Chemistry 5070 / 41 — October/November 2024

DifficultyEasy

Worked solution

Answer

L is the cathode (connected to the negative terminal).
M is the anode (connected to the positive terminal).

Material: carbon (or graphite) or platinum.

Both electrodes must be inert so they do not react with the molten electrolyte or the products formed at them. Carbon/graphite is the most commonly used material in school experiments.

Final answer

L = cathode, M = anode; material = carbon (graphite) or platinum

Detailed explanation

Walkthrough

The circuit diagram shows a power supply connected to two electrodes dipping into molten lead(II) bromide. The electrode connected to the negative terminal is the cathode (L), and the electrode connected to the positive terminal is the anode (M). This is a standard convention in electrolysis: cathode = negative, anode = positive.

The electrodes must be made of an inert material — one that will not react with the electrolyte or the products. The mark scheme accepts carbon (also called graphite) or platinum. Carbon is the most common choice in school laboratories because it is cheap and conducts electricity well. Platinum is inert but expensive.

Key Takeaways

In electrolysis, the cathode is the negative electrode and the anode is the positive electrode.
Electrodes for electrolysis must be inert — they do not participate in the reaction. Carbon/graphite and platinum are the standard inert electrode materials.

Common Mistakes

Confusing which electrode is which: remember cathode = negative (both start with 'c' and 'n' — no, that doesn't help; just remember: cathode attracts positive ions, so it must be negative).
Suggesting a reactive metal like copper or iron as the electrode material — these would react with the electrolyte or products and would not give the correct results.
Not naming the electrodes at all and just describing them.

Things to Be Careful About

The question asks for the name of the apparatus (cathode/anode) and the material they are made from — two separate pieces of information.
Carbon and graphite are acceptable alternatives; they are the same material (graphite is a form of carbon).
Platinum is also acceptable but less commonly used in school labs.

Techniques used

identify electrodes from circuit connectionsrecognise inert electrode materials for electrolysis

(b)

Describe the appearance of the products at L and at M.

appearance of product at L ______
appearance of product at M ______

DifficultyMedium-Easy

Worked solution

Answer

Appearance of product at L (cathode, negative electrode):
A shiny, silver-grey (or dark grey/black) solid or liquid deposit of lead metal. Since the electrolyte is molten and hot, the lead produced may appear as a silvery liquid or a grey solid on cooling.

Appearance of product at M (anode, positive electrode):
An orange-brown vapour (or gas) of bromine.

At the cathode: $\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}$ (lead metal is deposited).
At the anode: $2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-$ (bromine gas is released).

Final answer

At L: shiny silver/grey solid or liquid (lead); at M: orange-brown vapour (bromine)

Detailed explanation

Walkthrough

Molten lead(II) bromide, $\text{PbBr}_2$ , contains two types of ions: positive lead(II) ions ( $\text{Pb}^{2+}$ ) and negative bromide ions ( $\text{Br}^-$ ).

At the cathode (L, negative electrode):
Positive ions are attracted to the negative cathode. The $\text{Pb}^{2+}$ ions gain electrons (reduction) to form lead metal:

\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}

Lead is a dense, silvery-grey metal. At the high temperature of the molten electrolyte (melting point of $\text{PbBr}_2$ is about 373 °C), the lead produced may initially appear as a silvery liquid, but it quickly cools to a grey/silver solid deposit on the electrode.

At the anode (M, positive electrode):
Negative ions are attracted to the positive anode. The $\text{Br}^-$ ions lose electrons (oxidation) to form bromine molecules:

2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-

Bromine is a halogen that exists as a reddish-brown/orange-brown vapour at room temperature. At the elevated temperature of the experiment, it is released as an orange-brown gas/vapour above the molten electrolyte.

Key Takeaways

In electrolysis of a binary ionic compound, the metal is produced at the cathode and the non-metal (halogen) at the anode.
Lead metal appears as a shiny silver/grey solid or liquid; bromine appears as an orange-brown vapour.
The half-equations show reduction at the cathode (gain of electrons) and oxidation at the anode (loss of electrons).

