic solution, manganate(VII) ions, MnO4-, are reduced to Mn2+. Mn in MnO4- has oxidation state +7; in Mn2+ it is +2, so each MnO4- gains 5 electrons. Fe2+ is oxidised to Fe3+, losing 1 electron per ion. To combine the half-equations, the electrons must cancel.

Understanding the Question

You need the overall ionic equation for Fe2+ reducing MnO4- in acid. The question supplies the species involved: Fe2+, MnO4-, H+, and the products Fe3+, Mn2+ and water.

Approach

Write the two half-equations, balance atoms and charges, then multiply so electrons cancel.

Step-by-Step Reasoning

Reduction: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O. (Balance O with water, then H with H+, then charge with electrons.)
Oxidation: Fe2+ -> Fe3+ + e-.
Multiply the oxidation half-equation by 5: 5Fe2+ -> 5Fe3+ + 5e-.
Add the two half-equations; the 5e- cancel: MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+.

Key Takeaways

Redox equations in acid require balancing O with H2O, H with H+, and charge with electrons. The electron transfer must balance between the two half-reactions.

Common Mistakes

Forgetting H+ or H2O.
Using OH- instead of H+ (that would be alkaline conditions).
Wrong electron count (e.g. 3 electrons for MnO4- to Mn2+).
Not balancing charges.

Things to Be Careful About

MnO4- is purple; Mn2+ is pale pink/colourless, but that is not needed in the equation.
The equation must be ionic, so Fe2+, Fe3+, MnO4-, Mn2+ and H+ are written as ions; Cl- and K+ are spectator ions and are omitted.

Techniques used

balance redox half-equationscombine half-equationsbalance atoms and charges

(ii)

State here the $\text{Fe}^{2+} : \text{MnO}_4^-$ ratio from your equation in (i).

DifficultyEasy

Worked solution

Answer

$\text{Fe}^{2+} : \text{MnO}_4^- = 5 : 1$

Final answer

5 : 1

Detailed explanation

Background Concept

The coefficients in a balanced redox equation give the mole ratio in which the reactants combine. From the equation in (i), 5 Fe2+ ions react with 1 MnO4- ion.

Understanding the Question

Simply read the stoichiometric ratio from the balanced ionic equation.

Approach

Look at the coefficients of Fe2+ and MnO4- in the balanced equation.

Step-by-Step Reasoning

In MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+, the coefficient of Fe2+ is 5 and of MnO4- is 1, so the ratio is 5:1.

Key Takeaways

Always quote the ratio in the order asked: Fe2+ : MnO4-.

Common Mistakes

Quoting 1:5 instead of 5:1.
Using the ratio from an unbalanced equation.

Things to Be Careful About

The ratio is a pure number; no units.
If your equation in (i) were wrong, this ratio would carry your error forward (ecf).

Techniques used

read stoichiometric ratio from balanced equationstate mole ratio in required order

(iii)

Calculate the number of moles of $\text{MnO}_4^-$ used in the titration.

DifficultyEasy

Worked solution

Working

n(\text{MnO}_4^-) = 0.0200 \times \frac{15.0}{1000} = 3.00 \times 10^{-4}\ \text{mol}

Answer

$3.00 \times 10^{-4}\ \text{mol}$

Final answer

3.00 × 10^-4 mol

Detailed explanation

Background Concept

Concentration (mol dm^-3) × volume (dm^3) gives amount in mol. Convert cm^3 to dm^3 by dividing by 1000.

Understanding the Question

15.0 cm^3 of 0.0200 mol dm^-3 KMnO4 was used. Calculate the moles of MnO4-.

Approach

Use n = cV with V in dm^3.

Step-by-Step Reasoning

V = 15.0/1000 = 0.0150 dm^3. n = 0.0200 × 0.0150 = 3.00 × 10^-4 mol.

Key Takeaways

Always convert cm^3 to dm^3 before using n = cV.

Common Mistakes

Forgetting to divide by 1000.
Using 15.0 dm^3 instead of 0.0150 dm^3.
Significant figures: 0.0200 has 3 s.f., 15.0 has 3 s.f., so answer to 3 s.f.

Things to Be Careful About

KMnO4 is the source of MnO4-; 1 mol KMnO4 gives 1 mol MnO4-.
The answer should be in mol, not mol dm^-3.

Techniques used

calculate moles from concentration and volumeconvert cm^3 to dm^3

(iv)

Calculate the number of moles of $\text{Fe}^{2+}$ in $2.50 \text{ cm}^3$ of the partially-used-up solution.

