Module 3.2 Predicting Reactions of Metals

The most active metals tend to have which of the following properties?

• A higher ionisation energy and a higher electronegativity

• A lower ionisation energy and a higher electronegativity

• A higher ionisation energy and a lower electronegativity

• A lower ionisation energy and a lower electronegativity

Answer explanation

The most active metals tend to have a lower ionisation energy and a lower electronegativity. Active metals, especially those in Group 1 of the periodic table (alkali metals), have low ionisation energies, which means they lose their outer electrons easily. They also have low electronegativities, meaning they have a weak attraction for electrons in a chemical bond. This combination of properties makes these metals highly reactive.

A galvanic cell is constructed using zinc (Zn/Zn²⁺) and silver (Ag/Ag⁺) half-cells.

What is the voltage produced by this galvanic cell when measured under standard conditions and which species is being reduced?

• 0.04 V, Ag⁺ ions are being reduced

• 0.04 V, Zn atoms are being reduced

• 1.56 V, Ag⁺ ions are being reduced

• 1.56 V, Zn atoms are being reduced

Answer explanation

In a galvanic cell constructed with zinc (Zn/Zn²⁺) and silver (Ag/Ag⁺) half-cells, the voltage produced under standard conditions is calculated to be 1.56 V. This is determined using the standard reduction potentials: +0.80 V for the reduction of Ag⁺ to Ag and -0.76 V for the oxidation of Zn to Zn²⁺. The cell potential is found by subtracting the anode potential (zinc) from the cathode potential (silver), resulting in 1.56 V=(+0.80 V)−(−0.76 V). In this reaction, Ag⁺ ions are reduced to form silver metal at the cathode, making them the species being reduced.

What is the oxidation state of arsenic in the arsenate ion, AsO₄^3-?

• -2

• -3

• +4

• +5

Answer explanation

The oxidation state of arsenic in the arsenate ion (AsO4 3−) is +5. To determine this, use the fact that the oxidation states of oxygen are typically -2. In AsO4 3−​, there are four oxygen atoms contributing a total oxidation state of 4×(−2)=−8. To balance this and match the overall charge of the ion (-3), arsenic must have an oxidation state of +5. Therefore, the oxidation state of arsenic in AsO4 3− is +5.

A student electrically connected two galvanic half-cells to measure and compare their reduction potentials. The half-cells used were Ag(s)/Ag⁺ and Sn(s)/Sn²⁺.

Assuming the investigation was carried out under standard conditions, what is the maximum potential difference the student would expect to measure?

• 0.46 V

• 0.66 V

• 0.94 V

• 1.74 V

Answer explanation

To determine the maximum potential difference between the Ag/Ag⁺ and Sn/Sn²⁺ half-cells under standard conditions, use their standard reduction potentials:

• Ag⁺ + e⁻ → Ag with E∘=+0.80 V

• Sn²⁺ + 2e⁻ → Sn with E∘=−0.14 V

The cell potential is calculated using the formula: Ecell=Ecathode−Eanode

Here, silver (Ag⁺/Ag) will be the cathode (reduction) and tin (Sn/Sn²⁺) will be the anode (oxidation): Ecell = (+0.80 V) − (−0.14 V) = 0.94 V

Thus, the maximum potential difference the student would expect to measure is 0.94 V.

A student performed a practical investigation that involved placing samples of different metals, one of which was lead, into solutions containing the ions of other metals represented by the letters X, Y and Z. The results of this investigation are shown in the table.

Which of the following combinations of metal and metal ions would give the highest potential difference in an electrochemical cell constructed under standard conditions?

• Z and Pb²⁺

• Pb and X²⁺

• Z and X²⁺

• Y and Y²⁺

Answer explanation

The highest potential difference in an electrochemical cell is achieved between the metal that is most easily oxidised (Z in this case, since it reacts with X²⁺ and Pb²⁺) and the metal ion that is most easily reduced (X²⁺, since it reacts with the most number of metals). This combination maximises the difference in reduction potentials, resulting in the highest potential difference. Therefore, the correct answer is Z and X²⁺.

Standard solutions of nickel (II) and tin (II) are used to make the electrochemical cell. Note that the labels (X) and (Y) on the wire connecting the nickel and tin electrodes.

Which of the following describes the voltage produced by this cell and the direction of electron flow through the connecting wire?