Common Mistakes

Saying lead is produced at the anode or bromine at the cathode — the metal always goes to the cathode, the non-metal to the anode.
Describing bromine as a liquid — at the temperature of this experiment, bromine is a gas/vapour.
Not mentioning the colour: lead is silver/grey and bromine is orange/brown — these are the key observable features.
Writing the full equation instead of describing the appearance — the question asks for the appearance, not the equation.

Things to Be Careful About

The question asks for the appearance of the products, not the chemical names (though naming them helps). Focus on colour and state: shiny silver/grey solid/liquid at L, orange-brown vapour at M.
Lead(II) bromide melts at about 373 °C. At this temperature, lead (melting point 327 °C) may actually be liquid, so both "solid" and "liquid" are acceptable descriptions for the lead deposit.
Bromine vapour is toxic — in a real experiment, this would be done in a fume cupboard, but the question does not ask about safety.

Techniques used

predict products of electrolysis of a binary ionic compounddescribe physical appearance of products at each electrode

(c)

The teacher stops heating and allows the lead(II) bromide to cool.

Explain why the lamp goes out.

______

DifficultyMedium

Worked solution

Answer

When the teacher stops heating, the molten lead(II) bromide solidifies (cools to become a solid). In the solid state, the $\text{Pb}^{2+}$ and $\text{Br}^-$ ions are fixed in a rigid lattice and cannot move. Since electrical conduction in an ionic compound requires mobile ions to carry the charge through the substance, no current can flow through the solid lead(II) bromide. With no current flowing in the circuit, the lamp goes out.

In summary:

Heating stops → molten $\text{PbBr}_2$ solidifies.
Solid ionic compound → ions are fixed in place, not free to move.
No mobile ions → no electrical conductivity → no current → lamp goes out.

Final answer

The molten lead(II) bromide solidifies on cooling; ions are fixed in the lattice and cannot move, so current cannot flow and the lamp goes out.

Detailed explanation

Walkthrough

This question tests the understanding of why ionic compounds conduct electricity only when molten or dissolved, not when solid.

Step 1: What happens when heating stops?
The molten lead(II) bromide cools down and solidifies. Lead(II) bromide has a melting point of about 373 °C. Once the temperature drops below this, the substance becomes a solid.

Step 2: What is different about the solid state?
In the molten (liquid) state, the ions ( $\text{Pb}^{2+}$ and $\text{Br}^-$ ) are free to move around. This mobility allows them to carry electric charge through the electrolyte, completing the circuit and allowing current to flow — which is why the lamp was lit.

In the solid state, the ions are held in a fixed, rigid ionic lattice. They can vibrate about their positions but cannot move freely from one place to another. Without mobile charge carriers, the solid cannot conduct electricity.

Step 3: What is the consequence for the circuit?
With no conductivity through the electrolyte, the circuit is effectively broken. No electric current can flow, so the lamp goes out.

This is the key principle: ionic compounds conduct electricity only when the ions are free to move — i.e., in the molten state or in aqueous solution. In the solid state, they are insulators.

Key Takeaways

Ionic compounds conduct electricity only when molten or dissolved because the ions must be mobile to carry charge.
In the solid state, ions are fixed in a lattice and cannot move, so no current flows.
The lamp going out is direct evidence that the circuit has been broken by the loss of conductivity.

Common Mistakes

Saying "the electricity stops" without explaining why — must mention solidification and/or ions being fixed/cannot move.
Saying "ions are not present" in the solid — ions ARE present in solid lead(II) bromide, they just cannot move.
Saying "the power supply turns off" — the power supply is still on; the issue is the electrolyte stopping conduction.
Not mentioning that the compound solidifies — this is the trigger event that leads to the loss of conductivity.
Saying "electrons cannot flow" — in ionic conduction, it is ions that move, not electrons. Electrons flow in the external wires, but through the electrolyte it is ions carrying the charge.