DifficultyMedium-Easy

Worked solution

Working

From (i), $\text{Fe}^{2+} : \text{MnO}_4^- = 5 : 1$ .

n(\text{Fe}^{2+}) = 5 \times 3.00 \times 10^{-4} = 1.50 \times 10^{-3}\ \text{mol}

Answer

$1.50 \times 10^{-3}\ \text{mol}$

Final answer

1.50 × 10^-3 mol

Detailed explanation

Background Concept

The mole ratio from the balanced equation allows conversion from moles of titrant to moles of analyte.

Understanding the Question

Given the moles of MnO4- used, find moles of Fe2+ in the 2.50 cm^3 sample.

Approach

Multiply moles of MnO4- by the stoichiometric ratio 5/1.

Step-by-Step Reasoning

n(Fe2+) = 5 × n(MnO4-) = 5 × 3.00 × 10^-4 = 1.50 × 10^-3 mol.

Key Takeaways

Use the ratio from your own balanced equation; if that ratio is wrong, this value changes accordingly (ecf).

Common Mistakes

Dividing instead of multiplying (1/5 ratio).
Using the ratio from (ii) incorrectly.

Things to Be Careful About

The moles are for the 2.50 cm^3 sample, not for 1 dm^3.
Keep 3 significant figures.

Techniques used

apply mole ratio to find molesmultiply moles of titrant by stoichiometric factor

(v)

Calculate the $[\text{Fe}^{2+}]$ in the partially-used-up solution.

DifficultyMedium-Easy

Worked solution

Working

[\text{Fe}^{2+}] = \frac{1.50 \times 10^{-3}}{2.50/1000} = 0.600\ \text{mol dm}^{-3}

Answer

$0.600\ \text{mol dm}^{-3}$

Final answer

0.600 mol dm^-3

Detailed explanation

Background Concept

Concentration = amount / volume in dm^3. The 2.50 cm^3 sample contains 1.50 × 10^-3 mol Fe2+.

Understanding the Question

Find the concentration of Fe2+ in the partially-used solution from the moles in the small sample.

Approach

Convert the sample volume to dm^3, then divide moles by volume.

Step-by-Step Reasoning

V = 2.50/1000 = 0.00250 dm^3. [Fe2+] = 1.50 × 10^-3 / 0.00250 = 0.600 mol dm^-3.

Key Takeaways

This concentration is the same in the bulk solution because the sample is representative.

Common Mistakes

Forgetting to convert cm^3 to dm^3.
Inverting the fraction (dividing volume by moles).
Writing 0.6 instead of 0.600 (both are numerically correct, but 3 s.f. is safer).

Things to Be Careful About

Units: mol dm^-3.
ecf: if your moles in (iv) were different, use that value here.

Techniques used

calculate concentration from moles and volumeconvert sample volume to dm^3

(vi)

Calculate the mass of copper that could still be dissolved by $100 \text{ cm}^3$ of the partially-used-up solution.

mass of copper = ........................... g

DifficultyMedium

Worked solution

Working

Original solution: $100\ \text{cm}^3$ contains

n(\text{Fe}^{3+})_{\text{original}} = 1.50 \times \frac{100}{1000} = 0.150\ \text{mol}

Partially-used solution: $100\ \text{cm}^3$ contains

n(\text{Fe}^{2+}) = 0.600 \times \frac{100}{1000} = 0.0600\ \text{mol}

This $\text{Fe}^{2+}$ was formed from $\text{Fe}^{3+}$ , so remaining $\text{Fe}^{3+}$ :

n(\text{Fe}^{3+})_{\text{remaining}} = 0.150 - 0.0600 = 0.0900\ \text{mol}

From $2\text{Fe}^{3+} + \text{Cu} \rightarrow 2\text{Fe}^{2+} + \text{Cu}^{2+}$ :

n(\text{Cu}) = \frac{0.0900}{2} = 0.0450\ \text{mol}

m(\text{Cu}) = 0.0450 \times 63.5 = 2.86\ \text{g}

Answer

$2.86\ \text{g}$

Final answer

2.86 g

Detailed explanation

Background Concept

The PCB reaction consumes Fe3+ and produces Fe2+. The titration measures how much Fe2+ has been produced, which tells us how much Fe3+ has already been used up. The original Fe3+ concentration is known, so the remaining Fe3+ can be found by subtraction. Then the stoichiometry of the PCB reaction gives the mass of copper that can still react.

Understanding the Question

We have 100 cm^3 of partially-used solution. We know original [Fe3+] = 1.50 mol dm^-3 and from part (v) [Fe2+] = 0.600 mol dm^-3. We need the mass of Cu that the remaining Fe3+ can dissolve.

