Electrochemistry

Use the data in the table below to answer this question. The most powerful oxidising agent in the table is

• Mn²⁺(aq)

• Zn(s)

• MnO₄^-(aq)

• Zn²⁺(aq)

Answer explanation

MnO4- (aq) is the most powerful oxidising agent due to its high oxidation state of manganese, allowing it to readily accept electrons compared to the other options listed.

Which change to a hydrogen electrode has no effect on the electrode potential?

• the concentration of the hydrogen ions

• the pressure of the hydrogen

• the surface area of the platinum electrode

• the temperature of the acid

Answer explanation

The surface area of the platinum electrode does not affect the electrode potential because it does not change the concentration of reactants or products in the electrochemical reaction, unlike the other factors listed.

Use the data in the table below to answer this question. Which one of the following statements is not correct?

• Fe²⁺(aq) can reduce acidified MnO₄^-(aq) to Mn²⁺(aq)

• Cr₂O₇^2-(aq) can oxidise acidified Fe²⁺(aq) to Fe³⁺(aq)

• Zn(s) can reduce acidified Cr₂O₇^2-(aq) to Cr²⁺(aq)

• Fe²⁺(aq) can reduce acidified Cr³⁺(aq) to Cr²⁺(aq)

Answer explanation

Fe2+(aq) cannot reduce Cr3+(aq) to Cr2+(aq) because Cr3+ is already in a lower oxidation state than Cr2+. Therefore, this statement is incorrect.

Some electrode potential data are shown.

 

Zn²⁺(aq) + 2 e⁻ → Zn(s)

E^ɵ = − 0.76 V

Pb²⁺(aq) + 2 e⁻ → Pb(s)

E^ɵ = − 0.13 V

Which is a correct statement about this cell?

• Electrons travel in the external circuit from zinc to lead.

• The concentration of lead (II) ions increases.

• The maximum EMF of the cell is 0.89 V

• Zinc is deposited.

Answer explanation

In the cell, zinc has a lower reduction potential than lead, meaning zinc will oxidize and lose electrons. Therefore, electrons travel from zinc to lead in the external circuit.

The E^Θ values for two electrodes are shown.

Fe²⁺(aq) + 2 e^– → Fe(s) E^Θ= –0.44 V

Cu²⁺(aq) + 2 e^– → Cu(s) E^Θ= +0.34 V

What is the EMF of the cell Fe(s)|Fe²⁺(aq)||Cu²⁺(aq)|Cu(s)?

• +0.78 V

• +0.10 V

• −0.10 V

• −0.78 V

Answer explanation

The EMF of the cell is calculated using EΘ(cell) = EΘ(cathode) - EΘ(anode). Here, Cu is the cathode (+0.34 V) and Fe is the anode (-0.44 V). Thus, EMF = 0.34 - (-0.44) = 0.34 + 0.44 = +0.78 V.

Which ion cannot catalyse the reaction between iodide (I^–) and peroxodisulfate (S₂O₈^2–)?

Use the data below to help you answer this question.

• Co²⁺

• Cr²⁺

• Fe²⁺

• Fe³⁺

Answer explanation

Cr²⁺ cannot catalyse the reaction because it does not effectively participate in redox reactions with iodide and peroxodisulfate, unlike Co²⁺, Fe²⁺, and Fe³⁺, which can facilitate electron transfer.

The following cell has an EMF of +0.46 V. Which statement is correct about the operation of the cell?

• Metallic copper is oxidised by Ag⁺ ions.

• The silver electrode has a negative polarity.

• The silver electrode gradually dissolves to form Ag⁺ ions.

• Electrons flow from the silver electrode to the copper electrode         
   via an external circuit.

Answer explanation

In this cell, Ag+ ions oxidize metallic copper, meaning copper loses electrons while Ag+ gains them. This confirms that metallic copper is oxidized by Ag+ ions, making this statement correct.

In this question consider the data below. The e.m.f. of the cell Ag(s) | Ag⁺(aq) || Pb²⁺(aq) | Pb(s) is

• 0.93 V

• 0.67 V

• −0.67 V

• −0.93 V

Answer explanation

The cell reaction involves Ag and Pb. The standard reduction potentials are E°(Ag+/Ag) = +0.80 V and E°(Pb2+/Pb) = -0.13 V. The e.m.f. is calculated as E°cell = E°(Ag) - E°(Pb) = 0.80 - (-0.13) = -0.93 V.

In this question consider the data below. The e.m.f. of the cell Pt(s) | H₂(g) | H⁺(aq) || Ag⁺(aq) | Ag(s) would be increased by

• increasing the concentration of H⁺(aq).

• increasing the surface area of the Pt electrode.

• increasing the concentration of Ag⁺(aq).

• decreasing the pressure of H₂(g).

Answer explanation

Increasing the concentration of Ag+(aq) shifts the equilibrium towards the right, increasing the e.m.f. according to the Nernst equation. This enhances the cell's potential, making it the correct choice.

A disproportionation reaction occurs when a species M⁺ spontaneously undergoes simultaneous oxidation and reduction.

                2M⁺(aq) → M²⁺(aq) + M(s)

The table below contains E data for copper and mercury species. Using these data, which one of the following can be predicted?

• Both Cu(I) and Hg(I) undergo disproportionation.

• Only Cu(I) undergoes disproportionation.

• Only Hg(I) undergoes disproportionation.

• Neither Cu(I) nor Hg(I) undergoes disproportionation.

Answer explanation

Cu(I) has a standard reduction potential that allows it to undergo disproportionation, forming Cu(II) and elemental Cu. Hg(I) does not have a favorable potential for this reaction, thus only Cu(I) undergoes disproportionation.