Things to Be Careful About

The mark scheme awards two marks: one for stating the compound solidifies (M1), and one for explaining that current cannot flow because ions cannot move (M2). Both points are needed for full marks.
Use precise language: say "ions are fixed in a lattice and cannot move" rather than "ions stop moving" or "no ions."
This is a classic 5070 question linking the state of an ionic compound to its electrical conductivity. The same principle applies to aqueous solutions: solid NaCl does not conduct, but dissolved NaCl does.

Techniques used

relate state of ionic compound to electrical conductivityexplain circuit failure on cooling

Q214MMediumObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationUse of Techniques, Apparatus and Materials

A student titrates four samples of $0.800\text{ mol / dm}^3$ aqueous sodium hydroxide, $\text{NaOH(aq)}$ , with aqueous ethanedioic acid.

In titration 1 the student:

rinses and fills a burette with aqueous ethanedioic acid
uses a volumetric pipette to add $25.0\text{ cm}^3$ of $\text{NaOH(aq)}$ to a conical flask
adds thymolphthalein indicator to the conical flask
places the conical flask on a white tile
adds aqueous ethanedioic acid from the burette while swirling the flask, adding drop by drop near the end-point, until the solution just changes colour.

The student repeats the titration three more times.

(a)

Fig. 2.1 shows the burette readings for two of the titrations.

Fig. 2.1

Record the burette readings in Table 2.1.

Complete Table 2.1.

Table 2.1

	titration number
	1	2	3	4
final burette reading / $\text{cm}^3$			20.1	40.4
initial burette reading / $\text{cm}^3$				20.5
volume of ethanedioic acid added / $\text{cm}^3$			20.1
best titration results (✓)

DifficultyMedium-Easy

Worked solution

Answer

	titration number
	1	2	3	4
final burette reading / $\text{cm}^3$	21.1	41.8	20.1	40.4
initial burette reading / $\text{cm}^3$	0.2	21.4	0.0	20.5
volume of ethanedioic acid added / $\text{cm}^3$	20.9	20.4	20.1	19.9
best titration results (✓)

Final answer

Titration 1: final = 21.1, initial = 0.2, volume = 20.9; Titration 2: final = 41.8, initial = 21.4, volume = 20.4; Titration 3: initial = 0.0; Titration 4: volume = 19.9

Detailed explanation

Walkthrough

To complete Table 2.1 from the burette diagrams in Fig. 2.1:

Titration 1:
- Initial reading: The bottom of the meniscus is at $0.2\text{ cm}^3$ .
- Final reading: The bottom of the meniscus is at $21.1\text{ cm}^3$ .
- Volume added = $21.1 - 0.2 = 20.9\text{ cm}^3$ .
Titration 2:
- Initial reading: The bottom of the meniscus is at $21.4\text{ cm}^3$ .
- Final reading: The bottom of the meniscus is at $41.8\text{ cm}^3$ .
- Volume added = $41.8 - 21.4 = 20.4\text{ cm}^3$ .
Titration 3:
- Final reading is $20.1\text{ cm}^3$ and volume added is $20.1\text{ cm}^3$ , so initial reading = $20.1 - 20.1 = 0.0\text{ cm}^3$ .
Titration 4:
- Final reading is $40.4\text{ cm}^3$ and initial reading is $20.5\text{ cm}^3$ .
- Volume added = $40.4 - 20.5 = 19.9\text{ cm}^3$ .

All readings must be recorded to one decimal place, including trailing zeros like $0.0\text{ cm}^3$ .

Key Takeaways

Burette scales increase downwards.
Always read from the bottom of the meniscus.
All burette readings must be recorded to a consistent precision of one decimal place (e.g. $0.0$ , not just $0$ ).