Approach

Calculate original moles of Fe3+ in 100 cm^3.
Calculate moles of Fe2+ in 100 cm^3 (these are the Fe3+ already used).
Subtract to get remaining Fe3+.
Use the reaction stoichiometry 2Fe3+ : 1Cu to find moles of Cu.
Convert moles of Cu to mass using Ar(Cu) = 63.5.

Step-by-Step Reasoning

Original Fe3+ in 100 cm^3: 1.50 × 0.100 = 0.150 mol.
Fe2+ in 100 cm^3: 0.600 × 0.100 = 0.0600 mol.
Remaining Fe3+ = 0.150 - 0.0600 = 0.0900 mol.
From 2Fe3+ + Cu -> 2Fe2+ + Cu2+, 2 mol Fe3+ reacts with 1 mol Cu, so n(Cu) = 0.0900/2 = 0.0450 mol.
Mass = 0.0450 × 63.5 = 2.8575 ≈ 2.86 g.

Key Takeaways

In a redox reaction, the amount of product formed (Fe2+) equals the amount of reactant consumed (Fe3+). Titration of the product allows indirect determination of the remaining reactant.

Common Mistakes

Using the total Fe3+ instead of the remaining Fe3+.
Forgetting the 2:1 stoichiometry between Fe3+ and Cu.
Using Ar(Cu) = 63.5 but then rounding incorrectly.
Mixing up cm^3 and dm^3.

Things to Be Careful About

The 0.600 mol dm^-3 is the concentration of Fe2+ in the partially-used solution, not the concentration of remaining Fe3+.
The answer should be in grams, to 3 significant figures (2.86 g).
ecf: if your [Fe2+] from (v) were different, use that value here.

Techniques used

calculate moles from concentration and volumeuse stoichiometry to relate Fe3+ to Cuconvert moles to mass

(c)

When $\text{SiCl}_4$ vapour is passed over $\text{Si}$ at red heat, $\text{Si}_2\text{Cl}_6$ is formed. $\text{Si}_2\text{Cl}_6$ contains a $\text{Si-Si}$ bond.
The reaction of $\text{Si}_2\text{Cl}_6$ and $\text{Cl}_2$ re-forms $\text{SiCl}_4$ .

\text{Si}_2\text{Cl}_6(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{SiCl}_4(\text{g})

Use bond energy data from the Data Booklet to calculate $\Delta H^\ominus$ for this reaction.

$\Delta H^\ominus$ = ........................... $\text{kJ mol}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

Bonds broken: $1 \times \text{Si-Si} + 1 \times \text{Cl-Cl} = 222 + 244 = 466\ \text{kJ mol}^{-1}$

Bonds formed: $2 \times \text{Si-Cl} = 2 \times 359 = 718\ \text{kJ mol}^{-1}$

\Delta H^\ominus = 466 - 718 = -252\ \text{kJ mol}^{-1}

Answer

$-252\ \text{kJ mol}^{-1}$

Final answer

-252 kJ mol^-1

Detailed explanation

Background Concept

Bond enthalpy is the energy required to break one mole of a bond in the gaseous state. In a reaction, energy is absorbed to break bonds and released when new bonds form. The enthalpy change is approximately:

\Delta H = \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})

Understanding the Question

For Si2Cl6 + Cl2 -> 2SiCl4, identify which bonds are broken and which are formed, use the Data Booklet bond energies, and calculate ΔH.

Approach

Look at the structures: Si2Cl6 is Cl3Si-SiCl3, so it has one Si-Si bond and six Si-Cl bonds. Cl2 has one Cl-Cl bond. Each SiCl4 has four Si-Cl bonds, so two SiCl4 have eight Si-Cl bonds. The six Si-Cl bonds already present in Si2Cl6 remain unchanged; only two new Si-Cl bonds are formed. Therefore bonds broken: 1 Si-Si + 1 Cl-Cl; bonds formed: 2 Si-Cl.

Step-by-Step Reasoning

Bonds broken: one Si-Si (222) and one Cl-Cl (244), total 466 kJ mol^-1.
Bonds formed: two Si-Cl (2 × 359 = 718) kJ mol^-1.
ΔH = 466 - 718 = -252 kJ mol^-1. Negative because bonds formed are stronger than bonds broken.

Key Takeaways

When using bond energies, count only the bonds actually broken and formed, not all bonds in the molecules. Existing bonds that remain unchanged cancel out.

Common Mistakes

Counting all 8 Si-Cl bonds formed instead of only the 2 new ones.
Using wrong sign: ΔH = bonds broken - bonds formed.
Forgetting to include Cl-Cl bond in Si2Cl6 + Cl2.
Units: kJ mol^-1.