• Voltage = 0.38 V, Electrons flow from (X) to (Y)

• Voltage = 0.10 V, Electrons flow from (X) to (Y)

• Voltage = 0.38 V, Electrons flow from (Y) to (X)

• Voltage = 0.10 V, Electrons flow from (Y) to (X)

Answer explanation

To determine the voltage produced by the cell and the direction of electron flow, we need to compare the standard reduction potentials for nickel (II) and tin (II):

• Ni²⁺ + 2e⁻ → Ni with E∘=−0.25 V

• Sn²⁺ + 2e⁻ → Sn with E∘=−0.14 V

The more positive reduction potential is for Sn²⁺/Sn, which means tin will be reduced and act as the cathode, while nickel will be oxidised and act as the anode. Electrons flow from the anode to the cathode.

The cell potential Ecell​ is calculated as: Ecell = Ecathode−Eanode = (−0.14 V)−(−0.25 V) = 0.11 V

The voltage produced by the cell is approximately 0.10 V. Since electrons flow from the anode (nickel) to the cathode (tin), electrons flow from (X) to (Y) if X is connected to the nickel electrode and Y to the tin electrode.

Therefore, the correct answer is Voltage = 0.10 V, Electrons flow from (X) to (Y).

Two significant compounds of nitrogen are nitrogen dioxide (NO₂) and dinitrogen pentoxide (N₂O₅).

What are the oxidation numbers of nitrogen in these two compounds, respectively?

• +2 and +10

• -5 and -10

• +2 and +5

• +4 and +5

Answer explanation

The oxidation numbers of nitrogen in nitrogen dioxide (NO₂) and dinitrogen pentoxide (N₂O₅) are +4 and +5, respectively. For NO₂, where oxygen has an oxidation number of -2, the nitrogen must have an oxidation number of +4 to balance the overall charge to zero. In N₂O₅, with oxygen again at -2, the nitrogen must have an oxidation number of +5 to balance the two nitrogen atoms against the five oxygens, ensuring the molecule remains neutral. Thus, the correct oxidation numbers for nitrogen in these compounds are +4 and +5.

A student used a chemical data book to list a pattern of change in a property of some Group 1 elements on the periodic table but did not record the property’s units.

Which of the following alternatives identifies the property and gives a correct reason why the change occurs down Group 1?

• Atomic radius: Atoms get larger with the addition of each new shell of electrons

• Ionisation energy: Increases with addition of protons to the nuclei of each element’s atoms

• Electronegativity: Increases because the charge on the nucleus increases down a group

• Standard cell reduction potential: Increases because metal activity increases

Answer explanation

The correct property is atomic radius, with the reason being that atoms get larger with the addition of each new shell of electrons as you move down Group 1 in the periodic table. As each successive element in Group 1 has an additional electron shell compared to the one above it, the distance between the nucleus and the outermost electrons increases, leading to a larger atomic radius. The other options are incorrect: ionisation energy actually decreases down a group due to the increased distance between the nucleus and the outer electrons; electronegativity decreases down a group for similar reasons; and standard cell reduction potential does not follow a simple increasing pattern with metal activity down a group. Therefore, the correct answer is atomic radius: Atoms get larger with the addition of each new shell of electrons.

If the cell shown is kept at standard values, which is closest to the voltage created by the cell?

• 0.20 V

• 0.50 V

• 0.70 V

• 0.90 V

Answer explanation

To determine the voltage created by the cell with nickel and zinc electrodes, we use their standard reduction potentials:

• Zn²⁺ + 2e⁻ → Zn with E∘=−0.76 V

• Ni²⁺ + 2e⁻ → Ni with E∘=−0.25 V

The cell potential is calculated using the formula: Ecell = Ecathode−Eanode =(−0.25 V)−(−0.76 V) = 0.51 V

Therefore, the closest voltage created by the cell is 0.50 V.

The connecting wire that runs from the nickel electrode to the zinc electrode has two points on it labelled X and Y.

Which of the following is a correct statement about the electrodes and the direction of electron flow in the wire when the current is flowing?

• Nickel is oxidised and electrons flow from X to Y.

• Nickel is oxidised and electrons flow from Y to X.

• Zinc is oxidised and electrons flow from X to Y.

• Zinc is oxidised and electrons flow from Y to X.

Answer explanation

In a galvanic cell with nickel and zinc electrodes, zinc is the metal that gets oxidized, and nickel is reduced. Oxidation occurs at the anode, and reduction occurs at the cathode. Electrons flow from the anode to the cathode through the external circuit.

Since zinc is oxidized, the electrons will flow from the zinc electrode to the nickel electrode. If X is connected to the zinc electrode and Y to the nickel electrode, the correct statement is Zinc is oxidized, and electrons flow from X to Y.