Common Mistakes

Reading the scale upwards (e.g. reading $0.2$ as $0.8$ ).
Writing whole numbers without a decimal place (e.g. writing $0$ instead of $0.0$ ).
Incorrect subtraction when calculating the titre.

Things to Be Careful About

Ensure every numerical entry in the burette table has exactly 1 decimal place.

Techniques used

read burette scales to one decimal placecalculate titre volume by subtraction

(b)

Tick (✓) the two best titration results in Table 2.1.

(c)

Use the ticked (✓) titration results in Table 2.1 to calculate the average volume of aqueous ethanedioic acid needed to neutralise $25.0\text{ cm}^3$ of the aqueous sodium hydroxide.

volume = ______ $\text{cm}^3$

(d)

Calculate the number of moles of $\text{NaOH}$ in $25.0\text{ cm}^3$ of $0.800\text{ mol / dm}^3$ $\text{NaOH(aq)}$ .

number of moles = ______

(e)

One mole of ethanedioic acid is neutralised by two moles of sodium hydroxide.

Use your answers to (c) and (d) to calculate the concentration, in $\text{mol / dm}^3$ , of ethanedioic acid.

Give your answer to three significant figures.

concentration = ______ $\text{mol / dm}^3$

DifficultyMedium

Worked solution

Working

Moles of ethanedioic acid:

\text{moles} = \frac{0.0200}{2} = 0.0100\text{ mol}

Concentration of ethanedioic acid:

\text{concentration} = \frac{0.0100}{\frac{20.0}{1000}} = \frac{0.0100 \times 1000}{20.0} = 0.500\text{ mol / dm}^3

Answer

0.500

Final answer

0.500

Detailed explanation

Walkthrough

Use the mole ratio from the question: $1\text{ mol}$ of ethanedioic acid reacts with $2\text{ mol}$ of $\text{NaOH}$ . $n(\text{ethanedioic acid}) = \frac{n(\text{NaOH})}{2} = \frac{0.0200}{2} = 0.0100\text{ mol}$
Calculate concentration: $\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.0100}{\frac{20.0}{1000}} = 0.500\text{ mol / dm}^3$
The question asks for three significant figures: write $0.500\text{ mol / dm}^3$ .

Key Takeaways

$\text{Concentration} = \frac{\text{moles}}{\text{volume (dm}^3\text{)}}$ .
Always apply the stoichiometric ratio between acid and base.
Trailing zeros are required when expressing a value to a specified number of significant figures.

Common Mistakes

Forgetting to divide moles of $\text{NaOH}$ by $2$ .
Multiplying by $2$ instead of dividing.
Writing $0.5$ instead of $0.500$ (losing the mark for 3 significant figures).

Things to Be Careful About

Give the final answer to exactly 3 significant figures as instructed.

Techniques used

apply stoichiometry mole ratiocalculate concentration in mol / dm3round to specified significant figures

(f)

The formula of ethanedioic acid is $\text{C}_2\text{H}_2\text{O}_4\cdot n\text{H}_2\text{O}$ .

(i)

$100\text{ cm}^3$ of the aqueous ethanedioic acid contains $6.3\text{ g}$ of $\text{C}_2\text{H}_2\text{O}_4\cdot n\text{H}_2\text{O}$ .

Use your answer from (e) to calculate the relative formula mass, $M_r$ , of $\text{C}_2\text{H}_2\text{O}_4\cdot n\text{H}_2\text{O}$ .