Things to Be Careful About

Data Booklet values: Si-Si 222, Cl-Cl 244, Si-Cl 359 kJ mol^-1.
The answer is negative because the reaction is exothermic.
State symbols: all species are gaseous, but bond energies are for gaseous molecules anyway.

Techniques used

identify bonds broken and formeduse bond energy datacalculate enthalpy change from bond energies

(d)

Calcium forms three calcium silicides, $\text{Ca}_2\text{Si}$ , $\text{CaSi}$ and $\text{CaSi}_2$ . The first of these reacts with water as follows.

.........\text{Ca}_2\text{Si} + ........\text{H}_2\text{O} \rightarrow .......\text{Ca(OH)}_2 + ......\text{SiO}_2 + .....\text{H}_2

(i)

Balance this equation. You may find the use of oxidation numbers helpful.

DifficultyMedium-Easy

Worked solution

Answer

$\text{Ca}_2\text{Si} + 6\text{H}_2\text{O} \rightarrow 2\text{Ca(OH)}_2 + \text{SiO}_2 + 4\text{H}_2$

Final answer

Ca2Si + 6H2O -> 2Ca(OH)2 + SiO2 + 4H2

Detailed explanation

Background Concept

Balancing a chemical equation requires equal numbers of each atom on both sides. Oxidation numbers can help decide the coefficients when a redox reaction also produces hydrogen gas.

Understanding the Question

Balance the given skeleton equation for the reaction of calcium silicide with water.

Approach

Balance the atoms that appear in only one compound on each side first (Ca, Si), then balance O and H. Alternatively use oxidation numbers: Si is oxidised from -4 to +4 (8 e lost), so 8 H atoms must be reduced from +1 to 0, giving 4 H2.

Step-by-Step Reasoning

Ca: 2 on left, so 2 Ca(OH)2 on right.
Si: 1 on left, 1 SiO2 on right.
O: right has 2×2 (from Ca(OH)2) + 2 (from SiO2) = 6 O, so need 6 H2O on left.
H: left has 12 H; right has 4 H in 2Ca(OH)2 + 2y H in H2, so 4 + 2y = 12, y = 4. Thus 4 H2.
Check: Ca2Si + 6H2O -> 2Ca(OH)2 + SiO2 + 4H2.

Key Takeaways

Balancing by atoms is systematic: balance metals/metalloids first, then oxygen, then hydrogen. Oxidation numbers confirm the electron transfer: Si loses 8 e, 8 H atoms gain 1 e each.

Common Mistakes

Wrong coefficient for H2O (e.g. 4 or 5).
Forgetting to balance Ca(OH)2.
Writing H2O as OH- or H+.

Things to Be Careful About

The equation must be balanced in atoms and charge (neutral overall).
The product H2 is a gas; state symbols optional but can be added.

Techniques used

balance redox equation by atomsuse oxidation numbers to confirm electron transfer

(ii)

During this reaction, state

which element(s) have been oxidised, .....................................................................

which element(s) have been reduced. ......................................................................

DifficultyMedium-Easy

Worked solution

Answer

Oxidised: silicon

Reduced: hydrogen

Final answer

Oxidised: silicon; reduced: hydrogen

Detailed explanation

Background Concept

Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number). In Ca2Si, Ca is +2, so Si must be -4. In SiO2, Si is +4. In H2O, H is +1; in H2, H is 0.

Understanding the Question

From the balanced equation in (i), identify which elements change oxidation state and classify them as oxidised or reduced.

Approach

Assign oxidation numbers to each element in reactants and products, then compare.

Step-by-Step Reasoning

Si: -4 in Ca2Si -> +4 in SiO2. Increase of 8, so silicon is oxidised.
H: +1 in H2O -> 0 in H2. Decrease of 1 per H, so hydrogen is reduced.
Ca: +2 in Ca2Si and +2 in Ca(OH)2, unchanged.
O: -2 in H2O and -2 in SiO2/Ca(OH)2, unchanged.

Key Takeaways

Oxidation number changes identify redox. Here silicon is oxidised and hydrogen is reduced; Ca and O are unchanged.

Common Mistakes

Saying calcium is oxidised/reduced (it is unchanged).
Saying oxygen is reduced (it is unchanged).
Confusing oxidation and reduction.

Things to Be Careful About

The oxidation number of Si in Ca2Si is -4 because Ca is +2 (2 × +2 = +4; to neutralise, Si = -4).
H in metal hydrides can be -1, but here H is in water, so +1.
The mark scheme requires both: silicon oxidised AND hydrogen reduced.

Techniques used

assign oxidation numbersidentify oxidation and reduction

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