$M_r$ = ______

DifficultyMedium

Worked solution

Working

Concentration in $\text{g / dm}^3$ :

\text{mass in } 1000\text{ cm}^3 = 6.3 \times 10 = 63\text{ g / dm}^3

Relative formula mass, $M_r$ :

M_r = \frac{\text{concentration in g / dm}^3}{\text{concentration in mol / dm}^3} = \frac{63}{0.500} = 126

Answer

126

Final answer

126

Detailed explanation

Walkthrough

Method 1:

Find the concentration of ethanedioic acid in $\text{g / dm}^3$ $g / dm^{3}$ :
- $100\text{ cm}^3$ contains $6.3\text{ g}$ .
- Therefore, $1000\text{ cm}^3$ contains $6.3 \times 10 = 63\text{ g / dm}^3$ .
Calculate $M_r$ using the molar concentration from part (e) ( $0.500\text{ mol / dm}^3$ ): $M_r = \frac{\text{mass concentration (g / dm}^3\text{)}}{\text{molar concentration (mol / dm}^3\text{)}} = \frac{63}{0.500} = 126$

Alternative Method:

Moles of acid in $100\text{ cm}^3 = 0.500 \times \frac{100}{1000} = 0.0500\text{ mol}$ .
$M_r = \frac{\text{mass}}{\text{moles}} = \frac{6.3}{0.0500} = 126$ .

Key Takeaways

$M_r = \frac{\text{mass}}{\text{moles}} = \frac{\text{concentration (g / dm}^3\text{)}}{\text{concentration (mol / dm}^3\text{)}}$ .

Common Mistakes

Using $6.3\text{ g}$ directly with $0.500\text{ mol / dm}^3$ without adjusting for the volume ( $100\text{ cm}^3$ vs $1000\text{ cm}^3$ ).

Things to Be Careful About

Ensure the volume units for both mass and moles are aligned before calculating $M_r$ .

Techniques used

calculate concentration in g / dm3calculate relative formula mass from mass and moles

(ii)

Use your answer from (f)(i) to deduce the value of $n$ in $\text{C}_2\text{H}_2\text{O}_4\cdot n\text{H}_2\text{O}$ .

Give your answer to the nearest whole number.

[ $A_r$ : H, 1; C, 12; O, 16]

$n$ = ______

(g)

State why the conical flask is placed on a white tile before aqueous ethanedioic acid is added from the burette.

______

(h)

State why a measuring cylinder is not used to measure $25.0\text{ cm}^3$ of $\text{NaOH(aq)}$ in this experiment.

______

Q314MMediumExperimental ContextsUse of Techniques, Apparatus and MaterialsObservations and MeasurementsQualitative Analysis

A student investigates solid Y and solution Z.

(a)

Solid Y is a white powder.

The tests the student does on Y are shown in Table 3.1.

Some of the observations for these tests are also shown.

Table 3.1

	tests on solid Y	observations
1	Add excess dilute acid to Y in a boiling tube. The gas produced is tested using limewater.	colourless solution formed limewater becomes milky
2	Add dilute nitric acid to some of the solution from test 1. Then add aqueous barium nitrate.	white precipitate
3	Add aqueous sodium hydroxide drop by drop to some of the solution from test 1 until a change is seen. Then add excess aqueous sodium hydroxide.	white precipitate soluble in excess giving a colourless solution
4	Add aqueous ammonia drop by drop to some of the solution from test 1 until a change is seen. Then add excess aqueous ammonia.	white precipitate soluble in excess giving a colourless solution

(i)

Describe how the gas is passed through limewater in test 1.

You may draw a labelled diagram to help answer the question.

______

(ii)

Describe two other observations the student makes in test 1.

______
______

(iii)

Identify the gas produced in test 1.

______

(iv)

Y contains one anion. Use the observations from test 1 to identify this anion.

______

(v)

Use the observations from test 2 to identify the dilute acid used in test 1.

______

(vi)

Identify the cation in Y using the observations from tests 3 and 4.

______

(b)

Solution Z is colourless.

(i)

The student thinks that Z contains $\text{Cu}^{2+}$ ions.

State why the student is not correct.

______

(ii)

Solution Z contains $\text{K}^+$ ions.

Describe how to do a flame test on solution Z to confirm the identity of this cation.

______

(iii)

Solution Z contains ions of a Group VII element.

Describe a test and the possible results to identify which Group VII ion is present in Z.

______